Chapter 2: Conduction

Example no:2.1,Page no:2.18

In [6]:
#Thickness of insulation

#Variable declaration
A=1  #Area of heat transfer[sq metre]
Q=450  #Rate of heat loss/unit area[W/ sq mtre]
dT=400  #Temperature difference across insulation layer[K]
k=0.11 #k for asbestos[W/(m.K)]

#Calculation
#Q=(k* A*dT)/x
x1=(k*A*dT)/Q
X1=x1*1000 
#for fire clay insulation
k=0.84  #For fire clay insulation[W/(m.K)]
x=(k*A*dT)/Q 
X=x*1000 

#Result
print"Ans.(A).Thickness of asbestos is:",round(X1),"mm"
print"Ans.(B)Thickness of fire clay insulation is:",round(X),"mm"

Ans.(A).Thickness of asbestos is: 98.0 mm
Ans.(B)Thickness of fire clay insulation is: 747.0 mm

Example no:2.2,Page no:2.18

In [8]:
#Heat loss per metre

#Variable declaration
L=1.0 # Length of pipe[m]
r1=(50.0/2.0) # in mm
r1=r1/1000.0 # in m
r2=(25.0+3.0)/1000.0 # m


#Calculation
import math
rm1=(r2-r1)/math.log(r2/r1) 
k1=45.0 #W/(m.K)
R1=(r2-r1)/(k1*(2*math.pi*rm1*L)) # Thermal resistance of wall pipe[K/W]

#For inner lagging:
k2=0.08 #W/(m.K)
ri1=0.028 #m
ri2=(ri1+r1) # m
rmi1=(ri2-ri1)/math.log(ri2/ri1)
R2=(ri2-ri1)/(k2*2*math.pi*rmi1*L) #Thermal resistance of inner lagging [K/W]

#For  outer lagging:
k3=0.04 #W/(m.K)
ro1=0.053 #m
ro2=(ro1+0.04) # m
rmo1=(ro2-ro1)/math.log(ro2/ro1)
R3=(ro2-ro1)/(k3*2*math.pi*rmo1*L)  #Thermal resistance of outer lagging

R=R1+R2+R3
Ti=550.0 #K #inside
To=330.0 #K  # outside
dT=Ti-To  #Temperature difference
Q=dT/R


#Result
print"Rate of heat loss per metre of pipe,Q=",round(Q,1),"W/m"

Rate of heat loss per metre of pipe,Q= 62.7 W/m

Example no:2.3,Page no:2.19

In [7]:
#Heat Loss in pipe
import math
#Variable declaration
r1=44.0 #i [mm]
r1=r1/1000.0  #i[m]
r2=0.094   #i [m]
r3=0.124  #i [m]
T1=623.0  #iTemperature at outer surface of wall in[K]
T3=313.0  #iTemperature at outer surface of outer insulation  [K]
k1=0.087  #iThermal conductivity of insulation layer 1..in [W/m.K]
k2=0.064  #iThermal conductivity of insulation layer 2  [W/m.K]
l=1    #i Length of pipe   [m]

#Calculation
rm1=(r2-r1)/math.log(r2/r1)  #imath.log mean radius of insulation layer 1 [m]
rm2=(r3-r2)/math.log(r3/r2)  #imath.log mean radius of insulation layer 2[m]
#iPutting values in following eqn:
Q= (T1-T3)/((r2-r1)/(k1*2*math.pi*rm1*l)+(r3-r2)/(k2*2*math.pi*rm2*l)) 

#Result
print"Heat loss per meter pipe is",round(Q,1),"W/m(approx)"

Heat loss per meter pipe is 149.2 W/m(approx)

Example no:2.4,Page no:2.21

In [9]:
#Heat loss in interface

#Variable declaration
A=1 #Heat transfer area  [sq m]
x1=0.229  # thickness of fire brick in [m]
x2=0.115  # thickness of insulating brick in [m]
x3=0.229  # thickness of building brick in [m]
k1=6.05  #thermal conductivity of fir brick  [W/(m.K)]
k2=0.581  #thermal conductivity of insulating brick   [W/m.K]
k3=2.33  #thermal conductivity of building brick   [W/m.K]
T1=1223  # inside temperature [K]
T2=323 # Outside temperature[K]

#Calculation
dT=T1-T2  #Overall temp drop [K]
R1=(x1/k1*A) #thermal resistance 1
R2=(x2/k2*A) # Thermal resistance 2
R3=(x3/k3*A) #Thermal resistance 3
Q=dT/(R1+R2+R3) #w/SQ m
Ta=-((Q*R1)-T1)  #from  Q1=Q=(T1-Ta)/(x1/k1*A)
#Similarly
Tb=(Q*R3)+T2 

#Result
print"Interface temperature:\ni-Between  FB-IB=",round(Ta)," K  \nii-Between IB-PB=",round(Tb,1),"K"

Interface temperature:
i-Between  FB-IB= 1121.0  K  
ii-Between IB-PB= 587.8 K

Example no:2.5,Page no:2.23

In [10]:
#Heat loss per unit area
import math
#Variable declaration
A=1   #let [sq m]
x1=0.23    #thickness of fir brick layer[m]
x2=0.115    # [m]
x3=0.23    #[m]
T1=1213.0   #Temperature of furnace [K]
T2=318.0    #Temperature of furnace  [K]
dT=T1-T2   #[K]
k1=6.047    #W/(m.K) (fire brick)
k2=0.581    #W/(m.K) (insulating brick)
k3=2.33    #W/(m.K) (building brick)

#Calculation
Q_by_A=dT/((x1/k1)+(x2/k2)+(x3/k3))  #Heat lost per unit Area in Watt
Q_by_A=round(Q_by_A,1)
R1=(x1/k1)  #Thermal resistance
R2=(x2/k2)
R3=(x3/k3)
R1=round(R1,2)
R2=round(R2,1)
R3=round(R3,1)
Ta=T1-((dT*R1)/(R1+R2+R3))
Ta=round(Ta)
Tb=((dT*R3)/(R1+R2+R3))+T2


#Result
print"Heat loss per unit area is",Q_by_A,"W=",Q_by_A,"J/s\n"
print"Ta=",Ta,"K(APPROX) =Temperature at the interface between fire brick and insulating brick"
print"Tb=",round(Tb),"K Temperature at the interface between insulating and building brick"
print"\nNOTE:Tb is wrongly calculated in the book as 565 K"

Heat loss per unit area is 2674.2 W= 2674.2 J/s

Ta= 1108.0 K(APPROX) =Temperature at the interface between fire brick and insulating brick
Tb= 581.0 K Temperature at the interface between insulating and building brick

NOTE:Tb is wrongly calculated in the book as 565 K

Example no:2.7,Page no:2.26

In [11]:
#Heat loss in wall
#Part-(a)

#Variable declaration
A=1.0  # sq metre
x1=114.0 # mm
x1=x1/1000.0 # metre
k1=0.138 # W/(m.K)
R1= x1/(k1*A)
x2=229.0 #mm
x2= x2/1000.0  # metre
k2=1.38  # W/m.K
R2=x2/(k2*A)
dT=1033.0-349.0

#Calculation
#Heat loss
Q1=dT/(R1+R2)
#Part(b)
#contact resistance=cr
cr=0.09 #K/W
R=R1+R2+cr
Q=dT/R

#Result
print"Heat loss  from 1 sq metre wall=",round(Q1,1),"W"
print"Heat loss from 1 sq metre when resistance present=",round(Q,1),"W"

Heat loss  from 1 sq metre wall= 689.5 W
Heat loss from 1 sq metre when resistance present= 632.1 W

Example no:2.8,Page no:2.27

In [12]:
#Loss per area

#Variable declaration
x1=0.02  #[m]
x2=0.01  #[m]
x3=0.02  #[m]
k1=0.105  #W/(m.k)
k3=k1  #W/(m.K)
k2=0.041  #W/(m.K)
T1=303
T2=263

#Calculation
dT=T1-T2   #[K]
Q_by_A=dT/((x1/k1)+(x2/k2)+(x3/k3))
R=0.625  #K/W
Tx=293   #K
Rx=0.9524   #K/W
x=R*(T1-Tx)/(dT*Rx)
x=x*100  #mm

#Result
print"The temperature of 293 K will be reached at point",round(x,1),"mm from the outermost wall surface of the ice-box"

The temperature of 293 K will be reached at point 16.4 mm from the outermost wall surface of the ice-box

Example no:2.9,Page no:2.28

In [13]:
#Heat loss

#Variable declaration
import math
ID=50.0 #Internal diameter[mm]
dT=(573.0-303.0) 
r1=ID/2.0 #mm
r1=r1/1000.0 # metres
OD=150.0 #Outer diameter[mm]
r2=OD/2.0 # mm
r2=75.0/1000.0 # m
#Thermal conductivity
k=17.45 # W/(m.K) 

#Calculation
#Q/A=dT/(r2-r1)/k
A1=4*math.pi*(r1**2) 
A2=4*math.pi*(r2**2) 
A=math.sqrt(A1*A2)
Q=(A*k*dT)/(r2-r1)

#Result
print"Heat loss=Q=",round(Q),"W"

Heat loss=Q= 2220.0 W

Example no:2.10,Page no:2.29

In [14]:
#Heat Passed
import math
#Variable declaration
A= 1.0 #sq m
x1=0.15
x2=0.01
x4=0.15
T1=973.0  #[K]
T2=288.0  #[K]
dT=T1-T2  #[K]
#Thermal conductivities
k1=1.75  
k2=16.86
k3=0.033
k4=5.2

#Calculation
#in absence of air gap,sum of thermal resistances 
sR=(x1/k1*A)+(x2/k2*A)+(x4/k4*A)
round(sR,3)
sR=0.1153  #approximate
Q= dT/sR

#When heat loss,Q=1163,then new resistance =sR1
Q1=1163.0  #[W/sq m]
sR1=dT/Q1
#width of air gap be w then
w=(sR1-sR)*k3*A  # [m]
w=w*1000  #in [mm]

#Result
print"Heat lost per sq meter is",round(Q),"W/sq m"
print"Width of air gap is",round(w,1),"mm"

Heat lost per sq meter is 5941.0 W/sq m
Width of air gap is 15.6 mm

Example no:2.11,Page no:2.30

In [15]:
# Heat loss in Insulated pipe
import math

#Variable declaration
d1=300.0  #[mm]
r1=d1/2.0  # [mm]
r1=r1/1000.0  #[m]
r2=r1+0.05  #[m]
r3=r2+0.04  #[m]
x1=0.05  #[m]
x2=0.04  #[m]
k1=0.105  #W/(m.K)
k2=0.07   #W/(m.K)

#Calculation
rm1=  (r2-r1)/math.log(r2/r1)   # [m]
rm2=(r3-r2)/math.log(r3/r2)      #[m]
L=1  #let
A1=math.pi*rm1*L   #let L=1
R1=x1/(k1*A1) 
A2=math.pi*rm2*L
R2=x2/(k2*A2)
T1=623.0  #[K]
T2=323.0  #[K]
dT=T1-T2  #[K]

#Part a
Q_by_L= dT/(R1+R2)   #Heat loss

#Part b:
P=2*math.pi*(r1+x1+x2)  #[m]
Q_by_L_peri=Q_by_L/P  # [W/sq m]
R1=x1/(k1*A1) 
sR=0.871+0.827
dT1=dT*R1/sR


#Result
#Part a
print"Heat loss is:",round(Q_by_L,1),"W/m" 
#Part b:
print"Heat lost per sq meter of outer insulation is",round(Q_by_L_peri),"W/sq m"
print"Temperature between two layers of insulation=",round((T1-dT1)),"K"

  

Heat loss is: 176.3 W/m
Heat lost per sq meter of outer insulation is 117.0 W/sq m
Temperature between two layers of insulation= 469.0 K

Example no:2.12,Page no:2.31

In [16]:
#Heat loss in Composite brick
import math
#Variable declaration
x1=0.01  #[m]
x2=0.15  #[m]
x3=0.15  #[m]
T1=973.0  #[K]
T2=423.0  #[K]
dT=T1-T2 
#Thermal conductivities
k1=16.86  #[W/m.K]
k2=1.75  #[W/m.K]
k3=5.23  #[W/m.K]
k_air=0.0337  # [W/m.K]
A=1  #[sq m]

#Calculation
sigma_R=(x1/(k1*A)+x2/(k2*A)+x3/(k3*A))
Q=dT/sigma_R  #Heat flow in  [W
Tm= Q*x3/k3  #Temperature drop in magnesite brick
#Interface temperature=iT
iT=T2+Tm  #[K]
sigma_xbyk= A*dT/1163   #with air gap for reducing heat loss to 1163 per sq m
x_by_k=sigma_xbyk-sigma_R   #x/k for air
t=x_by_k*k_air
t=t*1000 

#Calculation
print"Width of the air gap is",round(t,2),"mm"

 

Width of the air gap is 12.06 mm

Example no:2.13,Page no:2.32

In [17]:
#Heat flow in a pipe
import math
#Variable declaration
L=1   #assume  [m]
k1=43.03   #[W/(m.K)
k2=0.07    #(W/m.K)
T1=423    #inside temperature [K]
T2=305     # [K]
r1=0.0525    #[mm]
r2=0.0575    #[m]
r3=0.1075 #[m]

#Calculation
#r3=r3/1000   #[m]
Q=(2*math.pi*L*(T1-T2))/(((math.log(r2/r1))/k1)+((math.log(r3/r2))/k2))      #Heat loss per metre 
#Part 2
#T=Temperature of outer surface
T=T1-(Q*math.log(r2/r1))/(k1*2*math.pi*L) 
#Part iii
id=0.105    #inside diametre in [m]
A=math.pi*id*1  #inside area in [sq m]
C=Q/(A*(T1-T2))    #conductance per length

#Result
print"Heat flow per metre of pipe is",round(Q,2),"W/m"
#Part 2
print"Temperature at outer surface of steel pipe:",round(T,2),"K" 
#Part iii
print"Conductance per m length based on inside area is",round(C,1),"W/K"

Heat flow per metre of pipe is 82.93 W/m
Temperature at outer surface of steel pipe: 422.97 K
Conductance per m length based on inside area is 2.1 W/K

Example no:2.15,Page no:2.35

In [18]:
#Thickness of insulation 
  
#Variable declaration
A=1  # [sq m]
x1=0.1    #m
x2=0.04
k1=0.7
k2=0.48

#Calculation
sigma=x1/(k1*A)+x2/(k2*A)   #K/W
#Q=4.42*dT
#Q=dT/sigma
#with rockwool insulation added,Q_dash=0.75*Q
k3=0.065     # W/(m.K)
#Q_dash=dT/sigma+x3/k3*A
#On solving Q and Q_dash we get
x3=((1/(0.75*4.42))-sigma)*k3    #[m]
x3=x3*1000     # [mm]

#Result
print"Thickness of rockwool insulation required=",round(x3,2),"mm"

