In [4]:

```
from __future__ import division
#Variable declaration:
H = 2.5 #Height of wall (m)
W = 4 #Width of wall (m)
h = 11 #Convective heat transfer coefficient (W/m^2.K)
T1 = 24 #Outside surface temperature (°C)
T3 = -15 #Outside air temperature (°C)
L = 7.62/10**3 #Insulation thickness (m)
k = 0.04 #Thermal conductivity of wool (W/m.K)
#Calculation:
A = H*W #Heat transfer area (m^2)
Q = h*A*(T1-T3) #Heat transfer rate (W)
Ri = L/(k*A) #Insuation resistance (K/W)
Rc = 1/(h*A) #Convective resitance (K/W)
R = Ri+Rc #Total resistance (K/W)
Qt = (T1-T3)/R #Revised heat transfer rate (Btu/h)
#Result:
print "1. The heat transfer rate without insulation is :",round(Q)," W ."
print "Or, the heat transfer rate without insulation is :",round(Q*3.412)," Btu/h ."
print "2. The revised heat transfer rate with insulation is :",round(Qt)," W ."
print "Or, the revised heat transfer rate with insulation is :",round(Qt*3.412)," Btu/h ."
print "There is a calculation mistake in book."
```

In [5]:

```
#Variable declaration:
#From example 19.1:
T1 = 24 #Outside surface temperature (°C)
Ri = 0.0191 #Insulation resistance (K/W)
Q = 1383 #Revised heat transfer rate (Btu/h)
#Calculation:
T2 = T1-Q*Ri #Temperature at outer surface of insulation (°C)
#Result:
print "The temperature at the outer surface of the insulation is :",round(T2,1)," °C ."
```

In [6]:

```
#Variable declaration:
#From example 19.1:
h = 11 #Convective heat transfer coefficient (W/m^2.K)
L = 7.62/10**3 #Insulation thickness (m)
k = 0.04 #Thermal conductivity of wool (W/m.K)
#Calculation:
Bi = h*L/k #Biot number
#Result:
print "The Biot nmuber is :",round(Bi,1)," ."
```

In [7]:

```
from __future__ import division
#Variable declaration:
k = 0.022 #Thermal conductivity of glass wool (Btu/h.ft.°F)
T1 = 400 #Inside wall temperature (°F)
T2 = 25 #Outside wall temperature (°C)
L = 3/12 #Length of insulation cover (ft)
#Calculation:
T_2 = T2*(9/5)+32 #Outside wall temperature in fahrenheit scale (°F)
QbyA = k*(T1-T_2)/L #Heat flux across the wall (Btu/h.ft^2)
#Result:
print "The heat flux across the wall is :",round(QbyA,1)," Btu/h.ft^2 ."
```

In [12]:

```
from __future__ import division
#Variable declaration:
w = 8 #Width of wall (m)
H = 3 #Height of wall (m)
h = 21 #Convective heat transfer coefficient between the air and the surface (W/m^2.K)
T1 = -18 #Outside surace of wall temperature (°C)
T3 = 26 #Surrounding air temperature (°C)
l1 = 80/100 #Reduction in cooling load
k = 0.0433 #Thermal conductivity of cork board insulation (W/m.K)
T = 12000 #Units Btu/h in 1 ton of refrigeration
#Calculation:
A = w*H #Heat transfer area (m^2) (part 1)
Q1 = h*A*(T1-T3) #Rate of heat flow in the absence of insulation (W)
Q2 = Q1*3.4123/T #Rate of heat flow in the absence of insulation (ton of refrigeration)
l2 = 1-l1 #Reduced cooling load (part 2)
Q3 = l2*Q1 #Heat rate with insulation (W)
Rt = (T1-T3)/Q3 #Total thermal resistance (°C/W)
R2 = 1/(h*A) #Convection thermal resistance (°C/W)
R1 = Rt-R2 #Insulation conduction resistance (°C/W)
L = R1*k*A #Required insulation thickness (m)
#Result:
print "1. The rate of heat flow through the rectangular wall without insulation is :",round(Q1/10**3,2)," kW ."
print "Or, the rate of heat flow through the rectangular wall without insulation in tons of refrigeration is :",round(Q2,1)," ton of refrigeration ."
if (Q1<0):
print " The negative sign indicates heat flow from the surrounding air into the cold room."
else :
print " The positive sign indicates heat flow from the surrounding air into the cold room."
print "2. The required thickness of the insulation board is :",round(L*10**3,2)," mm ."
```

