\n",
23 | "Determine the total heat transfer to the air and the temperature of the air leaving the duct heater.\n",
24 | "
\n",
25 | "
\n",
26 | "Determine the pressure drop across the element bank and the fan power requirement.\n",
27 | "
\n",
28 | "
\n",
29 | "Compare the average convection coefficient obtained in your analysis with the value for an isolated (single) element. Explain the difference between the results.\n",
30 | "
\n",
31 | "
\n",
32 | "What effect would increasing the longitudinal and transverse pitches to 30 mm have on the exit temperature of the air, the total heat rate, and the pressure drop?\n",
33 | "
\n",
34 | ""
35 | ]
36 | },
37 | {
38 | "cell_type": "code",
39 | "execution_count": null,
40 | "metadata": {},
41 | "outputs": [],
42 | "source": []
43 | },
44 | {
45 | "cell_type": "markdown",
46 | "metadata": {},
47 | "source": [
48 | "# Problem 3\n",
49 | "An aluminum transmission line with a diameter of\n",
50 | "20 mm has an electrical resistance of $R'_\\text{elec}=2.636\\times10^{-4}\\Omega/\\text{m}$\n",
51 | "and carries a current of 700 A. The line is\n",
52 | "subjected to frequent and severe cross winds, increasing the probability of contact between adjacent lines,\n",
53 | "thereby causing sparks and creating a potential fire hazard for nearby vegetation. The remedy is to insulate the\n",
54 | "line, but with the adverse effect of increasing the conductor operating temperature.\n",
55 | "\n",
56 | "
Calculate the conductor temperature when the air temperature is $20^\\circ\\text{C}$ and the cross flow wind velocity is $10\\text{m/s}$
\n",
57 | "
Calculate the conductor temperature for the same\n",
58 | "conditions, but with a $2\\text{mm}$-thick insulation having\n",
59 | "a thermal conductivity of $0.15\\text{W/(m.K)}$.
\n",
60 | "
Calculate and plot the temperatures of the bare and\n",
61 | "insulated conductors for wind velocities in the\n",
62 | "range from $2$ to $20\\text{m/s}$. Comment on features of\n",
63 | "the curves and the effect of the wind velocity on the\n",
64 | "conductor temperatures.
\n",
65 | ""
66 | ]
67 | },
68 | {
69 | "cell_type": "markdown",
70 | "metadata": {},
71 | "source": [
72 | "# Problem 4\n",
73 | "\n",
74 | "\n",
75 | "The top surface of a heated compartment consists of very smooth (A) and highly roughened (B) portions, and the surface is placed in an atmospheric airstream. \n",
76 | "In the interest of minimizing total convection heat transfer from the surface, which orientation, (1) or (2),\n",
77 | "is preferred? If $T_s=100^\\circ\\text{C}$, $T_\\infty= 20^\\circ\\text{C}$, and $u_\\infty=20\\text{ m/s}$, what is the convection heat transfer from the entire surface for this orientation?\n",
78 | "\n",
79 | "Calculate the value of the average heat transfer coefficient when the entire plate is rotated $90^\\circ$ so that half of the leading edge consists of a very smooth portion (A) and the other half consists of a highly roughened portion (B)."
80 | ]
81 | },
82 | {
83 | "cell_type": "code",
84 | "execution_count": null,
85 | "metadata": {},
86 | "outputs": [],
87 | "source": []
88 | }
89 | ],
90 | "metadata": {
91 | "kernelspec": {
92 | "display_name": "Python 3",
93 | "language": "python",
94 | "name": "python3"
95 | },
96 | "language_info": {
97 | "codemirror_mode": {
98 | "name": "ipython",
99 | "version": 3
100 | },
101 | "file_extension": ".py",
102 | "mimetype": "text/x-python",
103 | "name": "python",
104 | "nbconvert_exporter": "python",
105 | "pygments_lexer": "ipython3",
106 | "version": "3.8.5"
107 | }
108 | },
109 | "nbformat": 4,
110 | "nbformat_minor": 4
111 | }
112 |
--------------------------------------------------------------------------------
/Libraries/raw.githubusercontent.com/yvesdubief/UVM-ME144-Heat-Transfer/master/Libraries/Tables/Water1atm.csv:
--------------------------------------------------------------------------------
1 | Temperature (K),Pressure (MPA),Density (kg/m^3),Cp (J/g.K),viscosity (Pa.s),Thermal conductivity (W/m.K)
2 | 274.00,0.10132,999.89,4.2166,0.0017398,0.56269
3 | 275.00,0.10132,999.94,4.2135,0.0016818,0.56459
4 | 276.00,0.10132,999.96,4.2106,0.0016269,0.56649
5 | 277.00,0.10132,999.97,4.2079,0.0015748,0.56839
6 | 278.00,0.10132,999.97,4.2054,0.0015253,0.57029
7 | 279.00,0.10132,999.95,4.2031,0.0014783,0.57219
8 | 280.00,0.10132,999.91,4.2009,0.0014335,0.57409
9 | 281.00,0.10132,999.86,4.1990,0.0013909,0.57598
10 | 282.00,0.10132,999.79,4.1971,0.0013503,0.57788
11 | 283.00,0.10132,999.72,4.1954,0.0013115,0.57976
12 | 284.00,0.10132,999.62,4.1938,0.0012745,0.58165
13 | 285.00,0.10132,999.52,4.1924,0.0012392,0.58352
14 | 286.00,0.10132,999.40,4.1910,0.0012054,0.58539
15 | 287.00,0.10132,999.27,4.1898,0.0011730,0.58725
16 | 288.00,0.10132,999.13,4.1886,0.0011421,0.58910
17 | 289.00,0.10132,998.97,4.1876,0.0011124,0.59094
18 | 290.00,0.10132,998.80,4.1866,0.0010840,0.59278
19 | 291.00,0.10132,998.63,4.1857,0.0010567,0.59459
20 | 292.00,0.10132,998.44,4.1849,0.0010305,0.59640
21 | 293.00,0.10132,998.24,4.1842,0.0010053,0.59819
22 | 294.00,0.10132,998.03,4.1835,0.00098110,0.59997
23 | 295.00,0.10132,997.81,4.1829,0.00095784,0.60174
24 | 296.00,0.10132,997.58,4.1823,0.00093545,0.60349
25 | 297.00,0.10132,997.34,4.1818,0.00091389,0.60522
26 | 298.00,0.10132,997.09,4.1814,0.00089313,0.60694
27 | 299.00,0.10132,996.83,4.1810,0.00087312,0.60864
28 | 300.00,0.10132,996.56,4.1806,0.00085383,0.61032
29 | 301.00,0.10132,996.28,4.1803,0.00083521,0.61199
30 | 302.00,0.10132,995.99,4.1801,0.00081724,0.61363
31 | 303.00,0.10132,995.69,4.1798,0.00079990,0.61526
32 | 304.00,0.10132,995.39,4.1797,0.00078314,0.61687
33 | 305.00,0.10132,995.08,4.1795,0.00076694,0.61846
34 | 306.00,0.10132,994.75,4.1794,0.00075128,0.62002
35 | 307.00,0.10132,994.42,4.1793,0.00073613,0.62157
36 | 308.00,0.10132,994.08,4.1793,0.00072147,0.62310
37 | 309.00,0.10132,993.74,4.1792,0.00070728,0.62460
38 | 310.00,0.10132,993.38,4.1792,0.00069354,0.62609
39 | 311.00,0.10132,993.02,4.1793,0.00068023,0.62755
40 | 312.00,0.10132,992.65,4.1793,0.00066733,0.62900
41 | 313.00,0.10132,992.27,4.1794,0.00065482,0.63042
42 | 314.00,0.10132,991.89,4.1795,0.00064269,0.63181
43 | 315.00,0.10132,991.50,4.1796,0.00063092,0.63319
44 | 316.00,0.10132,991.10,4.1798,0.00061950,0.63455
45 | 317.00,0.10132,990.69,4.1799,0.00060842,0.63588
46 | 318.00,0.10132,990.28,4.1801,0.00059765,0.63719
47 | 319.00,0.10132,989.85,4.1803,0.00058720,0.63848
48 | 320.00,0.10132,989.43,4.1805,0.00057704,0.63975
49 | 321.00,0.10132,988.99,4.1808,0.00056716,0.64099
50 | 322.00,0.10132,988.55,4.1810,0.00055756,0.64221
51 | 323.00,0.10132,988.10,4.1813,0.00054823,0.64342
52 | 324.00,0.10132,987.65,4.1816,0.00053914,0.64460
53 | 325.00,0.10132,987.19,4.1819,0.00053031,0.64575
54 | 326.00,0.10132,986.72,4.1822,0.00052171,0.64689
55 | 327.00,0.10132,986.25,4.1826,0.00051333,0.64800
56 | 328.00,0.10132,985.77,4.1829,0.00050518,0.64909
57 | 329.00,0.10132,985.28,4.1833,0.00049724,0.65017
58 | 330.00,0.10132,984.79,4.1837,0.00048950,0.65122
59 | 331.00,0.10132,984.29,4.1840,0.00048197,0.65224
60 | 332.00,0.10132,983.78,4.1845,0.00047462,0.65325
61 | 333.00,0.10132,983.27,4.1849,0.00046745,0.65424
62 | 334.00,0.10132,982.76,4.1853,0.00046047,0.65521
63 | 335.00,0.10132,982.23,4.1858,0.00045366,0.65615
64 | 336.00,0.10132,981.71,4.1863,0.00044701,0.65708
65 | 337.00,0.10132,981.17,4.1867,0.00044053,0.65798
66 | 338.00,0.10132,980.63,4.1872,0.00043420,0.65887
67 | 339.00,0.10132,980.09,4.1878,0.00042802,0.65974
68 | 340.00,0.10132,979.54,4.1883,0.00042199,0.66058
69 | 341.00,0.10132,978.98,4.1888,0.00041610,0.66141
70 | 342.00,0.10132,978.42,4.1894,0.00041034,0.66222
71 | 343.00,0.10132,977.85,4.1900,0.00040472,0.66301
72 | 344.00,0.10132,977.28,4.1906,0.00039923,0.66378
73 | 345.00,0.10132,976.70,4.1912,0.00039387,0.66453
74 | 346.00,0.10132,976.12,4.1918,0.00038862,0.66527
75 | 347.00,0.10132,975.53,4.1924,0.00038349,0.66599
76 | 348.00,0.10132,974.93,4.1931,0.00037848,0.66669
77 | 349.00,0.10132,974.33,4.1938,0.00037358,0.66737
78 | 350.00,0.10132,973.73,4.1945,0.00036878,0.66803
79 | 351.00,0.10132,973.12,4.1952,0.00036409,0.66868
80 | 352.00,0.10132,972.50,4.1959,0.00035950,0.66931
81 | 353.00,0.10132,971.88,4.1966,0.00035501,0.66992
82 | 354.00,0.10132,971.26,4.1974,0.00035062,0.67052
83 | 355.00,0.10132,970.63,4.1982,0.00034632,0.67110
84 | 356.00,0.10132,969.99,4.1990,0.00034210,0.67167
85 | 357.00,0.10132,969.35,4.1998,0.00033798,0.67222
86 | 358.00,0.10132,968.71,4.2006,0.00033394,0.67275
87 | 359.00,0.10132,968.06,4.2015,0.00032998,0.67327
88 | 360.00,0.10132,967.40,4.2023,0.00032611,0.67378
89 | 361.00,0.10132,966.74,4.2032,0.00032231,0.67427
90 | 362.00,0.10132,966.08,4.2041,0.00031859,0.67474
91 | 363.00,0.10132,965.41,4.2051,0.00031495,0.67520
92 | 364.00,0.10132,964.74,4.2060,0.00031138,0.67565
93 | 365.00,0.10132,964.06,4.2070,0.00030787,0.67608
94 | 366.00,0.10132,963.37,4.2080,0.00030444,0.67649
95 | 367.00,0.10132,962.69,4.2090,0.00030107,0.67690
96 | 368.00,0.10132,961.99,4.2100,0.00029777,0.67729
97 | 369.00,0.10132,961.29,4.2111,0.00029453,0.67767
98 | 370.00,0.10132,960.59,4.2121,0.00029136,0.67803
99 | 371.00,0.10132,959.88,4.2132,0.00028824,0.67838
100 | 372.00,0.10132,959.17,4.2144,0.00028519,0.67872
101 | 373.00,0.10132,958.46,4.2155,0.00028219,0.67904
102 |
--------------------------------------------------------------------------------
/Libraries/HT_thermal_resistance.py:
--------------------------------------------------------------------------------
1 | """Object name: Resistance
2 | Function name: serial_sum(R,nori,nend), performs serial sum of a resistance object list from nori to nend
3 | Function name: parallel_sum(R,nori,nend), performs parallel sum of a resistance object list from nori to nend
4 | """
5 | ### definition of thermal resistance ###
6 | from sympy.interactive import printing
7 | printing.init_printing(use_latex='mathjax')
8 |
9 |
10 | from IPython.display import display,Image, Latex
11 | import numpy as np
12 | import math
13 | import scipy.constants as sc
14 |
15 | import sympy as sym
16 | #from sympy import *
17 |
18 | class Resistance(object):
19 | """ Defines thermal resistances for conduction, convection and radiation heat transfer.
20 | First define the object attached with class with the name used in the thermal circuit
21 | and the units, which can only be 'W', 'W/m' or 'W/m^2'
22 | Second use self.conduction, self.convection or self.radiation to calculate your
23 | resistance. Each mode requires different arguments:
24 |
25 | from Libraries import HT_thermal_resistance as res
26 | R = []
27 | R.append(res.Resistance("$label$", "units")) where units = 'W', 'W/m' or 'W/m^2'
28 |
29 | then
30 |
31 | For conduction, there are 3 options:
32 |
33 | - R.cond_plane(k, L, A = 1.0) for planar conduction: k is the thermal conductivity,
34 | L is the thickness of the wall, and A is the optional surface area (=1 by default)
35 | - R.cond_cylinder(k , ra, rb, L = 1.0, angle = 2.*math.pi) for conduction in a
36 | cylindrical shell between the radii ra (internal) and rb (external). L is the length
37 | of the shell (optional, default = 1) and angle is angular dimension of shell, also
38 | optional and set to a full revolution by default (2 pi)
39 | - R.cond_sphere(k, ra, rb, scale = 1.0) for conductuion within a spherical shell bounded by radii ra and rb
40 | ra < rb. The optional parameter scale allows to calculate the thermal resistance for a fraction
41 | of a spherical shell. For instance a cornea is about 1/3 of spherical shell, so scale = 1./3.
42 |
43 | Convection:
44 | - R.convection(h, A = 1.0), where h is the convection coefficient (W/m^2K) and A is
45 | the surface area (optional, default is unit surface aera 1 m^2)
46 |
47 | Radiation:
48 | - R.radiation(eps, T_s, T_sur, A = 1.0), where eps is the permissivity of the material, T_s
49 | the surface temperature, T_sur the far away surface temperature, A the surface area (optional,
50 | by default A is the unit surface area 1 m^2).
51 |
52 | Contact:
53 |
54 | - R.contact(R,A,R_name= "R_{t}",A_name = "A",T_a_name = "T_a",Tb_name = "T_b"), where R is the contact resistance, typically obtained from a table
55 | A is the surface area
56 | The minimum number of arguments are:
57 | R.contact(R,A)
58 |
59 | R.display_equation(index) displays the heat flux/rate equations for a given resistance. index is the number of
60 | your resistance (you specify)
61 |
62 | Outputs:
63 | - R[i].R the resistance of element i, R[i].h the convection or radiation coefficient.
64 |
65 | Functions include
66 | R_tot = res.serial_sum(R,first_resistance,last_resistance) sums serial resistance
67 | R_tot = res.parallel_sum(R,first_resistance,last_resistance) sums parallel resistance
68 |
69 |
70 |
71 | """
72 | def __init__(self,name,units):
73 | self.name = name
74 | self.units = units
75 | def cond_plane(self, k, L, A = 1.0):
76 | self.mode = "conduction"
77 | self.geometry = "planar"
78 | self.k = k
79 | if k <= 0.:
80 | print("problem with the definition of thermal conductivity")
81 | self.L = L
82 | self.A = A
83 | self.R = self.L / (self.k * self.A)
84 | def cond_cylinder(self, k , ra, rb, L = 1.0, angle = 2.*math.pi):
85 | self.mode = "conduction"
86 | self.geometry = "cylindrical"
87 | self.k = k
88 | if k <= 0.:
89 | print("problem with the definition of thermal conductivity")
90 | self.ra = ra
91 | self.rb = rb
92 | if ra*rb <= 0.:
93 | print("problem with the definition of radii")
94 | self.L = L
95 | self.angle = angle
96 | self.R = np.log(rb/ra)/(angle*L*k)
97 | def cond_sphere(self, k, ra, rb, scale = 1.0):
98 | self.mode = "conduction"
99 | self.geometry = "spherical"
100 | self.k = k
101 | if k <= 0.:
102 | print("problem with the definition of thermal conductivity")
103 | self.ra = ra
104 | self.rb = rb
105 | if ra*rb <= 0.:
106 | print("problem with the definition of radii")
107 | self.R = (1./r_a-1./r_b)/(scale*4.*math.pi*k)
108 | def convection(self, h, A = 1.0):
109 | self.mode = 'convection'
110 | self.geometry = "whatever"
111 | self.R = 1./(h*A)
112 | self.A = A
113 | self.h = h
114 | def radiation(self,eps,T_s,T_sur, A = 1.0):
115 | self.R = 1./(eps*sc.sigma*(T_s+T_sur)*(T_s**2+T_sur**2)*A)
116 | self.mode = 'radiation'
117 | self.geometry = "whatever"
118 | self.A = A
119 | self.h = eps*sc.sigma*(T_s+T_sur)*(T_s**2+T_sur**2)
120 | def contact(self, R, A=1.0):
121 | self.R = R/A
122 | self.geometry = 'whatever'
123 | self.mode = 'contact'
124 |
125 |
126 |
127 | ### summation of thermal resistance (R is a vector) ###
128 | def serial_sum(R,nori,nend):
129 | sum = 0.
130 | for i in range(nori,nend+1):
131 | sum += R[i].R
132 | return sum
133 |
134 | def parallel_sum(R,nori,nend):
135 | sum = 0.
136 | for i in range(nori,nend+1):
137 | sum += 1./R[i].R
138 | return 1./sum
139 |
140 |
141 |
--------------------------------------------------------------------------------
/Libraries/thermodynamics.py:
--------------------------------------------------------------------------------
1 | """ Object name: Fluid"""
2 | import sys
3 | sys.path.insert(0, 'Tables/')
4 | import numpy as np
5 | import scipy
6 | import scipy.optimize
7 | from scipy.constants import convert_temperature
8 | def C2K(T):
9 | return convert_temperature(T,'Celsius','Kelvin')
10 | def C2F(T):
11 | return convert_temperature(T,'Celsius','Fahrenheit')
12 | def F2K(T):
13 | return convert_temperature(T,'Fahrenheit','Kelvin')
14 | def F2C(T):
15 | return convert_temperature(T,'Fahrenheit','Celsius')
16 | def K2F(T):
17 | return convert_temperature(T,'Kelvin','Fahrenheit')
18 | def K2C(T):
19 | return convert_temperature(T,'Kelvin','Celsius')
20 | import scipy.constants as sc
21 |
22 | def interpolate_table(target,index,xquantity,yquantity):
23 | return yquantity[index] + \
24 | (yquantity[index+1]-yquantity[index])* \
25 | (target-xquantity[index])/(xquantity[index+1]-xquantity[index])
26 |
27 | class Fluid(object):
28 | """ How to:
29 | from NewLibraries import thermodynamics as thermo
30 |
31 | fluid_of_interest = thermo.Fluid(material,T) material can be air, water, argon and krypton (see below for ranges)
32 | and the temperature of the fluid T is in Kelvin.
33 | Outputs:
34 | The new object computes thermodynamic properties of air between -150 C and 400 C,
35 | water between 274K and 373K, argon between 100 and 700K and
36 | krypton between 150 and 700 K under 1 atm. Argon, krypton and water were obtained
37 | through http://webbook.nist.gov/chemistry/fluid/
38 | More fluids to be added in the future
39 | fluid_of_interest.beta thermal expansion coefficient
40 | fluid_of_interest.rho density
41 | fluid_of_interest.Cp specific heat
42 | fluid_of_interest.mu dynamic viscosity
43 | fluid_of_interest.k thermal conductivity
44 | fluid_of_interest.nu kinematic viscosity
45 | fluid_of_interest.alpha thermal diffusivity
46 | fluid_of_interest.Pr
47 |
48 |
49 | """
50 | def __init__(self,name,T,unit = "K",P = 101325.01):
51 | dirTables = 'https://raw.githubusercontent.com/yvesdubief/UVM-ME144-Heat-Transfer/master/Libraries/Tables/'
52 | self.name = name
53 | if unit == "C":
54 | T = C2K(T)
55 | elif unit == "F":
56 | T = F2K(T)
57 | self.T = T
58 | self.P = P
59 | if P != 101325.01:
60 | print("All available tables are for P=1ATM, reverting to P=101325.01Pa")
61 | self.P = 101325.01
62 | if self.name == 'water':
63 | if T < 274 or T > 373:
64 | print("Temperature is out of bounds for liquid water")
65 | return
66 | url = dirTables+'Water1atm.csv'
67 | Ttab,ptab,rhotab,Cptab,mutab,ktab = \
68 | np.genfromtxt(url, delimiter=',', skip_header = 1, unpack=True, dtype=float)
69 | Ntab = len(Ttab)
70 | Cptab *= 1e3
71 | nutab = mutab/rhotab
72 | alphatab = ktab/(rhotab*Cptab)
73 | Prtab = nutab/alphatab
74 | dTtab = Ttab[1] - Ttab[0]
75 | # compute beta from -rho(d rho/dT)
76 | betatab = -(1./rhotab)*np.gradient(rhotab)/dTtab
77 | i = int((T-Ttab[0])/dTtab)
78 | if (i == Ntab - 1):
79 | i == Ntab - 2
80 | elif self.name == 'argon':
81 | if T < 100 or T > 700:
82 | print("Temperature is out of bounds for argon")
83 | return
84 | url = dirTables + 'Argon1atm.csv'
85 | Ttab,ptab,rhotab,Cptab,mutab,ktab = \
86 | np.loadtxt(url, delimiter=',', skiprows = 1, unpack=True, dtype=float)
87 | Ntab = len(Ttab)
88 | Cptab *= 1e3
89 | nutab = mutab/rhotab
90 | alphatab = ktab/(rhotab*Cptab)
91 | Prtab = nutab/alphatab
92 | dTtab = Ttab[1] - Ttab[0]
93 | # compute beta from -rho(d rho/dT)
94 | betatab = -(1./rhotab)*np.gradient(rhotab)/dTtab
95 | i = int((T-Ttab[0])/dTtab)
96 | if (i == Ntab - 1):
97 | i == Ntab - 2
98 | elif self.name == 'krypton':
99 | if T < 150 or T > 740:
100 | print("Temperature is out of bounds for krypton")
101 | return
102 | url = dirTables + 'Krypton1atm.csv'
103 | Ttab,ptab,rhotab,Cptab,mutab,ktab = \
104 | np.loadtxt(url, delimiter=',', skiprows = 1, unpack=True, dtype=float)
105 | Ntab = len(Ttab)
106 | Cptab *= 1e3
107 | nutab = mutab/rhotab
108 | alphatab = ktab/(rhotab*Cptab)
109 | Prtab = nutab/alphatab
110 | dTtab = Ttab[1] - Ttab[0]
111 | # compute beta from -rho(d rho/dT)
112 | betatab = -(1./rhotab)*np.gradient(rhotab)/dTtab
113 | i = int((T-Ttab[0])/dTtab)
114 | if (i == Ntab - 1):
115 | i == Ntab - 2
116 | elif self.name == 'air':
117 | if T < C2K(-150.) or T > C2K(400.):
118 | print("Temperature is out of bounds of the table for air")
119 | return
120 | url = dirTables + 'Air1atm.csv'
121 | # url = 'https://raw.githubusercontent.com/yvesdubief/UVM-ME144-Heat-Transfer/master/Libraries/Tables/Air1atm.csv'
122 | # url = 'Tables/Air1atm.csv'
123 | Ttab,rhotab,Cptab,ktab,nutab,betatab,Prtab = \
124 | np.genfromtxt(url, delimiter=',', skip_header = 1, unpack=True, dtype=float)
125 | Ntab = len(Ttab)
126 | Ttab = C2K(Ttab)
127 | Cptab *= 1e3
128 | nutab *= 1e-6
129 | mutab = rhotab*nutab
130 | alphatab = ktab/(rhotab*Cptab)
131 | Prtab = nutab/alphatab
132 | i = 0
133 | while (Ttab[i] < T) and (i 0.):
101 | f_1 = 1./(-2.0*np.log10(e/3.71))**2
102 | else:
103 | f_1 = f_0
104 | f_guess = np.max([f_0,f_1])
105 | #f_guess = 0.04
106 | def f_tmp(x):
107 | y = (-2*np.log10((2.51/(Re*np.sqrt(x))) + (e/(3.71))) - 1.0/np.sqrt(x))
108 | return y
109 | y = scipy.optimize.fsolve(f_tmp, f_guess)
110 | self.f = y[0]
111 | self.dPdx = self.f*(self.L/self.D)*(self.rho*self.Um**2)/2.
112 |
113 | def laminar_isothermal(self):
114 | self.Nu = 3.66
115 |
116 | def laminar_isoflux(self):
117 | self.Nu = 4.36
118 |
119 | def Dittus_Boelter(self,mode,Pr,Re = 0.0):
120 | if Re == 0. and self.Re !=0:
121 | Re = self.Re
122 | else:
123 | print("Warning Reynolds number is not defined")
124 | if (mode == 'heating'):
125 | n = 0.4
126 | elif (mode == 'cooling'):
127 | n = 0.3
128 | else:
129 | print("Warning you have to specify mode='heating' or 'cooling'")
130 | self.Nu = 0.023*Re**(4./5.)*Pr**n
131 |
132 | def Sieder_Tate(self,Pr,mu,mu_s, Re = 0.0):
133 | if Re == 0. and self.Re !=0:
134 | Re = self.Re
135 | else:
136 | print("Warning Reynolds number is not defined")
137 | self.Nu = 0.027*Re**(4/5)*Pr**(1/3)*(mu/mu_s)**0.14
138 |
139 | def Gnielinski(self, Pr, f,Re = 0.0):
140 | if Re == 0. and self.Re !=0:
141 | Re = self.Re
142 | else:
143 | print("Warning Reynolds number is not defined")
144 | self.Nu = (f/8.)*(Re-1000.)*Pr/(1+12.7*(f/8.)**0.5*(Pr**(2./3.)-1.))
145 |
146 | def Skupinski(self,Pr, Re = 0.0):
147 | if Re == 0. and self.Re !=0:
148 | Re = self.Re
149 | else:
150 | print("Warning Reynolds number is not defined")
151 | self.Nu = 4.82+0.0185*(Re*Pr)**0.827
152 |
153 | def Seban(self,Pr, Re = 0.0):
154 | if Re == 0. and self.Re !=0:
155 | Re = self.Re
156 | else:
157 | print("Warning Reynolds number is not defined")
158 | self.Nu = 5.0+0.025*(Re*Pr)**0.8
159 |
160 | def log_mean_temperature(T_s,T_o,T_i):
161 | if (T_s < min(T_o,T_i)):
162 | DT_o = T_o-T_s
163 | DT_i = T_i-T_s
164 | elif (T_s > max(T_o,T_i)):
165 | DT_o = T_s-T_o
166 | DT_i = T_s-T_i
167 | return (DT_o-DT_i)/np.log(DT_o/DT_i)
168 |
169 |
170 | def T_mx_Ts_constant(T_s,T_mi,P,mdot,Cp,hbar,x):
171 | return T_s-(T_s-T_mi)*np.exp(-P*x*hbar/(mdot*Cp))
172 |
173 | def T_mo_T_infty(T_infty,T_mi,mdot,Cp,R_tot):
174 | return T_infty-(Tinfty-T_mi)*np.exp(-1/(mdot*Cp*Rtot))
175 |
176 | def L_given_other_params(T_infty,T_mo,T_mi,mdot,Cp,Rptot):
177 | return -mdot*Cp*Rptot*np.log((T_infty -T_mo)/(T_infty - T_mi))
178 |
179 |
--------------------------------------------------------------------------------
/Libraries/HT_natural_convection.py:
--------------------------------------------------------------------------------
1 | """
2 | Object name:
3 | - HorizontalCylinder
4 | - FlatPlate
5 | - VerticalEnclosure
6 | Functions: Gr(g,beta,DT,D,nu) gives the Grashoff number based on:
7 | gravity g, thermal expansion coefficient beta, Temperature difference DT,
8 | length scale D, viscosity nu
9 | Ra(g,beta,DT,D,nu,alpha) gives the Rayleigh number where alpha is the thermal conductivity.
10 | """
11 | import numpy as np
12 | import scipy
13 | import scipy.optimize
14 |
15 | class HorizontalCylinder(object):
16 | """ Natural convection about a horizontal cylinder
17 | from NewLibraries import HT_natural_convection as natconv
18 | cyl = natconv.HorizontalCylinder(correlation, Ra, Pr = 0.0)
19 | where correlation is "Morgan" or "Churchill-Chu"
20 | cyl = natconv.HorizontalCylinder("Morgan", Ra)
21 | cyl = natconv.HorizontalCylinder("Churchill-Chu", Ra, Pr = xx)
22 | """
23 |
24 | def __init__(self,correlation="Morgan", Ra=0.0, Pr = 0.0):
25 | self.correlation = correlation
26 | self.Ra = Ra
27 | if correlation == "Morgan":
28 | if (Ra <= 1e-2):
29 | C=0.675
30 | n=0.058
31 | elif (Ra <= 1e2):
32 | C=1.02
33 | n=0.148
34 | elif (Ra <= 1e4):
35 | C=0.85
36 | n=0.188
37 | elif (Ra <= 1e7):
38 | C=0.480
39 | n=0.250
40 | elif (Ra <= 1e12):
41 | C=0.125
42 | n=0.333
43 | self.Nu = C*Ra**n
44 | elif correlation == "Churchill-Chu":
45 | if Pr == 0.:
46 | print("Warning you must specify Pr for Churchill and Chu correlation")
47 | else:
48 | self.Nu = (0.60+(0.387*Ra**(1./6.))/(1.+(0.559/Pr)**(9./16.))**(8./27.))**2
49 | else:
50 | print("Warning wrong correlation name")
51 |
52 | class VerticalEnclosure(object):
53 | """ Natural convection about a horizontal cylinder
54 | from NewLibraries import HT_natural_convection as natconv
55 | cyl = natconv.HorizontalCylinder(correlation, Ra, Pr = 0.0)
56 | where correlation is "Morgan" or "Churchill-Chu"
57 | cyl = natconv.HorizontalCylinder("Morgan", Ra)
58 | cyl = natconv.HorizontalCylinder("Churchill-Chu", Ra, Pr = xx)
59 | """
60 |
61 | def __init__(self,Ra,Pr,H,L):
62 | self.Ra = Ra
63 | self.Pr = Pr
64 | self.H = H
65 | self.L = L
66 | if correlation == "Morgan":
67 | if (H/L) < 2.:
68 | if Ra*Pr/(0.2+Pr)> 1.e3:
69 | self.Nu = 0.18*(Pr/(0.2+Pr)*Ra)**0.29
70 | else:
71 | print('Ra is too low for this correlation')
72 | self.Nu = np.inf
73 | elif H/L < 10:
74 | if Ra < 1e10:
75 | self.Nu = 0.22*(Pr/(0.2+Pr)*Ra)**0.28*(H/L)**(-0.25)
76 | else:
77 | print('Ra is too high for this correlation')
78 | self.Nu = np.inf
79 | elif Ra < 1e4:
80 | print('Ra is too low for this correlation')
81 | self.Nu = np.inf
82 | elif Ra < 1e7:
83 | if Pr > 0.6 and Pr < 2e4:
84 | print('ok')
85 | self.Nu =0.42*Ra**0.25*Pr**0.012*(H/L)**(-0.3)
86 | else :
87 | print('Pr is out of bounds for this correlation')
88 | self.Nu = np.inf
89 | elif Ra < 1e9:
90 | if Pr > 0.6 and Pr < 20.:
91 | self.Nu =0.46*Ra**(1./3.)
