M237 HEAT TRANSFER SPREADSHEETS

CONDUCTION

Heat travels three ways:

* Conduction - heat transfer by spreading through solids.
* Convection - heat transfer by the movement of heated gasses and liquids.
* Radiation - heat in the form of radiation that travels through space at the speed of light.

Heat always travels from an area of higher temperature to an area of lower temperature.

Heat transfer (Q) is the flow rate of heat and is measured in Watts or Btu's per hour.

Spread Sheet Method:

1. Type in values for the input data.
2. Enter.
3. Answer: X = will be calculated.
4. Automatic calculations are bold type.

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Symbol Units Multiply

by
Q Btu/hr 0.2931
Q Btu/hr 3.93E-04
w Btu 778.2
K Btu/hr-ft-deg F 1.730
h Btu/hr-ft^2-F 5.5956
ρ lbm/ft^2 16.0185
Cp Btu/lbm-deg F 4187
μ lbm / sec-ft 3600
μ lbm / sec-ft 1.488
L ft 0.3048
L in 25.4
L in 0.0254
F lbf 4.4482
 w ft-lbf 1.3558
ν ft^2 / sec 0.0929

Heat transfer is measured in feet and meter units.

Use the above units table from left to right:
Input Data
Units = 42
Multiply by = 0.0254
Calculate
Units Obtained = 1.067

Use the above units table from right to left:

Input Data
Units Obtained = 22
Divide by = 5.5956
Calculate
Units = 3.932

Temperature is the intensity of heat: Input Data

T= 52 deg C
T = deg C + 273.2
T= 325.2 deg K

Input Data
T= 62 deg F
T = 5*(deg F - 32)/9
T= 16.67 deg C
Conduction Example
Input Data
Heat source on t1 side, Q = 12 Watts
High temperature, t1 = 100 deg C
Thermal conductivity, K = 111 W / m-K
Dimension in inches, L1 = 4.00 in
Dimension in inches, L2 = 4.00 in
Thickness in inches, X = 0.375 in
Calculate
Area, A = (L1*L2)*0.0254^2
Answer: A= 0.0103 m^2
Heat transfer, Q = K*A*(t1 - t2) / (X*0.0254)
Low temperature, t2 = t1 - (Q*X) / (K*A)
Answer: t2 = 99.90 deg C

CONDUCTIVITIES & DENSITIES

Properties at 68 deg F
MATERIAL ρ K
lbm/ft^3 Btu/hr-ft-F
Copper, pure 559 223
Brass, 70 Cu, 30 Zn 532 64
Silver, 99.9% pure 657 235
Duralumin, 3-5%Cu, trace Mg 174 95
Carbon Steel, 1.0%C 487 25
Bronze, 75 Cu, 25 Sn 541 15
Stainless Steel,18 Cr, 8 Ni 488 9.4
Concrete, stone, 1-2-4 mix 0.79
Glass, window 0.45
Brick, common building 0.40
Wood, fir 0.063
Wood, white pine 0.065
Glass Wool, 1.5 lb/ft^3 0.022

Heat Transmission Through Air Films and Solids

Conduction through wall and air films on each side of the wall.
Find the heat transfer rate Q and the inner and outer wall surface temperatures.
OVERALL HEAT TRANSFER COEFFICIENT U

The heat flow rate, Q, is the same through each layer in the diagram above.

