In [22]:

```
#Variable declaration:
w1 = 1.5 #Thicknessof fin (in)
L = 12.0 #Length of fin (in)
w2 = 0.1 #Thickness of fin(in)
#Calculation:
Af = 2*w1*L #Face area of fin (in^2)
At = Af + L*w2 #Total area of fin (in^2)
#Result:
print "The face area of the fin is :",round(Af)," in^2 ."
print "The face area of the fin is :",round(Af/12**2,2)," ft^2 ."
print "The total area of the fin is :",round(At,1)," in^2 ."
print "The total area of the fin is :",round(At/12**2,3)," ft^2 ."
```

In [3]:

```
from math import pi
#Variable declaration:
rf = 6.0/12.0 #Outside radius of fin (ft)
ro = 4.0/12.0 #Outside radius of pipe (ft)
t = 0.1/12.0 #Thickness of fin (ft)
#Calculation:
Af = 2*pi*(rf**2-ro**2) #Face area of fin (ft^2)
At = Af + 2*pi*rf*t #Total area of fin (ft^2)
#Result:
print "The total fin area is :",round(At,3)," ft^2 ."
```

In [4]:

```
from math import sqrt
#Variable declaration:
L = 3.0*0.0254 #Height of fin (m)
t = 1.0*0.0254 #Thickness of fin (m)
h = 15.0 #Heat transfer coefficient (W/m^2.K)
k = 300.0 #Thermal conductivity (W/m.K)
#Calculation:
Lc = L + t/2.0 #Corrected height of fin (m)
Ap = Lc*t #Profile area of fin (m^2)
x = sqrt((Lc**3*h)/(k*Ap)) #x-coordinate of figure 17.3
#From figure 17.3:
nf = 98.0 #Fin efficiency
#Result:
print "The fin efficiency is :",nf,"% ."
```

In [5]:

```
#Variable declaration:
#From example 17.4:
X = 0.1246 #X-coordinate of figure 17.3
#Calculation:
#Applying equation (A) from Table 17.3:
Y = 4.5128*X**3 - 10.079*X**2 - 31.413*X + 101.47
#Result:
print "The fin efficiency is :",round(Y,1),"% ."
```

In [6]:

```
from math import sqrt,atan
#Variable declaration:
w = 0.2/100.0 #Width of fin (m)
t = 0.2/100.0 #Thickness of fin (m)
L = 1.0/100.0 #Length of fin (m)
h = 16.0 #Heat transfer coefficient (W/m^2.K)
k = 400.0 #Thermal conductivity of fin (W/m.K)
Tc = 100.0 #Circuit temperature (°C)
Ta = 25.0 #Air temperature (°C)
#Calculation:
P = 4*w #Fin cross-section parameter (m)
Ac = w*t #Cross-sectional area of fin (m^2)
Lc = L+Ac/P #Corrected height of fin (m)
m = sqrt((h*P)/(k*Ac)) #Location of minimum temperature (m^-1)
Q = (sqrt(h*P*k*Ac))*(Tc-Ta)*atan(h)*(m*Lc) #Heat transfer from each micro-fin (W)
#Result:
print "The heat transfer from each micro-fin is :",round(Q,2)," W ."
```