In [9]:

```
#Variable declaration:
#From example 19.5:
Q = -4435.2 #Heat rate with insulation (W)
R2 = 0.00198 #Convection thermal resistance (°C/W)
T3 = 26 #Surrounding air temperature (°C)
h = 21 #Convective heat transfer coefficient between the air and the surface (W/m^2.K)
k = 0.0433 #Thermal conductivity of cork board insulation (W/m.K)
L = 0.00825 #Required insulation thickness (m)
#Calculation:
T2 = T3+Q*R2 #Interface temperature (°C) (part 1)
Bi = h*L/k #Biot number (part 2)
#Result:
print "1. The interface temperature is :",round(T2,2)," °C ."
print "2. The Biot number is :",round(Bi)," ."
print "3. Theoretical part."
```

In [11]:

```
from __future__ import division
from math import pi,log
#Variable declaration:
D2 = 0.5/10**3 #External diameter of needle (m)
h3 = 12 #Heat transfer coefficient (W/m^2.K)
L = 1 #Insulation thickness (m)
T1 = 95 #Reactant temperature (°C)
T3 = 20 #Ambient air temperature (°C)
k1 = 16 #Thermal conductivity of needle (W/m.K)
k3 = 0.0242 #Thermal conductivity of air (W/m.K)
D3 = 2/10**3 #Diameter of rubber tube (m)
#Calculation:
r2 = D2/2 #External radius of needle (m)
r3 = D3/2 #Radius of rubber tube (m)
Rt1 = 1/(h3*(2*pi*r2*L)) #Thermal resistance (°C/W)
Q1 = (T1-T3)/Rt1 #Rate of heat flow in the absence of insulation (W)
Bi = h3*D2/k1 #Biot number
Nu = h3*D2/k3 #Nusselt number
R2 = log(r3/r2) #Thermal resistance of needle (°C/W)
R3 = 1/(h3*(2*pi*r3*L)) #Thermal resistance of rubber tube (°C/W)
Rt2 = R2+R3 #Total thermal resistance (°C/W)
Q2 = (T1-T3)/Rt2 #Rate of heat loss (W)
#Result:
print "1. The rate of the heat loss from the hypodermic needle with the rubber insulation is :",round(Q1,2)," W ."
print " The rate of the heat loss from the hypodermic needle without the rubber insulation is :",round(Q2,2)," W ."
print "2. The Biot number is :",round(Bi,6)," ."
print " The nusselt number is :",round(Nu,3)," ."
```

In [15]:

```
from __future__ import division
from math import log, pi
#Variable declaration:
h = 140 #Convention heat transfer coefficient (W/m^2.K)
D1 = 10/10**3 #Rod diameter (m)
L = 2.5 #Rod length (m)
T1 = 200 #Surface temperature of rod (°C)
T2 = 25 #Fluid temperature (°C)
k = 1.4 #Thermal conductivity of bakellite (W/m.K)
l = 55/10**3 #Insulation thickness (m)
#Calculation:
Q1 = h*pi*D1*L*(T1-T2) #Rate of heat transfer for the bare rod (W) (part 1)
Bi = 2 #Critical Biot number (part 2)
D2 = Bi*k/h #Critical diameter associated with the bakelite coating (m)
r2 = D2/2 #Critical radius associated with the bakelite coating (m)
r1 = D1/2 #Rod radius (m)
R1 = log(r2/r1)/(2*pi*k*L) #Insulation conduction resistance (°C/W)
R2 = 1/(h*(2*pi*r2*L)) #Convection thermal resistance (°C/W)
Rt1 = R1+R2 #Total thermal resistance (°C/W)
Qc = (T1-T2)/Rt1 #Heat transfer rate at the critical radius (W)
r3 = r1+l #New radius associated with the bakelite coating after insulation (m) (part 3)
R3 = log(r3/r1)/(2*pi*k*L) #Insulation conduction bakelite resistance (°C/W)
R4 = 1/(h*(2*pi*r3*L)) #Convection bakelite thermal resistance (°C/W)
Rt2 = R3+R4 #Total bakelite thermal resistance (°C/W)
Q2 = (T1-T2)/Rt2 #Heat transfer rate at the bakelite critical radius (W)
Re = ((Q1-Q2)/Q1)*100 #Percent reduction in heat transfer rate relative to the case of a bare rod (%)
#Result:
print "1. The rate of heat transfer for the bare rod is :",round(Q1)," W ."
print "2. The critical radius associated with the bakelite coating is :",round(r2*10**3)," mm."
print " & the heat transfer rate at the critical radius is :",round(Qc)," W ."
print "3. The fractional reduction in heat transfer rate relative to the case of a bare rod is :",round(Re,1)," % ."
```

