In [1]:

```
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);
```

In [3]:

```
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);
```

In [5]:

```
# 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);
```

In [6]:

```
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);
```

In [7]:

```
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)
```

In [8]:

```
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);
```

In [10]:

```
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.
```

In [12]:

```
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.
```

In [13]:

```
# 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);
```

In [16]:

```
# 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.
```

In [14]:

```
# 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);
```

In [22]:

```
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[0] = nan
th2[0] = nan
th3[0] = 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";
```