Thermal Resistances of Layers

Heat transfer per unit area, Q / A = (t1 - t2) / (Xa / Ka)

Heat intensity = t1 -t2

Thermal resistance for layer a, Ra = Xa / Ka

Thermal resistance for layer o, Ro = 1/ ho

Overall heat transfer rate, Q / A = (t1 -t4) / (Ra + Rb + Rc)

Overall temperature difference, ∆T= t1 - t4

Overall heat transfer coefficient, U = 1 / (A * Σ Rn)

Heat transfer per unit area, Q / A = U*(∆T)

U.S. Units Input Data
Wall length, L = 300.00 in
Wall height, H = 48.00 in
Wall thickness, X = 7.00 in
Inside convective coefficient, ho = 2 Btu/hr-ft^2-F
to = 70.0 deg F
t5 = 36.0 deg F
Thickness, Xa = 1.00 in
Xb = 4.00 in
Xc = 0.75 in
Thermal conductivity, Ka = 0.065 Btu/hr-ft-F
Thermal conductivity, Kb = 0.400 Btu/hr-ft-F
Thermal conductivity, Kc = 0.280 Btu/hr-ft-F
Outside convective coefficient, h5 = 5.80 Btu/hr-ft^2-F

Calculate
Wall area, A = L*H / 144
A= 100.00 ft^2
Thermal Resistances
Ro = 1 / ho
Ro = 0.500
Ra = Xa / Ka
Ra = 1.2821
Rb = Xb / Kb
Rb = 0.8333
Rc = Xc / Kc
Rc = 0.2232
R5 = 1 / h5
R5 = 0.172
Overall heat transfer coefficient
1 / U = 1/ho + Xa / Ka + Xb / Kb + Xc / Kc + 1/h5
1 / U = Ro+R1+Ra+Rb+Rc+R5
1/U= 3.0110
Answer: U = 0.332
∆T = to - t5
Answer: ∆T = 34 deg F
Heat transfer per unit area, Q / A = U*(∆T)
Answer: Q / A = 11.29
Heat transfer, Q = U*A*(∆T)
Answer: Q = 1129 Btu/hr
 Answer: Q = 331 Watts
Surface temperature is found from Equation-1 above:
Heat transfer per unit area, Q / A = (to - t1) / (1 / ho)
t1 = to - (Q/A)*(1 / ho)
Answer: t1 = 64.4 deg F
Internal temperature follows:
Heat transfer per unit area, Q / A = (t1 - t2) / (Xa / Ka)
t2 = t1 - (Q/A)*(Xa / Ka)
Answer: t2 = 49.9 deg F

Input Data
Wall length, L = 12.00 in
Wall height, H = 12.00 in
Wall thickness, X = 0.5 ft
Inside temperature, t1 = 70 deg F
Outside temperature, t4 = 20 deg F
Wall material conductivity, K = 0.263 Btu/hr-ft-F
Inside convective coefficient, ha = 2.00 Btu/hr-ft^2-F
Outside convective coefficient, hc = 10.00 Btu/hr-ft^2-F
Calculate
Wall area, A = L*H / 144
A= 1.00 ft^2
Ra = 1 / ha
Ra = 0.50
Rb = X/K
Rb = 1.90
Rc = 1 / hc
Rc = 0.10
Overall thermal resistance, R = Ra + Rb + Rc
ΣR = 2.50
Overall temperature difference, ∆T = t1 - t4
∆T = 50.0 deg F

Heat transfer per unit area, Q / A = ∆T / ΣR

Q/A= 20.00 Btu / hr-ft^2
Heat transfer, Q = A*∆T / ΣR
Q= 20.00 Btu/hr
Q= 5.86 Watts
Internal temperatures are found from Equation-1 above:
 Heat transfer per unit area, Q / A = (t1 - t2) / (Ra)
t2 = t1 -(Q/A)*(Ra)
Answer: t2 = 60.00 deg F

Heat transfer per unit area, Q / A = (t3 - t4) / (Rc)

t3 = t4 + (Q/A)*(Rc)
Answer: t3 = 22.00 deg F

Two Dimensional Heat Transfer

FOR INFORMATION ONLY - NOT INCLUDED IN THE TEST

The steady state energy balance on the interior nodal point N is:

0 = Q1-N + Q2-N + Q3-N + Q4-N

L = Thickness into page.

