# Chapter 5 : Heat Transfer¶

### example 5.1 page number 171¶

In [1]:

import math
# Variables
A=5.*4   #in m2
T1=100.;   #in K
T2=30.;    #in K

# Calculations
delta_T=T1-T2;

x=0.25   #in m
k=0.70    #in W/mK
Q=k*A*(delta_T/x);

# Results
print "rate of heat loss = %f W"%(Q)

rate of heat loss = 3920.000000 W


### example 5.2 page number 171¶

In [2]:

import math
# Variables
d1=0.15   #in m
d2=0.16   #in m
l=1.   #in m

# Calculations
A1=3.14*d1*l;
A2=3.14*d2*l
Am=(A1-A2)/math.log (A1/A2);

T1=120.;   #in K
T2=119.8;   #in K

delta_T=T1-T2;
x=(d2-d1)/2;
k=50.   #in W/mK
Q=k*Am*(delta_T/x);

# Results
print "rate of heat loss per unit length = %f W/m"%(Q)

rate of heat loss per unit length = 973.062272 W/m


### example 5.3 page number 172¶

In [3]:

import math
# Variables
ri=0.5   #in m
ro=0.6;  #in m
A1=4*3.14*ri**2;
A2=4*3.14*ro**2;

# Calculations
Am=(A1*A2)**0.5;

Ti=140.;    #in K
To=50.;    #in K
delta_T=Ti-To;
x=0.1   #in m
k=0.12    #in W/mK

Q=k*Am*(delta_T/x);

# Results
print "Heat loss through sphere = %f W"%(Q)

Heat loss through sphere = 406.944000 W


### example 5.4 page number 173¶

In [4]:

import math
# Variables
x1=0.250;  #in m
k1=0.7;  #in W/mK
A1=1.;   #in m2
R1=x1/(k1*A1);   #in K/W

# Calculations and Results
#for the felt layer
x2=0.020;  #in m
k2=0.046;  #in W/mK
A2=1.;   #in m2
R2=x2/(k2*A2);   #in K/W
R=R1+R2;
print "Total resistance = %f K/W"%(R)

T1=110.;   #in K
T2=25.    #in K
delta_T=T1-T2;
Q=delta_T/R;
print "heat loss through wall = %f W/square m"%(Q)

Total resistance = 0.791925 K/W
heat loss through wall = 107.333333 W/square m


### example 5.5 page number 173¶

In [5]:

import math
# Variables
d1=0.15   #in m
d2=0.16   #in m
l=1.   #in m
A1=3.14*d1*l;
A2=3.14*d2*l
Am1=(A2-A1)/math.log (A2/A1);
x1=(d2-d1)/2.;
k1=50.   #in W/mK
R1=x1/(k1*Am1);

# Calculations and Results
#resistance by insulation
d2=0.16   #in m
d3=0.26   #in m
l=1.   #in m
A2=3.14*d2*l;
A3=3.14*d3*l
Am2=(A3-A2)/math.log (A3/A2);
x2=(d3-d2)/2.;
k2=0.08   #in W/mK
R2=x2/(k2*Am2);
R=R1+R2;

print "total resistance = %f K/W"%(R)

T1=120.;   #in K
T2=40.;   #in K
delta_T=T1-T2;
Q=delta_T/R;

print "heat loss = %f W/m"%(Q)

total resistance = 0.966583 K/W
heat loss = 82.765822 W/m


### example 5.6 page number 174¶

In [6]:

import math
# Variables
x1=0.1;   #in m
x2= 0.25;  #in m
k_rb=0.93;   #in W/mK
k_ib=0.116    #in W/mK
k_al=203.6    #in W/mK
A=0.1   #in m2

# Calculations and Results
#to find resistance without rivets
R=(1/A)*((x1/k_rb)+(x2/k_ib));
T1=225   #in K
T2=37    #in K
delta_T=T1-T2;
Q=delta_T/R;
print "heat transfer rate = %f W"%(Q)

#to find resistance with rivet
d=0.03   #in m
rivet_area= (3.14/4)*d**2;
R_r=(x1+x2)/(k_al*rivet_area);
area_norivet=A-rivet_area;
R_cl=(A/area_norivet)*R;
R_eq=1/(1/R_r+1/R_cl);
Q_new=delta_T/R_eq;

print "Rate of heat transfer with rivet = %f W"%(Q_new)
increase=((Q_new-Q)/Q)*100;
print "percentage increase in heat transfer rate = %f"%(increase)

heat transfer rate = 8.308660 W
Rate of heat transfer with rivet = 85.514415 W
percentage increase in heat transfer rate = 929.220242


### example 5.7 page number 187¶

In [4]:

import math

# variables
Cp = 4.178          # kJ/kg K for water
q = 1838.            # rate at which heat is transfered
A = .1005           # heat transfer area
dt1 = 80. - 24       # temperature diffference at hot end
dt2 = 36.-24         # temperature difference at cold end

# Calculations and Results
dtm = (56 + 12)/2.0
h = q/(A*dtm)
print "Heat transfer coefficient, h  = %.0f W/m**2 K"%h

dtm = (56 - 12)/math.log(56/12.)
h = q/(A*dtm)
print "h = %.0f W/m**2 K"%h

Heat transfer coefficient, h  = 538 W/m**2 K
h = 640 W/m**2 K


### example 5.8 page number 188¶

In [7]:

import math

# Variables
density=984.1     #in kg/cubic meter
v=3.   #in m/s
viscosity=485*10**-6;    #in Pa-s
k=0.657    #in W/mK
cp=4178.   #in J/kg K
d=0.016    #in m