Thickness of rockwool insulation required= 4.91 mm

Example no:2.16,Page no:2.36

In [19]:
#Reduction in heat loss in insulated pipe
d1=40.0    # Diameter of pipe[mm]
r1=(d1/2.0)/1000.0  #Outside radius in [m]
t1=20.0     #Insulation 1 thickness in [mm]
t1=t1/1000    #[m]
t2=t1          #Insulation 2 thickness in[m]
r2=r1+t1      #radius after 1st insulation  in [m]
r3=r2+t2      #Radius after second insulation in [m]

#Calculation
import math
#Since python does not handles symbolic constants,we will assume some values:
#(1)
#Let the layer M-1 be nearer to the surface
L=1.0          #[m]
T1=10.0        #Temperature of inner surface of pipe [K]
T2=5.0         #Temperature of outer surface of insulation [K]
k=1.0          #Thermal conductivity
k1=k         #For M-1 material
k2=3*k       #For material M-2
Q1=(T1-T2)/(math.log(r2/r1)/(2*math.pi*L*k1)+math.log(r3/r2)/(2*math.pi*L*k2))
#(2)
#Let the layer of material M-2 be nearer to the surface
Q2=(T1-T2)/(math.log(r2/r1)/(2*math.pi*L*k2)+math.log(r3/r2)/(2*math.pi*L*k1))
#For dummy variables unity...
#For any value of k,T1 and T2,Q1 is always less than Q2

#Result
print"M-1 near the surface is advisable(i.e Arrangement one will result in less heat loss)"
per_red=(Q2-Q1)*100/Q2
print"Percent reduction in heat loss is",round(per_red,1),"percent"
print"\nNOTE:Slight variation in answers due to less precise calculation in book.If performed manually,this answer stands to be correct"

M-1 near the surface is advisable(i.e Arrangement one will result in less heat loss)
Percent reduction in heat loss is 23.2 percent

NOTE:Slight variation in answers due to less precise calculation in book.If performed manually,this answer stands to be correct

Example no:2.17,Page no:2.37

In [20]:
#Heat loss in a pipe
import math

#Variable declaration
T1=523   #[K]
T2=323   #[K]
r1=0.05   #[m]
r2=0.055   #[m]
r3=0.105   #[m]
r4=0.155   #[m]
k1=50      #[W/(m.K)]
k2=0.06    #[W/(m.K)]
k3=0.12    #W/(m.K)

#Calculation
#CASE 1
Q_by_L1=2*math.pi*(T1-T2)/((math.log(r2/r1))/k1+(math.log(r3/r2))/k2+(math.log(r4/r3))/k3)   #[W/m]
#Case 2
Q_by_L2=2*math.pi*(T1-T2)/((math.log(r2/r1))/k1+(math.log(r3/r2))/k3+(math.log(r4/r3))/k2)
perct=(Q_by_L2-Q_by_L1)*100/Q_by_L1

#Result
#CASE 1
print"Heat loss=",round(Q_by_L1,1),"W/m"
#Case 2
print"If order is changed then heat loss=",round(Q_by_L2,2),"W/m" 
print"Loss of heat is increased by",round(perct,2)," percent by putting material with higher thermal conductivity near the pipe surface"

Heat loss= 89.6 W/m
If order is changed then heat loss= 105.76 W/m
Loss of heat is increased by 18.04  percent by putting material with higher thermal conductivity near the pipe surface

Example no:2.18,Page no:2.38

In [21]:
#Arrangements for heat loss

#Variable declaration
#Assume:
L=1.0    #[m]
r1=0.10    #[m]    Outside radius od pipe
ia=0.025    #inner insulaiton [m]
import math 
r2=r1+ia    #Outer radius of inner insulation
r3=r2+ia    #Outer radius of outer insulation

#Calculation
#CASE 1:'a' near the pipe surface
#let k1=1
k1=1.0     #Thermal conductivity of A[W/m.K]
#and k2=3k1=3
k2=3.0     #Thermal conductivity of B[W/m.K]
#Let dT=1
dT=1.0
Q1=dT/(math.log(r2/r1)/(2*math.pi*k1*L)+math.log(r3/r2)/(2*math.pi*k2*L))
#CASE 2:'b' near the pipe surface 
Q2=dT/(math.log(r2/r1)/(2*math.pi*k2*L)+math.log(r3/r2)/(2*math.pi*k1*L))

#Result
print"ANSWER-(i)\nQ1=",round(Q1,2),"W \nQ2=",round(Q2,2)," W "
print"Q1 is less than Q2.i.e arrangement A near the pipe surface and B as outer layer gives less heat loss\n"
percent=(Q2-Q1)*100/Q1     #percent reduction in heat loss
print"ANSWER-(ii) \nPercent reduction in heat loss (with near the pipe surface)=",round(percent,1),"%(approx)" 

ANSWER-(i)
Q1= 22.13 W 
Q2= 24.48  W 
Q1 is less than Q2.i.e arrangement A near the pipe surface and B as outer layer gives less heat loss

ANSWER-(ii) 
Percent reduction in heat loss (with near the pipe surface)= 10.6 %(approx)

Example no:2.19,Page no:2.39

In [22]:
#Insulation thickness

#Variable declaration
x1=0.224 # m
k1=1.3 # W/(m.K)
k2=0.346 # W/(m.K)
T1=1588.0 # K
T2= 299.0 # K
QA=1830.0 # W/ sq metre #heat loss

#Calculation
#Q/A=(T1-T2)/x1/k1+x2/k2
x2=k2*((T1-T2)*1/(QA)-(x1/k1))
x2=x2*1000 

#Result
print"Thickness of insulation=",round(x2),"mm"

Thickness of insulation= 184.0 mm

Example no:2.20,Page no:2.39

In [23]:
#Heat loss in furnace

#Variable declaration

#for clay
k1=0.533    #[W/(m.K)]
#for red brick
k2=0.7      #[W/m.K]


#Calculation

#Case 1
A=1     #Area
x1=0.125    #[m]
x2=0.5      #[m]
#Resistances
r1=x1/(k1*A) #Res of fire clay [K/W]
r2=x2/(k2*A)  #Res of red brick[K/W]
r=r1+r2
#Temperatures
T1=1373    #[K]
T2=323     #[K]
Q=(T1-T2)/r #[W/sq m]
Tdash=T1-Q*r1  #[K]

#Case2
# Heat loss must remain  unchanged,Thickness of red brick also reduces to its half
x3=x2/2   #[m]
r3=x3/(k2*A)  #[K/W]
Tdd= T2+(Q*r3)    #[K]
#Thickness of diatomite be x2,km be mean conductivity
Tm=(Tdash+Tdd)/2   #[K]
km=0.113+(0.00016*Tm)  #[W/(m.K]
x2=km*A*(Tdash-Tdd)/Q  #[m]
x2=x2*1000      #[mm]

#Result
print"Thickness of diatomite layer=",round(x2),"mm"
  

Thickness of diatomite layer= 93.0 mm

Example no:2.21,Page no:2.41

In [24]:
#Rate of heat loss in pipe

#Variable declaration
k1=0.7   #common brick   W/((m.K)
k2=0.48  #gypsum layer [W/(m.K)
k3=0.065 #Rockwool    [W/m.K]
#Heat loss with insulatiob will be 20% of without insulation
A=1      #sq m
x1=0.1   #[m]
x2=0.04 #[m]

#Calculation
R1=x1/(k1*A)   #K/W
R2=x2/(k2*A)   #K/W
R=R1+R2        #K/W
#R3=x3/(k3*A)
QbyQd=0.2
sigRbyRd=QbyQd
x3=(R/QbyQd-R)/15.4 #m
x3=x3*1000          #[mm]