Finite difference equations for each conductive flux:

Q1-N = K*L*(∆Y)*(T1 - TN) / (∆X)

Q2-N = K*L*(∆X)*(T2 - TN) / (∆Y)

Q3-N = K*L*(∆Y)*(T3 - TN) / (∆X)

Q4-N = K*L*(∆X)*(T4 - TN) / (∆Y)

If ∆X equals ∆Y:
0 = T1 +T2 + T3 + T4 -4*TN
 Node
1
2
3
4
5
6
g 7
8
9

Node
1
2
External Temperatures 3
4
5
6
7
8
9

0 = (T1 + T2)/ 2 + (h*∆X / k)*T∞ - (h*∆X / k + 1)*Tn

Input Data
Ambient temperature, T∞ = 90.0
Convective heat transfer coefficient, hc = 9.1
 Conductivity, k = 2.0
Grid spacing, ∆X = ∆Y = 1.5

0 = T1 + T4 + (T2 + T3)/ 2 + (h*∆X / k)*T∞ - (h*∆X / k + 3)*Tn

0 = (T1 + T2)/ 2 + T3 - 2*Tn

0 = T1 / a*(a + 1) + T2 / (b + 1) + T3 / (a + 1) + T4 / b*(b + 1)

- (1/a + 1/b)*Tn
END OF SECTION
e at the speed of light.
John Andrew LLC
of lower temperature. Copyright © 6/7/2006
DISCLAIMER: The materials contained in the online course are not
r Btu's per hour. intended as a representation or warranty on the part of John Andrew
LLC or any other person/organization named herein. The materials are
for general information only. They are not a substitute for competent
professional advice. Application of this information to a specific project
should be reviewed by a registered professional engineer. Anyone
making use of the information set forth herein does so at their own risk
and assumes any and all resulting liability arising therefrom.

Units
Obtained
W
hp
ft-lbf
W / m-C
W / m^2-C
kg / m^3
J/kg-C
lbm / hr-ft
kg / sec-m
m
mm
m
N
 N-m
m^2 / s

Input Data
T= 50 deg F
T = deg F + 460
T= 510 deg R

Input Data
T= 60 deg C
T = (9*deg C/5) + 32
T= 140 deg F
 K
W/m*C
386
111
407
164
43
26
16.3
1.37
0.78
0.69
0.109
0.112
0.038

mperatures.
gram above.

Equation 1
 Metric Un Input Data
Wall length, L = 300.00 in
Wall height, H = 48.00 in
Wall thickness, X = 6.00 in
Inside convective coefficient, ho = 9.38 Btu/hr-ft^2-F
to = 22.2 deg F
t5 = -1.0 deg F
Xa = 1.00 in
Xb = 0.66 in
Xc = 0.50 in
Ka = 0.10 Btu/hr-ft-F
Kb = 0.10 Btu/hr-ft-F
Kc = 0.52 Btu/hr-ft-F
Outside convective coefficient, h5 = 34.10 Btu/hr-ft^2-F

Calculate
Wall area, A = L*H*0.0929 / 144
A= 9.29 m^2
Thermal Resistances:
Ro = 1 / ho
Ro = 0.107
Ra = Xa / Ka
Ra = 0.2540
Rb = Xb / Kb
Rb = 0.1676
Rc = Xc / Kc
Rc = 0.0244
R5 = 1 / h5
R5 = 0.029
Overall heat transfer coefficient
b / Kb + Xc / Kc + 1/h5 1 / U = 1/ho + Xa / Ka + Xb / Kb + Xc / Kc + 1/h5
1 / U = Ro+R1+Ra+Rb+Rc+R5
1/U= 0.5820
Answer: U = 1.718
∆T = to - t5
Answer: ∆T = 23.2 deg C
Heat transfer per unit area, Q / A = U*(∆T)
Answer: Q / A = 39.86
Heat transfer, Q = U*A*(∆T)
Answer: Q = 370 Watts
 Surface temperature is found from Equation-1 above:
Heat transfer per unit area, Q / A = (to - t1) / (1 / ho)
t1 = to - (Q/A)*(1 / ho)
Answer: t1 = 18.0 deg F
Internal temperature follows:
Heat transfer per unit area, Q / A = (t1 - t2) / (Xa / Ka)
t2 = t1 - (Q/A)*(Xa / Ka)
Answer: t2 = 7.8 deg F
 Google apps are free

