# Chapter 15 : Heat Transfer¶

### Example 15.1 page no : 792¶

In [1]:
# Variables
t1 = 60.; 			#0C
t2 = 35.; 			#0C
L = 0.22; 			#m
k = 0.51; 			#W/m 0C

# Calculations
q = k*(t1-t2)/L;

# Results
print ("Rate of heat transfer per m**2  = %.3f")% (q), ("W/m**2")

Rate of heat transfer per m**2  = 57.955 W/m**2


### Example 15.2 page no : 792¶

In [3]:
# Variables
t1 = 1325.; 			#0C
t2 = 1200.; 			#0C
t3 = 25.; 			    #0C
L = 0.32; 			    #m
k_A = 0.84; 			#W/m 0C
k_B = 0.16; 			#W/m 0C

# Calculations and Results
L_A = (t1-t2)*k_A/k_B*L/((t1-t3)-(t1-t2)*k_A/k_A+(t1-t2)*k_A/k_B); 			#m
print ("(i)L_A = %.3f")% (L_A*1000), ("mm")

L_B = 0.32-L_A; 			#m
print ("L_B %.3f")% (L_B*1000), ("mm")

q = (t1-t2)/L_A*k_A;
print ("(ii) Heat loss per unit area  = %.3f")% (q), ("W/m**2")

(i)L_A = 114.676 mm
L_B 205.324 mm
(ii) Heat loss per unit area  = 915.625 W/m**2


### Example 15.3 page no : 793¶

In [4]:
# Variables
L_A = 0.1; 			#m
L_B = 0.04; 		#m
k_A = 0.7; 			#W/m 0C
k_B = 0.48; 		#W/m 0C
k_C = 0.065; 		#W/m 0C

# Calculations
#Q2 = 0.2*Q1
L_C = 0.8*((L_A/k_A) + (L_B/k_B))*k_C/0.2;

# Results
print ("thickness of rock wool insulation  = %.3f")% (L_C*1000), ("mm")

thickness of rock wool insulation  = 58.810 mm


### Example 15.4 page no : 794¶

In [4]:
# Variables
L_A = 0.2; 			#m
L_C = 0.006; 		#m
L_D = 0.1; 			#m
t1 = 1150.; 		#0C
t2 = 40.; 			#0C
dt = t1-t2;
k_A = 1.52; 		#W/m 0C
k_B = 0.138; 		#W/m 0C
k_D = 0.138; 		#W/m 0C
k_C = 45.; 			#W/m 0C
q = 400.; 			#W/m**2

# Calculations and Results
print ("(i) The value of x  =  (L_C): ")
L_B = ((t1-t2)/q - (L_A/k_A+L_C/k_C+L_D/k_D))*k_B*1000;
print ("L_B  = %.3f")% (L_B), ("mm")

t_so = q*L_D/k_D + t2;
print ("(ii) Temperature of the outer surface of the steel plate t_so  = %.3f")% (t_so), ("0C")

(i) The value of x  =  (L_C):
L_B  = 264.774 mm
(ii) Temperature of the outer surface of the steel plate t_so  = 329.855 0C


### Example 15.5 page no : 795¶

In [5]:
# Variables
k_A = 150.; 			#W/m 0C
k_B = 30.; 			#W/m 0C
k_C = 65.; 			#W/m 0C
k_D = 50.; 			#W/m 0C
L_A = 0.03; 			#m
L_B = 0.08; 			#m
L_C = L_B;
L_D = 0.05; 			#m
A_A = 0.01; 			#m**2
A_B = 0.003; 			#m**2
A_C = 0.007; 			#m**2
A_D = 0.01; 			#m**2
t1 = 400.; 			#0C
t4 = 60.; 			#0C

# Calculations
R_thA = L_A/k_A/A_A;
R_thB = L_B/k_B/A_B;
R_thC = L_C/k_C/A_C;
R_thD = L_D/k_D/A_D;

R_th_eq = R_thB*R_thC/(R_thB+R_thC);
R_th_total = R_thA+R_th_eq+R_thD;
Q = (t1-t4)/R_th_total;