92 | else :
93 | print('Pr is out of bounds for this correlation')
94 | self.Nu = np.inf
95 | else:
96 | print('Ra is too high, got nothing for you')
97 | self.Nu = np.inf
98 | class FlatPlate(object):
99 | """ Natural convection caused by a flat plate at temperature T_s
100 | in a fluid at given ambient temperature T_infty.
101 | Inputs:
102 | - Ra based on the |T_s - T_infty| and L=A_s/P_s where
103 | A_s and P_s are the area and the perimeter of the plate,
104 | respectively.
105 | - Pr
106 | - surface is either 'upper' or 'lower'
107 | - surfaceT is either 'cold' or 'hot'
108 | Output:
109 | - Average Nu_L
110 | Limits:
111 | from NewLibraries import HT_natural_convection as natconv
112 | plate = natconv.FlatPlate(Ra= Ra_L, Pr = Pr,
113 | surface = 'upper', surfaceT = 'hot')
114 | plate = natconv.FlatPlate(Ra= Ra_L, Pr = Pr,
115 | surface = 'lower', surfaceT = 'cold')
116 | - Ra > 1e4
117 | - If 1e4 <= Ra <= 1e7 then Pr>= 0.7
118 | - If 1e7 <= Ra <= 1e11 then all Pr
119 | plate = natconv.FlatPlate(Ra= Ra_L, Pr = Pr,
120 | surface = 'lower', surfaceT = 'hot')
121 | plate = natconv.FlatPlate(Ra= Ra_L, Pr = Pr,
122 | surface = 'upper', surfaceT = 'cold')
123 | - 1e4<= Ra <= 1e9, Pr>=0.7
124 |
125 | """
126 |
127 | def __init__(self,Ra,Pr,surface='none',surfaceT='none'):
128 | self.Ra = Ra
129 | self.Pr = Pr
130 | self.surface = surface
131 | self.surfaceT = surfaceT
132 | # print(self.surface,self.surfaceT)
133 | if (self.surface != 'upper') and (self.surface != 'lower'):
134 | print(self.surface)
135 | print("you must specify surface='upper' or 'lower'")
136 | if (self.surfaceT != 'hot') and (self.surfaceT != 'cold'):
137 | print("you must specify surfaceT='hot' or 'cold'")
138 | if ((self.surface == 'upper') and (self.surfaceT == 'hot')) or \
139 | ((self.surface == 'lower') and (self.surfaceT == 'cold')):
140 | if (self.Ra >= 1e4) and (self.Ra <= 1e7):
141 | if self.Pr < 0.7:
142 | print("Warning: For %s surface of %s plate, \
143 | the correlation is only valid for Pr>= 0.7 " %(self.surface,self.surfaceT))
144 | self.Nu = 0.54*self.Ra**(1/4)
145 | elif (self.Ra >= 1e7) : #and (self.Ra <= 1e11):
146 | self.Nu = 0.15*min(1e11,self.Ra)**(1/3)
147 | else:
148 | # print("Ra is too small")
149 | self.Nu = 0
150 | if ((self.surface == 'lower') and (self.surfaceT == 'hot')) or \
151 | ((self.surface == 'upper') and (self.surfaceT == 'cold')):
152 | if (self.Ra >= 1e4) : # and (self.Ra <= 1e9):
153 | self.Nu = 0.15*min(1e9,self.Ra)**(1/5)
154 | if self.Pr < 0.7:
155 | print("Warning: the correlation is only valid for Pr>= 0.7")
156 | else:
157 | # print("Ra is too small")
158 | self.Nu=0
159 |
160 |
161 | def Gr(g=9.81,beta=0.0,DT=0.0,D=0.0,nu=1.0):
162 | return (g*beta*DT*D**3)/(nu**2)
163 |
164 | def Ra(g=9.81,beta=0.0,DT=0.0,D=0.0,nu=1.0,alpha=1.0):
165 | return (g*beta*DT*D**3)/(nu*alpha)
166 |
167 |
168 |
169 |
--------------------------------------------------------------------------------
/.ipynb_checkpoints/ME144-HW1-checkpoint.ipynb:
--------------------------------------------------------------------------------
1 | {
2 | "cells": [
3 | {
4 | "cell_type": "raw",
5 | "metadata": {},
6 | "source": [
7 | "Text provided under a Creative Commons Attribution license, CC-BY. All code is made available under the FSF-approved MIT license. (c) Yves Dubief, 2016. NSF for support via NSF-CBET award #1258697."
8 | ]
9 | },
10 | {
11 | "cell_type": "markdown",
12 | "metadata": {},
13 | "source": [
14 | "The following cell should always be the first coding cell of your python notebooks"
15 | ]
16 | },
17 | {
18 | "cell_type": "code",
19 | "execution_count": 1,
20 | "metadata": {
21 | "collapsed": false
22 | },
23 | "outputs": [
24 | {
25 | "name": "stdout",
26 | "output_type": "stream",
27 | "text": [
28 | "Please enter your NETID (e.g. ydubief)ydubief\n",
29 | "ydubief\n"
30 | ]
31 | }
32 | ],
33 | "source": [
34 | "\n",
35 | "student_id = raw_input('Please enter your NETID (e.g. ydubief)')\n",
36 | "print(student_id)\n",
37 | "assignment_name = 'HW1_'+student_id\n"
38 | ]
39 | },
40 | {
41 | "cell_type": "code",
42 | "execution_count": 1,
43 | "metadata": {
44 | "collapsed": true
45 | },
46 | "outputs": [],
47 | "source": [
48 | "\"\"\"\n",
49 | "importing the necessary libraries, do not modify\n",
50 | "\"\"\"\n",
51 | "%matplotlib inline \n",
52 | "# plots graphs within the notebook\n",
53 | "%config InlineBackend.figure_format='svg' # not sure what this does, may be default images to svg format\n",
54 | "\n",
55 | "from IPython.display import display,Image, Latex\n",
56 | "from __future__ import division\n",
57 | "from sympy.interactive import printing\n",
58 | "printing.init_printing(use_latex='mathjax')\n",
59 | "\n",
60 | "\n",
61 | "from IPython.display import display,Image, Latex\n",
62 | "\n",
63 | "from IPython.display import clear_output\n",
64 | "\n",
65 | "import SchemDraw as schem\n",
66 | "import SchemDraw.elements as e\n",
67 | "\n",
68 | "import matplotlib.pyplot as plt\n",
69 | "import numpy as np\n",
70 | "import math\n",
71 | "import scipy.constants as sc\n",
72 | "\n",
73 | "import sympy as sym\n",
74 | "\n",
75 | "from IPython.core.display import HTML\n",
76 | "def header(text):\n",
77 | " raw_html = '
\n",
139 | "\n",
140 | "The rear window of an automobile is defogged by attaching a thin, transparent, film-type heating element to its inner surface. By electrically heating this element, a uniform heat flux may be established at the inner surface. \n",
141 | "\n",
142 | "(a)\tFor 4-mm-thick window glass, determine the electrical power required per unit window area to maintain an inner surface temperature of $15^\\circ C$ when the interior air temperature and convection coefficient are $T_{\\infty.i}= 25^\\circ C$ and $h_i=10 W/m^2.K$, while the exterior (ambient) air temperature and convection coefficient are $T_{\\infty.o}=-10^\\circ C$ and $h_o=65 W/m^2.K$.\n",
143 | "\n",
144 | "(b) In practice $T_{\\infty.o}$ and $h_o$ vary according to weather conditions and car speed. For values of $h_o=2,20,65,100 W/m^2.K$, determine and plot the electrical power requirement as a function of $T_{\\infty.o}$ for $-30\\leq T_{\\infty.o}\\leq 0^\\circ C$. From your results, what can you conclude about the need for heater operation at low values of ho? How is this conclusion affected by the value of $T_{\\infty.o}$? If h V n, where V is the vehicle speed and n is a positive exponent, how does the vehicle speed affect the need for heater operation?\n",
145 | "\n",
146 | "The thermal conductivity of this glass is $1.4 W/m.K$\n"
147 | ]
148 | },
149 | {
150 | "cell_type": "markdown",
151 | "metadata": {},
152 | "source": [
153 | "## Assumptions"
154 | ]
155 | },
156 | {
157 | "cell_type": "markdown",
158 | "metadata": {},
159 | "source": [
160 | "Steady state, 1D conduction, thermal resistance of the heating element is negligible. Negligible heat transfer by radiation."
161 | ]
162 | },
163 | {
164 | "cell_type": "markdown",
165 | "metadata": {},
166 | "source": [
167 | "## Parameters"
168 | ]
169 | },
170 | {
171 | "cell_type": "code",
172 | "execution_count": 2,
173 | "metadata": {
174 | "collapsed": false
175 | },
176 | "outputs": [],
177 | "source": [
178 | "L =0.004 #m\n",
179 | "\n",
180 | "k_glass = 1.4 #W/m.K thermal conductivity of glass\n",
181 | "\n",
182 | "T_inf_in = 25 #C\n",
183 | "T_inf_out = -10 #C\n",
184 | "h_in = 65.\n",
185 | "h_out = 65.\n",
186 | "T_s_i = 15 #C"
187 | ]
188 | },
189 | {
190 | "cell_type": "code",
191 | "execution_count": 23,
192 | "metadata": {
193 | "collapsed": false
194 | },
195 | "outputs": [
196 | {
197 | "name": "stdout",
198 | "output_type": "stream",
199 | "text": [
200 | "[NbConvertApp] Converting notebook Problem-Template0.ipynb to html\n",
201 | "[NbConvertApp] Writing 319091 bytes to ydubief.html\n"
202 | ]
203 | }
204 | ],
205 | "source": [
206 | "!ipython nbconvert --to html ME144-HW1.ipynb --output $assignment_name"
207 | ]
208 | },
209 | {
210 | "cell_type": "code",
211 | "execution_count": null,
212 | "metadata": {
213 | "collapsed": true
214 | },
215 | "outputs": [],
216 | "source": []
217 | }
218 | ],
219 | "metadata": {
220 | "kernelspec": {
221 | "display_name": "Python 2",
222 | "language": "python",
223 | "name": "python2"
224 | },
225 | "language_info": {
226 | "codemirror_mode": {
227 | "name": "ipython",
228 | "version": 2
229 | },
230 | "file_extension": ".py",
231 | "mimetype": "text/x-python",
232 | "name": "python",
233 | "nbconvert_exporter": "python",
234 | "pygments_lexer": "ipython2",
235 | "version": "2.7.11"
236 | }
237 | },
238 | "nbformat": 4,
239 | "nbformat_minor": 0
240 | }
241 |
--------------------------------------------------------------------------------
/.ipynb_checkpoints/HT-banks_of_tubes-checkpoint.ipynb:
--------------------------------------------------------------------------------
1 | {
2 | "cells": [
3 | {
4 | "cell_type": "code",
5 | "execution_count": 1,
6 | "metadata": {},
7 | "outputs": [],
8 | "source": [
9 | "%matplotlib inline \n"
10 | ]
11 | },
12 | {
13 | "cell_type": "markdown",
14 | "metadata": {},
15 | "source": [
16 | "This notebook deals with banks of cylinders in a cross flow. Cylinder banks are common heat exchangers where the cylinders may be heated by electricity or a fluid may be flowing within the cylinder to cool or heat the flow around the cylinders. The advantage of cylinder banks is the increase mixing in the fluid, thus the temperature downstream of the bank is likely to be quite homogeneous.\n",
17 | "\n",
18 | ""
19 | ]
20 | },
21 | {
22 | "cell_type": "markdown",
23 | "metadata": {},
24 | "source": [
25 | "The arrangement of cylinders may be aligned or staggered as shown in the following figures. The flow and geometrical parameters will be used in the derivation of temperature equations and Nusselt number correlation.\n",
26 | ""
27 | ]
28 | },
29 | {
30 | "cell_type": "markdown",
31 | "metadata": {},
32 | "source": [
33 | "This notebook should cover a wide variety of problems, providing that the assumption of isothermal boundary conditions on the tubes is (approximately) valid. The tube surface temperature is $T_s$. \n",
34 | "The flow and geometrical parameters of importance to solve this problem are:\n",
35 | "\n",
36 | "* Arithmetic mean of temperature between inlet $T_i$ and outlet $T_o$ of the bank. \n",
37 | "$$\n",
38 | "T_m = \\frac{T_i+T_o}{2}\n",
39 | "$$\n",
40 | "* Reynolds number based on the max velocity within the bank $V_\\text{max}$, the density and viscosity based on $T_m$:\n",
41 | "$$\n",
42 | "Re=\\frac{\\rho V_\\text{max}D}{\\mu}\n",
43 | "$$\n",
44 | "**Question: At what temperature should you estimate $\\rho$ and $\\mu$?** The energy of the flow comes from the inlet and the velocity $V_\\mathrm{max}$ is calculated from the inlet velocity. The density should therefore be estimated at $T_i$. The viscous forces however occur throughout the domain, so $\\mu$ should be estimated at $T_m$. In some cases $T_o$ is the quantity to be found. It is acceptable to use $\\mu(T_i)$, but you must verify that the temperature difference $\\Delta T=\\vert T_i-T_o\\vert$ is not too large. If it is, you must repeat the calculation iteratively with $\\mu(T_i)$ until $T_o$ converges.\n",
45 | "\n",
46 | "* Prandtl number $Pr$ based on $T_m$ \n",
47 | "* Surface Prandtl number $Pr_s$ based on $T_s$\n",
48 | "* Number of tubes in the transversal direction $N_T$, longitudinal direction $N_L$ and total $N=N_T\\times N_L$\n",
49 | "* The transversal $S_T$ and longitudinal $S_L$ separations between tubes in a row and between rows.\n",
50 | "* The type of tube arrangement: \n",
51 | " * Aligned\n",
52 | "$$\n",
53 | "V_\\text{max}=\\frac{S_T}{S_T-D}V_i\n",
54 | "$$\n",
55 | " * Staggered\n",
56 | "$$\n",
57 | "V_\\text{max}=\\frac{S_T}{2(S_D-D)}V_i\\text{ with }S_D=\\sqrt{S_L^2+\\left(\\frac{S_T}{2}\\right)^2}\n",
58 | "$$\n"
59 | ]
60 | },
61 | {
62 | "cell_type": "markdown",
63 | "metadata": {},
64 | "source": [
65 | "The Nusselt number correlation for a bank of tubes is a variation of the Zukauskas correlation:\n",
66 | "$$\n",
67 | "Nu = C_2C_1Re^mPr^{0.36}\\left(\\frac{Pr}{Pr_s}\\right)^{1/4}\n",
68 | "$$\n",
69 | "where $C_2$ depends on $N_L$. In the library, the function for this correlation is\n",
70 | "Nu_tube_banks(Re,Pr,Pr_s,S_L,S_T,N_L,arrangement) .\n",
71 | "\n",
72 | "The heat rate per unit length across the tube bank is\n",
73 | "$$\n",
74 | "q'=N\\overline{h}\\pi D \\Delta T_\\text{lm}\n",
75 | "$$\n",
76 | "where the temperature drop is the log-mean temperature difference\n",
77 | "$$\n",
78 | "\\Delta T_\\text{lm}=\\cfrac{(T_s-T_i)-(T_s-T_o)}{\\ln\\left(\\cfrac{T_s-T_i}{T_s-T_o}\\right)}\n",
79 | "$$\n",
80 | "which accounts for the exponential variation of temperature across the bank\n",
81 | "$$\n",
82 | "\\cfrac{T_s-T_o}{T_s-T_i}=\\exp\\left(-\\cfrac{\\pi D N \\overline{h}}{\\rho V_i N_T S_T C_p}\\right)\n",
83 | "$$\n",
84 | "where $\\rho$, $C_p$ and $V_i$ are inlet quantities if $T_o$ is unknown of the arthimetic mean temperature if available. Note that $N=N_L\\times N_T$ thus \n",
85 | "$$\n",
86 | "\\cfrac{T_s-T_o}{T_s-T_i}=\\exp\\left(-\\cfrac{\\pi D N_L \\overline{h}}{\\rho V_i S_T C_p}\\right)\n",
87 | "$$\n",
88 | "One may want to determine the number of tubes necessary to achieve a given $T_o$. The number of tubes in the transverse directions is typically dictated by the geometry of the system, so we are looking for $N_L$:\n",
89 | "$$\n",
90 | "N_L = \\cfrac{\\rho V_i S_T C_p}{\\pi D \\overline{h}} \\log\\left(\\cfrac{T_s-T_i}{T_s-T_o}\\right)\n",
91 | "$$\n"
92 | ]
93 | },
94 | {
95 | "cell_type": "markdown",
96 | "metadata": {},
97 | "source": [
98 | "The pressure loss through the tube bank is a critical component of the heat exchanger design. The presence of obstacles in the flow requires an increase in the mechanical energy necessary to drive the flow at a given flow rate. The pressure loss, given all parameters above, is\n",
99 | "$$\n",
100 | "\\Delta p = N_L\\,\\chi\\, f\\,\\frac{\\rho V_\\text{max}^2}{2}\n",
101 | "$$\n",
102 | "where the friction factor $f$ and the parameter $\\chi$ are given by the graphs below for the aligned (top) and staggered (bottom) arrangements. These graphs use two new quantities, the longitudnal and transverse pitches:\n",
103 | "$$\n",
104 | "P_L=\\frac{S_L}{D}\\text{ and } P_T=\\frac{S_T}{D}\n",
105 | "$$\n",
106 | "\n",
107 | ""
108 | ]
109 | },
110 | {
111 | "cell_type": "markdown",
112 | "metadata": {},
113 | "source": [
114 | "## Problem1\n",
115 | "A preheater involves the use of condensing steam on the inside of a bank of tubes to heat air that enters at $P_i=1 \\text{ atm}$ and $T_i=25^\\circ\\text{C}$. The air moves at $V_i=5\\text{ m/s}$ in cross flow over the tubes. Each tube is $L=1\\text{ m}$ long and has an outside diameter of $D=10 \\text{ mm}$. The bank consists of columns of 14 tubes in the transversal direction $N_T=14$ and $N_L$ rows in the direction of flow. The arrangement of tubes is aligned array for which $S_T=S_L=15\\text{ mm}$. What is the minimum value of $N_L$ needed to achieve an outlet temperature of $T_o=75^\\circ\\text{C}$? What is the corresponding pressure drop across the tube bank?\n"
116 | ]
117 | },
118 | {
119 | "cell_type": "code",
120 | "execution_count": 8,
121 | "metadata": {},
122 | "outputs": [],
123 | "source": [
124 | "import numpy as np\n",
125 | "from Libraries import thermodynamics as thermo\n",
126 | "from Libraries import HT_external_convection as extconv\n",
127 | "\n",
128 | "T_i = 25 #C\n",
129 | "T_o = 75 #C\n",
130 | "T_s = 100 #C\n",
131 | "V_i = 5 #m/s\n",
132 | "L = 1 #m\n",
133 | "D = 10e-3 #mm\n",
134 | "N_L = 14\n",
135 | "S_T = S_L = 15e-3 #m\n",
136 | "\n",
137 | "\n"
138 | ]
139 | },
140 | {
141 | "cell_type": "markdown",
142 | "metadata": {},
143 | "source": [
144 | "## Problem 2\n",
145 | "\n",
146 | "A preheater involves the use of condensing steam at $100^\\circ\\text{C}$ on the inside of a bank of tubes to heat air that enters at $1 \\text{ atm}$ and $25^\\circ\\text{C}$. The air moves at $5\\text{ m/s}$ in cross flow over the tubes. Each tube is $1\\text{ m}$ long and has an outside diameter of $10 \\text{ mm}$. The bank consists of 196 tubes in a square, aligned array for which $S_T=S_L=15\\text{ mm}$. What is the total rate of heat transfer to the air? What is the pressure drop associated with the airflow?"
147 | ]
148 | },
149 | {
150 | "cell_type": "markdown",
151 | "metadata": {},
152 | "source": [
153 | "## Problem 3"
154 | ]
155 | },
156 | {
157 | "cell_type": "markdown",
158 | "metadata": {},
159 | "source": [
160 | "\n",
161 | "An air duct heater consists of an aligned array of electrical heating elements in which the longitudinal and transverse pitches are $S_L=S_T= 24\\text{ mm}$. There are 3 rows of elements in the flow direction ($N_L=3$) and 4 elements per row ($N_T=4$). Atmospheric air with an upstream velocity of $12\\text{ m/s}$ and a temperature of $25^\\circ\\text{C}$ moves in cross flow over the elements, which have a diameter of $12\\text{ mm}$, a length of $250\\text{ mm}$, and are maintained at a surface temperature of $350^\\circ\\text{C}$.\n",
162 | "\n",
163 | "
\n",
164 | "Determine the total heat transfer to the air and the temperature of the air leaving the duct heater.\n",
165 | "
\n",
166 | "
\n",
167 | "Determine the pressure drop across the element bank and the fan power requirement.\n",
168 | "
\n",
169 | "
\n",
170 | "Compare the average convection coefficient obtained in your analysis with the value for an isolated (single) element. Explain the difference between the results.\n",
171 | "
\n",
172 | "
\n",
173 | "What effect would increasing the longitudinal and transverse pitches to 30 mm have on the exit temperature of the air, the total heat rate, and the pressure drop?\n",
174 | "
\n",
175 | ""
176 | ]
177 | },
178 | {
179 | "cell_type": "code",
180 | "execution_count": null,
181 | "metadata": {},
182 | "outputs": [],
183 | "source": [
184 | "\n",
185 | "\n"
186 | ]
187 | },
188 | {
189 | "cell_type": "code",
190 | "execution_count": null,
191 | "metadata": {},
192 | "outputs": [],
193 | "source": []
194 | }
195 | ],
196 | "metadata": {
197 | "kernelspec": {
198 | "display_name": "Python 3",
199 | "language": "python",
200 | "name": "python3"
201 | },
202 | "language_info": {
203 | "codemirror_mode": {
204 | "name": "ipython",
205 | "version": 3
206 | },
207 | "file_extension": ".py",
208 | "mimetype": "text/x-python",
209 | "name": "python",
210 | "nbconvert_exporter": "python",
211 | "pygments_lexer": "ipython3",
212 | "version": "3.8.5"
213 | }
214 | },
215 | "nbformat": 4,
216 | "nbformat_minor": 2
217 | }
218 |
--------------------------------------------------------------------------------
/.ipynb_checkpoints/HTpy-HW2-checkpoint.ipynb:
--------------------------------------------------------------------------------
1 | {
2 | "cells": [
3 | {
4 | "cell_type": "markdown",
5 | "metadata": {},
6 | "source": [
7 | "# HW2\n",
8 | "\n",
9 | "\n",
10 | "Before submitting your **HTpy-HW2.ipynb** file, perform the following\n",
11 | "* `Kernel>Restart & Run All`\n",
12 | "* Make sure that there are no errors\n",
13 | "\n",
14 | "The following includes problems that need coding and others that only need to perform simple operations (python as a calculator)."
15 | ]
16 | },
17 | {
18 | "cell_type": "code",
19 | "execution_count": 1,
20 | "metadata": {},
21 | "outputs": [],
22 | "source": [
23 | "from Libraries import thermodynamics as thermo\n",
24 | "import numpy as np\n",
25 | "import matplotlib.pyplot as plt\n",
26 | "import scipy.constants as csts"
27 | ]
28 | },
29 | {
30 | "cell_type": "markdown",
31 | "metadata": {},
32 | "source": [
33 | "## 1 Convection heat transfer coefficient\n",
34 | "A plate of metal of mass $m=4.25 kg$, specific heat $C_p= 2770 J/kg.K$ and surface area $0.4m\\times0.4m$ is cooled in ambient air at temperature $T_\\infty=25^\\circ C$. The plate is thin enough and its thermal conductivity high enough that the plate is assumed to be isothermal at any time. At a given time, the plate temperature is $T_s = 245^\\circ C$ and the rate of temperature cooling is\n",
35 | "$$\n",
36 | "\\frac{dT}{dt}=-0.028 K/s\n",
37 | "$$\n",
38 | "\n",
39 | "Find the convection heat transfer coefficient from conservation of energy applied to the plate."
40 | ]
41 | },
42 | {
43 | "cell_type": "markdown",
44 | "metadata": {},
45 | "source": [
46 | "### Solution\n",
47 | "Assumptions:\n",
48 | "* Plate is isothermal\n",
49 | "* Only heat transfer is convection heat transfer\n",
50 | "* Conservation of energy is applied to the plate\n",
51 | "* The plate is thin therefore the area exposed to convection is $A=2\\times0.4^2m^2$\n",
52 | "\n",
53 | "Conservation of energy applied to the metal plates writes:\n",
54 | "$$\n",
55 | "\\frac{d E_{st}}{dt}=\\dot{E}_{in} - \\dot{E}_{out} + \\dot{E}_g\n",
56 | "$$\n",
57 | "\n",
58 | "reduces to\n",
59 | "\n",
60 | "$$\n",
61 | "\\frac{d E_{st}}{dt}=- \\dot{E}_{out}\n",
62 | "$$\n",
63 | "\n",
64 | "where $\\dot{E}_{out}$ is the thermal energy lost by convection to the ambient air:\n",
65 | "$$\n",
66 | "\\dot{E}_{out}=q_{conv}=hA(T_s-T_\\infty).\n",
67 | "$$\n",
68 | "\n",
69 | "Since $T_s=T$, the conservation of energy becomes:\n",
70 | "$$\n",
71 | "\\frac{d}{dt}\\int_V\\rho C_pT dV=-hA(T-T_\\infty)\n",
72 | "$$\n",
73 | "\n",
74 | "$$\n",
75 | "mC_p\\frac{dT}{dt}=-hA(T-T_\\infty)\n",
76 | "$$\n",
77 | "\n",
78 | "Therefore,\n",
79 | "\n",
80 | "$$\n",
81 | "\\frac{dT}{dt}=-\\frac{hA}{mC_p}(T-T_\\infty)\n",
82 | "$$\n",
83 | "\n",
84 | "or\n",
85 | "\n",
86 | "$$\n",
87 | "h=-\\frac{mC_p}{A(T-T_\\infty)}\\frac{dT}{dt}\n",
88 | "$$\n",
89 | "\n"
90 | ]
91 | },
92 | {
93 | "cell_type": "code",
94 | "execution_count": 2,
95 | "metadata": {},
96 | "outputs": [
97 | {
98 | "name": "stdout",
99 | "output_type": "stream",
100 | "text": [
101 | "The convection heat transfer for the given conditions is 4.7 W/m^2.K\n"
102 | ]
103 | }
104 | ],
105 | "source": [
106 | "Cp = 2770. #J/kg.K\n",
107 | "m = 4.25 #kg\n",
108 | "dTdt = -0.028 #K/s\n",
109 | "T_infty = 25. #C\n",
110 | "T_s = 245. #C\n",
111 | "A = 2*0.4**2\n",
112 | "\n",
113 | "h = -m*Cp/(A*(T_s - T_infty))*dTdt\n",
114 | "print (\"The convection heat transfer for the given conditions is %.1f W/m^2.K\" %h)"
115 | ]
116 | },
117 | {
118 | "cell_type": "markdown",
119 | "metadata": {},
120 | "source": [
121 | "## 2 Radiation in a vacuum \n",
122 | "\n",
123 | "The same plate as above is now in a vacuum (no flow). Using all conditions defined in 1, determine the emissivity of the plate and the heat rate at which radiation is emitted from the plate. Assume that $T_{sur}=T_\\infty$. "
124 | ]
125 | },
126 | {
127 | "cell_type": "markdown",
128 | "metadata": {},
129 | "source": [
130 | "### Solution\n",
131 | "Assumptions:\n",
132 | "* Same as above, except no convection\n",
133 | "\n",
134 | "Using the same approache as above, with $\\dot{E}_{out}=q_{rad}=\\sigma\\varepsilon A(T_s^4-T_\\infty^4)$, the emissivity is\n",
135 | "$$\n",
136 | "m C_p\\frac{dT}{dt}=\\sigma\\varepsilon A(T_s^4-T_\\infty^4)\n",
137 | "$$\n",
138 | "$$\n",
139 | "\\varepsilon=-\\frac{mC_p}{\\sigma A(T_s^4-T_\\infty^4)}\\frac{dT}{dt}\n",
140 | "$$\n",
141 | "\n",
142 | "Emissive power of the plate is \n",
143 | "\n",
144 | "$$\n",
145 | "q_{rad}=\\varepsilon \\sigma A T_s^4\n",
146 | "$$"
147 | ]
148 | },
149 | {
150 | "cell_type": "code",
151 | "execution_count": 3,
152 | "metadata": {},
153 | "outputs": [
154 | {
155 | "name": "stdout",
156 | "output_type": "stream",
157 | "text": [
158 | "The Stefan-Boltzmann constant is available from scipy 5.67e-08 W/m^2.K^4\n"
159 | ]
160 | }
161 | ],
162 | "source": [
163 | "print('The Stefan-Boltzmann constant is available from scipy %.2e W/m^2.K^4' %csts.sigma)"
164 | ]
165 | },
166 | {
167 | "cell_type": "code",
168 | "execution_count": 5,
169 | "metadata": {},
170 | "outputs": [
171 | {
172 | "name": "stdout",
173 | "output_type": "stream",
174 | "text": [
175 | "The emissivity of the plate is 0.28\n",
176 | "The heat rate of radiation emitted by the plate is 370 W\n"
177 | ]
178 | }
179 | ],
180 | "source": [
181 | "Cp = 2770. #J/kg.K\n",
182 | "m = 4.25 #kg\n",
183 | "dTdt = -0.028 #K/s\n",
184 | "T_infty = 25. #C\n",
185 | "T_s = 245. #C\n",
186 | "A = 2*0.4**2\n",
187 | "eps = -m*Cp/(csts.sigma*A*(thermo.C2K(T_s)**4 - thermo.C2K(T_infty)**4))*dTdt\n",
188 | "print('The emissivity of the plate is %.2f' %eps)\n",
189 | "q_rad = eps*csts.sigma*A*thermo.C2K(T_s)**4\n",
190 | "print('The heat rate of radiation emitted by the plate is %.0f W' %q_rad)"
191 | ]
192 | },
193 | {
194 | "cell_type": "markdown",
195 | "metadata": {},
196 | "source": [
197 | "## 3 Surface temperature of steam line\n",
198 | "\n",
199 | "A steam line of diameter $D=0.1m$ and length $L=25m$ loses $18.405 kW$ to the ambient air. The air is at $T_\\infty=T_{sur}=25^\\circ C$, the convection coefficient of the air flow is $h=10W/m^2.K$ and the emissivity of the pipe is $\\varepsilon=0.8$.\n",
200 | "\n",
201 | "Find the surface temperature of the steam line, the annual energy loss and the annual cost assuming that the steam is generated by a furnace with an efficiency $\\eta_f = 0.9$ running at a cost of natural gas of $0.02\\$/MJ$.\n",
202 | "\n",
203 | "For the surface temperature, use `scipy.optimize.fsolve`"
204 | ]
205 | },
206 | {
207 | "cell_type": "markdown",
208 | "metadata": {},
209 | "source": [
210 | "### Solution\n",
211 | "Assumptions:\n",
212 | "* Heat transfer is steady and 1D\n",
213 | "* Conservation of energy is applied to the surface of the pipe. There is no energy generation in that control volume. \n",
214 | "\n",
215 | "The control volume is the external surface of the pipe of surface area $A=\\pi D\\times L=2.5m^2$. The steam loses heat to the outside at a rate $q=18.405kW$. This energy is transferred to the outside through radiation and convection. The conservation of energy applied to the surface of the pipe therefore reduces to:\n",
216 | "\n",
217 | "$$\n",
218 | "\\dot{E}_{in}-\\dot{E}_{out}=0\n",
219 | "$$\n",
220 | "\n",
221 | "with $\\dot{E}_{in}=q$ and $\\dot{E}_{out}=q_{conv}+q_{rad}$. For a surface temperature $T_s$, this equation becomes\n",
222 | "\n",
223 | "$$\n",
224 | "q-hA(T_s-T_\\infty)-\\sigma\\varepsilon A(T_s^4-T_\\infty^4)=0\n",
225 | "$$\n",
226 | "\n",
227 | "that we solve for $T_s$ using the function root finder solver fsolve from scipy.optimize.\n",
228 | "\n",
229 | "The annual energy loss $E_{loss}$ is q multiplied by a year. The energy consumed by the boiler to create the total energy loss is\n",
230 | "$$\n",
231 | "E_{furnace}=\\frac{E_{loss}}{\\eta_f}\n",
232 | "$$\n",
233 | "\n",
234 | "Warning: the cost is in dollars per MEGAJoule."