Step Easy Equation Solver:

1 Follow the example provide in
2 Written Equations
3 Next
4 Equation Mode
5 See Example
6 Select you variables > a,b,c,d
7 Continue
8 Insert your equatins below
9 - Empty -
10 - Empty -
11 - Empty -
12 - Empty -
13 Select -Empty- and enter first
14 Repeat
15 Select -Empty- and enter last

Final Equations
Node 1 0 = 900 + b + d - 4a
Node 2 0 = a + 700 + c - 4b
Node 3 0 = d + b + 500 - 4c
Node 4 0 = 700 + a + c - 4d

Solutions
Node 1 400
Node 2 350
Node 3 300
Node 4 350
Internal Temperatures
 External Temperatures

Original Equations
a 0 = 400 + 100 + T2 + T4 - 4*T1
b 0 = T1 + 100 + T3 + T5 - 4*T2
c 0 = T2 + 100 + 100 + T6 - 4*T3
d 0 = 400 + T1 + T5 + T7 - 4*T4
e 0 = T4 + T2 + T6 + T8 - 4*T5
f 0 = T5 + T3 + 100 + T9 - 4*T6
g 0 = 400 + T4 + T8 + 200 - 4*T7
h 0 = T7 + T5 + T9 + 200 - 4*T8
i 0 = T8 + T6 + 100 + 200 - 4*T9

Final Equations Solutions

a 0 = 500 + b + d - 4*a 236 deg C
b 0 = a + 100 + c + e - 4*b 166 deg C
c 0 = b + 200 + f - 4*c 129 deg C
d 0 = 400 + a + e + g - 4*d 277 deg C
e 0 = d + b + f + h - 4*e 200 deg C
f 0 = e + c + 100 + i - 4*f 148 deg C
g 0 = 600 + d + h - 4*g 271 deg C
h 0 = g + e + i + 200 - 4*h 209 deg C
i 0 = h + f + 300 - 4*i 164 deg C
Internal Temperatures

X / k)*T∞ - (h*∆X / k + 1)*Tn

)/ 2 + (h*∆X / k)*T∞ - (h*∆X / k + 3)*Tn

/ (b + 1) + T3 / (a + 1) + T4 / b*(b + 1)
P11
course are not
of John Andrew
The materials are
for competent
a specific project
neer. Anyone
o at their own risk
efrom.
on-1 above:
e apps are free

quation Solver: Linear Equation Solver

he example provide in this app.

ou variables > a,b,c,d,

ur equatins below

function : 123 : variable

+ - tap for + or hold for -

Empty- and enter first equation

Empty- and enter last equation

Final Equations Original Equations

900 + b + d - 4a <<<< Node 1 0 = 400 + 500 + T2 + T4 - 4*T1
a + 700 + c - 4b <<<< Node 2 0 = T1 + 500 + 200 + T3 - 4*T2
d + b + 500 - 4c <<<< Node 3 0 = T4 + T2 + 200 + 300 - 4*T3
700 + a + c - 4d <<<< Node 4 0 = 400 + T1 + T3 + 300 - 4*T4

Solutions
deg C
deg C
deg C
deg C
Temperatures
CONVECTION
Convection is heat transfer by the movement of heated gasses and liquids.