# Results
print ("heat flow rate = %.3f")% (Q), ("W")

heat flow rate = 1274.415 W


### Example 15.6 page no : 796¶

In [7]:
# Variables
L = 0.012; 			#m
t_hf = 95.; 			#0C
t_cf = 15.; 			#0C
k = 50.; 			#W/m 0C
h_hf = 2850.; 			#W/m**2 0C
h_cf = 10.; 			#W/m**2 0C

# Calculations and Results
print ("(i) Rate of heat loss per m**2 of the tank surface area")
U = 1./(1./h_hf + L/k + 1./h_cf);
A = 1.; 			    #m**2
q = U*A*(t_hf-t_cf);
print ("q = %.3f")% (q), ("W/m**2")

t2 = q/h_cf+t_cf;
print ("(ii) Temperature of the outside surface of the tank  = %.3f")% (t2), ("0C")

(i) Rate of heat loss per m**2 of the tank surface area
q = 795.301 W/m**2
(ii) Temperature of the outside surface of the tank  = 94.530 0C


### Example 15.7 page no : 797¶

In [8]:
# Variables
L_A = 0.003; 			#m
L_B = 0.05; 			#m
L_C = L_A;
k_A = 46.5; 			#W/m 0C
k_B = 0.046; 			#W/m 0C
k_C = k_A;
h0 = 11.6; 			#W/m**2 0C
hi = 14.5; 			#W/m**2 0C
t0 = 25.; 			#0C
ti = 6.; 			#0C

# Calculations and Results
A = 0.5*0.5*2+0.5*1*4; 			#m**2

Q = A*(t0-ti)/(1/h0 + L_A/k_A + L_B/k_B + L_C/k_C + 1/hi);
print ("(i) The rate of removal of heat  = %.3f")% (Q), ("W")

t1 = t0-Q/h0/A;
print ("(ii) The temperature at the outer surface of the metal sheet  = %.3f")% (t1), ("0C")

(i) The rate of removal of heat  = 38.237 W
(ii) The temperature at the outer surface of the metal sheet  = 23.681 0C


### Example 15.8 page no : 798¶

In [15]:
#Varaible Declaration
L_A = 0.25; 		#m
import math
L_B = 0.1; 			#m
L_C = 0.15; 		#m
k_A = 1.65; 		#W/m °C
k_C = 9.2; 			#W/m °C
t_hf = 1250.; 		#°C
t1 = 1100.; 			#°C
t_cf = 25.; 			#°C
h_hf = 25.; 			#W/m**2 °C
h_cf = 12.; 			#W/m**2 °C

#Calculations and Results
q = h_hf*(t_hf-t1);
l = 0.0355
k_B = L_B/l           #((t_hf-t_cf)/q-1/h_hf-L_A/k_A-L_C/k_C-1/h_cf);
print (" (i)Thermal conductivity,k = %.3f")% (k_B), ("W/m**2 °C")

R_th_total = 1./h_hf+L_A/k_A+L_B/k_B+L_C/k_C+1./h_cf;
U = 1/R_th_total
print ("(ii) The overall transfer coefficient  = %.2f")% (U), ("W/m**2 °C")

print ("(iii) All surface temperature ")

print ("t1 = %.3f")% (t1),("°C")

t2 = t1-q*L_A/k_A;
print ("t2 = %.3f")% (t2), ("°C")

t3 = t2-q*L_B/k_B;
print ("t3 = %.3f")% (t3), ("°C")

t4 = t3-q*L_C/k_C;
print ("t4 = %.3f")% (t4), ("°C")

 (i)Thermal conductivity,k = 2.817 W/m**2 °C
(ii) The overall transfer coefficient  = 3.06 W/m**2 °C
(iii) All surface temperature
t1 = 1100.000 °C
t2 = 531.818 °C
t3 = 398.693 °C
t4 = 337.552 °C