235 | ]
236 | },
237 | {
238 | "cell_type": "markdown",
239 | "metadata": {},
240 | "source": [
241 | "with \n",
242 | "$$\n",
243 | "\\dot{E}_{in}=q_{loss} = 18.405kW\n",
244 | "$$\n",
245 | "\n",
246 | "$$\n",
247 | "\\dot{E}_{out}=q_{conv}+q_{rad}\n",
248 | "$$"
249 | ]
250 | },
251 | {
252 | "cell_type": "code",
253 | "execution_count": null,
254 | "metadata": {},
255 | "outputs": [],
256 | "source": [
257 | "from scipy.optimize import fsolve\n",
258 | "D = 0.1\n",
259 | "L = 25.\n",
260 | "q = 18405.\n",
261 | "h = 10.\n",
262 | "T_infty = 25.\n",
263 | "eps = 0.8\n",
264 | "eta_f = 0.9\n",
265 | "Cost = 0.02\n",
266 | "A = np.pi*D*L\n",
267 | "\n",
268 | "\n",
269 | "# print('The steam line surface temperature is %.0f C' %T_s)\n",
270 | "# print('verification:',A*h*(T_s-T_infty) + eps*sccst.sigma*A*(C2K(T_s)**4-C2K(T_infty)**4))\n",
271 | "# cost = q*(365*24*3600)/0.9*C*1e-6\n",
272 | "# print('The cost of operation for a year is $%.f' %cost)"
273 | ]
274 | },
275 | {
276 | "cell_type": "markdown",
277 | "metadata": {},
278 | "source": [
279 | "## 4 Insulation\n",
280 | "\n",
281 | "Assuming that the surface temperature is $150^\\circ C$, find an insulation that works (i.e. an insulation that will not melt, for which you can retrieve the price and thermal conductivity). Derive the necessary equations that will give optimum savings.\n",
282 | "\n",
283 | "Hint: Derive the heat loss for a variable thickness of insulation, then apply this formula for a range of thicknesses. Compare savings of gas vs cost of insulation."
284 | ]
285 | },
286 | {
287 | "cell_type": "markdown",
288 | "metadata": {},
289 | "source": [
290 | "### Solution\n",
291 | "Assumptions:\n",
292 | "
"
322 | ]
323 | },
324 | {
325 | "cell_type": "code",
326 | "execution_count": null,
327 | "metadata": {},
328 | "outputs": [],
329 | "source": []
330 | },
331 | {
332 | "cell_type": "code",
333 | "execution_count": null,
334 | "metadata": {},
335 | "outputs": [],
336 | "source": []
337 | }
338 | ],
339 | "metadata": {
340 | "kernelspec": {
341 | "display_name": "Python 3",
342 | "language": "python",
343 | "name": "python3"
344 | },
345 | "language_info": {
346 | "codemirror_mode": {
347 | "name": "ipython",
348 | "version": 3
349 | },
350 | "file_extension": ".py",
351 | "mimetype": "text/x-python",
352 | "name": "python",
353 | "nbconvert_exporter": "python",
354 | "pygments_lexer": "ipython3",
355 | "version": "3.8.5"
356 | }
357 | },
358 | "nbformat": 4,
359 | "nbformat_minor": 4
360 | }
361 |
--------------------------------------------------------------------------------
/.ipynb_checkpoints/HW-01-checkpoint.ipynb:
--------------------------------------------------------------------------------
1 | {
2 | "cells": [
3 | {
4 | "cell_type": "markdown",
5 | "id": "0cc46860",
6 | "metadata": {},
7 | "source": [
8 | "# Homework Assignment 2\n",
9 | "\n",
10 | "Complete this notebook, export to pdf (`File>Download as> PDF (via LaTeX)`) and submit the pdf to BB"
11 | ]
12 | },
13 | {
14 | "cell_type": "markdown",
15 | "id": "85aecba7",
16 | "metadata": {},
17 | "source": [
18 | "## Pb 1: Air thermodynamic properties class\n",
19 | "\n",
20 | "The concept of class is central to most libraries. It consists of the definition of an object, which has specific properties, data and function.\n",
21 | "\n",
22 | "Zografos et al. (*Compt. Methods Appl. Mech Eng.*, 61:177-187, 1987) proposed polynomials to compute density, dynamic viscosity, thermal conductivity, and specific heat at constant temperature of air at atmospheric pressure as a function of the static temperature $T$ on the Kelvin scale as follows:\n",
23 | "* Density $\\rho$ in units of $\\mathrm{kg/m^3}$:\n",
24 | "\n",
25 | "$$\n",
26 | "\\rho=\\begin{cases}\n",
27 | "-2.440\\times10^{-2}T+5.9958\\,, & 100\\,\\mathrm{K}\\leq T<150\\,\\mathrm{K}\\\\\n",
28 | "345.57(T-2.6884)^{-1}\\,, & 150\\,\\mathrm{K}\\leq T\\leq 3000\\,\\mathrm{K}\n",
29 | "\\end{cases}\n",
30 | "$$\n",
31 | "* Dynamic viscosity $\\mu$ in units of $\\mathrm{N.s/m^2}=\\mathrm{kg/(m.s)}=\\mathrm{Pa.s}$:\n",
32 | "\n",
33 | "$$\n",
34 | "\\mu = 4.1130\\times10^{-6} + 5.0523\\times10^{-8}T - 1.4346\\times10^{-11}T^2+2.5914\\times10^{-15}T^3\n",
35 | "$$\n",
36 | "\n",
37 | "* Thermal conductivity $k$ in units of $\\mathrm{W/(m.K)}$:\n",
38 | "\n",
39 | "$$\n",
40 | "k= -7.488\\times10^{-3} + 1.7082\\times10^{-4}T - 2.3758\\times10^{-10}T^2 + 2.2012\\times10^{-10}T^3 \\\\\n",
41 | " + 9.4600\\times10^{-14}T^4 + 1.5797\\times10^{-17}T^5\n",
42 | "$$\n",
43 | "\n",
44 | "* Specific heat at constant pressure $c_p$ in unites of $\\mathrm{J/(kg.K)}$:\n",
45 | "\n",
46 | "$$\n",
47 | "c_p=1061.332-0.432819T + 1.02344\\times10^{-3}T^2 - 6.47474\\times10^{-7}T^3 + 1.3864\\times10^{-10}T^4\n",
48 | "$$\n",
49 | "\n",
50 | "Air density $\\rho$ of units $\\mathrm{kg/m^3}$ can be computed from \n",
51 | "\n",
52 | "The Prandtl number $Pr$ can be computed as\n",
53 | "\n",
54 | "$$\n",
55 | "Pr=\\frac{\\mu c_p}{k}\n",
56 | "$$\n",
57 | "\n",
58 | "and the thermal conductivity $\\alpha$ of units $\\mathrm{m^2/s}$ as\n",
59 | "\n",
60 | "$$\n",
61 | "\\alpha = \\frac{k}{\\rho c_p}\n",
62 | "$$\n",
63 | "\n",
64 | "Write a class storing all these variables for input temperature (float) and unit used for temperature (string, it can only be \"K\", \"C\", \"F\"). The class should convert the temperature to Kelvin if unit is either \"C\" or \"F\"). The class should give a warning if \n",
65 | "* the temperature is outside of the range,\n",
66 | "* the unit is not \"K\", \"C\", \"F\", e.g. \"c\"\n",
67 | "\n",
68 | "### Tasks\n",
69 | "* Complete the manual part of the class (text between \"\"\" xxxx \"\"\"), i.e. write appropriate units. \n",
70 | "* Complete the class to compute all quantities defined in the manual.\n",
71 | "* Apply your class to the examples below.\n",
72 | "* Define units for examples.\n",
73 | "* Plot density as function of temperature in the range $10^2\\leq T<3000$. Use log-scale for temperature. Axes must be labeled with their variables and units."
74 | ]
75 | },
76 | {
77 | "cell_type": "code",
78 | "execution_count": null,
79 | "id": "97650df2",
80 | "metadata": {},
81 | "outputs": [],
82 | "source": [
83 | "import numpy as np\n",
84 | "import math\n",
85 | "import matplotlib.pyplot as plt\n",
86 | "class Air:\n",
87 | " \"\"\"\n",
88 | " Returns an object containing the thermodynamic properties of air at atmospheric pressure\n",
89 | " Input: \n",
90 | " T: temperature (float)\n",
91 | " Tunit: unit of temperature (string). Can only be \"K\", \"C\", \"F\"\n",
92 | " output:\n",
93 | " self.rho: density [kg/m^3]\n",
94 | " self.mu: dynamic viscosity [write units here]\n",
95 | " self.k: thermal conductivity [write units here]\n",
96 | " self.cp: Specific heat at constant pressure [write units here]\n",
97 | " self.Pr: Prandtl number\n",
98 | " self.nu: kinematic viscosity [write units here]\n",
99 | " self.alpha: thermal diffusivity [write units here]\n",
100 | " \"\"\"\n",
101 | " def __init__(self,T,Tunit):\n",
102 | " if Tunit == \"C\":\n",
103 | " T += 273.15\n",
104 | " elif Tunit == \"F\":\n",
105 | " T = (T - 32)*5/9 +273.15\n",
106 | " elif Tunit != \"K\":\n",
107 | " print(\"Wrong input for unit of temperature\")\n",
108 | " print(\"Input should be 'K', 'C', or 'F'\")\n",
109 | " self.T = T\n",
110 | " self.mu = 4.1130E-6 + 5.0523E-8*T - 1.4346E-11*T**2 + 2.5914E-15*T**3\n",
111 | " \n"
112 | ]
113 | },
114 | {
115 | "cell_type": "code",
116 | "execution_count": null,
117 | "id": "0b87cbeb",
118 | "metadata": {},
119 | "outputs": [],
120 | "source": [
121 | "myair = Air(-200,\"F\")\n",
122 | "print(\"Temperature in K: %.2f\" %myair.T)\n",
123 | "print(\"Kinematic viscosity: %.2e [units], Thermal conductivity: %.4f [units]\" %(myair.nu,myair.k))"
124 | ]
125 | },
126 | {
127 | "cell_type": "code",
128 | "execution_count": null,
129 | "id": "f887dc6c",
130 | "metadata": {},
131 | "outputs": [],
132 | "source": [
133 | "myair = Air(-200,\"C\")"
134 | ]
135 | },
136 | {
137 | "cell_type": "code",
138 | "execution_count": null,
139 | "id": "5fdbb9f2",
140 | "metadata": {},
141 | "outputs": [],
142 | "source": [
143 | "myair = Air(2000,\"k\")"
144 | ]
145 | },
146 | {
147 | "cell_type": "code",
148 | "execution_count": null,
149 | "id": "8a8c3729",
150 | "metadata": {},
151 | "outputs": [],
152 | "source": [
153 | "myair = Air(2000,\"K\")\n",
154 | "print(\"Dynamic viscosity: %.2e [units], Thermal diffusivity: %.2e [units]\" %(myair.mu,myair.alpha))"
155 | ]
156 | },
157 | {
158 | "cell_type": "code",
159 | "execution_count": null,
160 | "id": "618ff68d",
161 | "metadata": {},
162 | "outputs": [],
163 | "source": [
164 | "T = np.logspace(2,np.log10(2999.),1000)\n",
165 | "rho = np.zeros_like(T)\n",
166 | "for i in range(T.shape[0]):\n",
167 | " # Your code here"
168 | ]
169 | },
170 | {
171 | "cell_type": "code",
172 | "execution_count": null,
173 | "id": "d67d9ede",
174 | "metadata": {},
175 | "outputs": [],
176 | "source": [
177 | "plt.figure(figsize=(3,2),dpi=100)\n",
178 | "# your line plot here\n",
179 | "plt.xlabel(\"$T\\;(\\mathrm{units})$\")\n",
180 | "plt.ylabel(r\"$\\rho\\;(\\mathrm{units})$\")\n",
181 | "plt.show()"
182 | ]
183 | },
184 | {
185 | "cell_type": "markdown",
186 | "id": "b4054fd7",
187 | "metadata": {},
188 | "source": [
189 | "## Pb 2 Window\n",
190 | "\n",
191 | "Set up equations and apply realistic numerical values to them to discuss heat losses of a single pane window, a single pane window with storm window and a double paned window with air trapped at a vacuum of $10^{-3} \\mathrm{torr}$ ina gap of $5\\mathrm{mm}$. Do not consider the effects of radiation for any of the window.\n",
192 | "\n",
193 | "\n",
194 | "\n",
195 | "#### Assumptions:\n",
196 | "* Heat transfer is steady and 1D\n",
197 | "* Radiation is ignored\n",
198 | "* Convection in the gap of the storm window is neglected\n",
199 | "* The temperature on the inside and outside surfaces of any window are isothermal and equal to $T_{si}=10^\\circ\\mathrm{C}$ and $T_{so}=-10^\\circ\\mathrm{C}$, respectively.\n",
200 | "* For the storm window, and the double pane window, the average temperature in the gap between the two glass pane is assumed to the average of $T_{si}$ and $T_{so}$ or $T_{mean}=0^\\circ\\mathrm{C}$.\n",
201 | "\n"
202 | ]
203 | },
204 | {
205 | "cell_type": "markdown",
206 | "id": "bd483448",
207 | "metadata": {},
208 | "source": [
209 | "#### Setup\n",
210 | "\n",
211 | "In this cell you are expected to:\n",
212 | "* Insert your sketches of the single pane and double pane windows showing heat fluxes, temperatures at boundaries and dimensions\n",
213 | "* Store in folder `MyFigures`. Change the names `your-single-pane-sketch` and `your-double-pane-sketch` in the HTML command below to include your figures to this cell.\n",
214 | "* Complete a **worded** write-up of the solutions (for both windows) including equations written in LaTeX\n",
215 | "\n",
216 | "\n",
217 | "\n",
218 | "\n",
219 | "*This is the type of introduction of your work that is expected in this class for any deliverable*\n",
220 | "\n",
221 | "The goal is to compute the heat loss from a model window under controlled conditions. We choose to fix the inside surface temperature to $10^\\circ\\mathrm{C}$ and the outside surface temperature to $-10^\\circ\\mathrm{C}$ for a window of height and width $h = 1.2\\;\\mathrm{m}$ and $w = 0.7\\;\\mathrm{m}$ with a surface aera $A=h\\times w$. The thickness of a residential window pane is $t_{glass}=3/32\\,\\mathrm{in}$ (https://ringerwindows.com/thick-glass-windows/) with a thermal conductivity $k=0.96\\;\\mathrm{W/m.K}$. The storm window has a $2.5\\,\\mathrm{in}$.\n",
222 | "\n",
223 | "The heat rate for a single window pane is...\n",
224 | "*Your derivation of the solution here*\n"
225 | ]
226 | },
227 | {
228 | "cell_type": "code",
229 | "execution_count": null,
230 | "id": "3a7346e0",
231 | "metadata": {},
232 | "outputs": [],
233 | "source": [
234 | "import numpy as np\n",
235 | "from Libraries import thermodynamics as thermo\n",
236 | "T_si = 10. #C\n",
237 | "T_so = -10. #C\n",
238 | "h = 1.2 #m\n",
239 | "w = 0.7 #m\n",
240 | "t_glass = 3./32.*0.0254 #m\n",
241 | "k_glass = 0.96\n",
242 | "A = h*w\n",
243 | "t_gap_storm = 2*0.0254 #m\n",
244 | "# getting thermodynamic properties of air for the storm window\n",
245 | "air_gap_storm = thermo.Fluid('air',thermo.C2K(0.))\n",
246 | "t_gap_vacuum = 0.005#m\n",
247 | "k_vacuum = xxxx #W/m.K (from graph)\n",
248 | "R_glass = \n",
249 | "R_gap_storm = \n",
250 | "R_gap_vacuum = \n",
251 | "R_total_single = \n",
252 | "R_total_storm = \n",
253 | "R_total_double = \n",
254 | "\n",
255 | "print(\"Thermal conductivity of air at 0 C: %.4f\" %air_gap_storm.k)\n",
256 | "print(\"Thermal resistance for the single window: %.4f K/W\" %R_total_single)\n",
257 | "print(\"Thermal resistance for the storm window: %.4f K/W\" %R_total_storm)\n",
258 | "print(\"Thermal resistance for the double pane window: %.4f K/W\" %R_total_double)\n",
259 | "\n",
260 | "DT = T_si - T_so\n",
261 | "q_single = DT/R_total_single\n",
262 | "q_storm = DT/R_total_storm\n",
263 | "q_double = DT/R_total_double\n",
264 | "print(\"Single window pane q: %.0f W\" %q_single)\n",
265 | "print(\"Storm window q: %.0f W\" %q_storm)\n",
266 | "print(\"Double pane window q: %.0f W\" %q_double)"
267 | ]
268 | },
269 | {
270 | "cell_type": "code",
271 | "execution_count": null,
272 | "id": "7d75b9d4",
273 | "metadata": {},
274 | "outputs": [],
275 | "source": []
276 | }
277 | ],
278 | "metadata": {
279 | "kernelspec": {
280 | "display_name": "Python 3 (ipykernel)",
281 | "language": "python",
282 | "name": "python3"
283 | },
284 | "language_info": {
285 | "codemirror_mode": {
286 | "name": "ipython",
287 | "version": 3
288 | },
289 | "file_extension": ".py",
290 | "mimetype": "text/x-python",
291 | "name": "python",
292 | "nbconvert_exporter": "python",
293 | "pygments_lexer": "ipython3",
294 | "version": "3.9.13"
295 | }
296 | },
297 | "nbformat": 4,
298 | "nbformat_minor": 5
299 | }
300 |
--------------------------------------------------------------------------------
/ME144-HW-04-11.ipynb:
--------------------------------------------------------------------------------
1 | {
2 | "cells": [
3 | {
4 | "cell_type": "markdown",
5 | "id": "cbbb79f1",
6 | "metadata": {},
7 | "source": [
8 | "# Homework Assignment due \n",
9 | "\n",
10 | "## Problem 1\n",
11 | "A thermos bottle has an inner diameter of $7\\;\\mathrm{mm}$ is $10\\;\\mathrm{cm}$ tall. The wall is composed of a sealed air chamber of thickness $1\\;\\mathrm{mm}$ sandwiched between two thin cylinders of stainless steal, $k_{stst}=15\\;\\mathrm{W/(m.K)}$, of thickness $2\\;\\mathrm{mm}$. The goal is to hold tea as close as possible to its ideal temperature of $80\\,^\\circ\\mathrm{C}$ for an hour in some of the weather we have had recently. For an ambient air at $T_\\infty=-20\\,^\\circ\\mathrm{C}$ and a wind of $10\\;\\mathrm{m/s}$, the convection coefficient around the bottle is $h = 28.7\\times10^3\\;\\mathrm{W/(m^2.K)}$. The radiation heat transfer is assumed to be negligible. All heat transfer calculations are assumed to be 1D.\n",
12 | "\n",
13 | "* 1.) Calculate the heat rate of the system if the air chamber is at atmospheric pressure $k_{air}=0.022\\;\\mathrm{W/(m.K)}$\n",
14 | "* 2.) Typical thermos bottles are designed with the air chamber vacuumed at about $10^{-3}\\;\\mathrm{torr}$ resulting in a thermal conductivity of $3\\times10^{-3}\\;\\mathrm{W/m.K}$. Calculate the new heat rate.\n",
15 | "* 3.) Describe briefly the microscopic phenomenon that causes vacuum air to reduce so significantly the heat rate.\n",
16 | "* 4.) Assuming that there is no heat loss from the bottom or the lid, give an estimate of the temperature loss of the tea per second in the vacuumed and non-vacuumed scenario. To do this you only need to apply the guiding principle of heat transfer (see equation sheet) and remember that thermal energy is $\\rho C_p T$, where the tea density is $1000\\;\\mathrm{kg/m^3}$ and $C_p= 4198.\\;\\mathrm{J/(kg.K)}$\n",
17 | "* 5.) Compute the heat loss from a lid made of $3 \\;\\mathrm{mm}$ of plastic with a conductivity of $5\\;\\mathrm{W/(m.K)}$. What is your conclusion regarding the ability of this thermos to hold your tea warm? If you can assume that heat loss rate is constant, how long would it take for your tea to be cold (decide what is cold based on your preference)?"
18 | ]
19 | },
20 | {
21 | "cell_type": "code",
22 | "execution_count": 1,
23 | "id": "843a3ff4",
24 | "metadata": {},
25 | "outputs": [],
26 | "source": [
27 | "import numpy as np\n",
28 | "import matplotlib.pyplot as plt\n",
29 | "from Libraries import thermodynamics as thermo\n",
30 | "from Libraries import HT_thermal_resistance as res\n",
31 | "im"
32 | ]
33 | },
34 | {
35 | "cell_type": "code",
36 | "execution_count": 3,
37 | "id": "14e6fddf",
38 | "metadata": {},
39 | "outputs": [],
40 | "source": [
41 | "L = 0.1 #m Height\n",
42 | "ri = 0.07 #m Inside radius\n",
43 | "t_sts = 0.002 #m Stainless steal wall thickness\n",
44 | "k_sts = 15 #W/m.K thermal conductivity of stainless steal wall\n",
45 | "t_gap = 0.001 #m \n",
46 | "k_gap_atm = 2.2e-2 #W/m.K air chamber without vacuum\n",
47 | "k_gap_vac = 3.0e-3 #W/m.K air chamber with vacuum\n",
48 | "h = 28000 #W/m^2.K convection coefficient\n",
49 | "T_i = 80 #C\n",
50 | "T_inf = -20 #C\n",
51 | "\n",
52 | "r1 = ri + t_sts\n",
53 | "r2 = r1 + t_gap\n",
54 | "ro = r2 + t_sts\n",
55 | "\n",
56 | "A_o = 2*np.pi*ro**2*L\n",
57 | "\n",
58 | "R = []\n",
59 | "# Conduction through Inner wall\n",
60 | "R.append(res.Resistance(\"$R_{cond,w1}$\", \"W\"))\n",
61 | "# Conduction through air gap \n",
62 | "R.append(res.Resistance(\"$R_{cond,a}$\", \"W\"))\n",
63 | "# Conduction through outer wall\n",
64 | "R.append(res.Resistance(\"$R_{cond,w2}$\", \"W\"))\n",
65 | "# Conduction through convection\n",
66 | "R.append(res.Resistance(\"$R_{conv}$\", \"W\"))\n",
67 | "\n"
68 | ]
69 | },
70 | {
71 | "cell_type": "markdown",
72 | "id": "0bcaacce",
73 | "metadata": {},
74 | "source": []
75 | },
76 | {
77 | "cell_type": "code",
78 | "execution_count": 4,
79 | "id": "6c2e5bdf",
80 | "metadata": {},
81 | "outputs": [],
82 | "source": [
83 | "\n",
84 | "R[0].cond_cylinder(k_sts,ra=ri,rb=r1,L=L)\n",
85 | "R[1].cond_cylinder(k_gap_atm,ra=r1,rb=r2,L=L)\n",
86 | "R[2].cond_cylinder(k_sts,ra=r2,rb=ro,L=L)\n",
87 | "R[3].convection(h,A=A_o)\n"
88 | ]
89 | },
90 | {
91 | "cell_type": "code",
92 | "execution_count": 5,
93 | "id": "c21e9015",
94 | "metadata": {},
95 | "outputs": [],
96 | "source": [
97 | "R_tot_atm = res.serial_sum(R,0,3)"
98 | ]
99 | },
100 | {
101 | "cell_type": "code",
102 | "execution_count": 9,
103 | "id": "f999a92d",
104 | "metadata": {},
105 | "outputs": [],
106 | "source": [
107 | "R[1].cond_cylinder(k_gap_vac,ra=r1,rb=r2,L=L)\n",
108 | "R_tot_vac = res.serial_sum(R,0,3)"
109 | ]
110 | },
111 | {
112 | "cell_type": "code",
113 | "execution_count": 10,
114 | "id": "0083d356",
115 | "metadata": {},
116 | "outputs": [
117 | {
118 | "data": {
119 | "text/latex": [
120 | "$\\displaystyle 7.33366778491278$"
121 | ],
122 | "text/plain": [
123 | "7.333667784912777"
124 | ]
125 | },
126 | "execution_count": 10,
127 | "metadata": {},
128 | "output_type": "execute_result"
129 | }
130 | ],
131 | "source": [
132 | "R_tot_vac"
133 | ]
134 | },
135 | {
136 | "cell_type": "code",
137 | "execution_count": 12,
138 | "id": "05887af4",
139 | "metadata": {},
140 | "outputs": [],
141 | "source": [
142 | "q_atm = 1/R_tot_atm*(T_i - T_inf)\n",
143 | "q_vac = 1/R_tot_vac*(T_i - T_inf)"
144 | ]
145 | },
146 | {
147 | "cell_type": "code",
148 | "execution_count": 14,
149 | "id": "316786a1",
150 | "metadata": {},
151 | "outputs": [
152 | {
153 | "data": {
154 | "text/latex": [
155 | "$\\displaystyle 13.635741750632$"
156 | ],
157 | "text/plain": [
158 | "13.635741750631993"
159 | ]
160 | },
161 | "execution_count": 14,
162 | "metadata": {},
163 | "output_type": "execute_result"
164 | }
165 | ],
166 | "source": [
167 | "q_vac"
168 | ]
169 | },
170 | {
171 | "cell_type": "code",
172 | "execution_count": null,
173 | "id": "083deacc",
174 | "metadata": {},
175 | "outputs": [],
176 | "source": [
177 | "print(\"Without vacuum, the heat rate is \")"
178 | ]
179 | },
180 | {
181 | "cell_type": "markdown",
182 | "id": "d5fb74e9",
183 | "metadata": {},
184 | "source": [
185 | "## Problem 2\n",
186 | "\n",
187 | "The key to a happy skier is to maintain the surface temperature of your skin larger than $20^\\circ\\mathrm{C}$. The objective is to determine the heat flux generated by a thin electrical heater (resistance) inserted on the top of the liner, i.e. between the liner and the sock. All calculations will assume 1D planar heat transfer and to be performed per unit surface area. The study is also limited to the heat transfer from the top of the foot only. The body is assumed to be at constant temperature $36^\\circ\\mathrm{C}$, or $309^\\circ\\mathrm{K}$, which is the inner boundary condition for a skin/fat layer of thickness $3\\;\\mathrm{mm}$ and $k_{skin}=3\\times10^{-1}\\;\\mathrm{W/(m.K)}$. The skin is covered by a sock of thickness $2.5\\;\\mathrm{mm}$ and $k_{sock}=7.5\\;\\mathrm{W/(m.K)}$. The liner of the boot is made of polymer fibers, thickness $5\\;\\mathrm{mm}$ and $k_\\text{liner}=6\\;\\mathrm{W/(m.K)}$ and the shell is carbon fiber, thickness $3\\;\\mathrm{mm}$ and $k_\\text{shell}=10\\;\\mathrm{W/(m.K)}$ with emissivity $\\varepsilon=0.9$. The air is at $T_\\infty= -20^\\circ\\mathrm{C}=253^\\circ\\mathrm{C}$ and the convection coefficient is $h=50\\;\\mathrm{W/(m^2.K)}$\n",
188 | "* 1.) Calculate the heat rate without the heating resistance. For radiation you will use the radiation coefficient heat transfer $h_r=\\varepsilon\\sigma(T_s+T_\\infty)(T_s^2+T_\\infty^2)$, where $\\sigma = 5.67\\times10^{-8}\\;\\mathrm{W/(m^2.K^4)}$ and $T_s$ is the boot surface temperature (outside surface). Take a guess for $T_s$ (between $T_0$ and $T_\\infty$), calculate the total flux. From the total flux, calculate $T_s$ and iterate once more. What is the skin surface temperature?\n",
189 | "* 2.) Modify your thermal circuit to include the thin heater between the sock and the liner and write the conservation of energy in terms of heat fluxes\n",
190 | "* 3.) What heat flux must the heating resistance generate to maintain the skin temperature at $25^\\circ\\mathrm{C}$ under the present conditions?\n",
191 | "* 4.) Calculate the boot outer surface temperature $T_s$.\n",
192 | "* 5.) Without any additional calculation, what is the most preferable material for a sock: wool, $k_\\text{wool}=7.5\\;\\mathrm{W/(m.K)}$, or cotton, $k_\\text{cotton}=17.5\\;\\mathrm{W/(m.K)}$?\n",
193 | "\n"
194 | ]
195 | },
196 | {
197 | "cell_type": "code",
198 | "execution_count": null,
199 | "id": "ff142774",
200 | "metadata": {},
201 | "outputs": [],
202 | "source": []
203 | },
204 | {
205 | "cell_type": "markdown",
206 | "id": "e100ec1d",
207 | "metadata": {},
208 | "source": [
209 | "## Problem 3\n",
210 | "\n",
211 | "To defog the rear window of an automobile, a very thin transparent heating element is attached to the inner surface of the window. A uniform flux $q''_{h}$ is provided to the heating element to maintain the inner surface of the window at a desired temperature $T_{s,i}$ . The window's thickness is $t=5mm$ and the thermal conductivity is $k=1.2\\;\\mathrm{W/m.K}$ . The interior temperature of the automobile is $T_{\\infty,i}=22^\\circ\\mathrm{C}$ and the convection coefficient is $h_i=15\\;\\mathrm{W/m^2.K}$ The outside ambient temperature is $t_{\\infty,o}=-5^\\circ C$ and the convection heat transfer is $h_o=100\\;\\mathrm{W/m^2.K}$. \n",
212 | "\n",
213 | "* 1.) Define your assumptions.\n",
214 | "* 2.) Sketch the thermal circuit of the system.\n",
215 | "* 3.) Write all equations necessary to solve this problem.\n",
216 | "* 4.) Determine the heat flux for $T_{s,i}=15^\\circ\\mathrm{C}$.\n",
217 | "\n",
218 | "The rear window is now tinted with an emissivity $\\varepsilon=0.8$. Radiation is assumed to take place only from the outer surface of the window to the outside. The sky temperature for radiation is $T_{sur}=T_{sky} = 5^\\circ\\mathrm{C}$ and the desired inner surface temperature is still $T_{s,i}=15^\\circ\\mathrm{C}$. \n",
219 | "* 5.) Sketch the new thermal circuit.\n",
220 | "* 6.) Write all equations necessary to solve this problem.\n",
221 | "* 7.) Determine the necessary heater heat flux to maintain. $T_{s,i}=15^\\circ\\mathrm{C}$"
222 | ]
223 | },
224 | {
225 | "cell_type": "code",
226 | "execution_count": null,
227 | "id": "cc5423a1",
228 | "metadata": {},
229 | "outputs": [],
230 | "source": []
231 | },
232 | {
233 | "cell_type": "markdown",
234 | "id": "95aade5f",
235 | "metadata": {},
236 | "source": [
237 | "## Problem 4\n",
238 | "\n",
239 | "\n",
240 | "A fluid at uniform temperature $T_i = 500^\\circ\\mathrm{K}$ is contained in an insulation blanket comprised of two semi-cylindrical shells of different materials. The outside ambient conditions are $T_\\infty = 300^\\circ\\mathrm{K}$ and $h = 25\\;\\mathrm{W/m^2.K}$. The thermal conductivities of the two materials are $k_A = 2\\;\\mathrm{W/m.K}$ and $k_B = 0.25\\;\\mathrm{W/m.K}$, respectively. The inner and outer diameters are $r_1 = 50\\;\\mathrm{mm}$ and $r_2 = 100\\;\\mathrm{mm}$.\n",
241 | "\n",
242 | "\n",
243 | "\n",
244 | "Assuming that the source of heating for the inside fluid stops suddenly. What is the initial rate of temperature change for a fluid of density $\\rho = 1000\\;\\mathrm{kg/m^3}$ and $C_p = 4000\\;\\mathrm{J/kg.K}$? If you have enough time left, show that the heat loss is $1040\\;\\mathrm{W/m}$.\n",
245 | "\n",
246 | "* 1.) Draw the thermal circuit.\n",
247 | "* 2.) Assuming that the source of heating for the inside fluid stops suddenly and the fluid temperature is spatially uniform (but varies in time). Derive the equation governing the rate of temperature change in the fluid ($dT/dt$)?\n",
248 | "* 3.) What is the initial rate of temperature change?\n",
249 | "* 4.) Bonus: Plot the evolution of temperature over time"
250 | ]
251 | },
252 | {
253 | "cell_type": "code",
254 | "execution_count": null,
255 | "id": "458e6b3c",
256 | "metadata": {},
257 | "outputs": [],
258 | "source": []
259 | }
260 | ],
261 | "metadata": {
262 | "kernelspec": {
263 | "display_name": "Python 3 (ipykernel)",
264 | "language": "python",
265 | "name": "python3"
266 | },
267 | "language_info": {
268 | "codemirror_mode": {
269 | "name": "ipython",
270 | "version": 3
271 | },
272 | "file_extension": ".py",
273 | "mimetype": "text/x-python",
274 | "name": "python",
275 | "nbconvert_exporter": "python",
276 | "pygments_lexer": "ipython3",
277 | "version": "3.10.10"
278 | }
279 | },
280 | "nbformat": 4,
281 | "nbformat_minor": 5
282 | }
283 |
--------------------------------------------------------------------------------
/HW-01.ipynb:
--------------------------------------------------------------------------------
1 | {
2 | "cells": [
3 | {
4 | "cell_type": "markdown",
5 | "id": "0cc46860",
6 | "metadata": {},
7 | "source": [
8 | "# Homework Assignment 2\n",
9 | "\n",
10 | "Complete this notebook, export to pdf (`File>Download as> PDF (via LaTeX)`) and submit the pdf to BB"
11 | ]
12 | },
13 | {
14 | "cell_type": "markdown",
15 | "id": "85aecba7",
16 | "metadata": {},
17 | "source": [
18 | "## Pb 1: Air thermodynamic properties class\n",
19 | "\n",
20 | "The concept of class is central to most libraries. It consists of the definition of an object, which has specific properties, data and function.\n",
21 | "\n",
22 | "Zografos et al. (*Compt. Methods Appl. Mech Eng.*, 61:177-187, 1987) proposed polynomials to compute density, dynamic viscosity, thermal conductivity, and specific heat at constant temperature of air at atmospheric pressure as a function of the static temperature $T$ on the Kelvin scale as follows:\n",
23 | "* Density $\\rho$ in units of $\\mathrm{kg/m^3}$:\n",
24 | "\n",
25 | "$$\n",
26 | "\\rho=\\begin{cases}\n",
27 | "-2.440\\times10^{-2}T+5.9958\\,, & 100\\,\\mathrm{K}\\leq T<150\\,\\mathrm{K}\\\\\n",
28 | "345.57(T-2.6884)^{-1}\\,, & 150\\,\\mathrm{K}\\leq T\\leq 3000\\,\\mathrm{K}\n",
29 | "\\end{cases}\n",
30 | "$$\n",
31 | "* Dynamic viscosity $\\mu$ in units of $\\mathrm{N.s/m^2}=\\mathrm{kg/(m.s)}=\\mathrm{Pa.s}$:\n",
32 | "\n",
33 | "$$\n",
34 | "\\mu = 4.1130\\times10^{-6} + 5.0523\\times10^{-8}T - 1.4346\\times10^{-11}T^2+2.5914\\times10^{-15}T^3\n",
35 | "$$\n",
36 | "\n",
37 | "* Thermal conductivity $k$ in units of $\\mathrm{W/(m.K)}$:\n",
38 | "\n",
39 | "$$\n",
40 | "k= -7.488\\times10^{-3} + 1.7082\\times10^{-4}T - 2.3758\\times10^{-10}T^2 + 2.2012\\times10^{-10}T^3 \\\\\n",
41 | " + 9.4600\\times10^{-14}T^4 + 1.5797\\times10^{-17}T^5\n",
42 | "$$\n",
43 | "\n",
44 | "* Specific heat at constant pressure $c_p$ in unites of $\\mathrm{J/(kg.K)}$:\n",
45 | "\n",
46 | "$$\n",
47 | "c_p=1061.332-0.432819T + 1.02344\\times10^{-3}T^2 - 6.47474\\times10^{-7}T^3 + 1.3864\\times10^{-10}T^4\n",
48 | "$$\n",
49 | "\n",
50 | "Air density $\\rho$ of units $\\mathrm{kg/m^3}$ can be computed from \n",
51 | "\n",
52 | "The Prandtl number $Pr$ can be computed as\n",
53 | "\n",
54 | "$$\n",
55 | "Pr=\\frac{\\mu c_p}{k}\n",
56 | "$$\n",
57 | "\n",
58 | "and the thermal conductivity $\\alpha$ of units $\\mathrm{m^2/s}$ as\n",
59 | "\n",
60 | "$$\n",
61 | "\\alpha = \\frac{k}{\\rho c_p}\n",
62 | "$$\n",
63 | "\n",
64 | "Write a class storing all these variables for input temperature (float) and unit used for temperature (string, it can only be \"K\", \"C\", \"F\"). The class should convert the temperature to Kelvin if unit is either \"C\" or \"F\"). The class should give a warning if \n",
65 | "* the temperature is outside of the range,\n",
66 | "* the unit is not \"K\", \"C\", \"F\", e.g. \"c\"\n",
67 | "\n",
68 | "### Tasks\n",
69 | "* Complete the manual part of the class (text between \"\"\" xxxx \"\"\"), i.e. write appropriate units. \n",
70 | "* Complete the class to compute all quantities defined in the manual.\n",
71 | "* Apply your class to the examples below.\n",
72 | "* Define units for examples.\n",
73 | "* Plot density as function of temperature in the range $10^2\\leq T<3000$. Use log-scale for temperature. Axes must be labeled with their variables and units."
74 | ]
75 | },
76 | {
77 | "cell_type": "code",
78 | "execution_count": null,
79 | "id": "97650df2",
80 | "metadata": {},
81 | "outputs": [],
82 | "source": [
83 | "import numpy as np\n",
84 | "import math\n",
85 | "import matplotlib.pyplot as plt\n",
86 | "class Air:\n",
87 | " \"\"\"\n",
88 | " Returns an object containing the thermodynamic properties of air at atmospheric pressure\n",
89 | " Input: \n",
90 | " T: temperature (float)\n",
91 | " Tunit: unit of temperature (string). Can only be \"K\", \"C\", \"F\"\n",
92 | " output:\n",
93 | " self.rho: density [kg/m^3]\n",
94 | " self.mu: dynamic viscosity [write units here]\n",
95 | " self.k: thermal conductivity [write units here]\n",
96 | " self.cp: Specific heat at constant pressure [write units here]\n",
97 | " self.Pr: Prandtl number\n",
98 | " self.nu: kinematic viscosity [write units here]\n",
99 | " self.alpha: thermal diffusivity [write units here]\n",
100 | " \"\"\"\n",
101 | " def __init__(self,T,Tunit):\n",
102 | " if Tunit == \"C\":\n",
103 | " T += 273.15\n",
104 | " elif Tunit == \"F\":\n",
105 | " T = (T - 32)*5/9 +273.15\n",
106 | " elif Tunit != \"K\":\n",
107 | " print(\"Wrong input for unit of temperature\")\n",
108 | " print(\"Input should be 'K', 'C', or 'F'\")\n",
109 | " self.T = T\n",
110 | " self.mu = 4.1130E-6 + 5.0523E-8*T - 1.4346E-11*T**2 + 2.5914E-15*T**3\n",
111 | " \n"
112 | ]
113 | },
114 | {
115 | "cell_type": "code",
116 | "execution_count": null,
117 | "id": "0b87cbeb",
118 | "metadata": {},
119 | "outputs": [],
120 | "source": [
121 | "myair = Air(-200,\"F\")\n",
122 | "print(\"Temperature in K: %.2f\" %myair.T)\n",
123 | "print(\"Kinematic viscosity: %.2e [units], Thermal conductivity: %.4f [units]\" %(myair.nu,myair.k))"
124 | ]
125 | },
126 | {
127 | "cell_type": "code",
128 | "execution_count": null,
129 | "id": "f887dc6c",
130 | "metadata": {},
131 | "outputs": [],
132 | "source": [
133 | "myair = Air(-200,\"C\")"
134 | ]
135 | },
136 | {
137 | "cell_type": "code",
138 | "execution_count": null,
139 | "id": "5fdbb9f2",
140 | "metadata": {},
141 | "outputs": [],
142 | "source": [
143 | "myair = Air(2000,\"k\")"
144 | ]
145 | },
146 | {
147 | "cell_type": "code",
148 | "execution_count": null,
149 | "id": "8a8c3729",
150 | "metadata": {},
151 | "outputs": [],
152 | "source": [
153 | "myair = Air(2000,\"K\")\n",
154 | "print(\"Dynamic viscosity: %.2e [units], Thermal diffusivity: %.2e [units]\" %(myair.mu,myair.alpha))"
155 | ]
156 | },
157 | {
158 | "cell_type": "code",
159 | "execution_count": null,
160 | "id": "618ff68d",
161 | "metadata": {},
162 | "outputs": [],
163 | "source": [
164 | "T = np.logspace(2,np.log10(2999.),1000)\n",
165 | "rho = np.zeros_like(T)\n",
166 | "for i in range(T.shape[0]):\n",
167 | " # Your code here"
168 | ]
169 | },
170 | {
171 | "cell_type": "code",
172 | "execution_count": null,
173 | "id": "d67d9ede",
174 | "metadata": {},
175 | "outputs": [],
176 | "source": [
177 | "plt.figure(figsize=(3,2),dpi=100)\n",
178 | "# your line plot here\n",
179 | "plt.xlabel(\"$T\\;(\\mathrm{units})$\")\n",
180 | "plt.ylabel(r\"$\\rho\\;(\\mathrm{units})$\")\n",
181 | "plt.show()"
182 | ]
183 | },
184 | {
185 | "cell_type": "markdown",
186 | "id": "b4054fd7",
187 | "metadata": {},
188 | "source": [
189 | "## Pb 2 Window\n",
190 | "\n",
191 | "Set up equations and apply realistic numerical values to them to discuss heat losses of a single pane window, a single pane window with storm window and a double paned window with air trapped at a vacuum of $10^{-3} \\mathrm{torr}$ ina gap of $5\\mathrm{mm}$. Do not consider the effects of radiation for any of the window.\n",
192 | "\n",
193 | "\n",
194 | "\n",
195 | "#### Assumptions:\n",
196 | "* Heat transfer is steady and 1D\n",
197 | "* Radiation is ignored\n",
198 | "* Convection in the gap of the storm window is neglected\n",
199 | "* The temperature on the inside and outside surfaces of any window are isothermal and equal to $T_{si}=10^\\circ\\mathrm{C}$ and $T_{so}=-10^\\circ\\mathrm{C}$, respectively.\n",
200 | "* For the storm window, and the double pane window, the average temperature in the gap between the two glass pane is assumed to the average of $T_{si}$ and $T_{so}$ or $T_{mean}=0^\\circ\\mathrm{C}$.\n",
201 | "\n"
202 | ]
203 | },
204 | {
205 | "cell_type": "markdown",
206 | "id": "bd483448",
207 | "metadata": {},
208 | "source": [
209 | "#### Setup\n",
210 | "\n",
211 | "In this cell you are expected to:\n",
212 | "* Insert your sketches of the single pane and double pane windows showing heat fluxes, temperatures at boundaries and dimensions\n",
213 | "* Store in folder `MyFigures`. Change the names `your-single-pane-sketch` and `your-double-pane-sketch` in the HTML command below to include your figures to this cell.\n",
214 | "* Complete a **worded** write-up of the solutions (for both windows) including equations written in LaTeX\n",
215 | "\n",
216 | "\n",
217 | "\n",
218 | "\n",
219 | "*This is the type of introduction of your work that is expected in this class for any deliverable*\n",
220 | "\n",
221 | "The goal is to compute the heat loss from a model window under controlled conditions. We choose to fix the inside surface temperature to $10^\\circ\\mathrm{C}$ and the outside surface temperature to $-10^\\circ\\mathrm{C}$ for a window of height and width $h = 1.2\\;\\mathrm{m}$ and $w = 0.7\\;\\mathrm{m}$ with a surface aera $A=h\\times w$. The thickness of a residential window pane is $t_{glass}=3/32\\,\\mathrm{in}$ (https://ringerwindows.com/thick-glass-windows/) with a thermal conductivity $k=0.96\\;\\mathrm{W/m.K}$. The storm window has a $2.5\\,\\mathrm{in}$.\n",
222 | "\n",
223 | "The heat rate for a single window pane is...\n",
224 | "*Your derivation of the solution here*\n"
225 | ]
226 | },
227 | {
228 | "cell_type": "code",
229 | "execution_count": null,
230 | "id": "3a7346e0",
231 | "metadata": {},
232 | "outputs": [],
233 | "source": [
234 | "import numpy as np\n",
235 | "from Libraries import thermodynamics as thermo\n",
236 | "T_si = 10. #C\n",
237 | "T_so = -10. #C\n",
238 | "h = 1.2 #m\n",
239 | "w = 0.7 #m\n",
240 | "t_glass = 3./32.*0.0254 #m\n",
241 | "k_glass = 0.96\n",
242 | "A = h*w\n",
243 | "t_gap_storm = 2*0.0254 #m\n",
244 | "# getting thermodynamic properties of air for the storm window\n",
245 | "air_gap_storm = thermo.Fluid('air',thermo.C2K(0.))\n",
246 | "t_gap_vacuum = 0.005#m\n",
247 | "k_vacuum = xxxx #W/m.K (from graph)\n",
248 | "R_glass = \n",
249 | "R_gap_storm = \n",
250 | "R_gap_vacuum = \n",
251 | "R_total_single = \n",
252 | "R_total_storm = \n",
253 | "R_total_double = \n",
254 | "\n",
255 | "print(\"Thermal conductivity of air at 0 C: %.4f\" %air_gap_storm.k)\n",
256 | "print(\"Thermal resistance for the single window: %.4f K/W\" %R_total_single)\n",
257 | "print(\"Thermal resistance for the storm window: %.4f K/W\" %R_total_storm)\n",
258 | "print(\"Thermal resistance for the double pane window: %.4f K/W\" %R_total_double)\n",
259 | "\n",
260 | "DT = T_si - T_so\n",
261 | "q_single = DT/R_total_single\n",
262 | "q_storm = DT/R_total_storm\n",
263 | "q_double = DT/R_total_double\n",
264 | "print(\"Single window pane q: %.0f W\" %q_single)\n",
265 | "print(\"Storm window q: %.0f W\" %q_storm)\n",
266 | "print(\"Double pane window q: %.0f W\" %q_double)"
267 | ]
268 | },
269 | {
270 | "cell_type": "markdown",
271 | "id": "a8f4fc81",
272 | "metadata": {},
273 | "source": [
274 | "## Pb 3\n",
275 | "A plate of metal of mass $m=4.25 kg$, specific heat $C_p= 2770 J/kg.K$ and surface area $0.4m\\times0.4m$ is cooled in ambient air at temperature $T_\\infty=25^\\circ C$. The plate is thin enough and its thermal conductivity high enough that the plate is assumed to be isothermal at any time. At a given time, the plate temperature is $T_s = 245^\\circ C$ and the rate of temperature cooling is\n",
276 | "$$\n",
277 | "\\frac{dT}{dt}=-0.028 K/s\n",
278 | "$$\n",
279 | "\n",
280 | "Find the convection heat transfer coefficient from conservation of energy applied to the plate."
281 | ]
282 | },
283 | {
284 | "cell_type": "code",
285 | "execution_count": null,
286 | "id": "344068f6",
287 | "metadata": {},
288 | "outputs": [],
289 | "source": []
290 | },
291 | {
292 | "cell_type": "markdown",
293 | "id": "3b915796",
294 | "metadata": {},
295 | "source": [
296 | "## Pb 4 Radiation in a vacuum \n",
297 | "\n",
298 | "The same plate as above is now in a vacuum (no flow). Using all conditions defined in 01.1, determine the emissivity of the plate and the heat rate at which radiation is emitted from the plate. Assume that $T_{sur}=T_\\infty$. "
299 | ]
300 | },
301 | {
302 | "cell_type": "code",
303 | "execution_count": null,
304 | "id": "ed0da580",
305 | "metadata": {},
306 | "outputs": [],
307 | "source": []
308 | },
309 | {
310 | "cell_type": "code",
311 | "execution_count": null,
312 | "id": "57e3221d",
313 | "metadata": {},
314 | "outputs": [],
315 | "source": []
316 | }
317 | ],
318 | "metadata": {
319 | "kernelspec": {
320 | "display_name": "Python 3 (ipykernel)",
321 | "language": "python",
322 | "name": "python3"
323 | },
324 | "language_info": {
325 | "codemirror_mode": {
326 | "name": "ipython",
327 | "version": 3
328 | },
329 | "file_extension": ".py",
330 | "mimetype": "text/x-python",
331 | "name": "python",
332 | "nbconvert_exporter": "python",
333 | "pygments_lexer": "ipython3",
334 | "version": "3.9.13"
335 | }
336 | },
337 | "nbformat": 4,
338 | "nbformat_minor": 5
339 | }
340 |
--------------------------------------------------------------------------------
/.ipynb_checkpoints/HT-ablation-BL-checkpoint.ipynb:
--------------------------------------------------------------------------------
1 | {
2 | "cells": [
3 | {
4 | "cell_type": "markdown",
5 | "metadata": {},
6 | "source": [
7 | "# External convection flow\n",
8 | "\n",
9 | "## Flat plate\n",
10 | "\n",
11 | "In the heating or cooling of a plate by a flow (or a flow by a plate), the quantities of interest are\n",
12 | "* Reynolds number and Prandtl number based on the thermodynamics of the **fluid** at **film temperature**.\n",
13 | "* Boundary layer thickness and thermal boundary layer thickness\n",
14 | "* Local and average coefficients of friction\n",
15 | "* Local and average oefficients of convection heat transfer obtained through a Nusselt number correlation\n"
16 | ]
17 | },
18 | {
19 | "cell_type": "markdown",
20 | "metadata": {},
21 | "source": [
22 | "# Design of a low temperature ablation experimental facility\n",
23 | "\n",
24 | "During atmopheric entry, the heat shield of a space vehicle experiences tremendous heat load and aggressive chemical reactions. On some materials, the formation of roughness patterns is observed. These patterns are caused by the local ablation of the matieral. Roughness may cause extra heat flux through increased turbulence that could lead to the failure of the heat shield. The mechanisms by which such ablation or erosion patterns are formed are poorly understood, leaving engineers with few if any ability to predict the outcome of the heat shield. \n",
25 | "\n",
26 | "Here we consider a simple system made out of a plate of wax of thickness $0.05\\text{ m}$ in order to study this system in a lab without worrying about high-temperatures.\n",
27 | "\n",
28 | "The local ablation velocity of wax $V_a(x)$ (i.e. the removal rate of solid wax) is derived from a conservation of energy:\n",
29 | "\n",
30 | "$$\n",
31 | "\\rho_{wax}h_lV_a(x) =\\begin{cases}\\max\\left(q''_\\text{conv}-q''_\\text{cond},0\\right)&\\mathrm{if}\\; T_s(x)\\geq T_m\\\\\n",
32 | "0& \\text{otherwise}\\end{cases}\n",
33 | "$$\n",
34 | "\n",
35 | "where $\\rho_{wax}=900\\,\\mathrm{kg}/\\mathrm{m}^3$ is the wax density, $h_l= 200\\;\\mathrm{kJ}/\\mathrm{kg}$ is the latent heat of wax and $q''_\\text{conv}$ and $q''_\\text{cond}$ are the *local* heat fluxes by convection and conduction, respectively. $T_s(x)$ an $T_m$ are the *local* surface temperature, and the surface temperature, respectively.\n",
36 | "\n",
37 | "The goal is to study the formation of erosion patterns on the surface of the wax plate, thus we are interested in maximumizing the ablation velocity. We will consider two fluids, air and water. \n",
38 | "\n",
39 | "For both case, the bottom of the wax plate is maintained at constant temperature $T_\\text{cool}=20^\\circ\\mathrm{C}$. \n",
40 | "In the following, you are asked to find the initial ablation velocity assuming that (a) the plate has not started to melt yet and (b) the flow-exposed surface is at melting temperature $T_\\text{melt}=50^\\circ\\mathrm{C}$. The thermal conductivity of wax is $k_{wax}=0.25\\mathrm{W}/\\mathrm{m.K}$ and radiation heat transfer is neglected."
41 | ]
42 | },
43 | {
44 | "cell_type": "markdown",
45 | "metadata": {},
46 | "source": [
47 | "2.1 The upstream air flow velocity is $10 \\,\\mathrm{m/s}$ and its upstream temperature is $150^\\circ\\mathrm{C}$. Plot the evolution of the initial ablation velocity in $\\mathrm{mm/s}$ and/or give its values for $x=0.01,0.1,0.2,0.5,1.0\\mathrm{m}$. The leading edge is a very thin, adiabatic rounded edge."
48 | ]
49 | },
50 | {
51 | "cell_type": "markdown",
52 | "metadata": {},
53 | "source": [
54 | "### Assumptions\n",
55 | "\n",
56 | "* Radiation is negligible\n",
57 | "* Leading edge does not trip boundary layer into turbulent regime\n",
58 | "* Heat transfer is predominantly vertical, we use the 1D approximation.\n",
59 | "* Both horizontal surfaces of the wax slab are isothermal.\n",
60 | "\n",
61 | "\n",
62 | "\n",
63 | "\n",
64 | "The ablation velocity is given by the equation above. The conduction heat flux within the wax slab is\n",
65 | "$$\n",
66 | "q''_{cond}=\\frac{k_{wax}}{t_{wax}}(T_m-T_{cool})\n",
67 | "$$\n",
68 | "\n",
69 | "The convection heat flux is defined as\n",
70 | "$$\n",
71 | "q''_{conv}(x)=h(x)(T_\\infty-T_m)\n",
72 | "$$\n",
73 | "and $h(x)$ is given by the correlation for external convection over a flat plate. \n",
74 | "\n",
75 | "The flow starts in the laminar regime and may transition to turbulence if $Re_{x=L}$ is larger than $Re_{c}\\approx 5\\times10^5$.\n",
76 | "\n",
77 | "Flow thermodynamics properties are determined at the film temperature:\n",
78 | "$$\n",
79 | "T_f=\\frac{T_m+T_\\infty}{2}\n",
80 | "$$"
81 | ]
82 | },
83 | {
84 | "cell_type": "code",
85 | "execution_count": 55,
86 | "metadata": {},
87 | "outputs": [
88 | {
89 | "data": {
90 | "image/png": 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\n",
91 | "text/plain": [
92 | ""
93 | ]
94 | },
95 | "metadata": {
96 | "needs_background": "light"
97 | },
98 | "output_type": "display_data"
99 | }
100 | ],
101 | "source": [
102 | "from Libraries import thermodynamics as thermo\n",
103 | "from Libraries import HT_external_convection as extconv\n",
104 | "import numpy as np\n",
105 | "import matplotlib.pyplot as plt\n",
106 | "import scipy.constants as csts\n",
107 | "\n",
108 | "# Upstream air properties:\n",
109 | "U_infty = 10. #m/s\n",
110 | "T_infty = 300. #C\n",
111 | "\n",
112 | "#Melting surface \n",
113 | "T_melt = 50. #C\n",
114 | "\n",
115 | "#Bottom temperature\n",
116 | "T_cool = 20. #C\n",
117 | "\n",
118 | "#Wax properties\n",
119 | "k_wax = 0.25 #W/m.K\n",
120 | "rho_wax = 900. #kg/m^3\n",
121 | "h_L_wax = 200.e3 #J/kg\n",
122 | "t_wax = 0.05 #m\n",
123 | "L_plate = 1. #m\n",
124 | "\n",
125 | "# conduction heat flux in the wax\n",
126 | "qpp_cond = k_wax*(T_melt - T_cool)/t_wax\n",
127 | "\n",
128 | "def Va(qpp, Ts):\n",
129 | " global T_melt, rho_wax, h_L_wax, qpp_cond\n",
130 | " if (Ts >= T_melt):\n",
131 | " Va = np.maximum.reduce([(qpp - qpp_cond)/(rho_wax*h_L_wax), 0.])\n",
132 | " else:\n",
133 | " Va = 0.\n",
134 | " return Va\n",
135 | "\n",
136 | "# convection heat flux\n",
137 | "T_f = (T_melt + T_infty)/2.\n",
138 | "airprop = thermo.Fluid('air',thermo.C2K(T_f))\n",
139 | "airconv = extconv.FlatPlate('mixed','isothermal',U_infty,airprop.nu,airprop.alpha,1.)\n",
140 | "\n",
141 | "npts = 500\n",
142 | "x = np.logspace(-3,0,npts)\n",
143 | "Va_x = np.zeros_like(x)\n",
144 | "for i in range(npts):\n",
145 | " xlocal = x[i]\n",
146 | " airconv.local(xlocal)\n",
147 | " hx = airconv.Nu_x*airprop.k/xlocal\n",
148 | " qpp_x = hx*(T_infty - T_melt)\n",
149 | " Va_x[i] = Va(qpp_x,T_melt)\n",
150 | "\n",
151 | "# plt.figure(figsize(3,2),dpi=150)\n",
152 | "plt.plot(x,Va_x*1e3)\n",
153 | "plt.xlabel(\"$x\\;(\\mathrm{m})$\")\n",
154 | "plt.ylabel(\"$Va\\;(\\mathrm{mm}/\\mathrm{s})$\")\n",
155 | "plt.show()"
156 | ]
157 | },
158 | {
159 | "cell_type": "code",
160 | "execution_count": 47,
161 | "metadata": {},
162 | "outputs": [
163 | {
164 | "data": {
165 | "text/latex": [
166 | "$\\displaystyle 4.5842204364201207e-05$"
167 | ],
168 | "text/plain": [
169 | "4.5842204364201207e-05"
170 | ]
171 | },
172 | "execution_count": 47,
173 | "metadata": {},
174 | "output_type": "execute_result"
175 | }
176 | ],
177 | "source": []
178 | },
179 | {
180 | "cell_type": "code",
181 | "execution_count": 43,
182 | "metadata": {},
183 | "outputs": [],
184 | "source": [
185 | "?extconv.FlatPlate"
186 | ]
187 | },
188 | {
189 | "cell_type": "markdown",
190 | "metadata": {},
191 | "source": [
192 | "2.2 The upstream water flow velocity is $1 \\mathrm{m/s}$ and its upstream temperature is $99^\\circ\\mathrm{C}$. Plot the evolution of the initial ablation velocity in $\\text{mm/s}$ and/or give its values for $x=0.01,0.1,0.2,0.5,1.0\\mathrm{m}$. Discuss which fluid is more likley to achieve the formation of the deepest erosion-driven roughness."
193 | ]
194 | },
195 | {
196 | "cell_type": "code",
197 | "execution_count": null,
198 | "metadata": {},
199 | "outputs": [],
200 | "source": []
201 | }
202 | ],
203 | "metadata": {
204 | "kernelspec": {
205 | "display_name": "Python 3",
206 | "language": "python",
207 | "name": "python3"
208 | },
209 | "language_info": {
210 | "codemirror_mode": {
211 | "name": "ipython",
212 | "version": 3
213 | },
214 | "file_extension": ".py",
215 | "mimetype": "text/x-python",
216 | "name": "python",
217 | "nbconvert_exporter": "python",
218 | "pygments_lexer": "ipython3",
219 | "version": "3.8.5"
220 | }
221 | },
222 | "nbformat": 4,
223 | "nbformat_minor": 4
224 | }
225 |
--------------------------------------------------------------------------------
/.ipynb_checkpoints/HTpy-Lecture9-1DConduction-checkpoint.ipynb:
--------------------------------------------------------------------------------
1 | {
2 | "cells": [
3 | {
4 | "cell_type": "markdown",
5 | "metadata": {},
6 | "source": [
7 | "# 1D Thermal Boundary Conduction: Boundary Conditions, Plans, Cylindrical and Spherical Conduction, Thermal Resistance"
8 | ]
9 | },
10 | {
11 | "cell_type": "markdown",
12 | "metadata": {},
13 | "source": [
14 | "Local heat flux is measured by the phenomenological Fourier's Law:\n",
15 | "$$\n",
16 | "\\vec{q}''=-k\\vec{\\nabla}T\n",
17 | "$$\n",
18 | "where $k$ is the local thermal conduction, W/(m.K), and the gradient of temperature $\\vec{\\nabla}T$. Since the latter is effectively a temperature difference, temperature may be expressed in Kelvin or Celsius.\n",
19 | "In the previous notebook, we established that in steady-state, planar, 1D conduction, the heat flux (and heat rate) in an homogenous material is proportional to the temperature difference between the boundaries of the material:\n",
20 | "$$\n",
21 | "q''=-\\frac{\\Delta T}{R''} \\text{ and } q=-\\frac{\\Delta T}{R}\n",
22 | "$$\n",
23 | "where the thermal resistances for heat flux and heat rate are\n",
24 | "$$\n",
25 | "R''=\\frac{L}{k}\\text{ and }R=\\frac{L}{kA}\n",
26 | "$$\n",
27 | "$A$ is the cross sectional area of the material.\n",
28 | "\n",
29 | "In this notebook, you will learn how to calculate heat flux with various boundary conditions and in different geometries. We will still limit heat transfer within the system of interest to pure conduction."
30 | ]
31 | },
32 | {
33 | "cell_type": "markdown",
34 | "metadata": {},
35 | "source": [
36 | "## 1. Boundary Conditions"
37 | ]
38 | },
39 | {
40 | "cell_type": "markdown",
41 | "metadata": {},
42 | "source": [
43 | "In the previous notebook, we saw that the solution of a heat transfer problem requires the knowledge of heat transfer at the boundaries of the domain of interest. This section defines the possible scenarios for boundary conditions.\n",
44 | "
"
45 | ]
46 | },
47 | {
48 | "cell_type": "markdown",
49 | "metadata": {},
50 | "source": [
51 | "### 1.1 Imposed Temperature (Isothermal)"
52 | ]
53 | },
54 | {
55 | "cell_type": "markdown",
56 | "metadata": {},
57 | "source": [
58 | "This is the first scenario in the above figure: $T_{s0}$ and and $T_{s3}$ are fixed, the flux is a consequence. This condition is called isothermal, and its mathematical name is the Dirichlet condition."
59 | ]
60 | },
61 | {
62 | "cell_type": "markdown",
63 | "metadata": {},
64 | "source": [
65 | "### 1.2 Imposed Flux"
66 | ]
67 | },
68 | {
69 | "cell_type": "markdown",
70 | "metadata": {},
71 | "source": [
72 | "If the wall is heated by an electric resistance for instance, the heat flux is fixed. The mathematical name is the Neumann condition. A special case is the adiabatic condition when $q_\\infty''=0$ (no heat transfer). In this case, the interface temperature $T_{s0}$ is governed by the imposed flux "
73 | ]
74 | },
75 | {
76 | "cell_type": "markdown",
77 | "metadata": {},
78 | "source": [
79 | "### 1.3 Convection Heat Transfer"
80 | ]
81 | },
82 | {
83 | "cell_type": "markdown",
84 | "metadata": {},
85 | "source": [
86 | "When a surface is cooled or warmed by a flowing fluid, the heat transfer is convection driven. There are three types of convection:\n",
87 | "
\n",
88 | "
Natural convection: the fluid motion is driven by buoyancy, i.e. the temperature gradients imposed by the boundary conditions.