Measuring Air Film Coefficient Input Data

Heat source, Q = 100 Watts
Surface area, A = 0.4 sq m
Inside air temp. thermocouple, t1 = 65.3 C
Inside surface temp. thermocouple, t2 = 20.0 C

Calculations
Heat convection, air layer, Q = h * A * (t1 - t2) Watts
h = Q / (A*(t1 - t2))
S.I. Answer: h = 5.52 W/m^2*C
h =(W/m^2*K)/5.596
U.S. Answer: h = 0.986 Btu/hr-ft^2*F

Boundary layer thickness = X mm

Air flow velocity = V m/s

Convective Heat Transfer Coefficient

Convective heat transfer coefficient, h = k*C*(Gr*Pr)^n / L

Description Length L Gr C n
Vertical Plate < 10^4 1.36 0.200
or 10^4<10^9 0.59 0.250
Vertical Cylinder 2*π*r 10^9<10^12 0.13 0.333

Horizontal Plate (S1 + S2) / 2 10^5<2*10^7 0.54 0.250

hot surface facing up 2*10^7<3*10^10 0.14 0.333

Horizontal Plate 3*10^5<3*10^10 0.27 0.250

hot surface facing down
Convective Heat Transfer Coefficient Calculation - Pick Horizontal or Vertical Plate:

Characteristic Length L Input Data-1

height = 8.00 in
width = 6.00 in
Exterior surface temperature, Ts = 146.0 deg C
Ambient (room) temperature, Ta = 36.0 deg C
Calculation P1
h, L=(height x width) / (height + width) = L = 3.43 in
Plate characteristic length, L = 0.09 m
Answer: Ts - Ta = 110 deg C
Th = (Ts + Ta) / 2 deg C
Answer: Th = 91 deg C

Air Properties at temperature, Th, from table below:

Input Data-2
Air conductivity, k = 0.03115 W / m*C
Air density, ρ = 0.964 kg / m^3
Air dynamic viscosity, μ = 2.15E-05 kg / sec*m
ary layer volume coefficient of expansion, β = 2.74E-03 1 / deg C
Gravitational constant, g = 9.81 m/sec^2
Prandtl number, Pr = 0.694 -
Grashof number, Gr = L^3*g*β*ρ^2*(Ts - Ta) / μ^2
Answer: Gr = 3.926E+06
Answer: Gr*Pr = 2.725E+06
C and n below are found in the table above for the Gr*Pr value calculated here.
Input Data-3
Gr*Pr = 10^3 < 10^9
C= 0.53
n= 0.25
Nusselt Equation, Air Film Coefficient, h = k*C*(Gr*Pr)^n / L
S.I. Answer: h = 7.702 W / m^2*C
U.S. Answer: h = (W/m^2*K)/5.5956
Btu/hr-ft^2*F
U.S. Answer: h = 1.376 Btu/hr-ft^2*F
CONDUCTIVITIES & DENSITIES
Properties at 68 deg F
MATERIAL ρ K K
lbm/ft^3 Btu/hr-ft-FW/m*C
Copper, pure 559 223 386
Brass, 70 Cu, 30 Zn 532 64 111
Silver, 99.9% pure 657 235 407
Duralumin, 3-5%Cu, trace Mg 174 95 164
Carbon Steel, 1.0%C 487 25 43
Bronze, 75 Cu, 25 Sn 541 15 26
Steel,18 Cr, 8 Ni 488 9.4 16.3
Concrete, stone, 1-2-4 mix 0.79 1.37
Glass, window 161 0.45 0.78
Brick, Diatomacious 480 0.40 0.69
Wood, fir 0.063 0.109
Wood, white pine 0.065 0.112
Glass Wool 1.5 0.022 0.038

FINS
Fins are used to increase heat transfer area
and provide a cooling effect.

However, if h*A / P*k is greater than 1.00

the fins will insulate and prevent heat flow.

h = air film convective coefficient

A = exposed surface area of the fins.
P = perimeter of the fins.
k = fin material conductivity coefficient.

Case 1. Fin, finite length, heat loss by convection at end.