In [16]:

```
from __future__ import division
from math import pi, log
#Variable declaration:
r1 = 1.1/100 #Inside radius of pipe (m)
r2 = 1.3/100 #Outside radius of pipe (m)
r3 = 3.8/100 #Outside radius of asbestos insulation (m)
L = 1 #Length of tube (m)
h1 = 190 #Heat transfer coefficient from ethylene glycol to the stainless steel pipe (W/m^2.K)
k2 = 19 #Thermal conductivity of pipe (W/m.K)
h2 = 14 #Outside heat transfer coefficient from the air to the surface of the insulation (W/m^2.K)
k3 = 0.2 #Thermal conductivity of asbestos (W/m.K)
T1 = 124 #Hot ethylene glycol temperature (°C)
T5 = 2 #Surrounding air temperature (°C)
k4 = 0.0242 #Thermal conductivity of air (W/m.K)
#Calculation:
A1 = 2*pi*r1*L #Inside surface area of pipe (m^2) (part1)
A2 = 2*pi*r2*L #Outside surface area of pipe (m^2)
A3 = 2*pi*r3*L #Outside surface area of asbestos insulation (m^2)
R1 = 1/(h1*A1) #Inside convection resistance (°C/W)
R2 = log(r2/r1)/(2*pi*k2*L) #Conduction resistance through the tube (°C/W)
R3 = 1/(h2*A2) #Outside convection resistance (°C/W)
Rt1 = R1+R2+R3 #Total resistance without insulation (°C/W)
Q1 = (T1 - T5)/Rt1 #Heat transfer rate without insulation (W)
R4 = log(r3/r2)/(2*pi*k3*L) #Conduction resistance associated with the insulation (°C/W) (part 2)
R5 = 1/(h2*A3) #Outside convection resistance (°C/W)
Rt2 = R1+R2+R4+R5 #Total rsistance with the insulation (°C/W)
Q2 = (T1-T5)/Rt2 #Heat transfer rate with the insulation (W)
U1 = 1/(Rt2*A1) #Overall heat transfer coefficient based on the inside area (W/m^2.K) (part 3)
U3 = 1/(Rt2*A3) #Overall heat transfer coefficient based on the outside area (W/m^2.K) (part 4)
T3 = T1-(R1+R2)*Q2 #Temperature at the steel–insulation interface (°C) (part 5)
Bi1 = h2*(2*r3)/k3 #Outside Biot number (part 6)
Bi2 = h1*(2*r1)/k2 #Inside Biot number
Nu = h1*(2*r1)/k4 #Nusselt number of the air
rlm = (r3-r2)/log(r3/r2) #Log mean radius of the insulation (m) (part 7)
#Result:
print "1. The rate of heat transfer without insulation is :",round(Q1,1)," W."
print "2. The rate of heat transfer with insulation is :",round(Q2,1)," W."
print "3. The overall heat transfer coefficient based on the inside area of the tube is :",round(U1,2)," W/m^2.K ."
print "4. The overall heat transfer coefficient based on the outside area of the insulation is :",round(U3,1)," W/m^2.K ."
print "5. The temperature, T3, at the steel–insulation interface is :",round(T3,1)," °C."
print "6. The inside Biot numbers is :",round(Bi2,2)," ."
print " The outside Biot numbers is :",round(Bi1,2)," ."
print " The Nusselt number is :",round(Nu,1)," ."
print "7. The log mean radius of insulation is :",round(rlm*100,2)," cm."
print "There is a printing mistake in book for unit in part 7."
```

In [17]:

```
from __future__ import division
from math import pi
#Variable declaration:
h1 = 800 #Heat transfer coefficient for steam condensing inside coil (Btu/h.ft^2.°F)
h2 = 40 #Heat transfer coefficient for oil outside coil (Btu/h.ft^2.°F)
h3 = 40 #Heat transfer coefficient for oil inside tank wal (Btu/h.ft^2.°F)
h4 = 2 #Heat transfer coefficient for outer tank wall to ambient air (Btu/h.ft^2.°F)
k1 = 0.039 #Thermal conductivity of insulation layer (Btu/h.ft.°F)
l1 = 2/12 #Thickness of insulation layer (ft)
D = 10 #Diameter of tank (ft)
H = 30 #Height of tank (ft)
k2 = 224 #Thermal conductivity of copper tube (Btu/h.ft.°F)
l2 = (3/4)/12 #Thickness of insulation layer (ft)
T1 = 120 #Temperature of tank (°F)
T2 = 5 #Outdoor temperature (°F)
#Calculation:
Uo1 = 1/(1/h3+(l1/k1)+1/h4) #Overall heat transfer coefficient for tank (Btu/h.ft^2.°F)
At = pi*(D+2*l1)*H #Surface area of tank (ft^2)
Q = Uo1*At*(T1-T2) #Heat transfer rate lost from the tank (Btu/h)
#From table 6.3:
l2 = 0.049/12 #Thickness of coil (ft)
A = 0.1963 #Area of 18 guage, 3/4-inch copper tube (ft^2/ft)
Uo2 = 1/(1/h2+(l2/k2)+1/h1) #Overall heat transfer coefficient for coil (Btu/h.ft^2.°F)
#From steam tables:
Tst = 240 #Temperature for 10 psia (24.7 psia) steam (°F)
Ac = Q/(Uo2*(Tst-T1)) #Area of tube (ft^2)
L = Ac/A #Lengt of tube (ft)
#Result:
print "The length ofcopper tubing required is :",round(L,1)," ft ."
```

In [3]:

```
from __future__ import division
from math import pi, log
from numpy import array,log as LOG
#Variable declaration:
#For 1-inch pipe schedule 40:
Di = 1.049/12 #Inside diameter (ft)
Do = 1.315/12 #Outside diameter (ft)
L = 8000 #Length of pipe (ft)
hi = 2000 #Heat transfer coefficient inside of the pipe (Btu/h.ft^2.°F)
ho = 100 #Outside heat transfer coefficient (Btu/h.ft.°F)
kl = 0.01 #Thermal conductivity of insulation (Btu/h.ft.°F)
T1 = 240 #Steam temperature (°F)
T2 = 20 #Air temperature (°F)
k = 24.8 #Thermal conductivity for steel (Btu/h.ft.°F)
Dxl = array([3/8,1/2,3/4,1])/12 #thickness(ft)
amt = array([1.51,3.54,5.54,8.36])/6 #Cost per feet($)
#Calculation:
D_ = (Do-Di)/log(Do/Di) #Log-mean diameter of the pipe (ft)
Dl = Do+2*(Dxl) #Insulation thickness (ft)
D_l = (Dl-Do)/LOG(Dl/Do) #Log mean diameter of pipe (ft)
Dxw = (Do-Di)/2 #Pipe thickness (ft)
Rw = Dxw/(k*pi*D_*L) #Wall resistance ((Btu/h.°F)^-1)
Ri = 1/(hi*pi*Di*L) #Inside steam convection resistance ((Btu/h.°F)^-1)
Rl = Dxl/(kl*pi*D_l*L) #Insulation resistance ((Btu/h.°F)^-1)
Ro = 1/(ho*pi*Dl*L) #Outside air convection resistance ((Btu/h.°F)^-1)
R = Ri+Rw+Rl+Ro #Total resistance ((Btu/h.°F)^-1)
Uo = 1/(R*pi*Dl*L) #Overall outside heat transfer coefficient (Btu/h.ft^2.°F)
Ui = 1/(R*pi*Di*L) #Overall inside heat transfer coefficient (Btu/h.ft^2.°F)
dT = T1-T2
Ai = pi*Di*L #Inside area (ft^2)
Q = Ui*Ai*dT #Energy loss (Btu/h)
def energyPerDollar(Q1,Q2,amt1,amt2):
return round((Q1-Q2)/(8000*(amt2-amt1)),1)
#Results:
print "Energy saved per dollar ingoing from 3/8 to 1/2 inch is :",energyPerDollar(Q[0],Q[1],amt[0],amt[1]),' Btu/h.$'
print "Energy saved per dollar ingoing from 1/2 to 3/4 inch is :",energyPerDollar(Q[1],Q[2],amt[1],amt[2]),' Btu/h.$'
print "Energy saved per dollar ingoing from 3/4 to 1 inch is :",energyPerDollar(Q[2],Q[3],amt[2],amt[3]),' Btu/h.$'
```

In [21]:

```
from __future__ import division
#Variable declaration:
ki = 0.44 #Thermal conductivity of insulation (Btu/h.ft.°F)
ho = 1.32 #Air flow coefficient (Btu/h.ft^2.°F)
OD = 2 #Outside diameter of pipe (in)
#Calculation:
rc = (ki/ho)*12 #Outer critical radius of insulation (in)
ro = OD/2 #Outside radius of pipe (in)
L = rc-ro #Critical insulation thickness (in)
#Result:
print "The outer critical radius of insulation is :",round(rc)," in ."
if ro<rc:
print "Since, ro<rc, the heat loss will increase as insulation is added."
else :
print "Sice, ro>rc, the heat loss will decrease as insulation is added."
```