\n",
89 | "
Forced convection: the flow is driven by an external force (fan, pressure gradient, gravity) in a way that is stronger than the buoyancy-driven flow.
\n",
90 | "
Mixed convection happens when the strength of the external force driven flow and that of the buoyancy driven flow are comparable.
\n",
91 | "
\n",
92 | "For now, convection heat transfer is modeled as\n",
93 | "$$\n",
94 | "q_\\text{conv}=hA\\left(T_{\\infty,0}-T_{s,0}\\right)\n",
95 | "$$\n",
96 | "and $h$ is the convection heat transfer coefficient, which is a function of the nature of the fluid, the flow characteristics, and the temperature scales involved. We will learn to determine $h$ later on. Until then, $h$ will be provided.\n",
97 | "\n",
98 | "The thermal resistance of convection heat transfer is\n",
99 | "$$\n",
100 | "R_\\text{conv}''=\\frac{1}{h}\\text{ and }R_\\text{conv}=\\frac{1}{hA}\n",
101 | "$$\n"
102 | ]
103 | },
104 | {
105 | "cell_type": "markdown",
106 | "metadata": {},
107 | "source": [
108 | "### 1.4 Radiation Heat Transfer"
109 | ]
110 | },
111 | {
112 | "cell_type": "markdown",
113 | "metadata": {},
114 | "source": [
115 | "The last mode of heat transfer is radiation. For now, we consider radiation between a small surface at temperature $T_\\text{s}$ and a vast surrounding at temperature $T_\\text{sur}$. The surface of interest is characterized by an emissivity $0\\leq\\varepsilon\\leq1$. Thermal radiation is energy emitted by matter and transported by electromagnetic waves. \n",
116 | "For now, we consider radiation from a small surface tThe radiation heat flux, or radiation heat transfer per unit area is\n",
117 | "$$\n",
118 | "q_\\text{rad}''=\\frac{q}{A}=\\varepsilon\\sigma\\left(T_\\text{s}^4-T_\\text{sur}^4\\right)\n",
119 | "$$\n",
120 | "where $\\sigma=5.67\\times10^8 \\text{W/(m}^2\\text{K}^4$ is the Stefan Boltzmann constant.\n",
121 | "\n",
122 | "You should note that the temperature should always be expressed in Kelvin when dealing with radiation. \n",
123 | "\n",
124 | "The radiation heat transfer coefficient is\n",
125 | "$$\n",
126 | "h_\\text{rad}=\\varepsilon\\sigma\\left(T_\\text{s}+T_\\text{sur}\\right)\\left(T_\\text{s}^2+T_\\text{sur}^2\\right)\n",
127 | "$$\n",
128 | "yielding\n",
129 | "$$\n",
130 | "q_\\text{rad}=h_\\text{rad}A\\left(T_\\text{s}-T_\\text{sur}\\right)\n",
131 | "$$\n",
132 | "and the thermal resitance of radiation heat transfer:\n",
133 | "$$\n",
134 | "R_\\text{rad}=\\frac{1}{h_\\text{rad}A}\n",
135 | "$$\n"
136 | ]
137 | },
138 | {
139 | "cell_type": "markdown",
140 | "metadata": {},
141 | "source": [
142 | "## 2. Cylindrical walls"
143 | ]
144 | },
145 | {
146 | "cell_type": "markdown",
147 | "metadata": {},
148 | "source": [
149 | "In many systems, heat transfer occurs between a fluid transported between two points: heating in building, mass transport in chemical and food processing plants, etc... Typically, ducts are cylindrical. Why?"
150 | ]
151 | },
152 | {
153 | "cell_type": "markdown",
154 | "metadata": {},
155 | "source": [
156 | "Consider an pipe of thickness $r_\\text{out}-r_\\text{in}$, where $r_\\text{in}$ and $r_\\text{out}$ are inner and outer radii, which are at constant temperatures $T_\\text{in}>T_\\text{out}$. To solve this problem, we proposed to\n",
157 | "\n",
158 | "
Assume 1D conduction heat transfer: The transport of heat is assumed to be purely radial, no heat escapes in other directions.
\n",
159 | "
Divide the system in $n$ shells of constant thickness. The cylindrical shell numbered $i$ is defined by the volume between the cylindrical surfaces $r_i-\\Delta r/2$ and $r_i+\\Delta r/2$. The cylindrical surface at the center of the cell has the radius\n",
160 | "$$\n",
161 | "r_i=r_\\text{in}+i.\\Delta r\\text{ with } \\Delta r=\\frac{r_\\text{out}-r_\\text{in}}{n-1}\n",
162 | "$$
\n",
163 | "\n",
164 | "Since the problem is 1D, the temperature is constant on any cylindrical surface of radius $r\\in\\left[r_\\text{in},r_\\text{out}\\right]$. \n",
165 | "\n",
166 | "Writing the conservation of energy between cylindrical shells results in\n",
167 | "$$\n",
168 | "-2\\pi\\left(r_i-\\frac{\\Delta r}{2}\\right)Lk\\frac{T_{i}-T_{i-1}}{\\Delta r}=\n",
169 | "-2\\pi\\left(r_i+\\frac{\\Delta r}{2}\\right)Lk\\frac{T_{i+1}-T_{i}}{\\Delta r}\n",
170 | "$$\n",
171 | "This equation states that the heat rate crossing surface $r_i-\\Delta r/2$ is equal to the heat rate crossing surface $r_i+\\Delta r/2$. You should recognise the surface area of the form $2\\pi r L$ and the Fourier's Law $-k(T_i-T_{i-1})/\\Delta r$. "
172 | ]
173 | },
174 | {
175 | "cell_type": "markdown",
176 | "metadata": {},
177 | "source": [
178 | "The conservation of energy or heat rate can be recast in a matrix friendly form:\n",
179 | "$$\n",
180 | "\\left(r_i-\\frac{\\Delta r}{2}\\right)T_{i-1}-2r_iT_{i}+\\left(r_i+\\frac{\\Delta r}{2}\\right)T_{i+1}=0\n",
181 | "$$\n",
182 | "Note that in this particular problem the temperature distribution does not depend upon the thermal conductivity of the cylinder.\n",
183 | "\n",
184 | "The boundary conditions are simply:\n",
185 | "$$\n",
186 | "T_0=T_\\text{in}\\text{ and }T_{n-1}=T_\\text{out}\n",
187 | "$$\n",
188 | "and the system in matrix form looks like this:"
189 | ]
190 | },
191 | {
192 | "cell_type": "markdown",
193 | "metadata": {},
194 | "source": [
195 | "$$\n",
196 | "\\left(\\begin{array}{ccccccc}\n",
197 | "1 & 0 & \\ldots & \\ldots & \\ldots & \\ldots & 0 \\\\\n",
198 | "r_{\\frac{1}{2}} & -2r_1 & r_{\\frac{3}{2}} & & & & \\vdots \\\\\n",
199 | "0 &\\ddots & \\ddots & \\ddots & & & \\vdots\\\\\n",
200 | "\\vdots & & r_{i-\\frac{1}{2}} & -2r_i & r_{i+\\frac{1}{2}} & & \\vdots\\\\\n",
201 | "\\vdots & & & \\ddots & \\ddots & \\ddots & 0\\\\\n",
202 | "\\vdots & & & & r_{n-\\frac{5}{2}} & -2r_{n-2} & r_{n-\\frac{3}{2}}\\\\ \n",
203 | "0 & \\ldots & \\ldots & \\ldots & \\ldots & 0 & 1\\\\\n",
204 | "\\end{array}\\right)\n",
205 | "\\left(\\begin{array}{c}\n",
206 | "T_0\\\\\n",
207 | "%T_1\\\\\n",
208 | "\\vdots \\\\\n",
209 | "T_{i-2} \\\\\n",
210 | "T_{i} \\\\\n",
211 | "T_{i+1} \\\\\n",
212 | "\\vdots \\\\\n",
213 | "%T_{n-2}\\\\\n",
214 | "T_{n-1}\\\\\n",
215 | "\\end{array}\\right)=\n",
216 | "\\left(\\begin{array}{c}\n",
217 | "T_\\text{in}\\\\\n",
218 | "0\\\\\n",
219 | "\\vdots \\\\\n",
220 | "\\vdots \\\\\n",
221 | "\\vdots \\\\\n",
222 | "0\\\\\n",
223 | "T_\\text{out}\\\\\n",
224 | "\\end{array}\\right)\n",
225 | "$$"
226 | ]
227 | },
228 | {
229 | "cell_type": "code",
230 | "execution_count": null,
231 | "metadata": {},
232 | "outputs": [],
233 | "source": [
234 | "%matplotlib inline \n",
235 | "# plots graphs within the notebook\n",
236 | "%config InlineBackend.figure_format='svg' # not sure what this does, may be default images to svg format\n",
237 | "\n",
238 | "import matplotlib.pyplot as plt #calls the plotting library hereafter referred as to plt\n",
239 | "import numpy as np #calls the numerical method library hereafter referred as to np for linear algebra\n",
240 | "\n",
241 | "##############################\n",
242 | "# Parameters #\n",
243 | "##############################\n",
244 | "r_in = 0.1 \n",
245 | "T_in = 10.\n",
246 | "T_out = 100.\n",
247 | "r_out = 10.\n",
248 | "n = 51\n",
249 | "\n",
250 | "##############################\n",
251 | "# Variable definition #\n",
252 | "##############################\n",
253 | "dr = float(r_out-r_in)/float(n-1) #shell thickness\n",
254 | "r = r_in + np.array([i*dr for i in range(n)]) #radius of the center of the shell\n",
255 | "A = np.zeros((n, n)) # pre-allocate [A] array\n",
256 | "b = np.zeros((n, 1)) # pre-allocate {b} column vector\n",
257 | "\n",
258 | "##################################\n",
259 | "# Matrix and vector definition #\n",
260 | "##################################\n",
261 | "A[0, 0] = 1.\n",
262 | "b[0, 0] = T_in\n",
263 | "A[n-1,n-1] = 1.\n",
264 | "b[n-1,0] = T_out\n",
265 | "\n",
266 | "for i in range(1, n-1):\n",
267 | " A[i, i-1] = r[i]-dr/2. # node-1\n",
268 | " A[i, i] = -2.*r[i] # node\n",
269 | " A[i, i+1] = r[i]+dr/2. # node+1\n",
270 | " \n",
271 | "#print('A \\n', A) # if you want to see what you matrix looks like\n",
272 | "#print('B \\n', b)\n",
273 | "\n",
274 | "#---- Solve using numpy.linalg.solve\n",
275 | "\n",
276 | "T = np.linalg.solve(A, b) # solve A*x = B for x\n",
277 | "\n",
278 | "#############################################\n",
279 | "# Compute linear temperature distribution #\n",
280 | "#############################################\n",
281 | "Tlin = np.zeros((n,1)) \n",
282 | "\n",
283 | "for i in range(n):\n",
284 | " Tlin[i] = T_in + (T_out - T_in)*(r[i] - r_in)/(r_out - r_in)\n",
285 | "\n",
286 | "plt.figure(figsize=(6,4), dpi=100)\n",
287 | "plt.plot(r,T, lw=2, label='Computed temperature')\n",
288 | "plt.plot(r,Tlin, lw=2, label='Linear temperature')\n",
289 | "plt.xlim([r_in,r_out])\n",
290 | "plt.ylim([T_out,T_in])\n",
291 | "plt.xlabel('$r$ (m)')\n",
292 | "plt.ylabel('$T$ ($^\\circ$C)')\n",
293 | "plt.legend()\n",
294 | "plt.show"
295 | ]
296 | },
297 | {
298 | "cell_type": "markdown",
299 | "metadata": {},
300 | "source": [
301 | "Play with different number of shells. Is the temperature distribution linear?\n",
302 | "To gain some insights on what type of distribution we are dealing with, we could try to plot in logarithmic-linear and log-log format. This investigation is motivated by the fact that many physical phenomena follow logarithmic or power laws. The former results in a line in a log-lin plot, and the latter in a log-log"
303 | ]
304 | },
305 | {
306 | "cell_type": "code",
307 | "execution_count": null,
308 | "metadata": {},
309 | "outputs": [],
310 | "source": [
311 | "plt.figure(figsize=(10,4), dpi=100)\n",
312 | "plt.subplot(1,2,1)\n",
313 | "plt.plot(r,T, lw=2, label='Computed temperature')\n",
314 | "plt.plot(r,Tlin, lw=2, label='Linear temperature')\n",
315 | "plt.xscale('log')\n",
316 | "plt.xlim([r_in,r_out])\n",
317 | "plt.ylim([T_out,T_in])\n",
318 | "plt.xlabel('$r$ (m)')\n",
319 | "plt.ylabel('$T$ ($^\\circ$C)')\n",
320 | "plt.legend()\n",
321 | "plt.subplot(1,2,2)\n",
322 | "plt.plot(r,T, lw=2, label='Computed temperature')\n",
323 | "plt.plot(r,Tlin, lw=2, label='Linear temperature')\n",
324 | "plt.xscale('log')\n",
325 | "plt.yscale('log')\n",
326 | "plt.xlim([r_in,r_out])\n",
327 | "plt.ylim([T_out,T_in])\n",
328 | "plt.xlabel('$r$ (m)')\n",
329 | "plt.ylabel('$T$ ($^\\circ$C)')\n",
330 | "plt.legend()\n",
331 | "plt.show\n"
332 | ]
333 | },
334 | {
335 | "cell_type": "markdown",
336 | "metadata": {},
337 | "source": [
338 | "Our educated guess payed off! The temperature distribution is obviously logarithmic, i.e.\n",
339 | "$$\n",
340 | "T(r)\\propto\\ln(r)\n",
341 | "$$\n",
342 | "To fit the boundary conditions, this profile ought to be:\n",
343 | "$$\n",
344 | "T(r)=T_\\text{in}+(T_\\text{out}-T_\\text{in})\\cfrac{\\ln\\left(\\cfrac{r}{r_\\text{in}}\\right)}{\\ln\\left(\\cfrac{r_\\text{out}}{r_\\text{in}}\\right)}\n",
345 | "$$\n",
346 | "and we can verify that this is the appropriate function:"
347 | ]
348 | },
349 | {
350 | "cell_type": "code",
351 | "execution_count": null,
352 | "metadata": {},
353 | "outputs": [],
354 | "source": [
355 | "\n",
356 | "Tlog = np.zeros((n,1))\n",
357 | "for i in range(n):\n",
358 | " Tlog[i] = T_in+(T_out-T_in)*np.log(r[i]/r_in)/np.log(r_out/r_in)\n",
359 | "plt.figure(figsize=(10,4), dpi=100)\n",
360 | "plt.subplot(1,2,1)\n",
361 | "plt.plot(r,T, lw=2, label='Computed temperature')\n",
362 | "plt.plot(r,Tlog,marker='o', mfc='none', label='Analytical solution')\n",
363 | "#plt.xscale('log')\n",
364 | "plt.xlim([r_in,r_out])\n",
365 | "plt.ylim([T_out,T_in])\n",
366 | "plt.xlabel('$r$ (m)')\n",
367 | "plt.ylabel('$T$ ($^\\circ$C)')\n",
368 | "plt.legend()\n",
369 | "plt.subplot(1,2,2)\n",
370 | "plt.plot(r,T, lw=2, label='Computed temperature')\n",
371 | "plt.plot(r,Tlog, marker='o', mfc='none', label='Analytical solution')\n",
372 | "plt.xscale('log')\n",
373 | "plt.xlim([r_in,r_out])\n",
374 | "plt.ylim([T_out,T_in])\n",
375 | "plt.xlabel('$r$ (m)')\n",
376 | "plt.ylabel('$T$ ($^\\circ$C)')\n",
377 | "plt.legend()\n",
378 | "plt.show"
379 | ]
380 | },
381 | {
382 | "cell_type": "markdown",
383 | "metadata": {},
384 | "source": [
385 | "Another way to derive the analytical solution is to consider the mathematical definition of Fourier's Law:\n",
386 | "$$\n",
387 | "q''=-k\\vec{\\nabla}T\\text{ which reduces to }q''=-k\\frac{dT}{dr}\\text{ in 1D radial conduction}\n",
388 | "$$\n",
389 | "Conservation of energy dictates that the heat rate be constant of any cylindrical surface of radius $r$. Owing to the expansion of the surface area with $r$, $A=2\\pi r L$ ($L$ is the length of the cylinder), the heat flux is not constant. \n",
390 | "\n",
391 | "Consequently, we have\n",
392 | "$$\n",
393 | "-(2\\pi r L)k\\frac{dT}{dr}=q\\text{ or }r\\frac{dT}{dr}=\\text{constant}=C\n",
394 | "$$\n",
395 | "which leads to:\n",
396 | "$$\n",
397 | "\\frac{dT}{dr}=\\frac{C}{r}\\Rightarrow T(r)=C\\ln(r)+B\n",
398 | "$$\n",
399 | "Consider the heat transfer between two cylindrical surfaces of length $L$ and radii $r_12.K. The pipe wall is 2 mm thick and its thermal conductivity is $k_\\text{p}=14$ W/m.K. The pipe is covered by insulation of thickness 10mm and thermal conductivity $k_\\text{ins}=0.05$ W/m.K. The oustide air is at $T_{\\infty,o}=23^\\circ$C and a convection heat transfer coefficient $h_\\text{o}=6$ W/m2.K. Calculate the heat gain in the fluid, and temperature distribution."
425 | ]
426 | },
427 | {
428 | "cell_type": "markdown",
429 | "metadata": {},
430 | "source": [
431 | "We break the system in cylindrical shells of constant thickness $\\Delta r$. According to Fourier's Law, the heat rates through the boundary at $r_i+\\Delta r/2$ and $r-\\Delta r_i/2$ must be equal and are respectively,\n",
432 | "$$\n",
433 | "-2\\pi\\left(r_i+\\Delta r/2\\right)k_{i-\\frac{1}{2}}\\frac{T_{i}-T_{i-1}}{\\Delta r}=\n",
434 | "-2\\pi\\left(r_i-\\Delta r/2\\right)k_{i+\\frac{1}{2}}\\frac{T_{i+1}-T_{i}}{\\Delta r}\n",
435 | "$$\n",
436 | "After a few simplifications, this identity reduces to\n",
437 | "$$\n",
438 | "\\left(r_i+\\Delta r/2\\right)k_{i-\\frac{1}{2}}T_{i-1}-\\left(\\left(r_i+\\Delta r/2\\right)k_{i-\\frac{1}{2}}+\\left(r_i-\\Delta r/2\\right)k_{i+\\frac{1}{2}}\\right)T_{i}+\\left(r_i-\\Delta r/2\\right)k_{i+\\frac{1}{2}}T_{i+1}=0\n",
439 | "$$\n",
440 | "At the boundaries, the temperature is defined by\n",
441 | "$$\n",
442 | "2\\pi r_0h_o\\left((T_{\\infty,\\text{out}}-T_0\\right)=-2\\pi\\left(r_0-\\Delta r/2\\right)k_\\frac{1}{2}\\frac{T_0-T_1}{\\Delta r}\n",
443 | "$$\n",
444 | "and\n",
445 | "$$\n",
446 | "-2\\pi\\left(r_{n-1}+\\Delta r/2\\right)k_\\frac{1}{2}\\frac{T_{n-2}-T_{n-1}}{\\Delta r}=2\\pi r_{n-1} h_\\text{in}\\left(T_{n-1}-T_{\\infty,\\text{in}}\\right)\n",
447 | "$$"
448 | ]
449 | },
450 | {
451 | "cell_type": "markdown",
452 | "metadata": {},
453 | "source": [
454 | "To summarize the system of equations writes:\n",
455 | "$$\n",
456 | "\\begin{split}\n",
457 | "&-\\left(r_0h_o+\\frac{\\left(r_0-\\Delta r/2\\right)k_\\frac{1}{2}}{\\Delta r}\\right)T_0+\\frac{\\left(r_0-\\Delta r/2\\right)k_\\frac{1}{2}}{\\Delta r}T_1=r_0h_0T_{\\infty,\\text{out}}\\\\\n",
458 | "&\\left(r_i+\\Delta r/2\\right)k_{i-\\frac{1}{2}}T_{i-1}-\\left(\\left(r_i+\\Delta r/2\\right)k_{i-\\frac{1}{2}}+\\left(r_i-\\Delta r/2\\right)k_{i+\\frac{1}{2}}\\right)T_{i}+\\left(r_i-\\Delta r/2\\right)k_{i+\\frac{1}{2}}T_{i+1}=0\\\\\n",
459 | "&\\frac{\\left(r_{n-1}+\\Delta r/2\\right)k_\\frac{1}{2}}{\\Delta r}T_{n-2}-\\left(h_\\text{in}r_{n-1}+\\frac{\\left(r_{n-1}+\\Delta r/2\\right)k_\\frac{1}{2}}{\\Delta r}T_{n-2}\\right)T_{n-1}=h_\\text{in}r_{n-1}T_{\\infty,\\text{in}}\n",
460 | "\\end{split}\n",
461 | "$$"
462 | ]
463 | },
464 | {
465 | "cell_type": "code",
466 | "execution_count": null,
467 | "metadata": {},
468 | "outputs": [],
469 | "source": [
470 | "%matplotlib inline\n",
471 | "%config InlineBackend.figure_format='svg'\n",
472 | "\n",
473 | "import matplotlib.pyplot as plt\n",
474 | "import numpy as np\n",
475 | "\n",
476 | "L = 12e-3 #wall thickness is 2.5 cm\n",
477 | "rout = 30e-3\n",
478 | "rin = 18e-3\n",
479 | "rinterface = 20e-3\n",
480 | "kins=0.05\n",
481 | "kpipe = 0.1\n",
482 | "hout = 6\n",
483 | "hin = 400\n",
484 | "n = 37 # Number of cells in the wall\n",
485 | "Tin = 6. #\n",
486 | "Tout = 23. #\n",
487 | "dr = float(L)/float(n-1) #cell size\n",
488 | "r = rout-np.array([i*dr for i in range(n)])\n",
489 | "A = np.zeros((n, n)) # pre-allocate [A] array\n",
490 | "b = np.zeros((n, 1)) # pre-allocate {b} column vector\n",
491 | "k = np.ones((n, 1))\n",
492 | "\n",
493 | "for i in range(n):\n",
494 | " if (r[i]-dr/2 > rinterface):\n",
495 | " k[i] = kins\n",
496 | " else:\n",
497 | " k[i] = kpipe\n",
498 | " \n",
499 | "A[0, 0] = -(rout*hout+(r[0]-dr/2)/dr*k[0])\n",
500 | "A[0, 1] = (r[0]-dr/2)/dr*k[0]\n",
501 | "b[0, 0] = -rout*hout*Tout\n",
502 | "A[n-1,n-2] = (r[n-1]+dr/2)/dr*k[n-1]\n",
503 | "A[n-1,n-1] = -(rin*hin+(r[n-1]+dr/2)/dr*k[n-1])\n",
504 | "b[n-1,0] = -rin*hin*Tin\n",
505 | "\n",
506 | "for i in range(1, n-1):\n",
507 | " A[i, i-1] = (r[i]+dr/2)/dr*k[i-1] # node-1\n",
508 | " A[i, i] = -(r[i]+dr/2)/dr*k[i-1]-(r[i]-dr/2)/dr*k[i] # node\n",
509 | " A[i, i+1] = (r[i]-dr/2)/dr*k[i] # node+1\n",
510 | " \n",
511 | "#print('A \\n', A)\n",
512 | "#print('B \\n', b)\n",
513 | "\n",
514 | "#---- Solve using numpy.linalg.solve\n",
515 | "\n",
516 | "T = np.linalg.solve(A, b) # solve A*x = B for x\n",
517 | "\n",
518 | "#print('T \\n', T)\n",
519 | "\n",
520 | "plt.figure(figsize=(6,4), dpi=100)\n",
521 | "plt.plot(r,T, lw=2, label='Wall temperature')\n",
522 | "#plt.xlim([0,L])\n",
523 | "#plt.ylim([To,Ti])\n",
524 | "plt.xlabel('$x$ (m)')\n",
525 | "plt.ylabel('$T$ ($^\\circ$C)')\n",
526 | "plt.legend()\n",
527 | "plt.show"
528 | ]
529 | },
530 | {
531 | "cell_type": "code",
532 | "execution_count": null,
533 | "metadata": {},
534 | "outputs": [],
535 | "source": []
536 | },
537 | {
538 | "cell_type": "code",
539 | "execution_count": null,
540 | "metadata": {},
541 | "outputs": [],
542 | "source": []
543 | }
544 | ],
545 | "metadata": {
546 | "kernelspec": {
547 | "display_name": "Python 3",
548 | "language": "python",
549 | "name": "python3"
550 | },
551 | "language_info": {
552 | "codemirror_mode": {
553 | "name": "ipython",
554 | "version": 3
555 | },
556 | "file_extension": ".py",
557 | "mimetype": "text/x-python",
558 | "name": "python",
559 | "nbconvert_exporter": "python",
560 | "pygments_lexer": "ipython3",
561 | "version": "3.8.5"
562 | }
563 | },
564 | "nbformat": 4,
565 | "nbformat_minor": 2
566 | }
567 |
--------------------------------------------------------------------------------
/Libraries/HT_external_convection.py:
--------------------------------------------------------------------------------
1 | """ Object name 1: FlatPlate
2 | Object name 2: CircularCylinder
3 | Object name 3: NoncircularCylinder
4 | Object name 4: BankofTubes
5 | """
6 |
7 | from sympy.interactive import printing
8 | printing.init_printing(use_latex='mathjax')
9 |
10 |
11 | from IPython.display import display,Image, Latex
12 | import numpy as np
13 | import math
14 | import scipy.constants as sc
15 | import sys
16 | try:
17 | import thermodynamics as thermo
18 | except ModuleNotFoundError:
19 | from Libraries import thermodynamics as thermo
20 | import sympy as sym
21 | #from sympy import *
22 |
23 | class FlatPlate(object):
24 | """ Definition of boundary layer thickness, friction coefficient, Nusselt number (both local and average)
25 | as a function of the regime.
26 | import HT_external_convection.py as extconv
27 |
28 | bl =extconv.FlatPlate(regime,thermal_bc,U_infty,nu,alpha,L,xi=0.0,Re_xc=5e5)
29 | where regime = 'laminar' or 'turbulent' or 'mixed',
30 | thermal_bc = 'isothermal', 'heat flux', 'unheated starting length',
31 | U_infty is the free stream velocity,
32 | nu the fluid viscosity,
33 | alpha the fluid thermal diffusivity,
34 | L length of the plate
35 | xi unheated started length (only applies of using unheated starting length)
36 | Re_xc critical Reynolds number for transition laminar to turbulence
37 |
38 | output: bl.Re_L Reynolds at the trailing edge of the plate (x=L)
39 |
40 | bl.local(x) calculates the local Re (bl.Re_x), Cf (bl.Cf_x), Nu (bl.Nu_x) and velocity
41 | thermal boundary layer thicknesses (bl.delta_x and bl.delta_Tx) at x based on thermal_bc
42 |
43 | bl.average(x) calculates the average Cf (bl.C_fave), Nu (bl.Nu_ave) over a length x from the leading edge
44 |
45 | """
46 | def __init__(self,regime,thermal_bc,U_infty,nu,alpha,L,xi=0.0,Re_xc=5e5):
47 | self.regime = regime
48 | self.thermal_bc = thermal_bc
49 | self.U_infty = U_infty
50 | self.nu = nu
51 | self.alpha = alpha
52 | self.Pr = self.nu/self.alpha
53 | self.L = L
54 | self.xi = xi
55 | self.Re_xc = Re_xc
56 | self.Re_L = self.L*self.U_infty/self.nu
57 | self.x_c = self.Re_xc*self.nu/self.U_infty
58 | if (self.regime != "laminar") and (self.regime != "turbulent") and (self.regime != "mixed"):
59 | print("Warning: regime is not properly defined")
60 | if self.thermal_bc != "isothermal" and self.thermal_bc != "heat flux" and self.thermal_bc != "unheated starting length":
61 | print("Warning: thermal boundary condition is not properly defined")
62 | if self.Re_L > self.Re_xc and self.regime == "laminar":
63 | print("Warning: The end plate Reynolds number is larger than the critical Reynolds number, consider 'mixed' regime instead")
64 | def local(self,x):
65 | self.x = x
66 | self.Re_x = self.U_infty*self.x/self.nu
67 | if x == 0.:
68 | self.delta_x = 0.
69 | self.delta_Tx = 0.
70 | self.Cf_x = 0.
71 | self.Nu_x = 0.
72 | else:
73 | if self.regime == "laminar":
74 | self.delta_x = 5.0*self.x/np.sqrt(self.Re_x)
75 | self.Cf_x = 0.664*np.power(self.Re_x,-1./2.)
76 | if self.thermal_bc == "isothermal":
77 | self.Nu_x = 0.332*np.power(self.Re_x,1./2.)*np.power(self.Pr,1./3.)
78 | elif self.thermal_bc == "heat flux":
79 | self.Nu_x = 0.453*np.power(self.Re_x,1./2.)*np.power(self.Pr,1./3.)
80 | elif self.thermal_bc == "unheated starting length":
81 | self.Re_xi = self.xi*self.U_infty/self.nu
82 | self.Nu_x = 0.332*np.power(self.Re_x,1./2.)*np.power(self.Pr,1./3.)/\
83 | np.power(1.-np.power(self.xi/self.x,3./4.),1./3.)
84 | elif self.regime == "turbulent":
85 | self.delta_x = 0.37*self.x*np.power(self.Re_x,-1./5.)
86 | self.Cf_x = 0.0592*np.power(self.Re_x,-1./5.)
87 | if self.thermal_bc == "isothermal":
88 | self.Nu_x = 0.0296*np.power(self.Re_x,4./5.)*np.power(self.Pr,1./3.)
89 | elif self.thermal_bc == "heat flux":
90 | self.Nu_x = 0.0296*np.power(self.Re_x,4./5.)*np.power(self.Pr,1./3.)
91 | elif self.thermal_bc == "unheated starting length":
92 | self.Re_xi = self.xi*self.U_infty/self.nu
93 | self.Nu_x = 0.0296*np.power(self.Re_x,4./5.)*np.power(self.Pr,1./3.)/\
94 | np.power(1.-np.power(self.xi/self.x,9./10.),1./9.)
95 | elif self.regime == "mixed":
96 | if self.x < self.x_c:
97 | self.delta_x = 5.0*self.x/np.sqrt(self.Re_x)
98 | self.Cf_x = 0.664*np.power(self.Re_x,-1./2.)
99 | if self.thermal_bc == "isothermal":
100 | self.Nu_x = 0.332*np.power(self.Re_x,1./2.)*np.power(self.Pr,1./3.)
101 | elif self.thermal_bc == "heat flux":
102 | self.Nu_x = 0.453*np.power(self.Re_x,1./2.)*np.power(self.Pr,1./3.)
103 | elif self.thermal_bc == "unheated starting length":
104 | self.Re_xi = self.xi*self.U_infty/self.nu
105 | self.Nu_x = 0.332*np.power(self.Re_x,1./2.)*np.power(self.Pr,1./3.)/\
106 | np.power(1.-np.power(self.xi/self.x,3./4.),1./3.)
107 | else:
108 | self.delta_x = 0.37*self.x*np.power(self.Re_x,-1./5.)
109 | self.Cf_x = 0.0592*np.power(self.Re_x,-1./5.)
110 | if self.thermal_bc == "isothermal":
111 | self.Nu_x = 0.0296*np.power(self.Re_x,4./5.)*np.power(self.Pr,1./3.)