#Result
print"Thickness of rockwool insulation",round(x3),"mm"

Thickness of rockwool insulation 59.0 mm

Example no:2.22,Page no:2.44

In [25]:
#Heat loss from insulated steel pipe

#Variable declaration
Ts=451.0       #Steam temperature in [K]
Ta=294.0       #Air temperature in [K]
Di=25.0    #Internal diameter of pipe  [mm]
Di=Di/1000               #[m]
od=33.0    #Outer diameter of pipe   [mm] 
od=od/1000               #[m]
hi=5678.0  #Inside heat transfer coefficient [W/(m**2.K)]
ho=11.36  #Outsideheat transfer coefficient  [W/(sq m.K)]

#Calculation
xw=(od-Di)/2     #Thickness of steel pipe  [m]
k2=44.97         #k for steel in W/(m.K)
k3=0.175         #k for rockwool in W/(m.K)
ti=38.0/1000            #thickness of insulation in [m]
r1=Di/2          #[m]
r2=od/2          #[m]
rm1=(r2-r1)/math.log(r2/r1)       #[m]
r3=r2+ti             #[m]
rm2=(r3-r2)/math.log(r3/r2)   #[m]
Dm1=2*rm1                #[m]
Dm2=2*rm2                #[m]
import math
#Rate of heat loss = dT/(sigma_R)
L=1              #[m]
R1=1/(hi*math.pi*Di*L)   #[K/W]
R1=round(R1,4)
R2=xw/(k2*math.pi*Dm1*L)
R2=round(R2,6)
R3=(r3-r2)/(k3*math.pi*Dm2*L)
R3=round(R3,3)
R3=1.086
Do=(od+2*ti)            #[mm]
R4=1/(ho*math.pi*Do*L)               #[m]
R4=round(R4,3)
sigma_R=R1+R2+R3+R4 
#Heat loss
dT=Ts-Ta                 #[K]
Q=dT/sigma_R             #Heat loss [W/m]

#Result
print"Rate of heat loss is",round(Q,2),"W/m"
print"\nNOTE:Slight variation in final answer due to mistake in calculation of sigma_R in textbook.\nIn book is is taken as 1.366\n "

Rate of heat loss is 116.63 W/m

NOTE:Slight variation in final answer due to mistake in calculation of sigma_R in textbook.
In book is is taken as 1.366

Example no:2.23,Page no:2.46

In [26]:
#Heat loss from furnace

#Variable declaration
T1=913.0     #[K]
T=513.0     #[K]
T2=313.0      #[K]

#Calculation

#Q=(T1-T)/(x/(k*A))
#Q=(T-T2)/(1/(h*A))
#x=2k/h
#Q=(T1-T2)/(x/(kA)+1/(h*A))
#Therefore,Q=hA/3*(T1-T2)
#With increase in thickness(100%)
#x1=4*k/h
#Q2=(T1-T2)/(x1/k*A+1/(h*A))
#Q2=(h*A)/5)*(T1-T2)
#Now
h=1.0     #Assume
A=1.0     #Assume for calculation
Q1=(h*A/3)*(T1-T2)
Q2=((h*A)/5)*(T1-T2)
percent=(Q1-Q2)*100/Q1      #Percent reduction in heat loss

#Result
print"Percentage reduction in heat loss is",percent,"%"

Percentage reduction in heat loss is 40.0 %

Example no:2.24,Page no:2.47

In [27]:
#Rate of heat loss

#Variable declaration
L=1.0#m
thp=2.0#Thickness of pipe  in mm
thi=10.0#Thickness of insulation  in mm
T1=373.0#K
T2=298.0#K
id=30.0#mm
r1=id/2#mm

#Calculation
r2=r1+thp#mm
r3=r2+thi#mm
#In S.I units
r1=r1/1000 #m
r2=r2/1000#m
r3=r3/1000#m
k1=17.44#W/(m.K)
k2=0.58#W/(m.K)
hi=11.63#W/(sq m.K)
ho=11.63#W/(sq m.K)
import math
Q=(2*math.pi*L*(T1-T2))/(1/(r1*hi)+(math.log(r2/r1))/k1+((math.log(r3/r2))/k2)+(1/(0.02*ho)))

#Result
print"Rate of heat loss,Q=",round(Q,1),"W"

Rate of heat loss,Q= 43.5 W

Example no:2.25,Page no:2.48

In [41]:
 #Thickness of insulation .
from scipy.optimize import fsolve
import math

#Variable declaration
h=8.5        #[W/sq m.K]
dT=175       #[K]
r2=0.0167            #[m]

#Calculation
Q_by_l=h*2*math.pi*r2*dT        #[W/m]
k=0.07           #For insulating material in  [W/m.K]
#for insulated pipe--50% reduction in heat loss
Q_by_l1=0.5*Q_by_l      #[w/m]
def f(r3):
    x=Q_by_l1-dT/((math.log(r3/r2))/(2*math.pi*k)+1/(2*math.pi*r3*h))
    return(x)
#by trial and error method we get:
r3=fsolve(f,0.05)
t=r3-r2         #thickness of insulation in [m]

#Result
print"Required thickness of insulation is",round(t[0],4),"m=",round(t[0]*1000,1)," mm or",round(t[0]*1000),"m"

Required thickness of insulation is 0.0188 m= 18.8  mm or 19.0 m

Example no:2.26,Page no:2.49

In [29]:
#Calculate heat loss per metre length
import math

#Variable declaration
id=0.1    #internal diameter in[m]
od=0.12    #outer diameter in [m]
T1=358    #Temperature of fluid   [K]
T2=298    #Temperature of surrounding   [K]
t=0.03    #thickness of insulation     [m]
k1=58    #[W/m.K]
k2=0.2    #W/(m.K) insulating material
h1=720    #inside heat transfer coeff [W/sq m .K]
h2=9      #W/sq m.K
r1=id/2    #[m]
r2=od/2    #[m]
r3=r2+t    #[m]

#Calculation
#Heat loss per meter=Q_by_L
Q_by_L=(T1-T2)/(1/(2*math.pi*r1*h1)+math.log(r2/r1)/(2*math.pi*k1)+math.log(r3/r2)/(2*math.pi*k2)+1/(2*math.pi*r3*h2))  #W/m

#Result
print"Heat loss per metre length of pipe=",round(Q_by_L,2),"W"

Heat loss per metre length of pipe= 114.49 W

Example no:2.27,Page no:2.50

In [30]:
#Mineral wool insulation
from scipy.optimize import fsolve

#Variable declaration
import math
T1=573           #[K]
T2=323          #[K]
T3=298          #[K]
h1=29            # Outside heat transfer coefficients [W/sq m.K]
h2=12           #[W/sq m.K]
r1=0.047             #Internal radius [m]
r2=0.05              #Outer radius[m]
k1=58                #[W/m.K]
k2=0.052             #[W/m.K]

#Calculation
#Q=(T1-T2)/(1/(r1*h1)+math.log(r2/r1)/k1+math.log(r3/r2)/k2)=(T2-T3)/(1/(r3*h2))
def f(r3):
    x=(T1-T2)/(1/(r1*h1)+math.log(r2/r1)/k1+math.log(r3/r2)/k2)-(T2-T3)/(1/(r3*h2))
    return(x)
	#by trial and error method :
r3=fsolve(f,0.05)
t=r3-r2             #Thickness of insulation in [m]
#Q=h2*2*math.pi*r3*L*(T2-T3)
Q_by_l=h2*2*math.pi*r3*(T2-T3)      #[W/m]