Heat flow rate, Q = (K*Ab*n*ΔT)*X1 / X2
Input Data
Temperature at fin base, T1 = 200 deg F
Room Temperature, T2 = 70 deg F
Fin width, w = 20 in
Fin thickness, t = 0.125 in
Fin material thermal conductivity, K = 174 W/m*C
Convective heat transfer coefficient, h = 5.7 W / m^2*C
Fin length, L = 0.750 in
Number of fins, N = 40
Calculations P3
Fin width, w = 0.508 m
Fin thickness, t = 0.0032 m
Fin base area, A = w*t m^2
A= 0.001613 m^2
Temperature difference, ΔT = T1 - T2
ΔT = 130 deg F
ΔT = 54.44 deg C
For thin fin, n = (2*h/K*t)^.5
n= 4.54
h / n*K = 0.0072
L= 0.019 m
n*L = 0.087
sinh(n*L) = 0.087
cosh(n*L) = 1.004
X1 = [(sinh(n*L) + (h / n*K)*cosh(n*L)]
X1 = 0.0006
X2 = [(cosh(n*L) + (h / n*K)*sinh(n*L)
X2 = 1.0044
Q = N*(K*A*n*ΔT)*(X1 / X2))
Answer: Q = 1.734 Watts
Case 2. Fin, finite length (L) insulated end.
tanh(n*L) = 0.08632
Heat flow rate, Q = N*(K*A*n*ΔT)*tanh(n*L)
Answer: Q = 239.66 Watts
Fin Efficiency
Ideal heat transfer occurs when the entire fin is at its base temperature.
Input Data
erature at fin base - Room Temperature, ΔT = 20 deg C
Fin material thermal conductivity, K = 4.00 W/ m*C
Convective heat transfer coefficient, h = 9.00 W / m^2*C
Fin width, w = 30.00 in
Fin thickness, t = 0.080 in
Fin length, L = 0.75 in
Calculations
Fin width, meters, w = 0.762 m
Fin thickness, meters, t = 0.0020 m
Fin base area, Ab = w*t m^2
Ab = 0.001548 m^2
Fin length, meters, L = 0.01905 m
Two Fin face areas, Af = 2*L*w
Af = 0.029032 m^2
Perimeter, P = 2*(w + t)
P= 1.5281 m

n= h*P/(K*Ab)
n= 47.12
n*L = 0.8977 P4

Fin efficiency, η = Actual Q / Ideal Q

Q = (h*P*K*Ab)^.5*(ΔT)*(tanh(n*L)) = (K*Ab*n)*(ΔT)*(tanh(n*L))
(h*P*K*A)^.5 = (K*A*n)

Ideal heat transfer, Qa = (h*P*K*A)^.5*ΔT*(tanh(n*L))

Qa = 4.174 Watts
Ideal heat transfer, Qi = (h*Af*ΔT)
Qi = 5.226 Watts
η= Qi / Qa
Answer: η = 79.9 %
Counter & Parallel Flow Heat Exchanger Tube Length Calculation

P5
Heat transfer, Q = U*A*∆Tm

Plane wall heat transfer coefficient, U = 1 / (1/ho + L/K + 1/h1

ylindrical wall heat transfer coefficient, U = 1 / (1/hi + [RoLn(Ro/Ri)]/K + Ri/Ro)

i and o refer to inside and outside tube surfaces.

Large temperature difference, ΔTbc = Ta - Td

ΔTbc = 20 From Input Data below.
Small temperature difference, ΔTad = Tb - Tc
ΔTad = 52 From Input Data below.
Logarithmic mean temp. difference, ΔTm = (ΔTbc - ΔTad)/ln(Tbc/ΔTad)
Answer: ΔTm = 33.5 deg C
The added resistance to heat transfer caused by corrosion is called fouling.
Fouling factor, R ranges between 0.0005 and 0.002. See manufactures data.
Fouling factor, R = (1/Udirty) - (1/Uclean)