In [24]:

```
from __future__ import division
#Variable declaration:
Lf = 6/12 #Length of firebrick (ft)
kf = 0.61 #Thermal conductivity of firebrick (Btu/h.ft.°F)
A = 480 #Surface area of wall (ft^2)
Lw = 8/12 #Length of rock wool (ft)
kw = 0.023 #Thermal conductivity of rock wool (Btu/h.ft.°F)
T1 = 1900 #Temperature of insulation of firebrick (°F)
T2 = 140 #Temperature of insulation of rock wool (°F)
#Calculation:
Rf = Lf/(kf*A) #Resistance of firebrick (h.°F/Btu)
Rw = Lw/(kw*A) #Resistance of rock wool (h.°F/Btu)
R = Rf+Rw #Total resitance (h.°F/Btu)
Q = (T1-T2)/R #Heat loss through the wall (Btu/h)
#Result:
print "The heat loss through the wall is :",round(Q)," Btu/h ."
```

In [25]:

```
from __future__ import division
#Variable declaration:
h1 = 1700 #Steam heat-transfer coefficient (Btu/h.ft^2.°F)
h2 = 2 #Air heat-transfer coefficient (Btu/h.ft^2.°F)
A = 1 #Area of base (ft^2) (assumption)
k1 = 26 #Thermal conductivity of steel (Btu/h.ft.°F)
k2 = 218 #Thermal conductivity of copper (Btu/h.ft.°F)
t = 0.375 #Thickness of steel sheet (in)
h3 = 2500 #Increased steam heat-transfer coefficient (Btu/h.ft^2.°F)
h4 = 12 #Increased air heat-transfer coefficient (Btu/h.ft^2.°F)
#Calculation:
R1 = 1/(h1*A) #Steam resistance (h.°F/Btu)
R2 = 1/(h2*A) #Air resistance (h.°F/Btu)
R3 = (t/12)/(k1*A) #Steel resistance (h.°F/Btu)
Rt1 = R1+R2+R3 #Total resistance (with steel) (h.°F/Btu)
R4 = (t/12)/(k2*A) #Copper resistance (h.°F/Btu) (part 1)
Rt2 = R1+R2+R4 #Total resistance (with copper) (h.°F/Btu)
R5 = 1/(h1*A) #New steam resistance (h.°F/Btu)
Rt3 = R5+R2+R3 #Total resistance after increasing the steam coefficient (h.°F/Btu)
R6 = 1/(h4*A) #Air resistance (h.°F/Btu)
Rt4 = R1+R6+R3 #Total resistance after increasing the air coefficient (h.°F/Btu)
#Result:
if (Rt1==Rt2):
print "1.The rate of heat transfer is essentially unaffected."
else :
print "1. The rate of heat transfer is essentially affected."
if (Rt1==Rt3):
print "2. The rate is again unaffected."
else :
print "2. The rate is again affected."
if (Rt1==Rt4):
print "3. The rate is unaffected for this case."
else :
print "3. The rate is affected for this case."
```

In [28]:

```
from __future__ import division
from math import pi, log, exp
#Variable declaration:
rfo = 12/2 #Outside radius of firebrick (ft)
rfi = 5.167 #Inside radius of firebrick (ft)
rso = 6.479 #Outside radius of sil-o-cel (ft)
rsi = 6.063 #Inside radius of fsil-o-cel (ft)
L = 30 #Length of incinerator (ft)
kf = 0.608 #Thermal conductivity of firebrick (Btu/h.ft.°F)
ks = 0.035 #Thermal conductivity of sil-o-cel (Btu/h.ft.°F)
#Calculation:
Rf= log(rfo/rfi)/(2*pi*L*kf) #Resistance of firebrick (h.ft.°F/Btu)
Rs= log(rso/rsi)/(2*pi*L*ks) #Resistance of sil-o-cel (h.ft.°F/Btu)
R = Rf+Rs #Total resistance (h.ft.°F/Btu)
ro = exp(R*(2*pi*L*ks))*rso #New outside radius of sil-o-cel (ft)
r= ro-rso #Extra thickness (ft)
#Result:
print "The extra thickness is :",round(r,3)," ft."
print "Or, the extra thickness is :",round(r*12,2)," in."
```