#Result
print"Thickness of insulation is",round(t*1000),"mm"
print"Rate of heat loss per unit length is",round(Q_by_l[0],1),"W/m"

Thickness of insulation is 32.0 mm
Rate of heat loss per unit length is 154.1 W/m

Example no:2.28,Page no:2.51

In [31]:
#Calculate heat loss per sq m and temperature of outside surface

#Variable declaration
A=1 #assume [sq m]
x1=0.006    #[m]
x2=0.075    #[m]
x3=0.2    #[m]
k1=39.0    #[W/m.K]
k2=1.1    #[W/m.K]
k3=0.66    #[W/m.K]
h0=65.0    #W/sq m .K
T1=900.0    #K
T2=300.0   #K

#Calculation
sigma_R=(x1/(k1*A)+x2/(k2*A)+x3/(k3*A)+1/(h0*A)) 

#To calculate heat loss/sq m area
Q=(T1-T2)/sigma_R    #[W/sq m]
#Q/A=T-T2/(1/h0), where T=Temp of outside surface
#So, T=T2+Q/(A*h0)
T=Q/(A*h0)+T2    #[K]

#Result
print"Heat loss per sq metre area is:",round(Q,1),"W/sq m" 
print"Temperature of outside surface of furnace is:",round(T,1),"K (",round(T-273,1),"degree C)"

Heat loss per sq metre area is: 1551.4 W/sq m
Temperature of outside surface of furnace is: 323.9 K ( 50.9 degree C)

Example no:2.29,Page no:2.52

In [32]:
#Determine necessary thickness of insulation brick

#Variable declaration
A=1.0    #Assume [sq m]
x1=0.003    #[m]
x3=0.008    #[m]
k1=30.0    #[W/m.K]
k2=0.7    #[W/m.K]
k3=40.0    #[W/m.K]
T1=363.0    #[K]
T=333.0    #[K]
T2=300.0    #[K]
h0=10.0    #W/sq m.K

#Calculation
#Q=(T1-T2)/(x1/(k1*A)+x2/(k2*A)+x3/(k3*A)+1/(h0*A))
#Also,Q=(T-T2)/(1/(h0*A))
#So, (T1-T2)/((x1/(k1*A)+x2/(k2*A)+x3/(k3*A)+1/(h0*A))=(T-T2)/(1/(h0*A))
#or,x2=k2*A((T1-T2)/((T-T2)*h0*A)-1/(h0*A)-x1/(k1*A)-x3/(k3*A))
x2=k2*A*((T1-T2)/((T-T2)*h0*A)-1/(h0*A)-x1/(k1*A)-x3/(k3*A))  #[m]

#Result
print"Thickness of insulating brick required is",round(x2*1000,1),"mm" 

Thickness of insulating brick required is 63.4 mm

Example no:2.30,Page no:2.53

In [51]:
#Heat flow through furnace wall

#Variable declaration
hi=75.0        #[W/sq m.K)
x1=0.2    #m
x2=0.1    #[m]
x3=0.1    #[m]
T1=1943    #[K]
k1=1.25    #W/m.K
k2=0.074    #/W/m.K
k3=0.555    #W/m.K
T2=343.0    #K
A=1.0    #assume [sq m]

#Calculation
sigma_R=1.0/(hi*A)+x1/(k1*A)+x2/(k2*A)+x3/(k3*A) 
#Heat loss per i sq m
Q=(T1-T2)/sigma_R    #[W]
#if T=temperature between chrome brick and koalin brick then 
#Q=(T1-T)/(1/(hi*A)+x1/(k1*A))
#or T=T1-(Q*(1/(hi*A)+x1/(k1*A)))
T=T1-(Q*(1.0/(hi*A)+x1/(k1*A)))     #[K]
#if Tdash=temperature at the outer surface of middle layer,then
#Q=(Tdash-T2)/(x3/(k1*A))
#or Tdash=T2+(Q*x3/(k3*A))
Tdash=T2+(Q*x3/(k3*A))    #[K]

#Result
print"Heat loss per sq m is:",round(Q,1),"W"
print"Temperature at inner surface of middle layer=",round(T,1),"K(",round(T-273,1),"degree C)"
print"Temperature at outer surface of middle layer=",round(Tdash,1),"K (",round(Tdash-273,1),"degree C)" 

Heat loss per sq m is: 938.5 W
Temperature at inner surface of middle layer= 1780.3 K( 1507.3 degree C)
Temperature at outer surface of middle layer= 512.1 K ( 239.1 degree C)

Example no:2.31,Page no:2.54

In [54]:
#Calculate:(a) Heat loss per unit length  
#(b)Reduction in heat loss
import math

#Variable declaration
hi=10    #W/sq m.K
h0=hi    #W/sq.m.K
r1=0.09    #m
r2=0.12    #m
t=0.05    #thickness of insulation [m]
k1=40    #W/m.K
k2=0.05    #W/m.K
T1=473    #K
T2=373    #K

#Calculation
Q_by_L=2*math.pi*(T1-T2)/(1/(r1*hi)+math.log(r2/r1)/k1+1/(r2*h0))     #W/m
#After addition of insulation:
r3=r2+t     #radius of outer surface of insulaiton
Q_by_L1=2*math.pi*(T1-T2)/(1/(r1*hi)+math.log(r2/r2)/k1+math.log(r3/r2)/k2+1/(r3*h0))     # W
Red=Q_by_L-Q_by_L1    #Reduciton in heat loss in [W/m]
percent_red=(Red/Q_by_L)*100    #% Reduction in heat loss

#Result
print"Ans (a) Heat loss=",round(Q_by_L,2),"W/m "
print"Ans (b) Percent reduction in heat loss is",round(percent_red,1),"%"

Ans (a) Heat loss= 321.94 W/m 
Ans (b) Percent reduction in heat loss is 77.5 %

Example no:2.32,Page no:2.55

In [33]:
#Determine: i-Heat flux across the layers and
#ii-Interfacial temperature between the layers

#Variable declaration
T1=798.0    #K
T2=298.0    #K
x1=0.02    #m
x2=x1    #m
k1=60.0    #W/m.K
k2=0.1    #W/m.K
hi=100.0    #W/sq m.K
h0=25.0    #W/sq m.K

#Calculation
Q_by_A=(T1-T2)/(1.0/hi+x1/k1+x2/k2+1.0/h0)     #W/sq m
#If Tis the interfacial temperature between steel plate and insulating material
#Q_by_A=(T-T2)/(x2/k2+1/h0)
T=Q_by_A*(x2/k2+1.0/h0)+T2

#Result
print"Ans (i)- Heat flux across the layers is",round(Q_by_A),"W/sq m" 
print"Ans-(ii)-Interfacial temperature between layers is",round(T,1),"K(",round(T-273,1),"degree C)"

Ans (i)- Heat flux across the layers is 1997.0 W/sq m
Ans-(ii)-Interfacial temperature between layers is 777.4 K( 504.4 degree C)

Example no:2.33,Page no:2.56

In [60]:
#Determine Temperature at the outer surface of wall and convective conductance on the outer wall
    
    
#Variable declaration    
T1=2273                 #Temperature of hot gas:[K]
T4=318                  #Ambient aur temperature[K]
Qr1_by_A=23260          #Heat flow by radiation from gases to inside surface of wall[W/sq m]
hi=11.63                #Heat transfer coefficient on inside wall:[W/sq m.K]
K=58                    #Thermal conductivity of wall[W.sq m/K]
Qr4_by_A=9300           #Heat flow by radiation from external surface to ambient[W/sq m]
T2=1273                 #Inside Wall temperature[K]
Qr1=Qr1_by_A    #W  for
A=1    #sq m


