# Chapter 9 : Heat transfer by the combined effect of conduction and convection¶

## Example 9.1 page : 159¶

In :
import math

# Variables
D  =  3./48;                                # diameter in ft
L  =  9./12;                                # Length of steam vessel in ft
T1  =  210.;                                # Vessel temperature in degF
T2  =  80.;                                 # Air temperature in degF
th0  =  T1-T2;                             # Temperature difference in degF
h  =  1.44;                                # Assumed heat coefficient in Btu/hr-ft**2-degF
C  =  math.pi*D;                               # Circumference of vessel in ft
A  =  math.pi*D*D/4;                           # Area of vessel in ft**2

# Calculations and Results
# For copper
k1  =  219;                                # Heat conductivity of copper in Btu/hr-ft-degF
m1  =  math.sqrt(h*C/(k1*A));                   # in  /ft
th1  =  th0*2/(math.exp(m1*L)+math.exp(-m1*L));
Tl1  =  round(th1+T2);                     # The temperaure at the free end in degF
print "Temperature at free end of the copper rod is %d degF "%(Tl1);

# For iron
k2  =  36;                                 # heat conductivity of copper in Btu/hr-ft-degF
m2  =  math.sqrt(h*C/(k2*A));                   # in  /ft
th2  =  th0*2/(math.exp(m2*L)+math.exp(-m2*L));
Tl2  =  th2+T2;                            # The temperaure at the free end in degF
print " Temperature at free end of the iron rod is %.2f degF "%(Tl2);

# For glass
k3  =  0.64;                                # Heat conductivity of copper in Btu/hr-ft-degF
m3  =  math.sqrt(h*C/(k3*A));                    # in  /ft
th3  =  th0*2/(math.exp(m3*L)+math.exp(-m3*L));
Tl3  =  th3+T2;                             # The temperaure at the free end in degF
print " Temperature at free end of the glass rod is %.2f degF "%(Tl3);

Temperature at free end of the copper rod is 196 degF
Temperature at free end of the iron rod is 151.80 degF
Temperature at free end of the glass rod is 80.03 degF


## Example 9.2 page : 163¶

In :
import math

# Variables
D  =  3./48;                                # diameter in ft
L  =  9./12;                                # Length of steam vessel in ft
T1  =  210.;                                # Vessel temperature in degF
T2  =  80.;                                 # Air temperature in degF
th0  =  T1-T2;                             # Temperature difference in degF
h  =  1.44;                                # Assumed heat coefficient in Btu/hr-ft**2-degF
C  =  math.pi*D;                               # Circumference of vessel in ft
A  =  math.pi*D*D/4;                           # Area of vessel in ft**2

# Calculations
k  =  36;                                  # heat conductivity of copper in Btu/hr-ft-degF
m  =  math.sqrt(h*C/(k*A));                     # in  /ft
q = k*A*m*th0*(math.exp(m*L)-math.exp(-m*L))/(math.exp(m*L)+math.exp(-m*L));
# Heat loss by iron rod in Btu/hr

# Results
print "The rate of heat loss by iron rod is %.1f Btu/hr"%(q);

The rate of heat loss by iron rod is 19.2 Btu/hr


## Example 9.3 page : 170¶

In :
# Variables
x  =  3./96;                             # Thickness of plate  in ft
k  =  220.;                              # thermal conductivity in Btu/hr-ft-degF
h1  =  480.;                             # Inner film coefficient in Btu/hr-ft**2-degF
h2  =  1250.;                            # Outer film coefficient in Btu/hr-ft**2-degF

# Calculations
U  =  1./((1/h1)+(x/k)+(1/h2));          # Overall heat transer coeeficient in Btu-hr-ft**2-degF

# Results
print "Overall heat transfer coefficient is %.f Btu/hr-ft**2-degF"%(U);

Overall heat transfer coefficient is 331 Btu/hr-ft**2-degF


## Example 9.4 page : 172¶

In :
import math

# Variables
r2  =  3./96;                            # Outer radius in ft
x  =  0.1/12;                           # Thickness of plate  in ft
r1  =  r2-x;                            # Outer radius in ft
k  =  200.;                              # thermal conductivity in Btu/hr-ft-degF
h1  =  280.;                             # Inner film coefficient in Btu/hr-ft**2-degF
h2  =  2000.;                            # Outer film coefficient in Btu/hr-ft**2-degF