Forced Convection - in Coiled Tube Heat Exchangers

Turbulent flow in coiled tube. Input Data
Temp. of water flowing in, Ta = 78 deg C
Temp. of water flowing out, Tb = 74 deg C
Tc = 22 deg C
Td = 58 deg C
Tube inside diameter, Di = 51 mm
Tube outside diameter, Do = 54 mm
Velocity of water in tube, V = 3.5 m/s

Water properties at temperature Tb deg C from the table above:

Water density, ρ = 994.7 kg / m^2
Cp = 4.183 k*J/kg*K
Water dynamic viscosity, μ = 6.82E-04 kg / m*s
Water conductivity, k = 0.6283 W / m*K
Prandtl number, Pr = 4.51 -
Factor, n = 0.40
Velocity of water in tube, V = 3 m/s

Calculations
Water bulk temperature, Tb = (Tin + Tout) / 2 deg C
Answer: Tb = 76 deg C
Reynolds Number, Re = V*D / ν
or, Re = V*D*ρ / μ
Answer: Re = 2.60E+05 Turbulent Re >4000

Convective heat transfer coefficient, h = 0.023*(Re^.8)*(Pr^n)*(k/d)

S.I. Answer: h= 11128 W / m^2*K
U.S. Answer: h=(W/m^2*K)/5.5956 Btu/hr-ft^2*F
U.S. Answer: h= 1989 Btu/hr-ft^2*F
 S.I. Answer from above: h = 11128 W / m^2*K
Large temperature difference, ΔTbc = Tb - Tc
Answer: ΔTbc= 52 deg C P6

Small temperature difference, ΔTad = Ta - Td

Answer: ΔTad = 20 deg C

Logarithmic temperature difference, ΔTm = (ΔTbc - ΔTad)/ln(Tbc/ΔTad)

Answer: ΔTm = 33.5 deg C

Overall heat transfer coefficient = Uo

Heat flow rate, Q = Uo*A*ΔTm
Heat flow rate thru inside tube wall, Qi = Uo*π*di*L*ΔTm
Heat flow rate thru outside tube wall, Qo = Uo*π*do*L*ΔTm
Uo = h *Ai / Ao

Tube inside area, Ai = π*di*L

Tube outside area, Ao = π*do*L
Overall heat transfer coefficient, Uo = h *di / do
Answer: Uo = 10510 W / m^2*K

Cp = Cp*1000 1000*J/kJ
Answer: Cp = 4183 J/kg*K

Disregarding tube fouling, determine the tube length required:

The tube length required, L = ρ*V*(di)^2*Cp*(Tout -Tin) / (4*Uo*do*ΔTm)
Answer: L = 15.37 m

This is the end of this spread sheet.

P7
RADIATION
Heat in the form of radiation travels through space at the speed of light.

Radiant heat energy is proportional to the 4th power of the absolute temperature.

Heat Radiation Upon a Surface

Fraction of heat energy absorbed, α = absorbtivity
Fraction of heat energy reflected, ρ = reflectivity
raction of heat energy passed thru., τ =transmissivity(transparent; solids, liquids, & gasses)

By definition, α + ρ + τ = 1.00

Black Body Radiation

An ideal black body absorbs all incident
radiation so that, α = 1.00
therefore, ρ = 0 and τ = 0

Input Data
t= 22 deg C
S.I. absolute temperature, Tk = t + 273.2 deg K
Answer: Tk = 295.2 deg K
Input Data
t= 70 deg F
U.S. absolute temperature, Tr = t + 459.8 deg R
Answer: Tr = 529.8 deg R

Radiation Example Input Data

Material surface emissivity, e = 0.91 -
Stefan-Boltzmann constant, σ = 5.67E-08 W / m^2-K^4
Dimension, L1 = 20.00 m
 Dimension, L2 = 3.00 m
Surface temperature, t1 = 38 deg C
Calculation
Area, A = L1*L2
Answer: A = 60.00 m^2
Absolute temperature. T = C + 273
T= 311 deg Abs
Radiant thermal flux reflected, Q = e*σ*T^4
Answer: Q = 482.7 Watts P1
Q is the radiant thermal flux per square foot reflected by the wall.