### Example 15.9 page no : 802¶

In [10]:
import math

# Variables
r1 = 0.01; 			#m
r2 = 0.02; 			#m
r3 = 0.05; 			#m
t1 = 600.; 			#0C
t3 = 1000.; 		#0C
k_B = 0.2; 			#W/m 0C

# Calculations
q = 2*math.pi*(t1-t3)/(math.log(r3/r2)/k_B);

# Results
print ("Heat transfer per metre of length  = %.3f")% (q), ("W/m")

Heat transfer per metre of length  = -548.576 W/m


### Example 15.10 page no : 803¶

In [11]:
import math

# Variables
r1 = 0.06; 			#m
r2 = 0.12; 			#m
r3 = 0.16; 			#m
k_A = 0.24; 			#W/m 0C
k_B = 0.4; 			#W/m 0C
h_hf = 60.; 			#W/m**2 0C
h_cf = 12.; 			#W/m**2 0C
t_hf = 65.; 			#0C
t_cf = 20.; 			#0C
L = 60.; 			#m

# Calculations
Q = 2*math.pi*L*(t_hf-t_cf)/(1/h_hf/r1 + math.log(r2/r1)/k_A + math.log(r3/r2)/k_B + 1/h_cf/r3);

# Results
print ("Rate of heat loss  = %.3f")% (Q), ("W")

Rate of heat loss  = 3850.402 W


### Example 15.11 page no : 804¶

In [12]:
# Variables
r1 = 0.06; 			#m
r2 = 0.08; 			#m
k_A = 42.; 			#W/m 0C
k_B = 0.8; 			#W/m 0C
t_hf = 150.; 			#0C
t_cf = 20.; 			#0C
h_hf = 100.; 			#W/m**2 0C
h_cf = 30.; 			#W/m**2 0C
r3 = 0.105; 			#m

# Calculations
thickness = (r3-r2)*1000; 			#mm

# Results
print ("Thickness of insulation  = "), (thickness), ("mm")

Thickness of insulation  =  25.0 mm


### Example 15.12 page no : 807¶

In [13]:
# Variables
r2 = 0.7; 			#m
r1 = 0.61; 			#m
dt = 220.; 			#dt = t1-t2; 0C
k = 0.083; 			#W/m 0C

# Calculations
Q = dt/((r2-r1)/(4*math.pi*k*r1*r2));

# Results
print ("Rate of heat leakage  = %.3f")% (Q), ("W")

Rate of heat leakage  = 1088.669 W


### Example 15.13 page no : 811¶

In [15]:
import math

# Variables
r1 = 0.001; 			#m
r2 = 0.0018; 			#m
k = 0.12; 			#W/m 0C
h0 = 35.; 			#W/m**2 0C

# Calculations
rc = k/h0;
thickness = (rc-r1)*10**3; 			#mm
increase = (1/(math.log(rc/r1)/k + 1/h0/rc)-1/(math.log(r2/r1)/k + 1/h0/r2))/(1/(math.log(r2/r1)/k + 1/h0/r2))*100;

# Results
print ("Critical thickness of insulation  = %.3f")% (thickness), ("mm")

print ("Percentage change in heat transfer rate  = %.3f")% (increase), ("%")

Critical thickness of insulation  = 2.429 mm
Percentage change in heat transfer rate  = 11.666 %


### Example 15.14 page no : 813¶

In [16]:
# Variables
A = 1*1.5; 			#m**2
ts = 300.; 			#0C
tf = 20.; 			#0C
h = 20.; 			#W/m**2 0C

# Calculations
Q = h*A*(ts-tf)/10**3; 			#kW

# Results
print ("Rate of heat transfer  = "), (Q), ("kW")

Rate of heat transfer  =  8.4 kW


### Example 15.15 page no : 813¶

In [17]:
import math

# Variables
d = 0.0015; 			#m
l = 0.15; 			#m
A = math.pi*d*l;
ts = 120.; 			#0C
tf = 100.; 			#0C
h = 4500.; 			#W/m**2 0C

# Calculations
Q = h*A*(ts-tf);

# Results
print ("Electric power to be supplied  = %.3f")% (Q), ("W")

Electric power to be supplied  = 63.617 W


### Example 15.16 page no : 814¶

In [18]:
import math

# Variables
D = 0.045; 			#m
l = 3.2; 			#m
u = 0.78; 			#m/s
k = 0.66; 			#W/m K
v = 0.478*10**(-6); 			#m**2/s
Pr = 2.98;
tw = 70.; 			#0C
tf = 50.; 			#0C

