# 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")
# 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")
# 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")
# 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")
# 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")
# 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")
# 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")
#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")
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")
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")
# 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")
# 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")
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), ("%")
# 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")
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")
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")
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")
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")
# 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")
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
# 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")
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")
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")
# 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")
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")
# Book answers are wrong. Please calculate them manually.
# 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")
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")
# 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")
# 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")
# 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")
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")
# 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), ("%")