112 | elif self.thermal_bc == "heat flux":
113 | self.Nu_x = 0.0296*np.power(self.Re_x,4./5.)*np.power(self.Pr,1./3.)
114 | elif self.thermal_bc == "unheated starting length":
115 | self.Re_xi = self.xi*self.U_infty/self.nu
116 | self.Nu_x = 0.0296*np.power(self.Re_x,4./5.)*np.power(self.Pr,1./3.)/\
117 | np.power(1.-np.power(self.xi/self.x,9./10.),1./9.)
118 |
119 | self.delta_Tx = self.delta_x*np.power(self.Pr,-1./3.)
120 | def average(self,x):
121 | self.x = x
122 | self.Re_x = self.U_infty*self.x/self.nu
123 | if x == 0.:
124 | print("The length cannot be zero")
125 | if self.regime == "laminar":
126 | self.Cf_ave = 1.328*np.power(self.Re_x,-1./2.)
127 | if self.thermal_bc == "isothermal" or self.thermal_bc == "heat flux":
128 | self.Nu_ave = 0.664*np.power(self.Re_x,1./2.)*np.power(self.Pr,1./3.)
129 | elif self.thermal_bc == "unheated starting length":
130 | p = 2.
131 | self.Re_xi = self.xi*self.U_infty/self.nu
132 | self.Nu_ave = 0.664*np.power(self.Re_x,1./2.)*np.power(self.Pr,1./3.)*\
133 | x/(x-self.xi)*np.power(1.-np.power(self.xi/x,(p+1.)/(p+2.)),p/(p+1.))
134 | elif self.regime == "turbulent":
135 | self.Cf_ave = 0.074*np.power(self.Re_x,-1./5.)
136 | if self.thermal_bc == "isothermal" or self.thermal_bc == "heat flux":
137 | self.Nu_ave = 0.037*np.power(self.Re_x,4./5.)*np.power(self.Pr,1./3.)
138 | elif self.thermal_bc == "unheated starting length":
139 | p = 8.
140 | self.Re_xi = self.xi*self.U_infty/self.nu
141 | self.Nu_ave = 0.664*np.power(self.Re_x,1./2.)*np.power(self.Pr,1./3.)*\
142 | x/(x-self.xi)*np.power(1.-np.power(self.xi/x,(p+1.)/(p+2.)),p/(p+1.))
143 | elif self.regime == "mixed":
144 | A = 0.037*np.power(self.Re_xc,4./5.)-0.664*np.power(self.Re_xc,1./2.)
145 |
146 | self.C_fave = 0.074*np.power(self.Re_x,-1./5.) - 2.*A/self.Re_x
147 | self.Nu_ave = (0.037*np.power(self.Re_x,4./5.) - A)*np.power(self.Pr,1./3.)
148 |
149 | class CircularCylinder(object):
150 | """ Nusselt correlations for cylinders
151 | import HT_external_convection.py as extconv
152 |
153 | bluff_body =extconv.CircularCylinder(correlation,Re,Pr,Pr_s = 0.0)
154 | where Re, Pr, and Pr_s are the Reynolds number, Prandtl number of the flow and surface Prandtl numbers, respectively. If using Hilpert of Churchill Bernstein correlations,
155 | Re and Pr must be defined at film temperature, Pr_s can be set to anything since it is not used.
156 | If using Zukauskas, Re and Pr are defined at temperature at infinity.
157 | correlation may be 'Hilpert', 'Churchill-Bernstein', 'Zukauskas'
158 | Example:
159 | bluff_body = extconv.CircularCylinder('Hilpert',Re,Pr)
160 | bluff_body = extconv.CircularCylinder('Churchill-Bernstein',Re,Pr)
161 | bluff_body = extconv.CircularCylinder('Zukauskas',Re,Pr,Pr_s = xx)
162 |
163 | Output: bluff_body.Nu average Nusselt number also bluff_body.correlation, bluff_body.Re, bluff_body.Pr, bluff_body.Pr_s
164 |
165 | bluff_body.correlation('Name of the correlation')
166 | Name of the correlation may be 'Hilpert', 'Churchill-Bernstein', 'Zukauskas'
167 |
168 |
169 | """
170 | def __init__(self,correlation,Re,Pr,Pr_s = 0.0):
171 | self.correlation = correlation
172 | self.Re = Re
173 | self.Pr = Pr
174 | self.Pr_s = Pr_s
175 | if correlation == "Zukauskas" and Pr_s == 0.0:
176 | print("Warning: Zukauskas correlation requires Pr_s")
177 | if self.correlation == "Hilpert":
178 | if self.Re < 0.4:
179 | print("Warning, Reynolds number too low for Hilpert Correlation")
180 | self.Nu = 0.
181 | elif self.Re < 4.:
182 | C = 0.989
183 | m = 0.33
184 | elif self.Re < 40:
185 | C = 0.911
186 | m = 0.385
187 | elif self.Re < 4000:
188 | C = 0.683
189 | m = 0.466
190 | elif self.Re < 40000.:
191 | C = 0.193
192 | m = 0.618
193 | elif self.Re <= 400000.:
194 | C = 0.027
195 | m = 0.805
196 | else :
197 | print("Warning Reynolds number is too high for the Hilpert Correlation")
198 | self.Nu = 0.
199 | if self.Re >= 0.4 and self.Re <= 400000.:
200 | self.Nu = C * self.Re**m * self.Pr**(1./3.)
201 | elif self.correlation == "Churchill-Bernstein":
202 | if (self.Re*self.Pr < 0.2):
203 | print("Warning: Product RePr lower than acceptable limit for Churchill Bernstein Correlation")
204 | self.Nu = 0.
205 |
206 | else:
207 | self.Nu = 0.3+(0.62*self.Re**(0.5)*self.Pr**(1./3.)) \
208 | /(1.+(0.4/self.Pr)**(2./3.))**(1./4.) \
209 | *(1.+(self.Re/282000.)**(5./8.))**(4./5.)
210 | elif self.correlation == "Zukauskas":
211 | if (self.Pr <= 10):
212 | n = 0.37
213 | else:
214 | n = 0.36
215 | if (self.Re < 1.) and (self.Re > 1.e6):
216 | print("Warning Reynolds number out of bounds for the Zukauskas Correlation")
217 | self.Nu = 0.
218 | else:
219 | if (self.Re < 40.):
220 | C = 0.75
221 | m = 0.4
222 | elif (self.Re < 1000.):
223 | C = 0.51
224 | m = 0.5
225 | elif (self.Re < 2.e5):
226 | C = 0.26
227 | m = 0.6
228 | else:
229 | C = 0.076
230 | m = 0.7
231 | self.Nu = C*self.Re**m*self.Pr**n*(self.Pr/self.Pr_s)**(1./4.)
232 |
233 | class NonCircularCylinder(object):
234 | """ Nusselt correlations for cylinders with non circular cross-sections.
235 | import HT_external_convection.py as extconv
236 |
237 | bluff_body =extconv.NonCircularCylinder(geometry,Re,Pr) where
238 | geometry = "angled square" square with stagnation point on one of its edges
239 | "square" square with stagnation point at the center of one of its faces
240 | "angled hexagon" hexagon with stagnation point on one of its edges
241 | "hexagon" hexagon with stagnation point at the center of one of its faces
242 | "thin plate" thin plate perpendicular to the flow
243 | Re: Reynolds number at film temperature
244 | Pr: Prandtl number at film temperature
245 |
246 | Output: bluff_body.Nu, bluff_body.Nu_front, bluff_body.Nu_back, the last two are for thin plate only
247 | also bluff_body.geometry, bluff_body.Re, bluff_body.Pr
248 |
249 | """
250 | def __init__(self,geometry,Re,Pr):
251 | self.geometry = geometry
252 | self.Re = Re
253 | self.Pr = Pr
254 | if self.geometry == "angled square":
255 | self.Nu_front = np.inf
256 | self.Nu_back = np.inf
257 | if self.Re < 6000:
258 | print("Warning, Reynolds number too low for Hilpert Correlation")
259 | self.Nu = np.inf
260 | elif self.Re <= 60000.:
261 | C = 0.304
262 | m = 0.59
263 | self.Nu = C * self.Re**m * self.Pr**(1./3.)
264 | else :
265 | print("Warning Reynolds number is too high for the Hilpert Correlation")
266 | self.Nu = np.inf
267 | elif self.geometry == "square":
268 | self.Nu_front = np.inf
269 | self.Nu_back = np.inf
270 | if self.Re < 5000:
271 | print("Warning, Reynolds number too low for Hilpert Correlation")
272 | self.Nu = np.inf
273 | elif self.Re <= 60000.:
274 | C = 0.158
275 | m = 0.66
276 | self.Nu = C * self.Re**m * self.Pr**(1./3.)
277 | else :
278 | print("Warning Reynolds number is too high for the Hilpert Correlation")
279 | self.Nu = np.inf
280 | elif self.geometry == "angled hexagon":
281 | self.Nu_front = np.inf
282 | self.Nu_back = np.inf
283 | if self.Re < 4500:
284 | print("Warning, Reynolds number too low for Hilpert Correlation")
285 | self.Nu = np.inf
286 | elif self.Re <= 90700.:
287 | C = 0.150
288 | m = 0.638
289 | self.Nu = C * self.Re**m * self.Pr**(1./3.)
290 | else :
291 | print("Warning Reynolds number is too high for the Hilpert Correlation")
292 | self.Nu = np.inf
293 | elif self.geometry == "hexagon":
294 | self.Nu_front = np.inf
295 | self.Nu_back = np.inf
296 | if self.Re < 5200:
297 | print("Warning, Reynolds number too low for Hilpert Correlation")
298 | self.Nu = np.inf
299 | elif self.Re <= 20400.:
300 | C = 0.164
301 | m = 0.638
302 | self.Nu = C * self.Re**m * self.Pr**(1./3.)
303 | elif self.Re <= 105000.:
304 | C = 0.039
305 | m = 0.78
306 | self.Nu = C * self.Re**m * self.Pr**(1./3.)
307 | else :
308 | print("Warning Reynolds number is too high for the Hilpert Correlation")
309 | self.Nu = np.inf
310 | elif self.geometry == "thin plate":
311 | self.Nu = np.inf
312 | if self.Re < 10000:
313 | print("Warning, Reynolds number too low for Hilpert Correlation")
314 | self.Nu_front = np.inf
315 | elif self.Re <= 50000.:
316 | C = 0.667
317 | m = 0.5
318 | self.Nu_front = C * self.Re**m * self.Pr**(1./3.)
319 | else :
320 | print("Warning Reynolds number is too high for the Hilpert Correlation for Nu_front")
321 | self.Nu_back = np.inf
322 | if self.Re < 7000:
323 | print("Warning, Reynolds number too low for Hilpert Correlation")
324 | self.Nu_back = np.inf
325 | elif self.Re <= 80000.:
326 | C = 0.191
327 | m = 0.667
328 | self.Nu_back = C * self.Re**m * self.Pr**(1./3.)
329 | else :
330 | print("Warning Reynolds number is too high for the Hilpert Correlation for Nu_front")
331 | self.Nu_back = np.inf
332 |
333 | class BankofTubes(object):
334 | """ Nusselt correlations for flow across banks of tubes
335 | import HT_external_convection.py as extconv
336 | bank = extconv.BankofTubes('aligned','air',T_i,T_s,T_o,"C",V_i,D,S_L,S_T,N_L=1)
337 | **Input:**
338 | arrangement = "aligned" tubes are aligned in row and column
339 | "staggered" tubes are staggered from one row to the next
340 | fluidspecie = 'air', 'water' or any other specie available in library thermodynamics
341 | T_i = inlet temperature
342 | T_s = tube surface temperature
343 | T_o = outlet temperature, provide a guess < T_s if unknown
344 | V_i: Inlet velocity
345 | S_L: tube center to tube center separation between two consecutive rows (perpendicular to the flow)
346 | S_T: tube center to tube center separation between two consecutive rows (aligned with the flow)
347 | N_L: number of rows aligned with flow if unknown give your best guess (default =1)
348 |
349 | Output:
350 | bank.T_m: arithmetic mean of T_i and T_o
351 | bank.Nu: average Nusselt number
352 | bank.Delta_T_lm: Log mean temperature difference
353 | bank.rho_i: inlet fluid density
354 | bank.mu_m: fluid viscosity at T_m
355 | bank.k_m: fluid thermal conductivity at T_m
356 | bank.Cp_m: Specific heat at T_m
357 | bank.Pr_m: Prandtl number at T_m
358 | bank.Pr_s: Prandtl number at T_s
359 | bank.Vmax: Max fluid velocity based on arrangement
360 | bank.Re: Reynolds number of the system
361 |
362 | bank.Nu: average Nusselt number
363 | bank.hbar: average heat transfer coefficient
364 |
365 | Functions:
366 | bank.heat_rate(N_T,N_L,L=1) creates bank.q, the heat rate per tube length is L is omitted
367 | or total heat rate if L is provided, based on the average convection coefficient
368 | bank.temperature_outlet_tube_banks(N_T,N_L) overwite bank.T_o (useful if T_o is unkown and you provided a
369 | guess. Rerun the object with new T_o to adjust Nu)
370 | bank.pressure_drop(N_L,f,chi) creates bank.Delta_p the pressure drop across the bank. f and chi must
371 | be extrapolated from graphs (see book or notebook)
372 |
373 |
374 | """
375 | def __init__(self,arrangement,fluidspecie,T_i,T_s,T_o,unit,V_i,D,S_L,S_T,N_L=1):
376 | self.arrangement = arrangement
377 | self.T_i = T_i
378 | self.T_s = T_s
379 | self.V_i = V_i
380 | if (T_o == T_s):
381 | print("T_o and T_s cannot be equal, setting T_o to (T_i+T_s)/2 (think of it as your first guess)")
382 | T_o = (T_i+T_s)/2.
383 | self.T_o = T_o
384 | T_m = (T_i + T_o)/2.
385 | self.T_m = T_m
386 | self.Delta_T_lm = ((T_s-T_i)-(T_s-T_o))/np.log((T_s-T_i)/(T_s-T_o))
387 | self.fluid = fluidspecie
388 | fluid_i = thermo.Fluid(fluidspecie,T_i,unit)
389 | fluid_m = thermo.Fluid(fluidspecie,T_m,unit)
390 | fluid_s = thermo.Fluid(fluidspecie,T_s,unit)
391 | self.rho_i = fluid_i.rho
392 | self.mu_m = fluid_m.mu
393 | self.k_m = fluid_m.k
394 | self.Cp_m = fluid_m.Cp
395 | self.Pr_m = fluid_m.nu/fluid_m.alpha
396 | self.Pr_s = fluid_s.nu/fluid_s.alpha
397 | self.S_L = S_L
398 | self.S_T = S_T
399 | self.N_L = N_L
400 | self.D = D
401 | if self.arrangement == 'aligned':
402 | self.Vmax = self.S_T*self.V_i/(self.S_T-D)
403 | elif self.arrangement == 'staggered':
404 | self.S_D = np.sqrt(self.S_L**2+(self.S_T/2.)**2)
405 | self.Vmax = self.S_T*V_i/(2.*(self.S_D-D))
406 | Re = self.rho_i*self.Vmax*self.D/self.mu_m
407 | self.Re = Re
408 | self.Nu = np.inf
409 | Corr_aligned = np.array([0.70,0.80,0.86,0.90,0.92,0.94,0.95,0.96,0.96,0.97,0.97,0.97,0.98,0.99,0.99,0.99,0.99,0.99,0.99])
410 | Corr_staggered = np.array([0.64,0.76,0.84,0.89,0.92,0.94,0.95,0.96,0.96,0.97,0.97,0.97,0.98,0.99,0.99,0.99,0.99,0.99,0.99])
411 | if (N_L < 20):
412 | if arrangement == 'aligned':
413 | Corr = Corr_aligned[N_L-1]
414 | elif arrangement == 'staggered':
415 | Corr = Corr_staggered[N_L-1]
416 | else:
417 | Corr = 1.
418 | if (Re < 10.):
419 | print('Warning: Re is out of bounds')
420 | if (Re >= 10.) and (Re <= 100.):
421 | if arrangement == 'aligned':
422 | C = 0.8
423 | m = 0.4
424 | elif arrangement == 'staggered':
425 | C = 0.9
426 | m = 0.4
427 | self.Nu = Corr*C*self.Re**m*self.Pr_m**(0.36)*(self.Pr_m/self.Pr_s)**(1./4.)
428 | elif (Re > 100.) and (Re <= 1000.):
429 | C = 0.51
430 | m = 0.
431 | self.Nu = Corr*C*self.Re**m*self.Pr_m**(0.36)*(self.Pr_m/self.Pr_s)**(1./4.)
432 | elif (Re > 1000.) and (Re <= 2.e5):
433 | if arrangement == 'aligned':
434 | if (S_T/S_L > 0.7):
435 | C = 0.27
436 | m = 0.63
437 | else:
438 | print('Warning: inefficient, S_T/S_L<0.7')
439 |
440 | elif arrangement == 'staggered':
441 | if (S_T/S_L < 2):
442 | C = 0.35*(S_T/S_L)**(1./5.)
443 | m = 0.6
444 | else:
445 | C = 0.40
446 | m = 0.6
447 | self.Nu = Corr*C*self.Re**m*self.Pr_m**(0.36)*(self.Pr_m/self.Pr_s)**(1./4.)
448 | elif (Re > 2e5) and (Re <= 2.e6):
449 | if arrangement == 'aligned':
450 | C = 0.021
451 | m = 0.84
452 | elif arrangement == 'staggered':
453 | C = 0.022
454 | m = 0.84
455 | self.Nu = Corr*C*self.Re**m*self.Pr_m**(0.36)*(self.Pr_m/self.Pr_s)**(1./4.)
456 | else:
457 | print('Warning: Re is out of bounds')
458 |
459 | self.hbar = self.Nu*self.k_m/self.D
460 | self.N_L_for_given_To = -np.log((self.T_s-self.T_o)/(self.T_s-self.T_i))/ \
461 | (np.pi*self.D*self.hbar)*(self.rho_i*self.V_i*self.S_T*self.Cp_m)
462 | if (self.N_L < 20) and (self.N_L_for_given_To >= 20):
463 | print("WARNING input N_L < 20 but N_L computed for input T_o >=20. \ Rerun your BankofTubes object with the appropriate N_L to calculate the correct Nu")
464 | if (self.N_L < 20) and (self.N_L_for_given_To >= 20):
465 | print("WARNING input N_L >= 20 but N_L computed for input T_o <20. \ Rerun your BankofTubes object with the appropriate N_L to calculate the correct Nu")
466 |
467 | def heat_rate(self,N_T,N_L,L=1):
468 | N = N_T*N_L
469 | if N_L > 20 and self.N_L < 20:
470 | print("WARNING: you chose N_L > 20 but your initial guess was < 20. \ Rerun your BankofTubes object with the appropriate N_L to calculate the correct Nu")
471 | elif N_L < 20 and self.N_L >= 20:
472 | print("WARNING: you chose N_L < 20 but your initial guess was > 20. \ Rerun your BankofTubes object with the appropriate N_L to calculate the correct Nu")
473 | self.q=N*self.hbar*np.pi*self.D*self.Delta_T_lm*L
474 |
475 |
476 | def temperature_outlet_tube_banks(self,N_T,N_L):
477 | if N_L >= 20 and self.N_L < 20:
478 | print("WARNING: you chose N_L > 20 but your initial guess was < 20. \ Rerun your BankofTubes object with the appropriate N_L to calculate the correct Nu")
479 | elif N_L < 20 and self.N_L >= 20:
480 | print("WARNING: you chose N_L < 20 but your initial guess was > 20. \ Rerun your BankofTubes object with the appropriate N_L to calculate the correct Nu")
481 | N = N_T*N_L
482 | self.T_o = self.T_s-(self.T_s-self.T_i)* \
483 | np.exp(-np.pi*self.D*N*self.hbar/(self.rho_i*self.V_i*N_T*self.S_T*self.Cp_m))
484 | def pressure_drop(self,N_L,f,chi):
485 | self.Delta_p = N_L*chi*(self.rho_i*self.Vmax**2/2)*f
486 |
487 |
488 |
489 |
490 |
491 |
--------------------------------------------------------------------------------
![\"my](\"ThermalCondAir.png\")
"
69 | ]
70 | },
71 | {
72 | "cell_type": "markdown",
73 | "metadata": {},
74 | "source": [
75 | "### Solution\n",
76 | "Assumptions:\n",
77 | "
\n",
75 | "The top surface of a heated compartment consists of very smooth (A) and highly roughened (B) portions, and the surface is placed in an atmospheric airstream. \n",
76 | "In the interest of minimizing total convection heat transfer from the surface, which orientation, (1) or (2),\n",
77 | "is preferred? If $T_s=100^\\circ\\text{C}$, $T_\\infty= 20^\\circ\\text{C}$, and $u_\\infty=20\\text{ m/s}$, what is the convection heat transfer from the entire surface for this orientation?\n",
78 | "\n",
79 | "Calculate the value of the average heat transfer coefficient when the entire plate is rotated $90^\\circ$ so that half of the leading edge consists of a very smooth portion (A) and the other half consists of a highly roughened portion (B)."
80 | ]
81 | },
82 | {
83 | "cell_type": "code",
84 | "execution_count": null,
85 | "metadata": {},
86 | "outputs": [],
87 | "source": []
88 | }
89 | ],
90 | "metadata": {
91 | "kernelspec": {
92 | "display_name": "Python 3",
93 | "language": "python",
94 | "name": "python3"
95 | },
96 | "language_info": {
97 | "codemirror_mode": {
98 | "name": "ipython",
99 | "version": 3
100 | },
101 | "file_extension": ".py",
102 | "mimetype": "text/x-python",
103 | "name": "python",
104 | "nbconvert_exporter": "python",
105 | "pygments_lexer": "ipython3",
106 | "version": "3.8.5"
107 | }
108 | },
109 | "nbformat": 4,
110 | "nbformat_minor": 4
111 | }
112 |
--------------------------------------------------------------------------------
/Libraries/raw.githubusercontent.com/yvesdubief/UVM-ME144-Heat-Transfer/master/Libraries/Tables/Water1atm.csv:
--------------------------------------------------------------------------------
1 | Temperature (K),Pressure (MPA),Density (kg/m^3),Cp (J/g.K),viscosity (Pa.s),Thermal conductivity (W/m.K)
2 | 274.00,0.10132,999.89,4.2166,0.0017398,0.56269
3 | 275.00,0.10132,999.94,4.2135,0.0016818,0.56459
4 | 276.00,0.10132,999.96,4.2106,0.0016269,0.56649
5 | 277.00,0.10132,999.97,4.2079,0.0015748,0.56839
6 | 278.00,0.10132,999.97,4.2054,0.0015253,0.57029
7 | 279.00,0.10132,999.95,4.2031,0.0014783,0.57219
8 | 280.00,0.10132,999.91,4.2009,0.0014335,0.57409
9 | 281.00,0.10132,999.86,4.1990,0.0013909,0.57598
10 | 282.00,0.10132,999.79,4.1971,0.0013503,0.57788
11 | 283.00,0.10132,999.72,4.1954,0.0013115,0.57976
12 | 284.00,0.10132,999.62,4.1938,0.0012745,0.58165
13 | 285.00,0.10132,999.52,4.1924,0.0012392,0.58352
14 | 286.00,0.10132,999.40,4.1910,0.0012054,0.58539
15 | 287.00,0.10132,999.27,4.1898,0.0011730,0.58725
16 | 288.00,0.10132,999.13,4.1886,0.0011421,0.58910
17 | 289.00,0.10132,998.97,4.1876,0.0011124,0.59094
18 | 290.00,0.10132,998.80,4.1866,0.0010840,0.59278
19 | 291.00,0.10132,998.63,4.1857,0.0010567,0.59459
20 | 292.00,0.10132,998.44,4.1849,0.0010305,0.59640
21 | 293.00,0.10132,998.24,4.1842,0.0010053,0.59819
22 | 294.00,0.10132,998.03,4.1835,0.00098110,0.59997
23 | 295.00,0.10132,997.81,4.1829,0.00095784,0.60174
24 | 296.00,0.10132,997.58,4.1823,0.00093545,0.60349
25 | 297.00,0.10132,997.34,4.1818,0.00091389,0.60522
26 | 298.00,0.10132,997.09,4.1814,0.00089313,0.60694
27 | 299.00,0.10132,996.83,4.1810,0.00087312,0.60864
28 | 300.00,0.10132,996.56,4.1806,0.00085383,0.61032
29 | 301.00,0.10132,996.28,4.1803,0.00083521,0.61199
30 | 302.00,0.10132,995.99,4.1801,0.00081724,0.61363
31 | 303.00,0.10132,995.69,4.1798,0.00079990,0.61526
32 | 304.00,0.10132,995.39,4.1797,0.00078314,0.61687
33 | 305.00,0.10132,995.08,4.1795,0.00076694,0.61846
34 | 306.00,0.10132,994.75,4.1794,0.00075128,0.62002
35 | 307.00,0.10132,994.42,4.1793,0.00073613,0.62157
36 | 308.00,0.10132,994.08,4.1793,0.00072147,0.62310
37 | 309.00,0.10132,993.74,4.1792,0.00070728,0.62460
38 | 310.00,0.10132,993.38,4.1792,0.00069354,0.62609
39 | 311.00,0.10132,993.02,4.1793,0.00068023,0.62755
40 | 312.00,0.10132,992.65,4.1793,0.00066733,0.62900
41 | 313.00,0.10132,992.27,4.1794,0.00065482,0.63042
42 | 314.00,0.10132,991.89,4.1795,0.00064269,0.63181
43 | 315.00,0.10132,991.50,4.1796,0.00063092,0.63319
44 | 316.00,0.10132,991.10,4.1798,0.00061950,0.63455
45 | 317.00,0.10132,990.69,4.1799,0.00060842,0.63588
46 | 318.00,0.10132,990.28,4.1801,0.00059765,0.63719
47 | 319.00,0.10132,989.85,4.1803,0.00058720,0.63848
48 | 320.00,0.10132,989.43,4.1805,0.00057704,0.63975
49 | 321.00,0.10132,988.99,4.1808,0.00056716,0.64099
50 | 322.00,0.10132,988.55,4.1810,0.00055756,0.64221
51 | 323.00,0.10132,988.10,4.1813,0.00054823,0.64342
52 | 324.00,0.10132,987.65,4.1816,0.00053914,0.64460
53 | 325.00,0.10132,987.19,4.1819,0.00053031,0.64575
54 | 326.00,0.10132,986.72,4.1822,0.00052171,0.64689
55 | 327.00,0.10132,986.25,4.1826,0.00051333,0.64800
56 | 328.00,0.10132,985.77,4.1829,0.00050518,0.64909
57 | 329.00,0.10132,985.28,4.1833,0.00049724,0.65017
58 | 330.00,0.10132,984.79,4.1837,0.00048950,0.65122
59 | 331.00,0.10132,984.29,4.1840,0.00048197,0.65224
60 | 332.00,0.10132,983.78,4.1845,0.00047462,0.65325
61 | 333.00,0.10132,983.27,4.1849,0.00046745,0.65424
62 | 334.00,0.10132,982.76,4.1853,0.00046047,0.65521
63 | 335.00,0.10132,982.23,4.1858,0.00045366,0.65615
64 | 336.00,0.10132,981.71,4.1863,0.00044701,0.65708
65 | 337.00,0.10132,981.17,4.1867,0.00044053,0.65798
66 | 338.00,0.10132,980.63,4.1872,0.00043420,0.65887
67 | 339.00,0.10132,980.09,4.1878,0.00042802,0.65974
68 | 340.00,0.10132,979.54,4.1883,0.00042199,0.66058
69 | 341.00,0.10132,978.98,4.1888,0.00041610,0.66141
70 | 342.00,0.10132,978.42,4.1894,0.00041034,0.66222
71 | 343.00,0.10132,977.85,4.1900,0.00040472,0.66301
72 | 344.00,0.10132,977.28,4.1906,0.00039923,0.66378
73 | 345.00,0.10132,976.70,4.1912,0.00039387,0.66453
74 | 346.00,0.10132,976.12,4.1918,0.00038862,0.66527
75 | 347.00,0.10132,975.53,4.1924,0.00038349,0.66599
76 | 348.00,0.10132,974.93,4.1931,0.00037848,0.66669
77 | 349.00,0.10132,974.33,4.1938,0.00037358,0.66737
78 | 350.00,0.10132,973.73,4.1945,0.00036878,0.66803
79 | 351.00,0.10132,973.12,4.1952,0.00036409,0.66868
80 | 352.00,0.10132,972.50,4.1959,0.00035950,0.66931
81 | 353.00,0.10132,971.88,4.1966,0.00035501,0.66992
82 | 354.00,0.10132,971.26,4.1974,0.00035062,0.67052
83 | 355.00,0.10132,970.63,4.1982,0.00034632,0.67110
84 | 356.00,0.10132,969.99,4.1990,0.00034210,0.67167
85 | 357.00,0.10132,969.35,4.1998,0.00033798,0.67222
86 | 358.00,0.10132,968.71,4.2006,0.00033394,0.67275
87 | 359.00,0.10132,968.06,4.2015,0.00032998,0.67327
88 | 360.00,0.10132,967.40,4.2023,0.00032611,0.67378
89 | 361.00,0.10132,966.74,4.2032,0.00032231,0.67427
90 | 362.00,0.10132,966.08,4.2041,0.00031859,0.67474
91 | 363.00,0.10132,965.41,4.2051,0.00031495,0.67520
92 | 364.00,0.10132,964.74,4.2060,0.00031138,0.67565
93 | 365.00,0.10132,964.06,4.2070,0.00030787,0.67608
94 | 366.00,0.10132,963.37,4.2080,0.00030444,0.67649
95 | 367.00,0.10132,962.69,4.2090,0.00030107,0.67690
96 | 368.00,0.10132,961.99,4.2100,0.00029777,0.67729
97 | 369.00,0.10132,961.29,4.2111,0.00029453,0.67767
98 | 370.00,0.10132,960.59,4.2121,0.00029136,0.67803
99 | 371.00,0.10132,959.88,4.2132,0.00028824,0.67838
100 | 372.00,0.10132,959.17,4.2144,0.00028519,0.67872
101 | 373.00,0.10132,958.46,4.2155,0.00028219,0.67904
102 |
--------------------------------------------------------------------------------
/Libraries/HT_thermal_resistance.py:
--------------------------------------------------------------------------------
1 | """Object name: Resistance
2 | Function name: serial_sum(R,nori,nend), performs serial sum of a resistance object list from nori to nend
3 | Function name: parallel_sum(R,nori,nend), performs parallel sum of a resistance object list from nori to nend
4 | """
5 | ### definition of thermal resistance ###
6 | from sympy.interactive import printing
7 | printing.init_printing(use_latex='mathjax')
8 |
9 |
10 | from IPython.display import display,Image, Latex
11 | import numpy as np
12 | import math
13 | import scipy.constants as sc
14 |
15 | import sympy as sym
16 | #from sympy import *
17 |
18 | class Resistance(object):
19 | """ Defines thermal resistances for conduction, convection and radiation heat transfer.
20 | First define the object attached with class with the name used in the thermal circuit
21 | and the units, which can only be 'W', 'W/m' or 'W/m^2'
22 | Second use self.conduction, self.convection or self.radiation to calculate your
23 | resistance. Each mode requires different arguments:
24 |
25 | from Libraries import HT_thermal_resistance as res
26 | R = []
27 | R.append(res.Resistance("$label$", "units")) where units = 'W', 'W/m' or 'W/m^2'
28 |
29 | then
30 |
31 | For conduction, there are 3 options:
32 |
33 | - R.cond_plane(k, L, A = 1.0) for planar conduction: k is the thermal conductivity,
34 | L is the thickness of the wall, and A is the optional surface area (=1 by default)
35 | - R.cond_cylinder(k , ra, rb, L = 1.0, angle = 2.*math.pi) for conduction in a
36 | cylindrical shell between the radii ra (internal) and rb (external). L is the length
37 | of the shell (optional, default = 1) and angle is angular dimension of shell, also
38 | optional and set to a full revolution by default (2 pi)
39 | - R.cond_sphere(k, ra, rb, scale = 1.0) for conductuion within a spherical shell bounded by radii ra and rb
40 | ra < rb. The optional parameter scale allows to calculate the thermal resistance for a fraction
41 | of a spherical shell. For instance a cornea is about 1/3 of spherical shell, so scale = 1./3.