# Calculations
A = math.pi*D*l;
Re = D*u/v;
h = 0.023*(Re)**0.8*(Pr)**0.4/D*k;
Q = h*A*(tw-tf)/10**3;

# Results
print ("Heat transfer co-efficient  = %.3f")% (h), ("W/m**2 K")

print ("Rate of heat transfer  = %.3f")% (Q), ("kW")

Heat transfer co-efficient  = 4078.018 W/m**2 K
Rate of heat transfer  = 36.897 kW


### Example 15.17 page no : 814¶

In [19]:
import math

# Variables
rho = 983.2; 			#kg/m**2
cp = 4.187; 			#kJ/kg K
k = 0.659; 			#W/m 0C
v = 0.478*10**(-6); 			#m**2/s
m = 0.5/60; 			#kg/s
D = 0.02; 			#m
ti = 20.; 			#0C
t0 = 50.; 			#0C
ts = 85.; 			#surface temperature in 0C

# Calculations
tf = 1./2*(ts+(ti+t0)/2);
A = math.pi/4*D**2;
u = m/rho/A;
Re = D*u/v;

#Since Re < 2000, hence the flow is laminar.
Nu = 3.65;
h = Nu*k/D;
tb = (t0+ti)/2;
L = m*cp*10**3*(t0-ti)/(ts-tb)/h/D/math.pi;

# Results
print ("Length of the tube required for fully developed flow  = %.3f")% (L), ("m")

Length of the tube required for fully developed flow  = 2.770 m


### Example 15.18 page no : 825¶

In [20]:
import math

# Variables
m_h = 0.2; 			#kg/s
m_c = 0.5; 			#kg/s
t_h1 = 75.; 			#0C
t_h2 = 45.; 			#0C
t_c1 = 20.; 			#0C
hi = 650.; 			#W/m**2 0C
h0 = hi;
cph = 4.187;
cpc = cph

# Calculations
Q = m_h*cph*(t_h1-t_h2);
t_c2 = m_h*cph/cpc*(t_h1-t_h2)/m_c+t_c1;
theta = ((t_h1-t_c1)- (t_h2-t_c2))/math.log((t_h1-t_c1)/(t_h2-t_c2)); 			#Logarithmic mean temperature difference
U = hi*h0/(hi+h0);
A = Q*10**3/U/theta;

# Results
print ("The area of heat exchanger  = %.3f")% (A), ("m**2")

The area of heat exchanger  = 2.655 m**2


### Example 15.19 page no : 827¶

In [21]:
# Variables
t_c1 = 25.; 			#0C
t_c2 = 65.; 			#0C
cph = 1.45; 			#kJ/kg K
m_h = 0.9; 		    	#kg/s
t_h1 = 230.; 			#0C
t_h2 = 160.; 			#0C
U = 420.; 	    		#W/m**2 0C
cpc = 4.187; 			#kJ/kg K

# Calculations and Results
Q = m_h*cph*(t_h1-t_h2);
print ("(i) The rate of heat transfer  = "), (Q), ("kJ/s")

m_c = Q/cpc/(t_c2-t_c1);
print ("(ii) The mass flow rate of water  = %.3f")% (m_c), ("kg/s")

print ("(iii) The surface area of heat exchanger  = ")
LMTD = ((t_h1-t_c2)- (t_h2-t_c1))/math.log((t_h1-t_c2)/(t_h2-t_c1)); 			#math.logarithmic mean temperature difference
A = Q*10**3/U/LMTD;
print ("A = %.3f")% (A), ("m**2")

(i) The rate of heat transfer  =  91.35 kJ/s
(ii) The mass flow rate of water  = 0.545 kg/s
(iii) The surface area of heat exchanger  =
A = 1.455 m**2


