# Variables
L = 0.02; #Thicness of stainless steel plate in m
T = [550,50]; #Temperatures at both the faces in degree C
k = 19.1; #Thermal Conductivity of stainless steel at 300 degree C in W/m.K
# Calculations
q = ((k*(T[0]-T[1]))/(L*1000)); #Heat transfered per uni area in kW/m**2
# Results
print 'The heat transfered through the material per unit area is %3.1f kW/m**2'%(q)
# Variables
L = 1.; #Length of the flat plate in m
w = 0.5; #Width of the flat plate in m
T = 30.; #Air stream temperature in degree C
h = 30.; #Convective heat transfer coefficient in W/m**2.K
Ts = 300.; #Temperature of the plate in degree C
# Calculations
A = (L*w) #Area of the plate in m**2
Q = (h*A*(Ts-T)/(1000)); #Heat transfer in kW
# Results
print 'Heat transfer rate is %3.2f kW'%(Q)
# Variables
T = 55. #Surface temperature in degree C
# Calculations
q = (5.6697*10**-8*(273+T)**4)/1000; #The rate at which the radiator emits radiant heat per unit area if it behaves as a black body in kW/m**2
# Results
print 'The rate at which the radiator emits radiant heat per unit area is %3.2f kW/m**2'%(q)
# Variables
k = 0.145; #Thermal conductivity of Firebrick in W/m.K
e = 0.85; #Emissivity
L = 0.145; #Thickness of the wall in m
Tg = 800.; #Gas temperature in degree C
Twg = 798.; #Wall temperature ion gas side in degree C
hg = 40.; #Film conductance on gas side in W/m**2.K
hc = 10.; #Film conductance on coolant side in W/m**2.K
F = 1.; #Radiation Shape factor between wall and gas
# Calculations
R1 = (((e*5.67*10**-8*F*((Tg+273)**4-(Twg+273)**4))/(Tg-Twg))+(1./hg)); #Thermal resistance inverse
R2 = (L/k); #Thermal resistance
R3 = (1./hc); #Thermal resistance
U = 1./((1./R1)+R2+R3); #Overall heat transfer coefficient in W/m**2.K
# Results
print 'Overall heat transfer coefficient is %3.3f W/m**2.K'%(U)
# Variables
D = 0.05; #Outside diameter of the pipe in m
e = 0.8; #Emmissivity
T = 30; #Room Temperature in degree C
Ts = 250; #Surface temperature in degree C
h = 10; #Convective heat transfer coefficient in W/m**2.K
# Calculations
q = ((h*3.14*D*(Ts-T))+(e*3.14*D*5.67*10**-8*((Ts+473)**4-(T+273)**4))); #Heat loss per unit length of pipe in W/m
# Results
print 'Heat loss per unit length of pipe is %3.1f W/m'%(q)
# Variables
A = 0.1; #Surface area of water heater in m**2
Q = 1000.; #Heat transfer rate in W
Twater = 40; #Temperature of water in degree C
h1 = 300; #Heat transfer coefficient in W/m**2.K
Tair = 40; #Temperature of air in degree C
h2 = 9; #Heat transfer coefficient in W/m**2.K
# Calculations
Tsw = (Q/(h1*A))+Twater; #Temperature when used in water in degree C
Tsa = (Q/(h2*A))+Tair; #Temperature when used in air in degree C
# Results
print 'Temperature when used in water is %3.1f degree C \n \
Temperature when used in air is %i degree C'%(Tsw,Tsa)