#Calculation
Qc1_by_A=hi*(T1-T2)    #W/sq m
Qc1=Qc1_by_A    #for A=1 sq m
    #Thermal resistance:
R=1.0/K    #K/W per sq m
#Now Q=(T2-T3)/R,i.e 
#External wall temp T3=T2-Q*R
#Q entering wall=
Q_enter=Qr1+Qc1    #W
T3=T2-Q_enter*R    #K
T3=673      #Approximate
#Heat loss due to convection:
Qc4_by_A=Q_enter-Qr4_by_A    #W/sq m
#Qc4_by_A=h0*(T3-T4)
#or  h0=Qc4_by_A/(T3-T4)
h0=Qc4_by_A/(T3-T4)    #W/sq m.K


#Result
print"Convective conductance is:",round(h0,1),"W/m^2.K"

Convective conductance is: 72.1 W/m^2.K

Example no:2.34,Page no:2.60

In [34]:
#Critical radius of insulation

#Variable declaration
T1=473    #[K]
T2=293    #[K]
k=0.17    #W/(m.K)
h=3    #W/(sq m.K)
h0=h    #W/sq m.K
rc=k/h    #m
r1=0.025    #Inside radius of insulaiton [mm]    

#Calculation
import math
q_by_l1=2*math.pi*(T1-T2)/(math.log(rc/r1)/k+1/(rc*h0))    #Heat transfer with insulation in W/m
#Without insulation:
q_by_l2=h*2*math.pi*r1*(T1-T2)    #W/m
inc=(q_by_l1-q_by_l2)*100/q_by_l2    #Increase of heat transfer
k=0.04    #Fibre glass insulaiton W/(sq m.K)
rc=k/h    #Critical radius of insulaiton

#Result
print"When covered with insulation:\nHeat loss=",round(q_by_l1,1),"W"
print"When without insulation,\nHeat loss=",round(q_by_l2,1),"W"
print"\nPercent increase =",round(inc,2),"percent" 
print"In this case the value of rc=",round(rc,4),"m is less than the outside radius of pipe (",r1,"m)"
print"So additon of any fibre glass would cause a decrease in the heat transfer"

When covered with insulation:
Heat loss= 105.7 W
When without insulation,
Heat loss= 84.8 W

Percent increase = 24.66 percent
In this case the value of rc= 0.0133 m is less than the outside radius of pipe ( 0.025 m)
So additon of any fibre glass would cause a decrease in the heat transfer

Example no:2.36,Page no:2.61

In [68]:
#Calculate the heat loss per metre of pipe and outer surface temperature

#Variable declaration
import math
k=1.0    #Thermal conductivity in [W/sq m.K]
h=8.0    #Het transfer coeff in W/sq m.K
rc=k/h  #Critical radius in m
T1=473.0    #K
T2=293.0    #K
r1=0.055    #Outer radius =inner radius in [m]

#Calculation
Q_by_L=2*math.pi*(T1-T2)/(math.log(rc/r1)/k+1.0/(rc*h))
#For outer surface
#Q_by_L=2*math.pi*(T-T2)/(1/rc*h)
# implies that, T=T2+Q_by_L/(rc*2*math.pi)
T=T2+Q_by_L/(rc*2*math.pi*h)    #K

#Result
print"Heat loss per meter of pipe is",round(Q_by_L),"W/m(approx)"
print"Outer surface temperature is: ",round(T,1),"K(",round(T-273,1),"degree C)"

Heat loss per meter of pipe is 621.0 W/m(approx)
Outer surface temperature is:  391.8 K( 118.8 degree C)

Example no:2.37,Page no:2.78

In [35]:
#Calculate the time required for a ball to attain a temperature of 423 K

#Variable declaration
k_steel=35.0    #W/m.K
Cp_steel=0.46    #kJ/(kg*K)
Cp_steel=Cp_steel*1000    #J/(kg*K)
h=10    #W/sq m.K
rho_steel=7800.0    #kg/cubic m
dia=50.0    #mm
dia=dia/1000    #m
R=dia/2      #radius in m

#Calculation
import math
A=4*math.pi*R**2  #Area  in sq m
V=A*R/3      #Volume in cubic meter
Nbi=h*(V/A)/k_steel
#As Nbi<0.10,internal temp gradient is negligible
T=423.0    #K
T0=723.0    #K
T_inf=373.0    #K
#(T-T_inf)/(T0-T_inf)=e**(-h*At/rho*Cp*V)
t=-rho_steel*Cp_steel*R*math.log((T-T_inf)/(T0-T_inf))/(3*h)     #s

#Result
print"Time required for a ball to attain a temperature of 423 K is",round(t),"s=",round(t/3600,2),"h"

Time required for a ball to attain a temperature of 423 K is 5818.0 s= 1.62 h

Example no:2.38,Page no:2.78

In [72]:
#Time take in Steel ball quenched

#Variable declaration
dia=50.0    #mm
dia=dia/1000    #m
r=dia/2    #radius in m
h=115.0    #W/sq m.K
rho=8000.0 #kg/cubic m
Cp=0.42    #kJ/kg.K
Cp=Cp*1000    #J/(kg*K)

#Calculation
import math
A=4*math.pi*r**2    #Area in sq m
V=A*r/3    #Volume in cubic m
T=423.0    #K
T_inf=363.0    #K
T0=723    #K
#(T-T_inf)/(T0-T_inf)=e**(-3ht/(rho*Cp*r))
t=-rho*Cp*r*math.log((T-T_inf)/(T0-T_inf))/(3*h)     #Time in seconds

#Result
print"Time taken by centre of ball to reach a temperature of 423 K is",round(t,2),"s (=",round(t/60,2),"minutes"

Time taken by centre of ball to reach a temperature of 423 K is 436.25 s (= 7.27 minutes

Example no:2.39,Page no:2.79

In [74]:
#HT in a Ball plunged in a medium

#Variable declaration
h=11.36    #W/sq m.K
k=43.3    #w/(m.K)
r=25.4    #radius in mm
r=r/1000    # radius in m

#Calculation
import math
A=4*math.pi*r**2    #Area of sphere [sq m]
V=A*r/3    #Volume in [cubic m]
rho=7849.0    #kg/cubic m
Cp=0.4606*10**3    #J/kg.K
t=1.0    #hour
t=t*3600    #seconds
T_inf=394.3    #[K]
T0=700.0    #[K]
# (T-T_inf)/(T0-T_inf)=e**(-3*h*t/rho*Cp*V)
T=T_inf+(T0-T_inf)*(math.exp((-h*A*t)/(rho*Cp*V))) 

#Result
print"Temperature of ball after 1 h=",round(T),"K (",round(T-273),"degree C)(APPROXIMATE)"

Temperature of ball after 1 h= 475.0 K ( 202.0 degree C)(APPROXIMATE)

Example no:2.40,Page no:2.80

In [36]:
#Slab temperature suddenly lowered

#Variable declaration
import math
rho=9000.0  #kg/cubic m
Cp=0.38   #kJ/(kg.K)
Cp=Cp*1000    #J/(kg.K)
k=370.0     #W/m.K
h=90.0      #W/sq m.K
l=400.0     #mm
l=l/1000     #length of copper slab
t=5.0/1000     #thickness in [m]

#Calculation
A=2*l**2     #Area of slab
V=t*l**2       #Volume in [cubic m]
L_dash=V/A    #[m]
#for slab of thickness 2x
#L_dash=x
L_dash=0.025     #[m]
Nbi=h*L_dash/k    #< 0.10
var=h*A/(rho*Cp*V)
#As Nbi<0.10,we can apply lumped capacity analysis
T=363.0    #[K]
T_inf=303.0    #[K]
T0=523.0    #[K]
t=-(math.log((T-T_inf)/(T0-T_inf)))/var