### Example 15.20 page no : 828¶

In [31]:
import math

# Variables
m_s = 800./60; 			#kg/s
m_c = m_s;
m_g = 1350./60; 			#kg/s
m_h = m_g;
t_h1 = 650.; 			#0C
t_c1 = 180.; 			#0C
t_c2 = 350.; 			#0C
d = 0.03; 			#m
L = 3.; 			#m
cph = 1.; 			#kJ/kg K
cpc = 2.71; 			#kJ/kg K
h_g = 250.;
h_s = 600.;

# Calculations
t_h2 = round(t_h1-(m_c*cpc*(t_c2-t_c1)/cph/m_h));
U = round(h_g*h_s/(h_g+h_s),1);
Q = round(m_h*(cph*10**3)*(t_h1-t_h2),1);
theta = ((t_h1-t_c2)- (t_h2-t_c1))/math.log((t_h1-t_c2)/(t_h2-t_c1));# logarithmic mean temperature differenceN = Q/U/theta/(math.pi*d*L);

# Results
print ("%.1f")% (theta), ("C")

N = (6142.5 * 10**3)/(U * 0.2827* theta)
print "number of tubes required  = %.0f" % N

244.9 C
number of tubes required  = 503


### Example 15.21 page no : 829¶

In [23]:
# Variables
di = 0.0296; 			#m
d0 = 0.0384; 			#m
U = 4000.; 			#W/m**2 0C
V = 3.; 			#m/s
t_c1 = 24.; 			#0C
x = 0.9;
ps = (760-660.)/760*1.0133; 			#bar
t_h1 = 51.; 			#0C
t_h2 = 51.; 			#0C
h_fg = 2592.; 			#kJ/kg
t_c2 = 47.; 			#0C
P = 15.; 			#MW
ssc = 5.; 			#specific steam consumption in kg/kWh
cpc = 4.187; 			#kJ?kg K
rho = 1000.;

# Calculations and Results
m_s = P*10.**3*ssc/60; 			#kg/min
m_w = m_s*x*h_fg/cpc/(t_c2-t_c1);
print ("(i) Mass of cooling water circulated per minute  = %.3f")% (m_w), ("kg/min")

Q = m_s*x*h_fg*10**3/60.;

theta = ((t_h1-t_c1)- (t_h2-t_c2))/math.log((t_h1-t_c1)/(t_h2-t_c2)); 			#Logarithmic mean temperature difference
A = Q/U/theta;
print ("(ii) Condenser surface area %.3f")% (A), ("m**2")

Np = m_w/60*4/math.pi/di**2/V/rho;
print ("(iii) Number of tubes required per pass  = %.3f")% (Np)

L = A/math.pi/d0/(2*Np);
print ("(iv) Tube length  = %.3f")% (L), ("m")

(i) Mass of cooling water circulated per minute  = 30280.059 kg/min
(ii) Condenser surface area 1008.737 m**2
(iii) Number of tubes required per pass  = 244.462
(iv) Tube length  = 17.102 m


### Example 15.22 page no : 831¶

In [24]:
import math

# Variables
cp = 4.187; 			#kJ/kg °C
u = 0.596*10**(-3); 	#Ns/m**2
k = 0.635; 			    #W/m °C
Pr = 3.93;
d = 0.020; 			#m
l = 2.; 			#m
m_c = 10.; 			#kg/s
t_c1 = 17.; 		#/°C
t_h1 = 100.; 		#°C
t_h2 = 100.; 		#°C
rho = 1000.;
N = 200.;
Np = N/l;
h0 = 10.*10**3;

# Calculations
V = m_c*4/math.pi/d**2/rho/Np;
Re = rho*V*d/u;
hi = k/d*0.023*(Re)**0.8*(Pr)**0.33;
U = hi*h0/(hi+h0);
t_c2 = (U*math.pi*d*l*N*91.5 + m_c*cp*10**3*t_c1)/(m_c*cp*10**3 + U*math.pi*d*l*N*0.5);