The total hemispherical heat radiation from

a black body surface (A) is, Eb = σ*T^4 W /m^2*K^4

Stefan-Boltzmann constant, σ = 5.670E-08 W /m^2*K^4

Stefan-Boltzmann constant, σ = 1.714E-09 Btu/h*ft^2*R

Gray body radiation, Eg = ε*σ*T^4 W /m^2*K^4

Gray body radiation factor = ε

figuration or Emissivity Factor, CF = E1 / E2

adiation transmitted from surface 1 = E1 W /m^2*K^4
Radiation transmitted to surface 2 = E2 W /m^2*K^4
 Surface 1 is smaller than surface 2.
If surface 2 completely surrounds
surface 1, Fa = 1.00

Total radiation transmitted, Et = ε*σ*CF*(T1^4-T1^4) W /m^2*K^4

Radiation Example
Horizontal duct under a house.
What is the heat loss per unit area of the duct?
Input Data
Duct diameter, D = 20 cm
Duct surface temperature, tduct = 85 C
Duct emissivity, ε = 0.8
Temperature under house, twalls = 20 C
Surrounding air temperature, tair = 18 C
Convective film coefficient, h = 5.5 W/ m^2*K 0.96 Btu/hr*ft^2*F
P2

1. Calculate heat lost by radiation: Calculations

Stefan-Boltzmann constant, σ = 5.670E-08 W /m^2*K^4
Tduct = tduct + 273
Answer: Tduct = 358 deg K
Twalls = twalls + 273
Answer: Twalls = 293 deg K

ce 2 completely surrounds surface 1

angement or Emissivity Factor, CF = 1.00 W/m^2*K^4
Heat transferred by radiation, Er = ε*σ*CF*(T1^4 - T2^4)
Answer: Er = 411 W/m^2*K^4

2. Determine heat lost by convection:

Convective heat transfer, Q / A = h*(tduct - tair)
Answer: Q / A = 368.5 W / m^2

3. Add to find total heat loss:

Qtotal / Area = Er + Q / A
Qtotal / Area = 779 W / m^2
This is the end of this spread sheet.

P3
& gasses)
MATH TOOLS

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Follow Steps>>
GOAL SEEK H = 10.00
Input V = 6.00
Horizontal force, H = 10.00 lbs TAN(A) = V/H
Vertical force, V = 6.00 lbs = 0.6000
 Calculation Angle A = ATAN(V/H)
Resultant force, R = ( H^2 + V^2 )^(1/2) = 0.5404
= 11.7 lbs A radians = 57.3*A
Angle, A = 57.30 * ATAN(V / H) Angle A = 57.3*A
= 30.97 deg = 30.97

GOAL SEEK method

Step-1 Select the green cell C38 containing a formula.
Step-2 Select: DATA > What-If Analysis > Goal Seek
Step-3 To value: 14, for example
Step-4 Pick cell containing value to be changed by Excel: C34 or C35 > OK

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Linear Thermal Expansion

Input Data Units or

Length, L = 120 in mm
Material Coefficient, α = 0.000012 in/in mm/mm
Temperature Change, Δt = 100 deg F deg C

efficients of Linear Expansion in the range 0 to 100C

Aluminum α = 0.0000238
Bronze α = 0.0000175
 Copper α = 0.0000165
Mild Steel α = 0.000012
Porcelain α = 0.000004

Calculations
Length change, ΔL = L * α * Δt
Answer: ΔL = 0.144 in or mm
 OLD EXCEL

Step Enter
1 10
26
lbs 3 Format
llbs 4 Format Cells
5 Number
number 6 Decimal Places
ATAN(V/H) 72
radians 8 OK
degrees 9 10.00
10 6.00
degrees