#Result
print"Time at which slab temperature becomes 363 K is",round(t,1),"s"
print"CALCULATION MISTAKE IN BOOK IN LAST LINE"

Time at which slab temperature becomes 363 K is 123.4 s
CALCULATION MISTAKE IN BOOK IN LAST LINE

Example no:2.41,Page no:2.80

In [77]:
#Flow over a flat plate
import math
#Variable declaration
rho=9000.0    #kg/cubic meter
Cp=0.38    #kJ/(kg.K)
Cp=Cp*1000    #J/kg.K
k=370.0    #W/(m.K)
T0=483.0    #K
T_inf=373.0    #K
delta_T=40.0    #K

#Calculation
T=T0-delta_T    #K
t=5.0    #time in [minutes]
t=t*60    #[seconds]
#A=2A.....Two faces
#V=A.2x
#2x=thickness of slab=30    mm=0.03    m
x=0.015    #[m]
th=2*x    #thickness of slab
h=-rho*Cp*x*math.log((T-T_inf)/(T0-T_inf))/t

#Result
print"Heat transfer coefficient is:",round(h,1),"W/(m^2.K)"

Heat transfer coefficient is: 77.3 W/(m^2.K)

Example no:2.42,Page no:2.81

In [79]:
#TIme required in Stainless steel rod immersed in water

#Variable declaration
rho=7800.0    #[kg per cubic m]
h=100.0    #W/(sq m.K)  Convective heat transfer coeff
Cp=460.0    #J/(kg.K)
k=40.0    #W/(m.K)
L=1.0    #[m] length ofrod
D=10.0    #mm    
D=D/1000    #diameter in[m]
R=D/2    #raidus in [m]

#Calculation
import math
#For cylindrical rod:
A=2*math.pi*R*L    #Area in [sq m]
V=math.pi*R**2*L    #Volume in [cubic m]
L_dash=V/A    #[m]
Nbi=h*L_dash/k    #Biot number
#N_bi<0.10,Hence lumped heat capavity is possible
T=473.0    #[K]    
T_inf=393.0    #[K]
T0=593.0    #[K]
t=-rho*Cp*V*math.log((T-T_inf)/(T0-T_inf))/(h*A)

#Result
print"Time required to reach temperature",T,"K is",round(t,1),"s"

Time required to reach temperature 473.0 K is 82.2 s

Example no:2.43,Page no:2.82

In [84]:
# Time constant of Chromel alumel thermocouple  

#Variable declaration
import math
rho=8600.0    #[kg/cubic m]
Cp=0.42    #kJ/(kg.K)
Cp=Cp*1000    #J/(kg.K)
dia=0.71    #[mm]
dia=dia/1000    #[dia in m]
R=dia/2    #radius [m]
h=600.0    #convective coeff W/(sq m.K)

#Calculation
#Let length =L=1
L=1.0         #[m]
A=2*math.pi*R*L 
V=math.pi*(R**2)*L 
tao=(rho*Cp*V)/(h*A) 
#at1
t=tao
#From  (T-T_inf)/(T0-T_inf)=e**(-t/tao)
ratio=math.exp(-t/tao)    #Ratio of thermocouple difference to initial temperature difference
print"Time constant of the thermocouple is",round(tao,3),"s" 
print"At the end of the time period t=tao=",round(tao,3),"s"
print"Temperature difference b/n the thermocouple and the gas stream would be",round(ratio,3),"of the initial temperature difference"
print"It should be reordered after",round(4*tao),"s"

Time constant of the thermocouple is 1.069 s
At the end of the time period t=tao= 1.069 s
Temperature difference b/n the thermocouple and the gas stream would be 0.368 of the initial temperature difference
It should be reordered after 4.0 s

Example no:2.44,Page no:2.83

In [37]:
#Thermocouple junction
import math

#Variable declaration
rho=8000.0    #kg/cubic m
Cp=420.0    #J/(kg.K)
h_hot=60.0    # for hot stream W/(sq m.K)    
dia=4.0    #[mm]
t=10.0 

#Calculation
r=dia/(2.0*1000)    #radius in [m]
#For sphere
V=(4.0/3.0)*math.pi*r**3    #Volume in [cubic m]
A=4*math.pi*r**2        #Volume in [sq m]
tao=rho*Cp*V/(h_hot*A)    # Time constant in [s]
ratio=math.exp(-t/tao)    # %e**(-t/tao)=(T-T-inf)/(T0-T_inf)
T_inf=573.0    #[K]
T0=313.0    #[K]
T=T_inf+ratio*(T0-T_inf)
#IN STILL AIR:
h_air=10.0    #W/(sq m .K)
tao_air=rho*Cp*V/(h_air*A)    #[s]
t_air=20.0    #[s]
ratio_air=math.exp(-t_air/tao_air)
T_inf_air=303.0    #[K]
T0_air=T 
T_air=T_inf_air+ratio_air*(T0_air-T_inf_air)

#Result
#ANS-[i]
print"Time constant of thermocouple is",round(tao,2),"s"
#ANS-[ii]
print"Temperature attained by junction 20 s after removing from the hot air stream is:",round(T_air),"K"

Time constant of thermocouple is 37.33 s
Temperature attained by junction 20 s after removing from the hot air stream is: 368.0 K

Example no:2.45,Page no:2.84

In [42]:
#Batch reactor time taken 
from scipy.optimize import fsolve
from scipy import integrate
import math

#Variable declaration
T_inf=390.0                #[K]
U=600.0                    #[W/sq m.K]
Ac=1.0                      #[sq m]
Av=10.0                       #Vessel area in [sq m]
m=1000.0                   #[kg]
Cp=3.8*10**3             #[J/kg.K]
To=290.0                   #[K]
T=360.0                    #[K]
h=8.5                   #[W/sq m.K]

#Calculation
#Heat gained from the steam=Rate of increase of internal energy
#U*A*(T_inf-T)=m*Cp*dT
def f(t):
    x=math.log((T_inf-To)/(T_inf-T))-U*Ac*t/(m*Cp) 
    return(x)
t=fsolve(f,1)           #[in s]
t=round(t)    #[in s]
Ts=290 
print"Time taken to heat the reactants over the same temperature range is",t,"s"
def g(T):
    t1=m*Cp/(U*Ac*(T_inf-T)-h*Av*(T-Ts))
    return(t1)
t1= integrate.quad(g,To,T)
def fx(Tmax):
    m=U*Ac*(T_inf-Tmax)-h*Av*(Tmax-Ts)
    return(m)
T_max=fsolve(fx,1)

#Result
print"In CASE 1\nTime  taken to heat the reactants =",t,"s .ie",round(t/3600,2),"h" 
print"In CASE 2 \nTime  taken to heat the reactants =",round(t1[0]),"s"
print"Maximum temperature at which temperature can be raised is",round(T_max,1),"K" 

Time taken to heat the reactants over the same temperature range is 7625.0 s
In CASE 1
Time  taken to heat the reactants = 7625.0 s .ie 2.12 h
In CASE 2 
Time  taken to heat the reactants = 8905.0 s
Maximum temperature at which temperature can be raised is 377.6 K

Example no:2.46,Page no:2.95

In [39]:
#Heat dissipation by aluminium rod  
import math 

#Variable declaration
dia=3.0    #[mm]
dia=dia/1000    #[m]
r=dia/2    #radius in[m]
k=150    #W/(m.K)
h=300    #W/(sq m.K)
T0=413    #[K]
T_inf=288    #[K]