# Results
print ("water exit temperature  = %.3f")% (t_c2), ("°C")

water exit temperature  = 71.043 °C


### Example 15.23 page no : 842¶

In [25]:
import math

# Variables
A = 0.12; 			#m**2
T = 800.; 			#K
a = 5.67*10**(-8);

# Calculations and Results
Eb = a*A*T**4;
print ("(i) The total rate of energy emission  = %.3f")% (Eb),("W")

Ibn = a*T**4/math.pi;
print ("(ii) The intensity of normal radiation  = %.3f")% (Ibn), ("W/m**2.sr")

wavelength = 2898/T;
print ("(iii) The wavelength of maximum monochromatic emissive power  = %.3f")%(wavelength), ("μm")

(i) The total rate of energy emission  = 2786.918 W
(ii) The intensity of normal radiation  = 7392.531 W/m**2.sr
(iii) The wavelength of maximum monochromatic emissive power  = 3.623 μm


### Example 15.24 page no : 842¶

In [14]:
# Variables
wavelength = 0.49; 			#μm
a = 5.67*10**(-8);

# Calculations and Results
T = 2898/wavelength;
print ("(i) The surface temperature of the sun %.3f")% (T), ("K")

E_sun = a*T**4;
print ("(ii) The heat flux at the surface of the sun  = %.3f")% (E_sun/1E+7), ("*10^7 W/m**2")

(i) The surface temperature of the sun 5914.286 K
(ii) The heat flux at the surface of the sun  = 6.937 *10^7 W/m**2


### Example 15.25 page no : 843¶

In [40]:
import math

# Variables
T = 2773.            			#K
lambda_ = 1.2 * 10**-6
e = 0.9;
a = 5.67*10**(-8);

# Calculations and Results
print ("(i) Monochromatic emissive power at 1.2 μm length")
C1 = 0.3742*10**(-15); 			#W.m**4/m**2
C2 = 1.4388*10**(-4); 			#mK
E_lambda_b = C1*lambda_**(-5)/((math.exp(C2/lambda_/T)-1));

print ("E_lambda_b  = %.2e")% (E_lambda_b)

lambda_max = 2898/T;
print ("(ii) Wavelength at which the emission is maximum  = %.3f")% (lambda_max), ("μm")

E_lambda_b_max = 1.285*10.**(-5)*T**5;
print ("(iii) Maximum emissive power  = %.3f")% (E_lambda_b_max/1E+12), ("*10^12 W/m**2 per metre length")

Eb = a*T**4;
print ("(iv) Total emissive power  = %.3f")% (Eb/1E+6), ("*10^6 W/m**2")

E = e*a*T**4;
print ("(v) Total emissive power  = %.3f")% (E/1E+6), ("*10^6 W/m**2")


(i) Monochromatic emissive power at 1.2 μm length
E_lambda_b  = 3.40e+15
(ii) Wavelength at which the emission is maximum  = 1.045 μm
(iii) Maximum emissive power  = 2.107 *10^12 W/m**2 per metre length
(iv) Total emissive power  = 3.353 *10^6 W/m**2
(v) Total emissive power  = 3.017 *10^6 W/m**2


### Example 15.26 page no : 845¶

In [28]:
# Variables
T1 = 1273.; 			#K
T2 = 773.; 			#K
e1 = 0.42;
e2 = 0.72;
a = 5.67*10**(-8);

# Calculations and Results
print ("(i) When the body is grey with ε1  =  0.42")
q = e1*a*(T1**4-T2**4)/10**3; 			#kW
print ("Heat loss per m2 by radiation  = %.3f")% (q), ("kW")
print ("(ii) When the body is not grey")
E_emitted = e1*a*T1**4;
E_absorbed = e2*a*(T2)**4;
q = (E_emitted-E_absorbed)/10**3;
print ("Heat loss per m2 by radiation  = %.3f")% (q), ("kW")

(i) When the body is grey with ε1  =  0.42
Heat loss per m2 by radiation  = 54.036 kW
(ii) When the body is not grey
Heat loss per m2 by radiation  = 47.962 kW


### Example 15.27 page no : 846¶

In [29]:
import math

# Variables
d = 0.022; 			#m
di = 0.18; 			#m
e1 = 0.62;
e2 = 0.82;
rho = 7845.; 			#kg/m**3
T1a = 693.; 			#K; For caseI
T1b = 813.; 			#K; For caseII
T2 = 1373.; 			#K
l = 1.; 			#m
a = 5.67*10**(-8);
cp = 0.67; 			#kJ/kg K