# Calculations
U  =  1/((r2/(h1*r1))+(r2*math.log(r2/r1)/k)+(1/h2));         # Overall heat transer coeeficient in Btu-hr-ft**2-degF

# Results
print "Overall heat transfer coefficient is %d Btu/hr-ft**2-degF"%(U);

Overall heat transfer coefficient is 184 Btu/hr-ft**2-degF


## Example 9.5 page : 176¶

In :
import math

# Variables
Tc1  =  120.;                                  # Inlet cold fluid temperature in degF
Tc2  =  310.;                                  # Outlet cold fluid temperature in degF
Th1  =  500.;                                  # Inlet hot fluid temperature in degF
Th2  =  400.;                                  # Outlet hot fluid temperature in degF

# Calculations
delt1  =  Th2-Tc1;                            # Maximum temperature difference in degF
delt2  =  Th1-Tc2;                            # Minimum temperature difference in degF
LMTD  =  (delt1-delt2)/math.log(delt1/delt2);      # Log mean temperature difference

# Results
print "The log mean temperature difference is %d degF"%(LMTD)

The log mean temperature difference is 232 degF


## Example 9.6 page : 178¶

In :
import math

# Variables
Tc1  =  120.;                                  # Inlet cold fluid temperature in degF
Tc2  =  310.;                                  # Outlet cold fluid temperature in degF
Th1  =  500.;                                  # Inlet hot fluid temperature in degF
Th2  =  400.;                                  # Outlet hot fluid temperature in degF

# Calculations
K  =  (Tc2-Tc1)/(Th2-Tc2);                    # Temperature ratio
R  =  (Th1-Th2)/(Tc2-Tc1);                    # Temperature ratio
delt1  =  Th2-Tc1;                            # Maximum temperature difference in degF
delt2  =  Th1-Tc2;                            # Minimum temperature difference in degF
LMTD  =  (delt1-delt2)/math.log(delt1/delt2);      # Log mean temperature difference
f  =  0.99;                                   # Correction factor as seen from figure
LMTDc  =  round(LMTD*f);                      # Corrected math.log mean temperature difference

# Results
print "Log mean temperature difference is %d degF"%(LMTDc);

Log mean temperature difference is 230 degF


## Example 9.7 page : 181¶

In :
import math

# Variables
Do = 1./12;                               # Outside diameter of the condenser in ft
Di = 0.902/12;                           # Outside diameter of the condenser in ft
Ts = 81.7;                               # Steam temperature in degF
Tw1 = 61.4;                              # Water inlet temperature in degF
Tw2 = 69.9;                              # Water outlet temperature in degF
k = 63.;                                  # Thermal conductivity in Btu/hr-ft-degF
v = 7.;                                   # average velocity in ft/sec
h1 = 1270.;                               # water side film coefficient i Btu/hr-ft**2-degF
h2 = 1000.;                               # Steam side film coefficient in Btu/hr-ft**2-degF

# Calculations
U = 1/((Do/(Di*h1))+(Do*math.log(Do/Di)/(2*k))+(1/h2));     # Heat transfer coefficient
LMTD = ((Ts-Tw1)-(Ts-Tw2))/math.log((Ts-Tw1)/(Ts-Tw2));   # Log mean temperature diff.
m = 731300;                                          # Saturated steam to be handled in lb/hr
L = 1097.4-49.7;                                     # Change in enthalpy in Btu/lb
q = m*L;                                             # Heat required in Btu/hr
A = q/(U*LMTD);                                      # Area of condenser in ft**2