#Calculation
A=math.pi*(r**2)    #Area in [sq m]
P=math.pi*dia      #[W/sq m.K]
Q=(T0-T_inf)*math.sqrt(h*P*k*A)   #Heat dissipated in [W]

#Result
print"Heat dissipated by the rod is",round(Q,3),"W"

Heat dissipated by the rod is 6.844 W

Example no:2.47,Page no:2.96

In [40]:
#Aluminium fin efficiency
#Variable declaration
k=200.0    #W/(m.K)
h=15.0    #W/(sq m.K)
T0=523.0    #[K]
T_inf=288.0    #[K]
theta_0=T0-T_inf    
dia=25.0    #diameter[mm]
dia=dia/1000    #diameter[m]
r=dia/2    #radius in [m]

#Calculation
import math
P=math.pi*dia    #[m]
A=math.pi*r**2    #[sq m]
#For insulated fin:
m=math.sqrt(h*P/(k*A))
L=100.0    #length of rod in [mm]
L=L/1000    #length of rod in [m]
Q=theta_0*math.tanh(m*L)*math.sqrt(h*P*k*A)    #Heat loss 
nf=math.tanh(m*L)/(m*L)    #Fin efficiency for  insulated fin
#At the end of the fin: theta/theta_0=(cosh[m(L-x)]/cosh(mL))
#at x=L, theta/theta_0=1/(cosh(mL)
T=T_inf+(T0-T_inf)*(1/math.cosh(m*L))    #[K]


#Result
#ANSWER-1
print"Heat loss by the insulated rod is",round(Q,1),"W"
#ANSWER-2
print"Fin efficiency is",round(nf*100,1),"percent"
#ANSWER-3
print"Temperature at the end of the fin is",round(T,1),"K"

Heat loss by the insulated rod is 26.6 W
Fin efficiency is 96.2 percent
Temperature at the end of the fin is 509.6 K

Example no:2.49,Page no:2.98

In [101]:
#Finding effective Pin fins
import math
#Variable declaration
k=300.0    #W/(m.K)
h=20.0    #W.(sq m.K)
P=0.05    #[m]
A=2.0    #[sq cm]
A=A/10000    #[sq m]
T0=503.0    #[K]
T_inf=303.0    #[K]


#Calculation
theta_0=T0-T_inf    #[K]
import math
m=math.sqrt(h*P/(k*A))

#CASE 1:  6 Fins of 100 mm length
L1=0.1    #Length of fin in [m]
Q=math.sqrt(h*P*k*A)*theta_0*math.tanh(m*L1)    #[W]
#For 6 fins
Q=Q*6    #for 6 fins [W]

#CASE 2:  10 fins of 60 mm length
L2=60.0    #[mm]
L2=L2/1000    #[m]
Q2=math.sqrt(h*P*k*A)*theta_0*math.tanh(m*L2)     #[W]
Q2=Q2*10    #For 10 fins



#Result
print"As,Q for 10 fins of 60 mm length(",round(Q2,2),"W) is more than Q for 6 fins of 100 mm length (",round(Q,2),"W)"
print"The agreement-->10 fins of 60 mm length is more effective"  

As,Q for 10 fins of 60 mm length( 117.66 W) is more than Q for 6 fins of 100 mm length ( 113.75 W)
The agreement-->10 fins of 60 mm length is more effective

Example no:2.50,Page no:2.98

In [103]:
#Metallic wall surrounded by oil and water   
#Variable declaration
h_oil=180.0    #W/(sq m.K)
h_air=15.0    #W/(sq m.K)
T_oil=353.0    #[K]
T_air=293.0    #[K]
delta_T=T_oil-T_air     #[K]
k=80.0    #Conductivity in  [W/(m.K)]
for_section=11.0*10**-3    #[m]
L=25.0    #[mm]
L=L/1000    #[m]
W=1.0    #[m] Width,..let
t=1.0    #[mm] 
t=t/1000    #[m]
A=W*t    #[m]
P=2*t
Af=2*L*W    #sq m
N=1.0

#Calculation

import math
Ab=for_section-A    #[sq m]
#CASE 1: Fin on oil side only
m=math.sqrt(h_oil*P/(k*A)) 
nf_oil=math.tanh(m*L)/(m*L)
Ae_oil=Ab+nf_oil*Af*N    #[sq m]
Q1=delta_T/(1/(h_oil*Ae_oil)+1/(h_air*for_section))    #[W]
#CASE 2: Fin on  air side only
m=math.sqrt(h_air*P/(k*A))
nf_air=math.tanh(m*L)/(m*L)
nf_air=0.928        #Approximation
Ae_air=Ab+nf_air*Af*N    #[sq m]
Q=delta_T/(1.0/(h_oil*for_section)+1.0/(h_air*Ae_air))    #[W]

#Result
print"In oil side,Q=",round(Q1,2),"W"
print"In air side,Q=",round(Q,2),"W"
print"From above results we see that more heat transfer takes place if fins are provided on the air side"

In oil side,Q= 9.75 W
In air side,Q= 35.56 W
From above results we see that more heat transfer takes place if fins are provided on the air side

Example no:2.51,Page no:2.100

In [5]:
#Brass wall having fins

#Variable declaration
k=75.0    #Thermal conductivity [W/(m.K)]
T_water=363.0    #[K]
T_air=303.0    #[K]    
dT=T_water-T_air    #delta T
h1=150.0    # for water[W/(sq m.K)]
h2=15.0       #for air [W/(sq m.K)]
W=0.5    #Width of wall[m]
L=0.025    #[m]

#Calculation
Area=W**2    #Base Area [sq m]
t=1.0    #[mm]
t=t/1000    #[m]
pitch=10.0    #[mm]
pitch=pitch/1000    #[m]
N=W/pitch    #[No of fins]


#Calculations
A=N*W*t    #Total cross-sectional area of fins   in [sq m]
Ab=Area-A    #[sq m]
Af=2*W*L    #Surface area of fins    [sq m]
from math import tanh
#CASE 1: HEAT TRANSFER WITHOUT FINS
A1=Area    #[sq m]
A2=A1    #[sq m]
Q=dT/(1.0/(h1*A1)+1.0/(h2*A2))         #[W]
print"Without fins,Q=",round(Q,2),"W"
#CASE 2: Fins on the water side
P=2*(t+W) 
A=0.5*10**-3 
m=math.sqrt(h1*P/(k*A))
nfw=math.tanh(m*L)/(m*L)   #Effeciency on water side
Aew=Ab+nfw*Af*N    #Effective area on the water side    [sq m]
Q=dT/(1.0/(h1*Aew)+1.0/(h2*A2))         #[W]
print"With fins on water side,Q=",round(Q,2),"W"
#CASE 3: FINS ON THE AIR SIDE
m=math.sqrt(h2*P/(k*A))
nf_air=tanh(m*L)/(m*L)    #Effeciency
Aea=Ab+nf_air*Af*N    #Effective area on air side
Q=dT/(1.0/(h1*A1)+1.0/(h2*Aea))         #[W]
print"With Fins on Air side,Q=",round(Q,1),"W"
#BOTH SIDE:
Q=dT/(1.0/(h1*Aew)+1.0/(h2*Aea))         #[W]

#Result
print"With Fins on both side,Q=",round(Q,1),"W" 

Without fins,Q= 204.55 W
With fins on water side,Q= 219.24 W
With Fins on Air side,Q= 800.3 W
With Fins on both side,Q= 1084.7 W