42 |
43 | Convection:
44 | - R.convection(h, A = 1.0), where h is the convection coefficient (W/m^2K) and A is
45 | the surface area (optional, default is unit surface aera 1 m^2)
46 |
47 | Radiation:
48 | - R.radiation(eps, T_s, T_sur, A = 1.0), where eps is the permissivity of the material, T_s
49 | the surface temperature, T_sur the far away surface temperature, A the surface area (optional,
50 | by default A is the unit surface area 1 m^2).
51 |
52 | Contact:
53 |
54 | - R.contact(R,A,R_name= "R_{t}",A_name = "A",T_a_name = "T_a",Tb_name = "T_b"), where R is the contact resistance, typically obtained from a table
55 | A is the surface area
56 | The minimum number of arguments are:
57 | R.contact(R,A)
58 |
59 | R.display_equation(index) displays the heat flux/rate equations for a given resistance. index is the number of
60 | your resistance (you specify)
61 |
62 | Outputs:
63 | - R[i].R the resistance of element i, R[i].h the convection or radiation coefficient.
64 |
65 | Functions include
66 | R_tot = res.serial_sum(R,first_resistance,last_resistance) sums serial resistance
67 | R_tot = res.parallel_sum(R,first_resistance,last_resistance) sums parallel resistance
68 |
69 |
70 |
71 | """
72 | def __init__(self,name,units):
73 | self.name = name
74 | self.units = units
75 | def cond_plane(self, k, L, A = 1.0):
76 | self.mode = "conduction"
77 | self.geometry = "planar"
78 | self.k = k
79 | if k <= 0.:
80 | print("problem with the definition of thermal conductivity")
81 | self.L = L
82 | self.A = A
83 | self.R = self.L / (self.k * self.A)
84 | def cond_cylinder(self, k , ra, rb, L = 1.0, angle = 2.*math.pi):
85 | self.mode = "conduction"
86 | self.geometry = "cylindrical"
87 | self.k = k
88 | if k <= 0.:
89 | print("problem with the definition of thermal conductivity")
90 | self.ra = ra
91 | self.rb = rb
92 | if ra*rb <= 0.:
93 | print("problem with the definition of radii")
94 | self.L = L
95 | self.angle = angle
96 | self.R = np.log(rb/ra)/(angle*L*k)
97 | def cond_sphere(self, k, ra, rb, scale = 1.0):
98 | self.mode = "conduction"
99 | self.geometry = "spherical"
100 | self.k = k
101 | if k <= 0.:
102 | print("problem with the definition of thermal conductivity")
103 | self.ra = ra
104 | self.rb = rb
105 | if ra*rb <= 0.:
106 | print("problem with the definition of radii")
107 | self.R = (1./r_a-1./r_b)/(scale*4.*math.pi*k)
108 | def convection(self, h, A = 1.0):
109 | self.mode = 'convection'
110 | self.geometry = "whatever"
111 | self.R = 1./(h*A)
112 | self.A = A
113 | self.h = h
114 | def radiation(self,eps,T_s,T_sur, A = 1.0):
115 | self.R = 1./(eps*sc.sigma*(T_s+T_sur)*(T_s**2+T_sur**2)*A)
116 | self.mode = 'radiation'
117 | self.geometry = "whatever"
118 | self.A = A
119 | self.h = eps*sc.sigma*(T_s+T_sur)*(T_s**2+T_sur**2)
120 | def contact(self, R, A=1.0):
121 | self.R = R/A
122 | self.geometry = 'whatever'
123 | self.mode = 'contact'
124 |
125 |
126 |
127 | ### summation of thermal resistance (R is a vector) ###
128 | def serial_sum(R,nori,nend):
129 | sum = 0.
130 | for i in range(nori,nend+1):
131 | sum += R[i].R
132 | return sum
133 |
134 | def parallel_sum(R,nori,nend):
135 | sum = 0.
136 | for i in range(nori,nend+1):
137 | sum += 1./R[i].R
138 | return 1./sum
139 |
140 |
141 |
--------------------------------------------------------------------------------
/Libraries/thermodynamics.py:
--------------------------------------------------------------------------------
1 | """ Object name: Fluid"""
2 | import sys
3 | sys.path.insert(0, 'Tables/')
4 | import numpy as np
5 | import scipy
6 | import scipy.optimize
7 | from scipy.constants import convert_temperature
8 | def C2K(T):
9 | return convert_temperature(T,'Celsius','Kelvin')
10 | def C2F(T):
11 | return convert_temperature(T,'Celsius','Fahrenheit')
12 | def F2K(T):
13 | return convert_temperature(T,'Fahrenheit','Kelvin')
14 | def F2C(T):
15 | return convert_temperature(T,'Fahrenheit','Celsius')
16 | def K2F(T):
17 | return convert_temperature(T,'Kelvin','Fahrenheit')
18 | def K2C(T):
19 | return convert_temperature(T,'Kelvin','Celsius')
20 | import scipy.constants as sc
21 |
22 | def interpolate_table(target,index,xquantity,yquantity):
23 | return yquantity[index] + \
24 | (yquantity[index+1]-yquantity[index])* \
25 | (target-xquantity[index])/(xquantity[index+1]-xquantity[index])
26 |
27 | class Fluid(object):
28 | """ How to:
29 | from NewLibraries import thermodynamics as thermo
30 |
31 | fluid_of_interest = thermo.Fluid(material,T) material can be air, water, argon and krypton (see below for ranges)
32 | and the temperature of the fluid T is in Kelvin.
33 | Outputs:
34 | The new object computes thermodynamic properties of air between -150 C and 400 C,
35 | water between 274K and 373K, argon between 100 and 700K and
36 | krypton between 150 and 700 K under 1 atm. Argon, krypton and water were obtained
37 | through http://webbook.nist.gov/chemistry/fluid/
38 | More fluids to be added in the future
39 | fluid_of_interest.beta thermal expansion coefficient
40 | fluid_of_interest.rho density
41 | fluid_of_interest.Cp specific heat
42 | fluid_of_interest.mu dynamic viscosity
43 | fluid_of_interest.k thermal conductivity
44 | fluid_of_interest.nu kinematic viscosity
45 | fluid_of_interest.alpha thermal diffusivity
46 | fluid_of_interest.Pr
47 |
48 |
49 | """
50 | def __init__(self,name,T,unit = "K",P = 101325.01):
51 | dirTables = 'https://raw.githubusercontent.com/yvesdubief/UVM-ME144-Heat-Transfer/master/Libraries/Tables/'
52 | self.name = name
53 | if unit == "C":
54 | T = C2K(T)
55 | elif unit == "F":
56 | T = F2K(T)
57 | self.T = T
58 | self.P = P
59 | if P != 101325.01:
60 | print("All available tables are for P=1ATM, reverting to P=101325.01Pa")
61 | self.P = 101325.01
62 | if self.name == 'water':
63 | if T < 274 or T > 373:
64 | print("Temperature is out of bounds for liquid water")
65 | return
66 | url = dirTables+'Water1atm.csv'
67 | Ttab,ptab,rhotab,Cptab,mutab,ktab = \
68 | np.genfromtxt(url, delimiter=',', skip_header = 1, unpack=True, dtype=float)
69 | Ntab = len(Ttab)
70 | Cptab *= 1e3
71 | nutab = mutab/rhotab
72 | alphatab = ktab/(rhotab*Cptab)
73 | Prtab = nutab/alphatab
74 | dTtab = Ttab[1] - Ttab[0]
75 | # compute beta from -rho(d rho/dT)
76 | betatab = -(1./rhotab)*np.gradient(rhotab)/dTtab
77 | i = int((T-Ttab[0])/dTtab)
78 | if (i == Ntab - 1):
79 | i == Ntab - 2
80 | elif self.name == 'argon':
81 | if T < 100 or T > 700:
82 | print("Temperature is out of bounds for argon")
83 | return
84 | url = dirTables + 'Argon1atm.csv'
85 | Ttab,ptab,rhotab,Cptab,mutab,ktab = \
86 | np.loadtxt(url, delimiter=',', skiprows = 1, unpack=True, dtype=float)
87 | Ntab = len(Ttab)
88 | Cptab *= 1e3
89 | nutab = mutab/rhotab
90 | alphatab = ktab/(rhotab*Cptab)
91 | Prtab = nutab/alphatab
92 | dTtab = Ttab[1] - Ttab[0]
93 | # compute beta from -rho(d rho/dT)
94 | betatab = -(1./rhotab)*np.gradient(rhotab)/dTtab
95 | i = int((T-Ttab[0])/dTtab)
96 | if (i == Ntab - 1):
97 | i == Ntab - 2
98 | elif self.name == 'krypton':
99 | if T < 150 or T > 740:
100 | print("Temperature is out of bounds for krypton")
101 | return
102 | url = dirTables + 'Krypton1atm.csv'
103 | Ttab,ptab,rhotab,Cptab,mutab,ktab = \
104 | np.loadtxt(url, delimiter=',', skiprows = 1, unpack=True, dtype=float)
105 | Ntab = len(Ttab)
106 | Cptab *= 1e3
107 | nutab = mutab/rhotab
108 | alphatab = ktab/(rhotab*Cptab)
109 | Prtab = nutab/alphatab
110 | dTtab = Ttab[1] - Ttab[0]
111 | # compute beta from -rho(d rho/dT)
112 | betatab = -(1./rhotab)*np.gradient(rhotab)/dTtab
113 | i = int((T-Ttab[0])/dTtab)
114 | if (i == Ntab - 1):
115 | i == Ntab - 2
116 | elif self.name == 'air':
117 | if T < C2K(-150.) or T > C2K(400.):
118 | print("Temperature is out of bounds of the table for air")
119 | return
120 | url = dirTables + 'Air1atm.csv'
121 | # url = 'https://raw.githubusercontent.com/yvesdubief/UVM-ME144-Heat-Transfer/master/Libraries/Tables/Air1atm.csv'
122 | # url = 'Tables/Air1atm.csv'
123 | Ttab,rhotab,Cptab,ktab,nutab,betatab,Prtab = \
124 | np.genfromtxt(url, delimiter=',', skip_header = 1, unpack=True, dtype=float)
125 | Ntab = len(Ttab)
126 | Ttab = C2K(Ttab)
127 | Cptab *= 1e3
128 | nutab *= 1e-6
129 | mutab = rhotab*nutab
130 | alphatab = ktab/(rhotab*Cptab)
131 | Prtab = nutab/alphatab
132 | i = 0
133 | while (Ttab[i] < T) and (i
"
19 | ]
20 | },
21 | {
22 | "cell_type": "markdown",
23 | "metadata": {},
24 | "source": [
25 | "The arrangement of cylinders may be aligned or staggered as shown in the following figures. The flow and geometrical parameters will be used in the derivation of temperature equations and Nusselt number correlation.\n",
26 | "
"
27 | ]
28 | },
29 | {
30 | "cell_type": "markdown",
31 | "metadata": {},
32 | "source": [
33 | "This notebook should cover a wide variety of problems, providing that the assumption of isothermal boundary conditions on the tubes is (approximately) valid. The tube surface temperature is $T_s$. \n",
34 | "The flow and geometrical parameters of importance to solve this problem are:\n",
35 | "\n",
36 | "* Arithmetic mean of temperature between inlet $T_i$ and outlet $T_o$ of the bank. \n",
37 | "$$\n",
38 | "T_m = \\frac{T_i+T_o}{2}\n",
39 | "$$\n",
40 | "* Reynolds number based on the max velocity within the bank $V_\\text{max}$, the density and viscosity based on $T_m$:\n",
41 | "$$\n",
42 | "Re=\\frac{\\rho V_\\text{max}D}{\\mu}\n",
43 | "$$\n",
44 | "**Question: At what temperature should you estimate $\\rho$ and $\\mu$?** The energy of the flow comes from the inlet and the velocity $V_\\mathrm{max}$ is calculated from the inlet velocity. The density should therefore be estimated at $T_i$. The viscous forces however occur throughout the domain, so $\\mu$ should be estimated at $T_m$. In some cases $T_o$ is the quantity to be found. It is acceptable to use $\\mu(T_i)$, but you must verify that the temperature difference $\\Delta T=\\vert T_i-T_o\\vert$ is not too large. If it is, you must repeat the calculation iteratively with $\\mu(T_i)$ until $T_o$ converges.\n",
45 | "\n",
46 | "* Prandtl number $Pr$ based on $T_m$ \n",
47 | "* Surface Prandtl number $Pr_s$ based on $T_s$\n",
48 | "* Number of tubes in the transversal direction $N_T$, longitudinal direction $N_L$ and total $N=N_T\\times N_L$\n",
49 | "* The transversal $S_T$ and longitudinal $S_L$ separations between tubes in a row and between rows.\n",
50 | "* The type of tube arrangement: \n",
51 | " * Aligned\n",
52 | "$$\n",
53 | "V_\\text{max}=\\frac{S_T}{S_T-D}V_i\n",
54 | "$$\n",
55 | " * Staggered\n",
56 | "$$\n",
57 | "V_\\text{max}=\\frac{S_T}{2(S_D-D)}V_i\\text{ with }S_D=\\sqrt{S_L^2+\\left(\\frac{S_T}{2}\\right)^2}\n",
58 | "$$\n"
59 | ]
60 | },
61 | {
62 | "cell_type": "markdown",
63 | "metadata": {},
64 | "source": [
65 | "The Nusselt number correlation for a bank of tubes is a variation of the Zukauskas correlation:\n",
66 | "$$\n",
67 | "Nu = C_2C_1Re^mPr^{0.36}\\left(\\frac{Pr}{Pr_s}\\right)^{1/4}\n",
68 | "$$\n",
69 | "where $C_2$ depends on $N_L$. In the library, the function for this correlation is\n",
70 | "Nu_tube_banks(Re,Pr,Pr_s,S_L,S_T,N_L,arrangement) .\n",
71 | "\n",
72 | "The heat rate per unit length across the tube bank is\n",
73 | "$$\n",
74 | "q'=N\\overline{h}\\pi D \\Delta T_\\text{lm}\n",
75 | "$$\n",
76 | "where the temperature drop is the log-mean temperature difference\n",
77 | "$$\n",
78 | "\\Delta T_\\text{lm}=\\cfrac{(T_s-T_i)-(T_s-T_o)}{\\ln\\left(\\cfrac{T_s-T_i}{T_s-T_o}\\right)}\n",
79 | "$$\n",
80 | "which accounts for the exponential variation of temperature across the bank\n",
81 | "$$\n",
82 | "\\cfrac{T_s-T_o}{T_s-T_i}=\\exp\\left(-\\cfrac{\\pi D N \\overline{h}}{\\rho V_i N_T S_T C_p}\\right)\n",
83 | "$$\n",
84 | "where $\\rho$, $C_p$ and $V_i$ are inlet quantities if $T_o$ is unknown of the arthimetic mean temperature if available. Note that $N=N_L\\times N_T$ thus \n",
85 | "$$\n",
86 | "\\cfrac{T_s-T_o}{T_s-T_i}=\\exp\\left(-\\cfrac{\\pi D N_L \\overline{h}}{\\rho V_i S_T C_p}\\right)\n",
87 | "$$\n",
88 | "One may want to determine the number of tubes necessary to achieve a given $T_o$. The number of tubes in the transverse directions is typically dictated by the geometry of the system, so we are looking for $N_L$:\n",
89 | "$$\n",
90 | "N_L = \\cfrac{\\rho V_i S_T C_p}{\\pi D \\overline{h}} \\log\\left(\\cfrac{T_s-T_i}{T_s-T_o}\\right)\n",
91 | "$$\n"
92 | ]
93 | },
94 | {
95 | "cell_type": "markdown",
96 | "metadata": {},
97 | "source": [
98 | "The pressure loss through the tube bank is a critical component of the heat exchanger design. The presence of obstacles in the flow requires an increase in the mechanical energy necessary to drive the flow at a given flow rate. The pressure loss, given all parameters above, is\n",
99 | "$$\n",
100 | "\\Delta p = N_L\\,\\chi\\, f\\,\\frac{\\rho V_\\text{max}^2}{2}\n",
101 | "$$\n",
102 | "where the friction factor $f$ and the parameter $\\chi$ are given by the graphs below for the aligned (top) and staggered (bottom) arrangements. These graphs use two new quantities, the longitudnal and transverse pitches:\n",
103 | "$$\n",
104 | "P_L=\\frac{S_L}{D}\\text{ and } P_T=\\frac{S_T}{D}\n",
105 | "$$\n",
106 | "
\n",
107 | "
"
108 | ]
109 | },
110 | {
111 | "cell_type": "markdown",
112 | "metadata": {},
113 | "source": [
114 | "## Problem1\n",
115 | "A preheater involves the use of condensing steam on the inside of a bank of tubes to heat air that enters at $P_i=1 \\text{ atm}$ and $T_i=25^\\circ\\text{C}$. The air moves at $V_i=5\\text{ m/s}$ in cross flow over the tubes. Each tube is $L=1\\text{ m}$ long and has an outside diameter of $D=10 \\text{ mm}$. The bank consists of columns of 14 tubes in the transversal direction $N_T=14$ and $N_L$ rows in the direction of flow. The arrangement of tubes is aligned array for which $S_T=S_L=15\\text{ mm}$. What is the minimum value of $N_L$ needed to achieve an outlet temperature of $T_o=75^\\circ\\text{C}$? What is the corresponding pressure drop across the tube bank?\n"
116 | ]
117 | },
118 | {
119 | "cell_type": "code",
120 | "execution_count": 8,
121 | "metadata": {},
122 | "outputs": [],
123 | "source": [
124 | "import numpy as np\n",
125 | "from Libraries import thermodynamics as thermo\n",
126 | "from Libraries import HT_external_convection as extconv\n",
127 | "\n",
128 | "T_i = 25 #C\n",
129 | "T_o = 75 #C\n",
130 | "T_s = 100 #C\n",
131 | "V_i = 5 #m/s\n",
132 | "L = 1 #m\n",
133 | "D = 10e-3 #mm\n",
134 | "N_L = 14\n",
135 | "S_T = S_L = 15e-3 #m\n",
136 | "\n",
137 | "\n"
138 | ]
139 | },
140 | {
141 | "cell_type": "markdown",
142 | "metadata": {},
143 | "source": [
144 | "## Problem 2\n",
145 | "\n",
146 | "A preheater involves the use of condensing steam at $100^\\circ\\text{C}$ on the inside of a bank of tubes to heat air that enters at $1 \\text{ atm}$ and $25^\\circ\\text{C}$. The air moves at $5\\text{ m/s}$ in cross flow over the tubes. Each tube is $1\\text{ m}$ long and has an outside diameter of $10 \\text{ mm}$. The bank consists of 196 tubes in a square, aligned array for which $S_T=S_L=15\\text{ mm}$. What is the total rate of heat transfer to the air? What is the pressure drop associated with the airflow?"
147 | ]
148 | },
149 | {
150 | "cell_type": "markdown",
151 | "metadata": {},
152 | "source": [
153 | "## Problem 3"
154 | ]
155 | },
156 | {
157 | "cell_type": "markdown",
158 | "metadata": {},
159 | "source": [
160 | "
\n",
161 | "An air duct heater consists of an aligned array of electrical heating elements in which the longitudinal and transverse pitches are $S_L=S_T= 24\\text{ mm}$. There are 3 rows of elements in the flow direction ($N_L=3$) and 4 elements per row ($N_T=4$). Atmospheric air with an upstream velocity of $12\\text{ m/s}$ and a temperature of $25^\\circ\\text{C}$ moves in cross flow over the elements, which have a diameter of $12\\text{ mm}$, a length of $250\\text{ mm}$, and are maintained at a surface temperature of $350^\\circ\\text{C}$.\n",
162 | "![\"my](\"https://github.com/yvesdubief/UVM-ME144-Heat-Transfer/blob/master/Figures/HW1-ThermalCondAir.png?raw=true\")
\n",
194 | "\n",
195 | "#### Assumptions:\n",
196 | "* Heat transfer is steady and 1D\n",
197 | "* Radiation is ignored\n",
198 | "* Convection in the gap of the storm window is neglected\n",
199 | "* The temperature on the inside and outside surfaces of any window are isothermal and equal to $T_{si}=10^\\circ\\mathrm{C}$ and $T_{so}=-10^\\circ\\mathrm{C}$, respectively.\n",
200 | "* For the storm window, and the double pane window, the average temperature in the gap between the two glass pane is assumed to the average of $T_{si}$ and $T_{so}$ or $T_{mean}=0^\\circ\\mathrm{C}$.\n",
201 | "\n"
202 | ]
203 | },
204 | {
205 | "cell_type": "markdown",
206 | "id": "bd483448",
207 | "metadata": {},
208 | "source": [
209 | "#### Setup\n",
210 | "\n",
211 | "In this cell you are expected to:\n",
212 | "* Insert your sketches of the single pane and double pane windows showing heat fluxes, temperatures at boundaries and dimensions\n",
213 | "* Store in folder `MyFigures`. Change the names `your-single-pane-sketch` and `your-double-pane-sketch` in the HTML command below to include your figures to this cell.\n",
214 | "* Complete a **worded** write-up of the solutions (for both windows) including equations written in LaTeX\n",
215 | "\n",
216 | "![\"my](\"MyFigures/your-single-pane-sketch.png\")
\n",
217 | "![\"my](\"MyFigures/your-double-pane-sketch.png\")
\n",
218 | "\n",
219 | "*This is the type of introduction of your work that is expected in this class for any deliverable*\n",
220 | "\n",
221 | "The goal is to compute the heat loss from a model window under controlled conditions. We choose to fix the inside surface temperature to $10^\\circ\\mathrm{C}$ and the outside surface temperature to $-10^\\circ\\mathrm{C}$ for a window of height and width $h = 1.2\\;\\mathrm{m}$ and $w = 0.7\\;\\mathrm{m}$ with a surface aera $A=h\\times w$. The thickness of a residential window pane is $t_{glass}=3/32\\,\\mathrm{in}$ (https://ringerwindows.com/thick-glass-windows/) with a thermal conductivity $k=0.96\\;\\mathrm{W/m.K}$. The storm window has a $2.5\\,\\mathrm{in}$.\n",
222 | "\n",
223 | "The heat rate for a single window pane is...\n",
224 | "*Your derivation of the solution here*\n"
225 | ]
226 | },
227 | {
228 | "cell_type": "code",
229 | "execution_count": null,
230 | "id": "3a7346e0",
231 | "metadata": {},
232 | "outputs": [],
233 | "source": [
234 | "import numpy as np\n",
235 | "from Libraries import thermodynamics as thermo\n",
236 | "T_si = 10. #C\n",
237 | "T_so = -10. #C\n",
238 | "h = 1.2 #m\n",
239 | "w = 0.7 #m\n",
240 | "t_glass = 3./32.*0.0254 #m\n",
241 | "k_glass = 0.96\n",
242 | "A = h*w\n",
243 | "t_gap_storm = 2*0.0254 #m\n",
244 | "# getting thermodynamic properties of air for the storm window\n",
245 | "air_gap_storm = thermo.Fluid('air',thermo.C2K(0.))\n",
246 | "t_gap_vacuum = 0.005#m\n",
247 | "k_vacuum = xxxx #W/m.K (from graph)\n",
248 | "R_glass = \n",
249 | "R_gap_storm = \n",
250 | "R_gap_vacuum = \n",
251 | "R_total_single = \n",
252 | "R_total_storm = \n",
253 | "R_total_double = \n",
254 | "\n",
255 | "print(\"Thermal conductivity of air at 0 C: %.4f\" %air_gap_storm.k)\n",
256 | "print(\"Thermal resistance for the single window: %.4f K/W\" %R_total_single)\n",
257 | "print(\"Thermal resistance for the storm window: %.4f K/W\" %R_total_storm)\n",
258 | "print(\"Thermal resistance for the double pane window: %.4f K/W\" %R_total_double)\n",
259 | "\n",
260 | "DT = T_si - T_so\n",
261 | "q_single = DT/R_total_single\n",
262 | "q_storm = DT/R_total_storm\n",
263 | "q_double = DT/R_total_double\n",
264 | "print(\"Single window pane q: %.0f W\" %q_single)\n",
265 | "print(\"Storm window q: %.0f W\" %q_storm)\n",
266 | "print(\"Double pane window q: %.0f W\" %q_double)"
267 | ]
268 | },
269 | {
270 | "cell_type": "code",
271 | "execution_count": null,
272 | "id": "7d75b9d4",
273 | "metadata": {},
274 | "outputs": [],
275 | "source": []
276 | }
277 | ],
278 | "metadata": {
279 | "kernelspec": {
280 | "display_name": "Python 3 (ipykernel)",
281 | "language": "python",
282 | "name": "python3"
283 | },
284 | "language_info": {
285 | "codemirror_mode": {
286 | "name": "ipython",
287 | "version": 3
288 | },
289 | "file_extension": ".py",
290 | "mimetype": "text/x-python",
291 | "name": "python",
292 | "nbconvert_exporter": "python",
293 | "pygments_lexer": "ipython3",
294 | "version": "3.9.13"
295 | }
296 | },
297 | "nbformat": 4,
298 | "nbformat_minor": 5
299 | }
300 |
--------------------------------------------------------------------------------
/ME144-HW-04-11.ipynb:
--------------------------------------------------------------------------------
1 | {
2 | "cells": [
3 | {
4 | "cell_type": "markdown",
5 | "id": "cbbb79f1",
6 | "metadata": {},
7 | "source": [
8 | "# Homework Assignment due \n",
9 | "\n",
10 | "## Problem 1\n",
11 | "A thermos bottle has an inner diameter of $7\\;\\mathrm{mm}$ is $10\\;\\mathrm{cm}$ tall. The wall is composed of a sealed air chamber of thickness $1\\;\\mathrm{mm}$ sandwiched between two thin cylinders of stainless steal, $k_{stst}=15\\;\\mathrm{W/(m.K)}$, of thickness $2\\;\\mathrm{mm}$. The goal is to hold tea as close as possible to its ideal temperature of $80\\,^\\circ\\mathrm{C}$ for an hour in some of the weather we have had recently. For an ambient air at $T_\\infty=-20\\,^\\circ\\mathrm{C}$ and a wind of $10\\;\\mathrm{m/s}$, the convection coefficient around the bottle is $h = 28.7\\times10^3\\;\\mathrm{W/(m^2.K)}$. The radiation heat transfer is assumed to be negligible. All heat transfer calculations are assumed to be 1D.\n",
12 | "\n",
13 | "* 1.) Calculate the heat rate of the system if the air chamber is at atmospheric pressure $k_{air}=0.022\\;\\mathrm{W/(m.K)}$\n",
14 | "* 2.) Typical thermos bottles are designed with the air chamber vacuumed at about $10^{-3}\\;\\mathrm{torr}$ resulting in a thermal conductivity of $3\\times10^{-3}\\;\\mathrm{W/m.K}$. Calculate the new heat rate.\n",
15 | "* 3.) Describe briefly the microscopic phenomenon that causes vacuum air to reduce so significantly the heat rate.\n",
16 | "* 4.) Assuming that there is no heat loss from the bottom or the lid, give an estimate of the temperature loss of the tea per second in the vacuumed and non-vacuumed scenario. To do this you only need to apply the guiding principle of heat transfer (see equation sheet) and remember that thermal energy is $\\rho C_p T$, where the tea density is $1000\\;\\mathrm{kg/m^3}$ and $C_p= 4198.\\;\\mathrm{J/(kg.K)}$\n",
17 | "* 5.) Compute the heat loss from a lid made of $3 \\;\\mathrm{mm}$ of plastic with a conductivity of $5\\;\\mathrm{W/(m.K)}$. What is your conclusion regarding the ability of this thermos to hold your tea warm? If you can assume that heat loss rate is constant, how long would it take for your tea to be cold (decide what is cold based on your preference)?"