In [21]:

```
from __future__ import division
from math import sqrt
#Variable declaration:
h1 = 13.0 #Air-side heat transfer coefficient (W/m^2.K)
A = 1.0 #Base wall area (m^2)
L = 2.5/100 #Length of steel fins (m)
L2 = 1.5/10**3 #Length of steel wall (m)
k = 13.0 #Thermal conductivity of fin (W/m.K)
k1 = 38.0 #Thermal conductivity of steel wall (W/m.K)
h2 = 260.0 #Water side heat transfer coefficient (W/m^2.K)
T4 = 19.0 #Air temperature (°C)
T1 = 83.0 #Water temperature (°C)
t = 1.3/10**3 #Thickness of steel fins (m)
w = 1.0 #Width of wall (m)
S = 1.3/100 #Fin pitch(m)
#Calculation:
R1 = 1/(h1*A) #Air resistance (°C/W) (part 1)
R2 = L2/(k1*A) #Conduction resistance (°C/W)
R3 = 1/(h2*A) #Water resistance (°C/W)
Rt = (R1+R3) #Total resistance (°C/W) (part 2)
Q = (T1-T4)/Rt #Total heat transfer (W)
Nf = 1/S #Number of fins (part 3)
Lbe = w - Nf*t #Unfinned exposed base surface
Abe = w*Lbe #Exposed base surface area (m^2)
Lc = L+t/2 #Corrected length (m)
Ap = Lc*t #Profile area (m^2)
Af = 2*w*Lc #Fin surface area (m^2)
Bi = h1*(t/2)/k1 #Biot number
a = sqrt(Lc**3*h1/(k*Ap)) #Abscissa of the fin efficiency
#From figure 17.3:
nf = 0.88 #Fin efficiency
Rb = 1/(h1*Abe) #Air thermal resistance of base wall (°C/W)
Rf = 1/(h1*Nf*Af*nf) #Air thermal resistance of fins (°C/W)
RT1 = 1/(1/Rb+1/Rf) #Total outside resistance of the fin array (°C/W)
Rt3 = RT1+R3 #Total resistance on air side fins (°C/W)
Qt = (T1-T4)/round(Rt3,5) #Heat transfer rate on air side fins (W)
I = (Qt/Q - 1)*100 #Percent increase in heat transfer rate to air side fins (W)
A = sqrt(Lc**3*h2/(k1*Ap)) #Abscissa of the new fin efficiency (part 4)
#From figure 17.3:
nf2 = 38.0 #New fin efficiency
Rb2 = 1/(h2*Abe) #Thermal resistance of base wall (°C/W)
Rf2 = 1/(h2*Nf*Af*nf2) #Thermal resistance of fins (°C/W)
Rt4 = 1/(1/Rb2+1/Rf2) #Total resistance of the finned surface (°C/W)
Rt5 = R1+Rt4 #Total resistance on water side fins (°C/W)
QT1 = (T1-T4)/Rt5 #Heat transfer rate on water side fins (W)
I2 = (QT1/Q - 1)*100 #Percent increase in heat transfer rate to water side fins (W)
#Result:
if (R2<R1 or R2<R3):
print "1. The conduction resistance may be neglected."
else:
print "1. The conduction resistance can not be neglected."
print "2. The rate of heat transfer from water to air is :",round(Q,1)," W ."
print "3. The percent increase in steady-state heat transfer rate by adding fins to the air side of the plane wall is :",round(I,1)," % ."
print "4. The percent increase in steady-state heat transfer rate by adding fins to the water side of the plane wall is :",round(I2,1)," % ."
print "____There is a calculation mistake in book in calculating Qt(83-19/0.0214 = 2999), hence slight differences in answer______"
```

In [10]:

```
from __future__ import division
from math import pi,sqrt
#Variable declaration:
Do = 2.5/100 #Outside diameter of tube (m)
t = 1/10**3 #Thickness of fin (m)
T = 25 #Fluid temperature (°C)
Tb = 170 #Surface temperature (°C)
h = 130 #Heat transfer coefficient (W/m^2.K)
k = 200 #Thermal conductivity of fin (W/m.K)
rf = 2.75/100 #Outside radius of fin (m)
#Calculation:
ro = Do/2 #Radius of tube (m)
Ab = 2*pi*ro*t #Area of the base of the fin (m^2)
Te = Tb-T #Excess temperature at the base of the fin (K)
Q1 = h*Ab*Te #Total heat transfer rate without the fin (W)
Bi = h*(t/2)/k #Biot number
L = rf-ro #Fin height (m)
rc = rf+t/2 #Corrected radius (m)
Lc = L+t/2 #Corrected height (m)
Ap = Lc*t #Profile area (m^2)
Af = 2*pi*(rc**2-ro**2) #Fin surface area (m^2)
Qm = h*Af*Te #Maximum fin heat transfer rate (W)
A = sqrt(Lc**3*h/(k*Ap)) #Abscissa of fin efficiency
C = rf/ro #Curve parameter of fin efficiency
#From figure 17.4:
nf = 0.86 #Fin efficiency
Qf = nf*Qm #Fin heat transfer rate (W)
R = Te/Qf #Fin resistance (K/W)
#Result:
print "1. The heat transfer rate without the fin is :",round(Q1,2)," W ."
print "Or, the heat transfer rate without the fin is :",round(Q1*3.412)," Btu/h ."
print "2. The corrected length is :",round(Lc,4)," m ."
print "3. The outer radius is :",round(rc,3)," m ."
print "4. The maximum heat transfer rate from the fin is :",round(Qm,2)," W ."
print "5. The fin efficiency is :",round(nf*100)," % ."
print "6. The fin heat transfer rate is :",round(Qf)," W ."
print "Or, the fin heat transfer rate is :",round(Qf*3.412)," Btu/h ."
print "7. The fin thermal resistance is :",round(R,2)," K/W ."
```

