Chapter 17: Fins and Extended Surfaces

ILLUSTRATIVE EXAMPLE 17.1, Page number: 358

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 ."

The face area of the fin is : 36.0  in^2 .
The face area of the fin is : 0.25  ft^2 .
The total area of the fin is : 37.2  in^2 .
The total area of the fin is : 0.258  ft^2 .

ILLUSTRATIVE EXAMPLE 17.3, Page number: 359

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 ."

The total fin area is : 0.899  ft^2 .

ILLUSTRATIVE EXAMPLE 17.4, Page number: 364

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,"% ."

The fin efficiency is : 98.0 % .

ILLUSTRATIVE EXAMPLE 17.5, Page number: 365

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),"% ."

The fin efficiency is : 97.4 % .

ILLUSTRATIVE EXAMPLE 17.6, Page number: 365

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 ."

The heat transfer from each micro-fin is : 0.15  W .

ILLUSTRATIVE EXAMPLE 17.8, Page number: 366

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______"

1. The conduction resistance may be neglected.
2. The rate of heat transfer from water to air is : 792.4  W .
3. The percent increase in steady-state heat transfer rate by adding fins to the air side of the plane wall is : 276.7  % .
4. The percent increase in steady-state heat transfer rate by adding fins to the water side of the plane wall is : 5.0  % .
____There is a calculation mistake in book in calculating Qt(83-19/0.0214 = 2999), hence slight differences in answer______

ILLUSTRATIVE EXAMPLE 17.10, Page number: 369

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 ."

1. The heat transfer rate without the fin is : 1.48  W .
Or, the heat transfer rate without the fin is : 5.0  Btu/h .
2. The corrected length is : 0.0155  m .
3. The outer radius is : 0.028  m .
4. The maximum heat transfer rate from the fin is : 74.35  W .
5. The fin efficiency is : 86.0  % .
6. The fin heat transfer rate is : 64.0  W .
Or, the fin heat transfer rate is : 218.0  Btu/h .
7. The fin thermal resistance is : 2.27  K/W .

ILLUSTRATIVE EXAMPLE 17.11, Page number: 370

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."

The fin effectiveness is : 43.2  .
Hence, the use of the fin is justified.

ILLUSTRATIVE EXAMPLE 17.12, Page number: 370

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)," ."

1. The total surface area for heat transfer is : 0.465  m^2 .
2. The exposed tube base total heat transfer rate is : 1332.4  W .
Or, the exposed tube base total heat transfer rate is : 4546.0  Btu/h .
3. The overall efficiency of the surface is : 88.3  % .
4. The overall surface effectiveness is : 5.22  .

ILLUSTRATIVE EXAMPLE 17.13, Page number: 374

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)," ."

1. The heat transfer rate without the fin is : 18.4  W .
2. The maximum heat transfer rate from the fin is : 110.0  W .
3. The fin efficiency is : 99.0  % .
   The fin thermal resistance is : 2.1  °C/W .
   The fin effectiveness is : 5.9  .

ILLUSTRATIVE EXAMPLE 17.14, Page number: 375

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."

The fin effectiveness is : 5.92  .
Hence, the use of the fin is justified.

ILLUSTRATIVE EXAMPLE 17.15, Page number: 375

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."

The fin efficiency is : 97.0  % .
The fin thermal resistance is : 3.6  °C/W .
The fin effectiveness is : 11.05  .
The maximum heat transfer rate from a single fin is : 51.53  W .
Since Ef = FCP>2, the use of the fin is justified.

ILLUSTRATIVE EXAMPLE 17.16, Page number: 376

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)," ."

The rate of heat transfer per unit length of tube is : 6815.5  W .
Or, the rate of heat transfer per unit length of tube is : 6.82  kW .
The overall fin efficiency is : 97.3  % .
The overall fin effectiveness is : 6.03  .

ILLUSTRATIVE EXAMPLE 17.17, Page number: 377

In [19]:
#Variable declaration:
print 'Analytical Solution'

Analytical Solution

ILLUSTRATIVE EXAMPLE 17.18, Page number: 379

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)," % ."

The fin efficiency is : 74.7  % .