18 | ]
19 | },
20 | {
21 | "cell_type": "code",
22 | "execution_count": 1,
23 | "id": "843a3ff4",
24 | "metadata": {},
25 | "outputs": [],
26 | "source": [
27 | "import numpy as np\n",
28 | "import matplotlib.pyplot as plt\n",
29 | "from Libraries import thermodynamics as thermo\n",
30 | "from Libraries import HT_thermal_resistance as res\n",
31 | "im"
32 | ]
33 | },
34 | {
35 | "cell_type": "code",
36 | "execution_count": 3,
37 | "id": "14e6fddf",
38 | "metadata": {},
39 | "outputs": [],
40 | "source": [
41 | "L = 0.1 #m Height\n",
42 | "ri = 0.07 #m Inside radius\n",
43 | "t_sts = 0.002 #m Stainless steal wall thickness\n",
44 | "k_sts = 15 #W/m.K thermal conductivity of stainless steal wall\n",
45 | "t_gap = 0.001 #m \n",
46 | "k_gap_atm = 2.2e-2 #W/m.K air chamber without vacuum\n",
47 | "k_gap_vac = 3.0e-3 #W/m.K air chamber with vacuum\n",
48 | "h = 28000 #W/m^2.K convection coefficient\n",
49 | "T_i = 80 #C\n",
50 | "T_inf = -20 #C\n",
51 | "\n",
52 | "r1 = ri + t_sts\n",
53 | "r2 = r1 + t_gap\n",
54 | "ro = r2 + t_sts\n",
55 | "\n",
56 | "A_o = 2*np.pi*ro**2*L\n",
57 | "\n",
58 | "R = []\n",
59 | "# Conduction through Inner wall\n",
60 | "R.append(res.Resistance(\"$R_{cond,w1}$\", \"W\"))\n",
61 | "# Conduction through air gap \n",
62 | "R.append(res.Resistance(\"$R_{cond,a}$\", \"W\"))\n",
63 | "# Conduction through outer wall\n",
64 | "R.append(res.Resistance(\"$R_{cond,w2}$\", \"W\"))\n",
65 | "# Conduction through convection\n",
66 | "R.append(res.Resistance(\"$R_{conv}$\", \"W\"))\n",
67 | "\n"
68 | ]
69 | },
70 | {
71 | "cell_type": "markdown",
72 | "id": "0bcaacce",
73 | "metadata": {},
74 | "source": []
75 | },
76 | {
77 | "cell_type": "code",
78 | "execution_count": 4,
79 | "id": "6c2e5bdf",
80 | "metadata": {},
81 | "outputs": [],
82 | "source": [
83 | "\n",
84 | "R[0].cond_cylinder(k_sts,ra=ri,rb=r1,L=L)\n",
85 | "R[1].cond_cylinder(k_gap_atm,ra=r1,rb=r2,L=L)\n",
86 | "R[2].cond_cylinder(k_sts,ra=r2,rb=ro,L=L)\n",
87 | "R[3].convection(h,A=A_o)\n"
88 | ]
89 | },
90 | {
91 | "cell_type": "code",
92 | "execution_count": 5,
93 | "id": "c21e9015",
94 | "metadata": {},
95 | "outputs": [],
96 | "source": [
97 | "R_tot_atm = res.serial_sum(R,0,3)"
98 | ]
99 | },
100 | {
101 | "cell_type": "code",
102 | "execution_count": 9,
103 | "id": "f999a92d",
104 | "metadata": {},
105 | "outputs": [],
106 | "source": [
107 | "R[1].cond_cylinder(k_gap_vac,ra=r1,rb=r2,L=L)\n",
108 | "R_tot_vac = res.serial_sum(R,0,3)"
109 | ]
110 | },
111 | {
112 | "cell_type": "code",
113 | "execution_count": 10,
114 | "id": "0083d356",
115 | "metadata": {},
116 | "outputs": [
117 | {
118 | "data": {
119 | "text/latex": [
120 | "$\\displaystyle 7.33366778491278$"
121 | ],
122 | "text/plain": [
123 | "7.333667784912777"
124 | ]
125 | },
126 | "execution_count": 10,
127 | "metadata": {},
128 | "output_type": "execute_result"
129 | }
130 | ],
131 | "source": [
132 | "R_tot_vac"
133 | ]
134 | },
135 | {
136 | "cell_type": "code",
137 | "execution_count": 12,
138 | "id": "05887af4",
139 | "metadata": {},
140 | "outputs": [],
141 | "source": [
142 | "q_atm = 1/R_tot_atm*(T_i - T_inf)\n",
143 | "q_vac = 1/R_tot_vac*(T_i - T_inf)"
144 | ]
145 | },
146 | {
147 | "cell_type": "code",
148 | "execution_count": 14,
149 | "id": "316786a1",
150 | "metadata": {},
151 | "outputs": [
152 | {
153 | "data": {
154 | "text/latex": [
155 | "$\\displaystyle 13.635741750632$"
156 | ],
157 | "text/plain": [
158 | "13.635741750631993"
159 | ]
160 | },
161 | "execution_count": 14,
162 | "metadata": {},
163 | "output_type": "execute_result"
164 | }
165 | ],
166 | "source": [
167 | "q_vac"
168 | ]
169 | },
170 | {
171 | "cell_type": "code",
172 | "execution_count": null,
173 | "id": "083deacc",
174 | "metadata": {},
175 | "outputs": [],
176 | "source": [
177 | "print(\"Without vacuum, the heat rate is \")"
178 | ]
179 | },
180 | {
181 | "cell_type": "markdown",
182 | "id": "d5fb74e9",
183 | "metadata": {},
184 | "source": [
185 | "## Problem 2\n",
186 | "\n",
187 | "The key to a happy skier is to maintain the surface temperature of your skin larger than $20^\\circ\\mathrm{C}$. The objective is to determine the heat flux generated by a thin electrical heater (resistance) inserted on the top of the liner, i.e. between the liner and the sock. All calculations will assume 1D planar heat transfer and to be performed per unit surface area. The study is also limited to the heat transfer from the top of the foot only. The body is assumed to be at constant temperature $36^\\circ\\mathrm{C}$, or $309^\\circ\\mathrm{K}$, which is the inner boundary condition for a skin/fat layer of thickness $3\\;\\mathrm{mm}$ and $k_{skin}=3\\times10^{-1}\\;\\mathrm{W/(m.K)}$. The skin is covered by a sock of thickness $2.5\\;\\mathrm{mm}$ and $k_{sock}=7.5\\;\\mathrm{W/(m.K)}$. The liner of the boot is made of polymer fibers, thickness $5\\;\\mathrm{mm}$ and $k_\\text{liner}=6\\;\\mathrm{W/(m.K)}$ and the shell is carbon fiber, thickness $3\\;\\mathrm{mm}$ and $k_\\text{shell}=10\\;\\mathrm{W/(m.K)}$ with emissivity $\\varepsilon=0.9$. The air is at $T_\\infty= -20^\\circ\\mathrm{C}=253^\\circ\\mathrm{C}$ and the convection coefficient is $h=50\\;\\mathrm{W/(m^2.K)}$\n",
188 | "* 1.) Calculate the heat rate without the heating resistance. For radiation you will use the radiation coefficient heat transfer $h_r=\\varepsilon\\sigma(T_s+T_\\infty)(T_s^2+T_\\infty^2)$, where $\\sigma = 5.67\\times10^{-8}\\;\\mathrm{W/(m^2.K^4)}$ and $T_s$ is the boot surface temperature (outside surface). Take a guess for $T_s$ (between $T_0$ and $T_\\infty$), calculate the total flux. From the total flux, calculate $T_s$ and iterate once more. What is the skin surface temperature?\n",
189 | "* 2.) Modify your thermal circuit to include the thin heater between the sock and the liner and write the conservation of energy in terms of heat fluxes\n",
190 | "* 3.) What heat flux must the heating resistance generate to maintain the skin temperature at $25^\\circ\\mathrm{C}$ under the present conditions?\n",
191 | "* 4.) Calculate the boot outer surface temperature $T_s$.\n",
192 | "* 5.) Without any additional calculation, what is the most preferable material for a sock: wool, $k_\\text{wool}=7.5\\;\\mathrm{W/(m.K)}$, or cotton, $k_\\text{cotton}=17.5\\;\\mathrm{W/(m.K)}$?\n",
193 | "\n"
194 | ]
195 | },
196 | {
197 | "cell_type": "code",
198 | "execution_count": null,
199 | "id": "ff142774",
200 | "metadata": {},
201 | "outputs": [],
202 | "source": []
203 | },
204 | {
205 | "cell_type": "markdown",
206 | "id": "e100ec1d",
207 | "metadata": {},
208 | "source": [
209 | "## Problem 3\n",
210 | "\n",
211 | "To defog the rear window of an automobile, a very thin transparent heating element is attached to the inner surface of the window. A uniform flux $q''_{h}$ is provided to the heating element to maintain the inner surface of the window at a desired temperature $T_{s,i}$ . The window's thickness is $t=5mm$ and the thermal conductivity is $k=1.2\\;\\mathrm{W/m.K}$ . The interior temperature of the automobile is $T_{\\infty,i}=22^\\circ\\mathrm{C}$ and the convection coefficient is $h_i=15\\;\\mathrm{W/m^2.K}$ The outside ambient temperature is $t_{\\infty,o}=-5^\\circ C$ and the convection heat transfer is $h_o=100\\;\\mathrm{W/m^2.K}$. \n",
212 | "\n",
213 | "* 1.) Define your assumptions.\n",
214 | "* 2.) Sketch the thermal circuit of the system.\n",
215 | "* 3.) Write all equations necessary to solve this problem.\n",
216 | "* 4.) Determine the heat flux for $T_{s,i}=15^\\circ\\mathrm{C}$.\n",
217 | "\n",
218 | "The rear window is now tinted with an emissivity $\\varepsilon=0.8$. Radiation is assumed to take place only from the outer surface of the window to the outside. The sky temperature for radiation is $T_{sur}=T_{sky} = 5^\\circ\\mathrm{C}$ and the desired inner surface temperature is still $T_{s,i}=15^\\circ\\mathrm{C}$. \n",
219 | "* 5.) Sketch the new thermal circuit.\n",
220 | "* 6.) Write all equations necessary to solve this problem.\n",
221 | "* 7.) Determine the necessary heater heat flux to maintain. $T_{s,i}=15^\\circ\\mathrm{C}$"
222 | ]
223 | },
224 | {
225 | "cell_type": "code",
226 | "execution_count": null,
227 | "id": "cc5423a1",
228 | "metadata": {},
229 | "outputs": [],
230 | "source": []
231 | },
232 | {
233 | "cell_type": "markdown",
234 | "id": "95aade5f",
235 | "metadata": {},
236 | "source": [
237 | "## Problem 4\n",
238 | "\n",
239 | "![](\"DataFigures/pipe.png\")
\n",
240 | "A fluid at uniform temperature $T_i = 500^\\circ\\mathrm{K}$ is contained in an insulation blanket comprised of two semi-cylindrical shells of different materials. The outside ambient conditions are $T_\\infty = 300^\\circ\\mathrm{K}$ and $h = 25\\;\\mathrm{W/m^2.K}$. The thermal conductivities of the two materials are $k_A = 2\\;\\mathrm{W/m.K}$ and $k_B = 0.25\\;\\mathrm{W/m.K}$, respectively. The inner and outer diameters are $r_1 = 50\\;\\mathrm{mm}$ and $r_2 = 100\\;\\mathrm{mm}$.\n",
241 | "\n",
242 | "![](\"DataFigures/ABC.png\")
\n",
243 | "\n",
244 | "Assuming that the source of heating for the inside fluid stops suddenly. What is the initial rate of temperature change for a fluid of density $\\rho = 1000\\;\\mathrm{kg/m^3}$ and $C_p = 4000\\;\\mathrm{J/kg.K}$? If you have enough time left, show that the heat loss is $1040\\;\\mathrm{W/m}$.\n",
245 | "\n",
246 | "* 1.) Draw the thermal circuit.\n",
247 | "* 2.) Assuming that the source of heating for the inside fluid stops suddenly and the fluid temperature is spatially uniform (but varies in time). Derive the equation governing the rate of temperature change in the fluid ($dT/dt$)?\n",
248 | "* 3.) What is the initial rate of temperature change?\n",
249 | "* 4.) Bonus: Plot the evolution of temperature over time"
250 | ]
251 | },
252 | {
253 | "cell_type": "code",
254 | "execution_count": null,
255 | "id": "458e6b3c",
256 | "metadata": {},
257 | "outputs": [],
258 | "source": []
259 | }
260 | ],
261 | "metadata": {
262 | "kernelspec": {
263 | "display_name": "Python 3 (ipykernel)",
264 | "language": "python",
265 | "name": "python3"
266 | },
267 | "language_info": {
268 | "codemirror_mode": {
269 | "name": "ipython",
270 | "version": 3
271 | },
272 | "file_extension": ".py",
273 | "mimetype": "text/x-python",
274 | "name": "python",
275 | "nbconvert_exporter": "python",
276 | "pygments_lexer": "ipython3",
277 | "version": "3.10.10"
278 | }
279 | },
280 | "nbformat": 4,
281 | "nbformat_minor": 5
282 | }
283 |
--------------------------------------------------------------------------------
/HW-01.ipynb:
--------------------------------------------------------------------------------
1 | {
2 | "cells": [
3 | {
4 | "cell_type": "markdown",
5 | "id": "0cc46860",
6 | "metadata": {},
7 | "source": [
8 | "# Homework Assignment 2\n",
9 | "\n",
10 | "Complete this notebook, export to pdf (`File>Download as> PDF (via LaTeX)`) and submit the pdf to BB"
11 | ]
12 | },
13 | {
14 | "cell_type": "markdown",
15 | "id": "85aecba7",
16 | "metadata": {},
17 | "source": [
18 | "## Pb 1: Air thermodynamic properties class\n",
19 | "\n",
20 | "The concept of class is central to most libraries. It consists of the definition of an object, which has specific properties, data and function.\n",
21 | "\n",
22 | "Zografos et al. (*Compt. Methods Appl. Mech Eng.*, 61:177-187, 1987) proposed polynomials to compute density, dynamic viscosity, thermal conductivity, and specific heat at constant temperature of air at atmospheric pressure as a function of the static temperature $T$ on the Kelvin scale as follows:\n",
23 | "* Density $\\rho$ in units of $\\mathrm{kg/m^3}$:\n",
24 | "\n",
25 | "$$\n",
26 | "\\rho=\\begin{cases}\n",
27 | "-2.440\\times10^{-2}T+5.9958\\,, & 100\\,\\mathrm{K}\\leq T<150\\,\\mathrm{K}\\\\\n",
28 | "345.57(T-2.6884)^{-1}\\,, & 150\\,\\mathrm{K}\\leq T\\leq 3000\\,\\mathrm{K}\n",
29 | "\\end{cases}\n",
30 | "$$\n",
31 | "* Dynamic viscosity $\\mu$ in units of $\\mathrm{N.s/m^2}=\\mathrm{kg/(m.s)}=\\mathrm{Pa.s}$:\n",
32 | "\n",
33 | "$$\n",
34 | "\\mu = 4.1130\\times10^{-6} + 5.0523\\times10^{-8}T - 1.4346\\times10^{-11}T^2+2.5914\\times10^{-15}T^3\n",
35 | "$$\n",
36 | "\n",
37 | "* Thermal conductivity $k$ in units of $\\mathrm{W/(m.K)}$:\n",
38 | "\n",
39 | "$$\n",
40 | "k= -7.488\\times10^{-3} + 1.7082\\times10^{-4}T - 2.3758\\times10^{-10}T^2 + 2.2012\\times10^{-10}T^3 \\\\\n",
41 | " + 9.4600\\times10^{-14}T^4 + 1.5797\\times10^{-17}T^5\n",
42 | "$$\n",
43 | "\n",
44 | "* Specific heat at constant pressure $c_p$ in unites of $\\mathrm{J/(kg.K)}$:\n",
45 | "\n",
46 | "$$\n",
47 | "c_p=1061.332-0.432819T + 1.02344\\times10^{-3}T^2 - 6.47474\\times10^{-7}T^3 + 1.3864\\times10^{-10}T^4\n",
48 | "$$\n",
49 | "\n",
50 | "Air density $\\rho$ of units $\\mathrm{kg/m^3}$ can be computed from \n",
51 | "\n",
52 | "The Prandtl number $Pr$ can be computed as\n",
53 | "\n",
54 | "$$\n",
55 | "Pr=\\frac{\\mu c_p}{k}\n",
56 | "$$\n",
57 | "\n",
58 | "and the thermal conductivity $\\alpha$ of units $\\mathrm{m^2/s}$ as\n",
59 | "\n",
60 | "$$\n",
61 | "\\alpha = \\frac{k}{\\rho c_p}\n",
62 | "$$\n",
63 | "\n",
64 | "Write a class storing all these variables for input temperature (float) and unit used for temperature (string, it can only be \"K\", \"C\", \"F\"). The class should convert the temperature to Kelvin if unit is either \"C\" or \"F\"). The class should give a warning if \n",
65 | "* the temperature is outside of the range,\n",
66 | "* the unit is not \"K\", \"C\", \"F\", e.g. \"c\"\n",
67 | "\n",
68 | "### Tasks\n",
69 | "* Complete the manual part of the class (text between \"\"\" xxxx \"\"\"), i.e. write appropriate units. \n",
70 | "* Complete the class to compute all quantities defined in the manual.\n",
71 | "* Apply your class to the examples below.\n",
72 | "* Define units for examples.\n",
73 | "* Plot density as function of temperature in the range $10^2\\leq T<3000$. Use log-scale for temperature. Axes must be labeled with their variables and units."
74 | ]
75 | },
76 | {
77 | "cell_type": "code",
78 | "execution_count": null,
79 | "id": "97650df2",
80 | "metadata": {},
81 | "outputs": [],
82 | "source": [
83 | "import numpy as np\n",
84 | "import math\n",
85 | "import matplotlib.pyplot as plt\n",
86 | "class Air:\n",
87 | " \"\"\"\n",
88 | " Returns an object containing the thermodynamic properties of air at atmospheric pressure\n",
89 | " Input: \n",
90 | " T: temperature (float)\n",
91 | " Tunit: unit of temperature (string). Can only be \"K\", \"C\", \"F\"\n",
92 | " output:\n",
93 | " self.rho: density [kg/m^3]\n",
94 | " self.mu: dynamic viscosity [write units here]\n",
95 | " self.k: thermal conductivity [write units here]\n",
96 | " self.cp: Specific heat at constant pressure [write units here]\n",
97 | " self.Pr: Prandtl number\n",
98 | " self.nu: kinematic viscosity [write units here]\n",
99 | " self.alpha: thermal diffusivity [write units here]\n",
100 | " \"\"\"\n",
101 | " def __init__(self,T,Tunit):\n",
102 | " if Tunit == \"C\":\n",
103 | " T += 273.15\n",
104 | " elif Tunit == \"F\":\n",
105 | " T = (T - 32)*5/9 +273.15\n",
106 | " elif Tunit != \"K\":\n",
107 | " print(\"Wrong input for unit of temperature\")\n",
108 | " print(\"Input should be 'K', 'C', or 'F'\")\n",
109 | " self.T = T\n",
110 | " self.mu = 4.1130E-6 + 5.0523E-8*T - 1.4346E-11*T**2 + 2.5914E-15*T**3\n",
111 | " \n"
112 | ]
113 | },
114 | {
115 | "cell_type": "code",
116 | "execution_count": null,
117 | "id": "0b87cbeb",
118 | "metadata": {},
119 | "outputs": [],
120 | "source": [
121 | "myair = Air(-200,\"F\")\n",
122 | "print(\"Temperature in K: %.2f\" %myair.T)\n",
123 | "print(\"Kinematic viscosity: %.2e [units], Thermal conductivity: %.4f [units]\" %(myair.nu,myair.k))"
124 | ]
125 | },
126 | {
127 | "cell_type": "code",
128 | "execution_count": null,
129 | "id": "f887dc6c",
130 | "metadata": {},
131 | "outputs": [],
132 | "source": [
133 | "myair = Air(-200,\"C\")"
134 | ]
135 | },
136 | {
137 | "cell_type": "code",
138 | "execution_count": null,
139 | "id": "5fdbb9f2",
140 | "metadata": {},
141 | "outputs": [],
142 | "source": [
143 | "myair = Air(2000,\"k\")"
144 | ]
145 | },
146 | {
147 | "cell_type": "code",
148 | "execution_count": null,
149 | "id": "8a8c3729",
150 | "metadata": {},
151 | "outputs": [],
152 | "source": [
153 | "myair = Air(2000,\"K\")\n",
154 | "print(\"Dynamic viscosity: %.2e [units], Thermal diffusivity: %.2e [units]\" %(myair.mu,myair.alpha))"
155 | ]
156 | },
157 | {
158 | "cell_type": "code",
159 | "execution_count": null,
160 | "id": "618ff68d",
161 | "metadata": {},
162 | "outputs": [],
163 | "source": [
164 | "T = np.logspace(2,np.log10(2999.),1000)\n",
165 | "rho = np.zeros_like(T)\n",
166 | "for i in range(T.shape[0]):\n",
167 | " # Your code here"
168 | ]
169 | },
170 | {
171 | "cell_type": "code",
172 | "execution_count": null,
173 | "id": "d67d9ede",
174 | "metadata": {},
175 | "outputs": [],
176 | "source": [
177 | "plt.figure(figsize=(3,2),dpi=100)\n",
178 | "# your line plot here\n",
179 | "plt.xlabel(\"$T\\;(\\mathrm{units})$\")\n",
180 | "plt.ylabel(r\"$\\rho\\;(\\mathrm{units})$\")\n",
181 | "plt.show()"
182 | ]
183 | },
184 | {
185 | "cell_type": "markdown",
186 | "id": "b4054fd7",
187 | "metadata": {},
188 | "source": [
189 | "## Pb 2 Window\n",
190 | "\n",
191 | "Set up equations and apply realistic numerical values to them to discuss heat losses of a single pane window, a single pane window with storm window and a double paned window with air trapped at a vacuum of $10^{-3} \\mathrm{torr}$ ina gap of $5\\mathrm{mm}$. Do not consider the effects of radiation for any of the window.\n",
192 | "\n",
193 | "![\"my](\"BLablation.png\")
\n",
63 | "\n",
64 | "The ablation velocity is given by the equation above. The conduction heat flux within the wax slab is\n",
65 | "$$\n",
66 | "q''_{cond}=\\frac{k_{wax}}{t_{wax}}(T_m-T_{cool})\n",
67 | "$$\n",
68 | "\n",
69 | "The convection heat flux is defined as\n",
70 | "$$\n",
71 | "q''_{conv}(x)=h(x)(T_\\infty-T_m)\n",
72 | "$$\n",
73 | "and $h(x)$ is given by the correlation for external convection over a flat plate. \n",
74 | "\n",
75 | "The flow starts in the laminar regime and may transition to turbulence if $Re_{x=L}$ is larger than $Re_{c}\\approx 5\\times10^5$.\n",
76 | "\n",
77 | "Flow thermodynamics properties are determined at the film temperature:\n",
78 | "$$\n",
79 | "T_f=\\frac{T_m+T_\\infty}{2}\n",
80 | "$$"
81 | ]
82 | },
83 | {
84 | "cell_type": "code",
85 | "execution_count": 55,
86 | "metadata": {},
87 | "outputs": [
88 | {
89 | "data": {
90 | "image/png": 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VbmaznMPiBJad1g7AK/t6m1yJmVlzOSxO4NyuuQBs2320yZWYmTWXw+IEVi7qQIKtu440uxQzs6ZyWJxAe0uWpQvaeWG3w8LMZjeHRRWvXzzXPQszm/UcFlW8vmsuL+w+QmGo2OxSzMyaxmFRxYVL59NfKPJz9y7MbBZzWFSx+uwFADy540BT6zAzayaHRRUrFs5hXluOJ7YfbHYpZmZN47CoQhKrz17Apu0Hml2KmVnTOCxq8JZzTuPZ1w5xoNd3zTOz2clhUYN3nLeICHj0hb3NLsXMrCkcFjVYvWwBna05fvTzPc0uxcysKRwWNchlM1x27kIeeX43EdHscszMGs5hUaMrV53BLw4cY9MO/yrKzGafhoWFpDWStkjaKunmMbZL0pdL25+UdHHZtpckPSXpCUkbGlVzuff+8hLyWfGvm3qa8fFmZk3VkLCQlAVuAa4CVgEfkbQq1ewq4LzSYy1wW2r7FRHxpojorne9Y5nfnufyNyzm/zzZw1DRQ1FmNrs0qmdxCbA1IrZFxABwN3B1qs3VwD9E4nFggaQzG1RfTT508VJ2Hurn4Wd3NrsUM7OGalRYLAW2ly3vKK2rtU0AD0raKGnteB8iaa2kDZI27N69ewrKHu3KVWdw1vw2vvbYS1P+3mZm01mjwkJjrEuP5Zyozdsi4mKSoapPSXrnWB8SEXdERHdEdHd1dU2+2nHkshmuvewcfrx1L5t7fKLbzGaPRoXFDuDssuVlQPpM8bhtImL4eRdwL8mwVlN87LJzmNeW468ffL5ZJZiZNVyjwmI9cJ6klZJagGuAdak264CPl34VdRlwMCJeldQhqRNAUgfwHuDpBtVdYX57nhvfdS7fe24X61/a16wyzMwaqiFhEREF4CbgAeBZ4F8iYrOkGyXdWGp2H7AN2Ap8Ffjt0vozgH+XtAn4KfBvEXF/I+oez/W/upIl89r47LefZtA3RTKzWUCn6ozk7u7u2LChflMyHtz8Gmv/cSN/+N7z+dQVr6/b55iZNYqkjeNNT/AM7kl6zy8v4f0XncmX/u/z/OyV/c0ux8ysrhwWJ+HPPngRS+a38amv/4w9R/qbXY6ZWd04LE7C/Dl5brv2Lew7OsD1f7eew32DzS7JzKwuHBYn6cKl87ntYxfz7KuHuOFrGzjaX2h2SWZmU85hMQXefcEZ/PWHV7P+pX189KuPe0jKzE45DospcvWblvKV67rZsvMwH7rtUZ599VCzSzIzmzIOiyl05aoz+MZ/uYzegSGuvuXHfP0nL/tmSWZ2SnBYTLGLl5/Gd3/vHVy68nT++N6n+djf/oSX9hxtdllmZifFYVEHi+a28rXrL+ELH7iQJ7cf5L1feoS/vP85Dh7zr6XMbGZyWNRJJiOuu+wcHvr05ay5cAm3/uAF3vmX3+fWH2zlYK9Dw8xmFl/uo0E29xzkiw9s4QdbdjOnJcuHu8/m+ret4JyFHc0uzcwMOPHlPhwWDba55yB/+6MXWbeph0IxuHTl6XzoLct430VnMrc11+zyzGwWc1hMQ68d7OObG7fzzY07eGlvL+35LO86v4srV53Buy9YzII5Lc0u0cxmGYfFNBYR/OyV/dzzs1/w0DM72XW4n2xG/MqK07j8DYt567kLufCseeSyPr1kZvXlsJghisVg044DPPTMTh5+dhdbdh4GoLM1x6WvO51LVy7kzcsXcOHS+bTls02u1sxONQ6LGWrX4T4e37aPx17Yy2Mv7OGlvb0AZDPigiWdvOnsBaxetoALzuzkvMWdtLc4QMxs8hwWp4hdh/vYtP0gT2zfz6btB9m0/QCHSxculOCc0+dw/pJOzl8yj/PP6OT1i+dyzsI57oWYWU1OFBb++c0MsrizjStXtXHlqjOAZNjqpb1HeX7nYZ577TBbXjvMlp2HeeiZnRTL/gY4c34b5yycw4qFHaxY1MGKhXNYfnoHSxe0M689h6QmHZGZzRQOixkskxGv65rL67rmsubCM0fW9w0OsXXXEbbtOcrLe47y4t6jvLy3l4ee2cneowOj3qOjJcuZC9o5a0E7Sxe0ceb85PVZ89s4Y34bXZ2tdLY6UMxmO4fFKagtn+XCpfO5cOn8im2H+gZ5ZW8vL+/tpefAMXoOHqPnwDFePdjHMz0H2XNkoGKfllyGrrmtdHW2sqj0PPKY20JXZyund7Ry2pw889ryZDIOFrNTjcNilpnXlh83SCDplbx2sI+eg8fYdaifPUf62X249DjSz479vTyxfT97jw4w1ukuCea35zltTgsL5ox+Pm1OngVly/Pa8nS25UqPPC05/zzYbLpyWNgobflscl5j0YkvQ1IYKrKvd2AkSPb3DrD/6CAHegfY3zvI/t4BDvQOsvNQH1teO8yB3gGODgyd8D1bcxk62/LMKwuQ4TBJguX4ckdrjjktWea0JM8drTk6WrLMac3Rns+Sde/GbEo5LGxSctkMizvbWNzZVvM+/YUhDvYOjoTJ4b4Ch/sGRz0f6itwqGzda4f6Rrb1Vgmbcm35DB0tOea0ZpPnVLAMP7fns7Tls7TlM6Net+aztOWytLcky22549va8llacxmfx7FZxWFhDdOay7J4XpbF82oPmHKFoSJH+gsjwXF0oEBv/xC9A6OXjw4v9xc4NjB6ec+R/rL9hjg2WHsAlZOSnlBbWaiMLJcFSksuS0s2Q2s+kzznMrTkktctpdetuezI61FtcuO0zx5v7x6UNYrDwmaMXDZTOucxddfNKhaD/kKRY4ND9I08ivQVhugbGEqeB4sj64fb9Q8O0VdI1h8bOP462VZk75EBBgpFBoaK9A8OJc+F4si6qZrelMtoVNDksxnyWZHPZshlM7RkRS6bGWmXy6jUJkMuq9HtMxnyOZHPjLEtmyGf0ch+LcPrRvYV+VyGfOk9cplkXTYjctnScyYJt3zZcka4hzZDOCxsVstkRHtLtqGz3yOCQjGS4CiFx0ChSH9h6HiglK0fCZ1CcfT2QpGBoaFU+6BQLDI4VGRwKCiUngeHihztL4x8bqGYrCuUtg2/HhhKtg0VGzdZdyRURsIlCZpcRmSzx0Omol0p1I6vy5Ttc3z98PtVtBv1Gcn6rJIrJGQyIqvjz8PrchmRKS1nMxx/Pdy2tD1X9jrddvg5l/6ckdeMfOZ0ClKHhVmDSRr5i7yjtdnVjK1YDAaLlYFzPFCKDBZKbcrCZ7j9wFCRYgSFoSR4CsVk/XAQjX4urR8qtSsWk21D6bbFsvc6HorHBiPVPt2u7DPK3q+RgThZEqMDKhVK2VKwZMrCZ1FHK/9y41unvBaHhZlVyGREaybLqXyLlYjj4TE4VKRYhKHSumIcD5Ri2brhoBmrbbEYI+tG75u0LQ6vjxO3Heuzh8bcl5HXxeEQjKCzTv9op/B/BTOz8UnJ+ZRcFl8/rQaeBWVmZlU1LCwkrZG0RdJWSTePsV2Svlza/qSki2vd18zM6qshYSEpC9wCXAWsAj4iaVWq2VXAeaXHWuC2CexrZmZ11KiexSXA1ojYFhEDwN3A1ak2VwP/EInHgQWSzqxxXzMzq6NGhcVSYHvZ8o7Sulra1LIvAJLWStogacPu3btPumgzM0s0KizGmlmS/pHzeG1q2TdZGXFHRHRHRHdXV9cESzQzs/E06qezO4Czy5aXAT01tmmpYV8zM6ujRvUs1gPnSVopqQW4BliXarMO+HjpV1GXAQcj4tUa9zUzszpqSM8iIgqSbgIeALLAXRGxWdKNpe23A/cB7wO2Ar3A9Sfat9pnbty4cY+klydR7iJgzyT2m8l8zLODj3l2OJljPme8DYqpuvzlKULShojobnYdjeRjnh18zLNDvY7ZM7jNzKwqh4WZmVXlsKh0R7MLaAIf8+zgY54d6nLMPmdhZmZVuWdhZmZVOSzMzKyqWRsWJ3PJ9JmqhmO+tnSsT0p6VNLqZtQ5lWq9vL2kX5E0JOk3GllfPdRyzJLeJekJSZsl/bDRNU61Gv5vz5f0r5I2lY75+mbUOVUk3SVpl6Snx9k+9d9fETHrHiST+14AXkdyOZFNwKpUm/cB3yW5NtVlwE+aXXcDjvlXgdNKr6+aDcdc1u57JBNDf6PZdTfg33kB8AywvLS8uNl1N+CY/zvwF6XXXcA+oKXZtZ/EMb8TuBh4epztU/79NVt7FidzyfSZquoxR8SjEbG/tPg4yXW4ZrJaL2//O8C3gF2NLK5OajnmjwL3RMQrABEx04+7lmMOoFOSgLkkYVFobJlTJyIeITmG8Uz599dsDYuTuWT6TDXR4/kkyV8mM1nVY5a0FPggcHsD66qnWv6d3wCcJukHkjZK+njDqquPWo75fwO/RHIR0qeA34uIYmPKa4op//5q1FVnp5uTuWT6TFXz8Ui6giQs3l7XiuqvlmP+EvCZiBhK/uic8Wo55hzwFuA/AO3AY5Iej4jn611cndRyzO8FngDeDZwLPCTpRxFxqM61NcuUf3/N1rA4mUumz1Q1HY+kNwJ3AldFxN4G1VYvtRxzN3B3KSgWAe+TVIiIbzekwqlX6//tPRFxFDgq6RFgNTBTw6KWY74e+PNIBvS3SnoRuAD4aWNKbLgp//6arcNQJ3PJ9Jmq6jFLWg7cA1w3g//KLFf1mCNiZUSsiIgVwDeB357BQQG1/d/+DvAOSTlJc4BLgWcbXOdUquWYXyHpSSHpDOB8YFtDq2ysKf/+mpU9iziJS6bPVDUe8+eAhcCtpb+0CzGDr9hZ4zGfUmo55oh4VtL9wJNAEbgzIsb8CeZMUOO/8xeAv5f0FMkQzWciYsZeulzSPwHvAhZJ2gH8CZCH+n1/+XIfZmZW1WwdhjIzswlwWJiZWVUOCzMzq8phYWZmVTkszMysKoeFmZlV5bAwM7OqHBZmdSapXdIPJWUnsE+LpEckzcqJszb9OCzM6u+3SC4JPlTrDqVLbT8M/Ke6VWU2AQ4Ls5Mk6fuSriy9/lNJX041uZbkekxIWiHpOUl3Snpa0tcl/ZqkH0v6uaRLyvb7dmlfs6bz5T7MTpKkdwKfB75KcmOhXx/uRZQubPdKRCwpLa8guV7Pm4HNJBfB20RySfhfB66PiA+U2maB1yKiq5HHYzYWj4eanaSIeKR0B7ZPA+9KDTctAg6kdnkxIp4CkLQZeDgionSRuxVl7zskaUBSZ0QcrutBmFXhYSizkyTpIuBMoH+ML/VjQFtqXX/Z62LZcpHKP+Bagb4pKtVs0hwWZiehdF/jr5Pc8/iopPeWby/d0zwrKR0Ytbz3QmB3RAxOSbFmJ8FhYTZJpRsH3QP8t4h4luSeCf9jjKYPMrlb1F5Bcl8Cs6bzCW6zOpP0ZuDTEXHdBPe7B/ijiNhSn8rMaueehVmdRcT/A74/0Ul5wLcdFDZduGdhZmZVuWdhZmZVOSzMzKwqh4WZmVXlsDAzs6ocFmZmVpXDwszMqvr/G9VjThsb5YcAAAAASUVORK5CYII=\n",
91 | "text/plain": [
92 | "![](\"Notebook2-1D-cell.jpg\")