In [11]:

```
#Variable declaration:
#From example 17.10:
Qf = 64 #Fin heat transfer rate (W)
Q1 = 1.48 #Total heat transfer rate without the fin (W)
#Calculation:
E = Qf/Q1 #Fin effectiveness
#Result:
print "The fin effectiveness is :",round(E,1)," ."
if E>2:
print "Hence, the use of the fin is justified."
```

In [13]:

```
from __future__ import division
from math import pi
#Variable declaration:
w = 1 #Length of tube (m)
S = 10/10**3 #Fin patch (m)
#From example 17.10:
t = 1/10**3 #Thickness of fin (m)
ro = 0.0125 #Radius of tube (m)
Af = 3.94*10**-3 #Fin surface area (m^2)
Tb = 145 #Excess temperature at the base of the fin (K)
h = 130 #Heat transfer coefficient (W/m^2.K)
Qf = 64 #Fin heat transfer rate (W)
#Calculation:
Nf = w/S #Number of fins in tube length
wb = w-Nf*t #Unfinned base length (m)
Ab = 2*pi*ro*wb #Unfinned base area (m^2)
At =Ab+Nf*Af #Total transfer surface area (m^2)
Qt = h*(2*pi*ro*w*Tb) #Total heat rate without fins (W)
Qb = h*Ab*Tb #Heat flow rate from the exposed tube base (W)
Qft = Nf*Qf #Heat flow rate from all the fins (W)
Qt2 = Qb+Qft #Total heat flow rate (W)
Qm = h*At*Tb #Maximum heat transfer rate (W)
no = Qt2/Qm #Overall fin efficiency
Eo = Qt2/Qt #Overall effectiveness
Rb = 1/(h*Ab) #Thermal resistance of base (K/W)
Rf = 1/(h*Nf*Af*no) #Thermal resistance of fins (K/W)
#Result:
print "1. The total surface area for heat transfer is :",round(At,3)," m^2 ."
print "2. The exposed tube base total heat transfer rate is :",round(Qb,1)," W ."
print "Or, the exposed tube base total heat transfer rate is :",round(Qb*3.412)," Btu/h ."
print "3. The overall efficiency of the surface is :",round(no*100,1)," % ."
print "4. The overall surface effectiveness is :",round(Eo,2)," ."
```

In [15]:

```
from __future__ import division
from math import sqrt
#Variable declaration:
w = 1 #Width of single of fin (m)
t = 2/10**3 #Fin base thickness (m)
l = 6/10**3 #Fin length thickness (m)
T1 = 250 #Surface temperature (°C)
T2 = 20 #Ambient air temperature (°C)
h = 40 #Surface convection coefficient (W/m^2.K)
k = 240 #Thermal conductivity of fin (W/m.K)
#Calculation:
Ab = t*w #Base area of the fin (m^2)
Te = T1-T2 #Excess temperature at the base of the fin (K)
Qw = h*Ab*Te #Heat transfer rate without a fin (W)
Af = 2*w*(sqrt(l**2-(t/2)**2)) #Fin surface area (m^2)
Qm = h*Af*Te #Maximum heat transfer rate (m^2)
Bi = h*(t/2)/k #Biot number
Lc = l #Corrected length (m)
Ap = l*t/2 #Profile area (m^2)
A = sqrt((Lc**3*h)/k*Ap) #Abscissa for the fin efficiency figure
#From figure 17.4:
nf = 0.99 #Fin efficiency
Qf = nf*Qm #Fin heat transfer rate (W)
R = Te/Qf #Fin thermal resistance (K/W)
E = Qf/Qw #Fin effectiveness
#Result:
print "1. The heat transfer rate without the fin is :",round(Qw,1)," W ."
print "2. The maximum heat transfer rate from the fin is :",round(Qm,-1)," W ."
print "3. The fin efficiency is :",round(nf*100)," % ."
print " The fin thermal resistance is :",round(R,1)," °C/W ."
print " The fin effectiveness is :",round(E,1)," ."
```

In [16]:

```
#Variable declaration:
#From example 17.13:
Qf = 108.9 #Fin heat transfer rate (W)
Qw = 18.4 #Total heat transfer rate without the fin (W)
#Calculation:
E = Qf/Qw #Fin effectiveness
#Result:
print "The fin effectiveness is :",round(E,2)," ."
if E>2:
print "Hence, the use of the fin is justified."
```

In [17]:

```
from __future__ import division
from math import pi,sqrt
#Variable declaration:
Do = 50/10**3 #Outside diameter of tube (m)
t = 4/10**3 #Thickness of fin (m)
T = 20 #Fluid temperature (°C)
Tb = 200 #Surface temperature (°C)
h = 40 #Heat transfer coefficient (W/m^2.K)
k = 240 #Thermal conductivity of fin (W/m.K)
l = 15/10**3 #Length of fin (m)
#Calculation:
ro = Do/2 #Radius of tube (m)
rf = ro+l #Outside radius of fin (m)
Ab = 2*pi*ro*t #Area of the base of the fin (m^2)
Te = Tb-T #Excess temperature at the base of the fin (K)
Q1 = h*Ab*Te #Total heat transfer rate without the fin (W)
Bi = h*(t/2)/k #Biot number
L = rf-ro #Fin height (m)
rc = rf+t/2 #Corrected radius (m)
Lc = L+t/2 #Corrected height (m)
Ap = Lc*t #Profile area (m^2)
Af = 2*pi*(rc**2-ro**2) #Fin surface area (m^2)
Qm = h*Af*Te #Maximum fin heat transfer rate (W)
A = sqrt(Lc**3*h/(k*Ap)) #Abscissa of fin efficiency
C = rf/ro #Curve parameter of fin efficiency
#From figure 17.4:
nf = 0.97 #Fin efficiency
Qf = nf*Qm #Fin heat transfer rate (W)
R = Te/Qf #Fin resistance (K/W)
E = Qf/Q1 #Fin effectiveness
#Result:
print "The fin efficiency is :",round(nf*100)," % ."
print "The fin thermal resistance is :",round(R,1)," °C/W ."
print "The fin effectiveness is :",round(E,2)," ."
print "The maximum heat transfer rate from a single fin is :",round(Qm,2)," W ."
if E>2:
print "Since Ef = FCP>2, the use of the fin is justified."
```

In [2]:

```
from __future__ import division
from math import pi,sqrt
#Variable declaration:
Nf = 125 #Array of fins per meter
w = 1 #Length of fin (m)
#From example 17.15:
t = 4/10**3 #Thickness of fin (m)
Do = 50/10**3 #Outside diameter of tube (m)
Af = 7.157*10**-3 #Fin surface area (m^2)
h = 40 #Heat transfer coefficient (W/m^2.K)
DTb = 180 #Excess temperature at the base of the fin (K)
Qf = 50 #Fin heat transfer rate (W)
#Calculation:
ro = Do/2 #Radius of tube (m)
wb = w-Nf*t #Unfinned exposed base length (m)
Ab = 2*pi*ro*wb #Area of the base of the fin (m^2)
At = Ab+Nf*Af #Total heat transfer surface area (m^2)
Qw = h*(2*pi*ro*w)*DTb #Heat rate without fin (W)
Qb = h*Ab*DTb #Heat rate from the base (W)
Qft = Nf*Qf #Heat rate from the fin (W)
Qt = Qb+Qft #Total heat rate (W)
Qm = h*At*DTb #Maximum heat transfer rate (W)
n = Qt/Qm #Overall fin efficiency
E = Qt/Qw #Overall fin effectiveness
Rb = 1/(h*Ab) #Thermal resistance of base (°C/W)
Rf = 1/(h*Nf*Af*n) #Thermal resistance of fin (°C/W)
#Result:
print "The rate of heat transfer per unit length of tube is :",round(Qt,1)," W ."
print "Or, the rate of heat transfer per unit length of tube is :",round(Qt/10**3,2)," kW ."
print "The overall fin efficiency is :",round(n*100,1)," % ."
print "The overall fin effectiveness is :",round(E,2)," ."
```

In [19]:

```
#Variable declaration:
print 'Analytical Solution'
```

In [20]:

```
from __future__ import division
#Variable declaration:
#From example 17.18:
T = 250 #Base temperature of fin (°F)
h = 15 #Convection coefficient of heat transfer (Btu/h.ft.°F)
w = 1 #Base width of fin (ft)
t = 1 #Thickness of fin (in)
H = 1/8 #Height of fin (in)
l = 1 #Length of fin (in)
Q = 357.2 #Heat transfer rate (Btu/h.ft)
#Calculation:
A = (l*w+t*w+H*w)/12 #Heat transfer area of fin (ft^2)
Qm = h*A*(T-70) #Maximum heat transfer rate (Btu/h.ft)
n = Q/Qm*100 #Fin efficiency
#Result:
print "The fin efficiency is :",round(n,1)," % ."
```