# Results
print "The area of steam condenser is %d ft**2"%(A);

# book answer is rounded. kindly check.

The area of steam condenser is 94927 ft**2


## Example 9.8 page : 182¶

In :
import math

# Variables
Tc1  =  139.7;                                  # Inlet cold fluid temperature in degF
Tc2  =  59.5;                                   # Outlet cold fluid temperature in degF
Th1  =  108.7;                                 # Inlet hot fluid temperature in degF
Th2  =  97.2;                                   # Outlet hot fluid temperature in degF

# Calculations and Results
delt1  =  Tc1-Th2;                         # Maximum temperature difference in degF
delt2  =  Th1-Tc2;                         # Minimum temperature difference in degF
LMTD  =  round((delt1-delt2)/math.log(delt1/delt2));
print "  The math.log mean temperature difference is %d degF"%(LMTD);

m  =  18210.;                              # Flow rate through tubes
q  =  m*(Th2-Tc2);                        # Heat loss in Btu/hr
A  =  48.1;                               # Area in ft**2
U  =  q/(A*LMTD);                         # Overall heat transfer coefficient
print "  The overall heat transfer coefficient is %d Btu/hr-ft**2-degF "%(U);

# To calcalute using equations estabilished by correlation
Ts  =  113.;                               # Average tube temperature in degF
Tf  =  (123.9+Ts)/2;                      # Film temperature in degF
# At this temperature thermal properties are considered
p1  =  61.7/32.2;                              # Density in slug/ft**3
u1  =  1.38/32.2;                              # Vismath.cosity in slug/ft-hr
Cp1  =  1*32.2;                                # Btu/slug/ft
k1  =  0.366;                                  # Thermal conductivity in Btu/hr-ft-degF
D1  =  0.375/12;                               # Diameter in ft
v1  =  7610.;                                   # Velocity in ft/sec
Nre1  =  v1*D1*p1/u1;                          # Reynolds number
Npr1  =  u1*Cp1/k1;                            # Prandtls number
Nnu1  =  0.33*Nre1**0.6*Npr1**(1./3);             # Nusselt number
h1  =  Nnu1*k1/D1;                             # Heat transfer coefficient
print "  The outer heat transfer coefficient is %d Btu/hr-ft**2-degF "%(h1);

# Taking the thermal properties at 78.3 degF
p2  =  62.2/32.2;                              # Density in slug/ft**3
u2  =  2.13/32.2;                              # Vismath.cosity in slug/ft-hr
Cp2  =  1*32.2;                                # Heat capacity in Btu/slug/ft
k2  =  0.348;                                  # Thermal conductivity in Btu/hr-ft-degF
D2  =  0.277/12;                               # Diameter in ft
v2  =  7140;                                   # Velocity in ft/sec
Nre2  =  v2*D2*p2/u2;                          # Reynolds number
Npr2  =  u2*Cp2/k2;                            # Prandtls number
Nnu2  =  0.023*Nre2**0.8*Npr2**(0.4);            # Nusselt number
h2  =  Nnu2*k2/D2;                             # Heat transfer coefficient
print "  The inner heat transfer coefficient is %d Btu/hr-ft**2-degF"%(h2);

k3  =  58;
U1  =  1/((D1/(D2*h2))+(D1*math.log(D1/D2)/(2*k3))+(1/h1));   # Heat transfer coefficient
print "  The overall heat transfer coefficient accordind to estabilished correlation  is %d Btu/hr-ft**2-degF "%(U1);

# note : rounding off error.

  The math.log mean temperature difference is 46 degF
The overall heat transfer coefficient is 310 Btu/hr-ft**2-degF
The outer heat transfer coefficient is 1567 Btu/hr-ft**2-degF
The inner heat transfer coefficient is 631 Btu/hr-ft**2-degF
The overall heat transfer coefficient accordind to estabilished correlation  is 349 Btu/hr-ft**2-degF


## Example 9.9 page : 188¶

In :
# Variables
To1 = 140.;                                # inlet temperature of oil in degF
To2 = 90.;                                 # Outlet temperature of oil in degf
Cpo = 0.5;                                # Specific heat capacity in Btu/lb-degf
Tw1 = 60.;                                 # Inlet tempearture of water in degF
Tw2 = 80.;                                 # Outlet temperature of water in degF
mo = 2000.;                                # Mass flow rate of oil in lb/hr

# Calculations
q = mo*Cpo*(To1-To2);                     # Heat transferred in Btu/hr
Cpw = 1;                                  # Heat capacity of water in Btu/hr
mw = q/(Cpw*(Tw2-Tw1));                   # Mass flow rate in lb/hr
E1 = (Tw1-Tw2)/(Tw1-To2);                 # Effective ratio
# Seeing the effective ratio and mass flow rate ratio, from the graph we get UA
UA = 1.15*mo*Cpo;

# Results
print "The product of overall heat transfer and total area is %d Btu/hr-degF"%(UA);

The product of overall heat transfer and total area is 1150 Btu/hr-degF


## Example 9.9 page : 191¶

In :
# Variables
t = 2.;                                  # Thickness of wall in ft
To = 100.;                               # Initial temperature of wall in degF
Tg = 1000.;                              # Temperature of hot gases math.exposed in degF
k = 8.;                                  # Thermal conductivity in Btu/hr-ft-degF
p = 162.;                                # density in lb/ft**-3
Cp = 0.3;                               # Heat capacity in Btu/lb-degF
h = 1.6;                                # Heat transfer coefficient in Btu/hr-ft**-2-degF
a = k/(p*Cp);                           # Thermal diffusivity

# Considering the values of a and 4at/L**2 and hl/2k, the value of Phis, Phic and Si can be obtained
Phis = 0.37;
Phic = 0.41;
Si = 0.62;

# Calculations and Results
Ta = Tg+(To-Tg)*Phis;                   # Temperature of surface in degF
print "The temperature of surface is %d degF  "%(Ta);
Tc = Tg+(To-Tg)*Phic;                   # Temperature of center plane in degF
print "The temperature of surface is %d degF  "%(Tc);
A = 10;                                  # area of wall through which heat is absorbed
q = p*Cp*t*A*Si*(To-Tg);                 # Heat absorbed in Btu/hr
print "The heat absorbed by wall is %d Btu"%(q);

# note : book answer is rounded off.

The temperature of surface is 667 degF
The temperature of surface is 631 degF
The heat absorbed by wall is -542376 Btu


## Example 9.10 page : 189¶

In :
# Variables
To1 = 160.;                               # inlet temperature of oil in degF
Cpo = 0.5;                               # Specific heat capacity in Btu/lb-degf
Tw1 = 60.;                                # Inlet temperature of water in degF
mo = 1000.;                               # Mass flow rate of oil in lb/hr
mw = 2500.;                               # Mass flow rate of water in lb/hr
Cpw = 1.;                                 # Heat capacity of water in Btu/hr

# Calculations and Results
X = mo*Cpo/(mw*Cpw);                     # Ratio of flow rates
UA = 1.15*mo*Cpo;
B = UA/mo*Cpo;

# from the graph, we can locate the point of A and B And corresponding effectiveness ratio
E = 0.86;                                # Effectiveness ratio
To2 = To1-E*(To1-Tw1);                   # Outlet temperature of oil in degF
print "The outlet temperature of oil is %d degF "%(To2);

q = mo*Cpo*(To1-To2);                     # Heat transferred in Btu/hr
Tw2 = Tw1+(q/(mw*Cpw));                   # Outlet temperature of oil in degF
print " The outlet tempearture of water is %.1f degF"%(Tw2);


The outlet temperature of oil is 74 degF
The outlet tempearture of water is 77.2 degF


## Example 9.11 page : 193¶

In :
import math
from numpy import zeros, nan

# To compute the temprature distribution
h = 1.;                                        # Heat transfer coefficient in Btu/hr-ft**2-degF
x = 1.;                                        # Assumed thickness in ft
k = 1.;                                        # Thermal conductivity in Btu/hr-ft-degF
N = h*x/k;
t0 = 600.;
t4 = 200.;
t1 = [500, 550, 550, 525, 525, 512.5, 512.5, 512.5, 506.2, 506.2, 506.2, 506.2, 503.1, 503.1];
t2 = [450, 450, 450, 450, 425, 425, 425, 412.5, 412.5, 412.5, 406.3, 406.3, 406.3, 403.1];
t3 = [350, 350, 325, 325 ,325, 325, 312.5, 312.5, 312.5 ,306.3, 306.3, 303.1, 303.1, 303.1];
th1 = zeros(14)
th2 = zeros(14)
th3 = zeros(14)
th1 = nan
th2 = nan
th3 = nan

# Assumed temperatures in degF for points 1 2 & 3 respectively
for i in range(0,14):
th1[i] = th1[i]+t0+t2[i]-2*t1[i];
th2[i] = th2[i]+t1[i]+t3[i]-2*t2[i];
th3[i] = th3[i]+t2[i]+t4-2*t3[i];
print "Assuming t1 = %.1f degF  t2 = %.1fdegF  t3 = %.1fdegF  th1 = %.1f  degF  th2 = %.1f  degF  th3 = %.1f  degF  "%(t1[i],t2[i],t3[i],th1[i],th2[i],th3[i]);

print "This way assumption must be continued till all sink strengths are zero";

Assuming t1 = 500.0 degF  t2 = 450.0degF  t3 = 350.0degF  th1 = nan  degF  th2 = nan  degF  th3 = nan  degF
Assuming t1 = 550.0 degF  t2 = 450.0degF  t3 = 350.0degF  th1 = -50.0  degF  th2 = 0.0  degF  th3 = -50.0  degF
Assuming t1 = 550.0 degF  t2 = 450.0degF  t3 = 325.0degF  th1 = -50.0  degF  th2 = -25.0  degF  th3 = 0.0  degF
Assuming t1 = 525.0 degF  t2 = 450.0degF  t3 = 325.0degF  th1 = 0.0  degF  th2 = -50.0  degF  th3 = 0.0  degF
Assuming t1 = 525.0 degF  t2 = 425.0degF  t3 = 325.0degF  th1 = -25.0  degF  th2 = 0.0  degF  th3 = -25.0  degF
Assuming t1 = 512.5 degF  t2 = 425.0degF  t3 = 325.0degF  th1 = 0.0  degF  th2 = -12.5  degF  th3 = -25.0  degF
Assuming t1 = 512.5 degF  t2 = 425.0degF  t3 = 312.5degF  th1 = 0.0  degF  th2 = -25.0  degF  th3 = 0.0  degF
Assuming t1 = 512.5 degF  t2 = 412.5degF  t3 = 312.5degF  th1 = -12.5  degF  th2 = 0.0  degF  th3 = -12.5  degF
Assuming t1 = 506.2 degF  t2 = 412.5degF  t3 = 312.5degF  th1 = 0.1  degF  th2 = -6.3  degF  th3 = -12.5  degF
Assuming t1 = 506.2 degF  t2 = 412.5degF  t3 = 306.3degF  th1 = 0.1  degF  th2 = -12.5  degF  th3 = -0.1  degF
Assuming t1 = 506.2 degF  t2 = 406.3degF  t3 = 306.3degF  th1 = -6.1  degF  th2 = -0.1  degF  th3 = -6.3  degF
Assuming t1 = 506.2 degF  t2 = 406.3degF  t3 = 303.1degF  th1 = -6.1  degF  th2 = -3.3  degF  th3 = 0.1  degF
Assuming t1 = 503.1 degF  t2 = 406.3degF  t3 = 303.1degF  th1 = 0.1  degF  th2 = -6.4  degF  th3 = 0.1  degF
Assuming t1 = 503.1 degF  t2 = 403.1degF  t3 = 303.1degF  th1 = -3.1  degF  th2 = 0.0  degF  th3 = -3.1  degF
This way assumption must be continued till all sink strengths are zero