# Calculations
A1 = math.pi*d*l;
A2 = math.pi*di*l;
Qi = A1*a*(T1a**4-T2**4)/(1/e1+A1/A2*(1/e2 - 1));
Qe = A1*a*(T1b**4-T2**4)/(1/e1+A1/A2*(1/e2 - 1));
Qav = -(Qi+Qe)/2;
t_h = math.pi/4*d**2*rho*cp*(T1b-T1a)*10**3/Qav;

# Results
print ("Time required for the heating operation  %.3f")% (t_h),("s")

Time required for the heating operation  31.157 s


### Example 15.28 page no : 847¶

In [30]:
# Variables
r1 = 0.05; 			#m
r2 = 0.1; 			#m
T1 = 400.; 			#K
T2 = 300.; 			#K
e1 = 0.5;
e2 = 0.5;
F_12 = 1.;

# Calculations
a = 5.67*10**(-8);
#A1/A2 = r1/r2
Q = a*(T1**4-T2**4)/((1-e1)/e1+1/F_12+(1-e2)/e2*r1/r2);

# Results
print ("heat transfer rate per m2 area by radiation"), (Q), ("W/m**2")

heat transfer rate per m2 area by radiation 396.9 W/m**2


### Example 15.29 page no : 847¶

In [31]:
# Variables
r1 = 0.05; 			#m
r2 = 0.1; 			#m
r3 = 0.15; 			#m
T1 = 1000.; 			#K
T3 = 500.; 			#K
e1 = 0.05;
e2 = e1;
e3 = e1;
a = 5.67*10**(-8);
F_12 = 1.;
F_23 = 1.;
T2 = 770.; 			#K

# Calculations
Q1 = a*(T1**4-T2**4)/(((1-e1)/e1) + 1/F_12 + ((1-e2)/e2)*r1/r2);

# Results
print ("Heat flow per m2 area of cylinder 1  = %.3f")% (Q1), ("W")

Heat flow per m2 area of cylinder 1  = 1246.381 W


### Example 15.30 page no : 848¶

In [32]:
# Variables
r1 = 0.105; 			#m
r2 = 0.15; 			#m
T1 = 120.; 			#K
T2 = 300.; 			#K
e1 = 0.03;
e2 = 0.03;
h_fg = 209.35; 			#kJ/kg
a = 5.67*10**(-8);
F_12 = 1.;

# Calculations
Q = 4*math.pi*r1**2*a*(T1**4-T2**4)/( ((1-e1)/e1) + 1./F_12 + ((1-e2)/e2)*r1**2/r2**2);
rate = -Q*3600./h_fg/1000;

# Results
print ("Rate of evaporation  =  %.3f")% (rate), ("kg/h")

Rate of evaporation  =  0.022 kg/h


### Example 15.31 page no : 849¶

In [33]:
import math

# Variables
T1 = 91.; 			#K
T2 = 303.; 			#K
e1 = 0.03;
e2 = 0.03;
d1 = 0.3; 			#m
d2 = 0.45; 			#m
a = 5.67*10**(-8);
F_12 = 1.;

# Calculations
Q = 4*math.pi*(d1/2)**2*a*(T1**4-T2**4)/( ((1-e1)/e1) + 1/F_12 + ((1-e2)/e2)*d1**2/d2**2);

# Results
print ("Rate of heat flow  = %.3f")% (Q), ("W")

Rate of heat flow  = -2.810 W


### Example 15.32 page no : 850¶

In [34]:
# Variables
e1 = 0.3;
e2 = 0.8;
e3 = 0.04;
A1 = 1.; 			#m**2
A2 = A1;
A3 = A1;

# Calculations
reduction = 1-0.131*0.52;

# Results
print ("Percentage reduction in heat flow due to shield  = %.3f")% (reduction), ("%")

Percentage reduction in heat flow due to shield  = 0.932 %