# Calculations and Results
Re=(density*v*d)/viscosity;
Pr=(cp*viscosity)/k;

#dittus boelter equation
h=0.023*Re**0.8*Pr**0.3*(k/d);
print "heat transfer coefficient = %f W/sq meter K"%(h)

#Sieder Tate equation
viscosity_w=920*10**-6.
h1=0.023*Re**0.8*Pr**(1./3)*(k/d)*(viscosity/viscosity_w)**0.14;
print "heat transfer coefficient = %f W/sq meter K"%(h1)

heat transfer coefficient = 12964.257508 W/sq meter K
heat transfer coefficient = 12306.258209 W/sq meter K


### example 5.9 page number 191¶

In [8]:

import math
# Variables
T_sun = 5973   #in degree C
d = 1.5*10**13    #in cm
R = 7.1*10**10;  #in cm

# Calculations
T_earth = ((R/(2*d))**0.5)*T_sun;

# Results
print "Temperature of earth = %f C"%(T_earth-273)

Temperature of earth = 17.576884 C


### example 5.10 page number 191¶

In [9]:

import math
# Variables
R=7*10**10;   #in cm
Ts=6000;   #in K

# Calculations
l=1.5*10**13;   #in m
To=((R**2/(4*l**2))**0.25)*Ts;

# Results
print "temperature of earth = %f K"%(To)

temperature of earth = 289.827535 K


### example 5.11 page number 192¶

In [10]:

import math
# Variables
R=6.92*10**5    #in km
l=14.97*10**7   #in km
Ts=6200;    #in K

# Calculations
To=(R**2/l**2)**0.25*Ts;

# Results
print "Equilibrium temperature = %f K"%(To)

Equilibrium temperature = 421.535191 K


### example 5.12 page number 192¶

In [11]:

import math
# Variables
view_factor=0.5;
R=6.92*10**5    #in km
l=14.97*10**7   #in km
Ts=6200;    #in K

# Calculations
To=(view_factor*(R**2/l**2))**0.25*Ts;

# Results
print "Equilibrium temperature = %f K"%(To)

Equilibrium temperature = 354.467431 K


### example 5.13 page number 193¶

In [12]:

import math
# Variables
view_factor=0.25;
R=7.1*10**10    #in cm
l=1.5*10**13   #in cm
Ts=5973;    #in K
alpha=0.2;
epsilon=0.1;

# Calculations
ratio=alpha/epsilon;
To=(ratio*view_factor*(R**2/l**2))**0.25*Ts;

# Results
print "Equilibrium temperature = %f K"%(To)

Equilibrium temperature = 345.556097 K


### example 5.14 page number 193¶

In [13]:

import math
# Variables
R=7*10**10;   #in cm
l=1.5*10**13;  #in cm
sigma=5.3*10**-5;   #in erd/s(cm2)(K)4
T=6000;    #in K

# Calculations
S=(R/l)**2*(sigma)*(T**4)*60;

# Results
print "solar constant = %f J/sq cm min"%(S/10**7)

solar constant = 8.975232 J/sq cm min


### example 5.15 page number 207¶

In [7]:

import math
# Variables
F = 5000.   #in kg/hr
xF = 0.01
xL = 0.02;

# Calculations and Results
L = F*xF/xL;
V = F-L;
print "L = %f Kg/hr V = %f kg/hr"%(L,V)

TF= 303  #in K
hF = 125.9   #in KJ/kg
T1 = 373.2   #in K
Hv = 2676.1   #in kJ/kg
hL = 419.04;   #in kJ/kg
Ts = 383.2   #in K
Hs = 2691.5   #in kJ/kg
hs = 461.30   #in kJ/kg

S = (F*hF-L*hL-V*Hv)/(hs-Hs);
print "amount of steam = %f kg steam/h"%(S)

q = S*(Hs - hs);
q = q*1000/3600   #conversion to Watt
U = q/(69.9*10);
print "heat transfer coefficient = %f W/sq m K"%(U)

L = 2500.000000 Kg/hr V = 2500.000000 kg/hr
amount of steam = 3187.315039 kg steam/h
heat transfer coefficient = 2824.809251 W/sq m K


### example 5.16 page number 208¶

In [15]:

import math
from numpy import *
# Variables
b1 = 6000*125.79+3187.56*2691.5-3187.56*461.30;    #data from previous problem
b2 = 6000;
A = array([[419.04, 2676.1],[1, 1]])

# Calculations and Results
b = array([[b1],[b2]]);
x = linalg.solve(A,b)
#x = x*b
L = x[0];
V = x[1];

print "L = %f kg/hrV = %f kg/hr"%(L,V)

F = 6000   #in kg/hr
xF = 0.01;
xL = F*xF/L;
print "percentage increase in outlet concentration = %f"%(xL*100)

L = 3629.927289 kg/hrV = 2370.072711 kg/hr
percentage increase in outlet concentration = 1.652926


### example 5.17 page number 209¶

In [6]:

import math
# Variables
Hv=2635.3    #kJ/kg
hL=313.93    #in kJ/kg

# Calculations and Results
S=(2500*313.93+2500*2635.3-5000*125.79)/(2691.5-461.30);
print "steam flow rate = %f kg steam/hr"%(S)

q = S*(2691.5 - 461.30);
q = q*1000./3600    #in W
U = 2833.13;   #in W/m2 K
delta_T = 383.2-348.2;   #in K
A = q/(U*delta_T);

print "Area = %f sq meter"%(A)
print "in this case a condensor and vaccum pump should be used"

# Note : there is mistake in calculation in Book. Please calculate manually.

steam flow rate = 3024.000090 kg steam/hr
Area = 18.892462 sq meter
in this case a condensor and vaccum pump should be used

In [ ]: