# Chapter 07:Principles of forced convection¶

## Ex7.1:pg-296¶

In :
import math

print"Introduction to heat transfer by S.K.Som, Chapter 7, Example 1"
#Engine oil at temprature,Tinf=60Â°C with a velocity of Uinf=1m/s flows over plate of length(L)=5m whose temprature(Tw)=30Â°C
Tw=30;
L=5;
Uinf=1;
Tinf=60;
#The properties at a film temprature of 45Â°C are as follows density(rho=870kg/m**3),Prandtl number(Pr=2850),conductivity(k=0.145W/(m*Â°C)),kinematic viscosity(nu=250*10**-6m**2/s).
rho=870;
Pr=2850;
k=0.145;
nu=250*10**-6;
print"First we check from reynolds no. that the flow is laminar or tubulent"
#Reynolds number is given by Re=(Uinf*L)/nu
print"Reynold number is"
Re=(Uinf*L)/nu
print"Re=",Re
print"which is less than critical reynolds number,So the flow is laminar."
#NuL is the average nusselt number
print"The average nusselt number over the entire length under the situation is given by NuL=0.664*Re**0.5*Pr**(1/3)"
NuL=0.664*Re**0.5*Pr**(1/3)
print"NuL=",NuL
#Heat flux is given by h=(k/L)*NuL
print"Heat flux in W/(m**2*K) is"
h=(k/L)*NuL
print"h=",h
#The rate of heat transfer per unit width is Q=h*A*(Tinf-Tw)
#Since unit width is considerd so B=1
#Area(A)=L*B
B=1;
A=L*B;
print"The rate of heat transfer per unit width in W is"
Q=h*A*(Tinf-Tw)
print"Q=",Q

Introduction to heat transfer by S.K.Som, Chapter 7, Example 1
First we check from reynolds no. that the flow is laminar or tubulent
Reynold number is
Re= 20000.0
which is less than critical reynolds number,So the flow is laminar.
The average nusselt number over the entire length under the situation is given by NuL=0.664*Re**0.5*Pr**(1/3)
NuL= 93.9037805416
Heat flux in W/(m**2*K) is
h= 2.72320963571
The rate of heat transfer per unit width in W is
Q= 408.481445356


## Ex7.2:pg-298¶

In :
import math

print"Introduction to heat transfer by S.K.Som, Chapter 7, Example 2"
#Atmospheric air at temprature,Tinf=300K and with a free stream Velocity Uinf=30m/s flows over a flat plate parallel to a side of length(L)=2m.
Tinf=300;
Uinf=30;
L=2;
#It is maintained at a uniform temprature of Tw=400K.
Tw=400;
#The properties of air at the film temprature of 350K are Prandtl number(Pr=0.705),conductivity(k=0.026W/(m*Â°C)),kinematic viscosity(nu=16.5*10**-6m**2/s)
Pr=0.705;
k=0.026;
nu=16.5*10**-6;
#We first find the location x(for reynolds number,Re=5*10**5) where the transition occurs
#Rex is reynolds number
print"The location x in m where the transition occurs"
Rex=5*10**5;
x=(nu*Rex)/Uinf
print"x=",x
#The average Nusselt number for the laminar zone is given by Nux=0.664*Re**0.5*Pr**(1/3)
print"The average Nusselt number for the laminar zone is"
Nux=0.664*Rex**0.5*Pr**(1/3)
print"Nux=",Nux
#Heat flux is given by h=(k/x)*Nux
print"Heat flux in W/(m**2*K) is"
h=(k/x)*Nux
print"h=",h
#Reynolds number is given by ReL=(Uinf*L)/nu
print"The reynolds number at L=2m is"
ReL=(Uinf*L)/nu
print"ReL=",ReL
#The average heat transfer coefficient over L=2m is determined from hbarL=(k/L)*(0.037*(ReL)**(4/5)-871)*Pr**(1/3)
print"The average heat transfer coefficient over L=2m in W/(m**2*K)"
hbarL=(k/L)*(0.037*(ReL)**(4/5)-871)*Pr**(1/3)
print"hbarL=",hbarL
#The rate of heat transfer per unit width is Q=h*A*(Tinf-Tw)
#Since unit width is considerd so B=1
#Area(A)=L*B
B=1;
A=L*B;
print"The rate of heat transfer per unit width in W is"
Q=hbarL*A*(Tw-Tinf)
print"Q=",Q

Introduction to heat transfer by S.K.Som, Chapter 7, Example 2
The location x in m where the transition occurs
x= 0.275
The average Nusselt number for the laminar zone is
Nux= 469.518902708
Heat flux in W/(m**2*K) is
h= 44.3908780742
The reynolds number at L=2m is
ReL= 3636363.63636
The average heat transfer coefficient over L=2m in W/(m**2*K)
hbarL= -11.322519
The rate of heat transfer per unit width in W is
Q= -2264.5038


## Ex7.3:pg-314¶

In :
import math

print"Introduction to heat transfer by S.K.Som, Chapter 7, Example 3"
#Air at a pressure of 101kPa and temprature,Tinf=20Â°C flows with a velocity(Uinf) of 5m/s over a flat plate whose temprature is kept constant at Tw=140Â°C.
Tw=140;
Tinf=20;
Uinf=5;
#The properties at the film temprature of 80Â°C are Prandtl number(Pr=0.706),Conductivity(k=0.03W/(m*Â°C)),kinematic viscosity(nu=2*10**-5m**2/s)
Pr=0.706;
k=0.03;
nu=2*10**-5;
#ReL is reynolds number and L is length of flat plate
print"(a)When the air flows parallel to the long side we have L=5 and the Reynolds no. becomes"
L=5;
ReL=(Uinf*L)/nu
print"ReL=",ReL
print"which is greater than critical Reynolds number."
#Thus we have combined laminar and tubulent flow.
# So The average heat transfer coefficient over L=5m is determined from hbarL=(k/L)*(0.037*(ReL)**(4/5)-871)*Pr**(1/3)
print"The average heat transfer coefficient over L=5m in W/(m**2*K)"
hbarL=(k/L)*(0.037*(ReL)**(4/5)-871)*Pr**(1/3)
print"hbarL=",hbarL
#The rate of heat transfer per unit width is Q=h*A*(Tinf-Tw)
#Since width is 1m so B=1
#Area(A)=L*B
B=1;
A=L*B;
#Q is the rate of heat transfer
print"The rate of heat transfer per unit width in W is"
Q=hbarL*A*(Tw-Tinf)
print"Q=",Q
#When the air flow is parallel to the 1m side we have L=1
print"(b)When the air flow is parallel to the 1m side we have L=1 an the Reynolds no. becomes "
L=1;
ReL=(Uinf*L)/nu
print"which is less than critical Reynolds number."
print"ReL=",ReL
#Thus we have laminar flow
#Heat flux is given by h=(k/L)*0.664*ReL**0.5*Pr**(1/3)
print"Heat flux in W/(m**2*K) is"
h=(k/L)*0.664*ReL**0.5*Pr**(1/3)
print"h=",h
#The rate of heat transfer per unit width is Q=h*A*(Tinf-Tw)
#Now width is 5m so B=5
#Area(A)=L*B
B=5;
A=L*B;
#Q is the rate of heat transfer
print"The rate of heat transfer per unit width in W is"
Q=h*A*(Tw-Tinf)
print"Q=",Q

Introduction to heat transfer by S.K.Som, Chapter 7, Example 3
(a)When the air flows parallel to the long side we have L=5 and the Reynolds no. becomes
ReL= 1250000.0
which is greater than critical Reynolds number.
The average heat transfer coefficient over L=5m in W/(m**2*K)
hbarL= -5.225778
The rate of heat transfer per unit width in W is
Q= -3135.4668
(b)When the air flow is parallel to the 1m side we have L=1 an the Reynolds no. becomes
which is less than critical Reynolds number.
ReL= 250000.0
Heat flux in W/(m**2*K) is
h= 9.96
The rate of heat transfer per unit width in W is
Q= 5976.0


## Ex7.4:pg-322¶

In :
import math

print"Introduction to heat transfer by S.K.Som, Chapter 7, Example 4"
#Castor oil at temprature,Tinf=36Â°C flows over a heated plate of length,L=6m and breadth,B=1m at velocity,Uinf=0.06m/s
Tinf=36;
L=6;
B=1;
Uinf=0.06;
#For a surface temprature at Tw=96Â°C
Tw=96;
#The properties at film temprature 66Â°C conductivity(k=0.21W/(m*K)),kinematic viscosity(nu=6*10**-5m**2/s),Thermal diffusivity(alpha=7.22*10**-8 m**2/s)
nu=6*10**-5;
k=0.21;
alpha=7.22*10**-8;
#ReL is reynolds number
print"(a)Reynolds number is"
ReL=(Uinf*L)/nu
print"ReL=",ReL
#Therefore the boundary layer is laminar over the entire plate.
#delta is the boundary layer thickness
print"The boundary layer thickness in m is"
delta=(5*L)/(ReL)**0.5
print"delta=",delta
#Pr is prandtl number.
print"Prandtl no. is"
Pr=nu/alpha
print"Pr=",Pr
#deltaT is thermal boundary layer thickness
print"The thermal boundary layer thickness in m is"
deltaT=delta/(Pr**(1/3))#NOTE:Answer in the book is incorrect(calculation mistake)
print"deltaT=",deltaT
#NuL is the nusselt number
print"(b)Since the prandtl number is high So Nusselt no. is"
NuL=0.339*(ReL)**0.5*Pr**(1/3)
print"NuL=",NuL
#Heat flux is given by hL=(k/L)*NuL
print"Heat flux in W/(m**2*K) is"
hL=(k/L)*NuL
print"hL=",hL
#hbarL is the average heat flux over length L
print"hbarL in W/(m**2*K) is"
hbarL=2*hL
print"hbarL=",hbarL
#The rate of heat transfer is Q=h*A*(Tinf-Tw)
#Area(A)=L*B
A=L*B;
print"(c)The rate of heat transfer in W is"
Q=hbarL*A*(Tw-Tinf)
print"Q=",Q

Introduction to heat transfer by S.K.Som, Chapter 7, Example 4
(a)Reynolds number is
ReL= 6000.0
The boundary layer thickness in m is
delta= 0.387298334621
Prandtl no. is
Pr= 831.024930748
The thermal boundary layer thickness in m is
deltaT= 0.387298334621
(b)Since the prandtl number is high So Nusselt no. is
NuL= 26.2588270873
Heat flux in W/(m**2*K) is
hL= 0.919058948055
hbarL in W/(m**2*K) is
hbarL= 1.83811789611
(c)The rate of heat transfer in W is
Q= 661.7224426


## Ex7.5:pg-322¶

In :
import math

print"Introduction to heat transfer by S.K.Som, Chapter 7, Example 5"
#A flat plate of width B=1m is maintained at a uniform surface temprtaure(Tw)=225Â°C
Tw=225;
B=1;
#Heating is done by rectangular modules of thickness t=10mm and length l=40mm.
t=10;
l=40;
#atmospheric air at temprature,Tinf=25Â°C flows over the plate at velocity(Uinf)=30m/s.
Tinf=25;
Uinf=30;
#The thermophysical properties of module are conductivity(km=5.2W/(m*K)),specific heat(cp=320J/(kg/K)),density(rho=2300kg/m**3).
km=5.2;
cp=320;
rho=2300;
#Assume the air properties at the film temprature of 125Â°C conductivity(ka=0.031W/(m*K)),kinematic viscosity(nu=22*10**-6m**2/s),Prandtl number(Pr=0.7)
ka=0.031;
nu=22*10**-6;
Pr=0.7;
#Module is placed at a distance of 800mm from the leading edge
#The distance from leading edge to the centre-line of the module,L=800+20=820mm.
L=0.0082;#in metre
#ReL is the reynolds number
print"Reynolds number is"
ReL=(Uinf*L)/nu
print"ReL=",ReL
print"Therefore the flow is turbulent over the module "
#The local heat transfer coefficient at L is calculated using hL=(k/L)*0.0296*(ReL)**(4/5)*(Pr)**(1/3)
print"The local heat transfer coefficient at L in W/(m**2*K)is"
hL=(ka/L)*0.0296*(ReL)**(4/0.5)*(Pr)**(1/0.3)
print"hL=",hL
#We consider that the local heat transfer coefficient at L=0.82m remains the same over the module which extends from L=0.80m to 0.84m
#If qm be the power generation in W/m**2 within the module ,we can write from energy balance qm*(t/0.1000)*(l/0.1000)*(B)=hbarL*(t/0.1000)*(B)*(Tw-Tinf)
print"The required power generation in W/m**3 is"
qm=(hL*(l/0.1000)*(B)*(Tw-Tinf))/((t/0.1000)*(l/0.1000)*(B))
print"qm=",qm

Introduction to heat transfer by S.K.Som, Chapter 7, Example 5
Reynolds number is
ReL= 11181.8181818
Therefore the flow is turbulent over the module
The local heat transfer coefficient at L in W/(m**2*K)is
hL= 8.32911955901e+30
The required power generation in W/m**3 is
qm= 1.6658239118e+31


## Ex7.6:pg-327¶

In :
import math

print"Introduction to heat transfer by S.K.Som, Chapter 7, Example 6"
#An aircraft is moving at a velocity of Uinf=150m/s in air at an altitude where the pressure is 0.7bar and the temprature is Tinf=-5Â°C.
Tinf=-5;
Uinf=150;
#The top surface of the wing absorbs solar radiation at a rate of Qr=900W/m**2.
Qr=900;
#Considering the wing as a flat plate of length(L)=2m and to be of solid construction with a single uniform surface temprature .
L=2;
#The properties of air at 268K and 0.7 bar are conductivity(k=0.024W/(m*K)),kinematic viscosity(nu=2*10**-5m**2/s),Prandtl number(Pr=0.72)
k=0.024;
nu=2*10**-5;
Pr=0.72;
#ReL is reynolds number
print"Reynolds number is"
ReL=Uinf*L/nu
print"ReL=",ReL
#Rec is critical reynolds number
print"Since ReL>Rec(=5*10**5) the flow is approximated as turbulent over the entire surface of the wing "
#Nusselt number is given by Nux=0.0308*ReL**(4/5)*Pr**(1/3)
Nux=0.0308*ReL**(4/5)*Pr**(1/3);
print"Nux=",Nux
#NubarL is average nusselt number over length L
print"Nusselt number is "
NubarL=(5/4)*Nux
print"NubarL=",NubarL
#Average heat transfer coefficient is given by hbarL=(k/L)*NubarL
print"Average heat transfer coefficient in W/(m**2*K) is"
hbarL=(k/L)*NubarL
print"hbarL=",hbarL
#From an energy balance the airfoil at steady state,Qr*As=2*hbarL*As*(Tw-Tinf) where Qr=radiation flux,As=upper or lower surface area.
#Therefore we can write Surface temprature of wing, Tw=Tinf+(Qr/(2*hbarL))
print"Surface temprature of wing in kelvin is"
Tw=(273+Tinf)+(Qr/(2*hbarL))
print"Tw=",Tw

Introduction to heat transfer by S.K.Som, Chapter 7, Example 6
Reynolds number is
ReL= 15000000.0
Since ReL>Rec(=5*10**5) the flow is approximated as turbulent over the entire surface of the wing
Nux= 0.0308
Nusselt number is
NubarL= 0.0308
Average heat transfer coefficient in W/(m**2*K) is
hbarL= 0.0003696
Surface temprature of wing in kelvin is
Tw= 1217800.46753


## Ex7.7:pg-331¶

In :
import math

print"Introduction to heat transfer by S.K.Som, Chapter 7, Example 7"
#A fine wire having a diameter(D)=0.04mm is placed in an air stream at temprature,Tinf=25Â°C having a flow velocity of Uinf=60m/s perpendicular to wire.
D=0.04;
Tinf=25;
Uinf=60;
#An electric current is passed through the wire ,raising its surface temprature to Tw=50Â°C
Tw=50;
#For air at the film temprature of 37.5Â°C,conductivity(k=0.027 W/(m*K)),kinematic viscosity(nu=17*10**-6m**2/s) and Prandtl number(Pr=0.71)
k=0.027;
nu=17*10**-6;
Pr=0.71;
#Re is reynolds number
print"Reynolds number is"
Re=Uinf*(D*10**-3)/nu
print"Re=",Re
#C and n are constants
#The values of C and n are found  for Re=141 are C=0.683 and n=0.466
#NuD is nusselt number
print"Nusselt number is"
NuD=(0.683)*Re**0.466*Pr**(1/3)
print"NuD=",NuD
#hbar is the average Heat transfer coefficient
print"The average Heat transfer coefficient in W/(m**2*K) is"
hbar=(k/(D*10**-3))*NuD
print"hbar=",hbar
#Heat transfer per unit length(qL) is given by pi*D*hbar*(Tw-Tinf)
print"Heat transfer per unit length in W/m is"
qL=math.pi*(D*10**-3)*hbar*(Tw-Tinf)
print"qL=",qL
#NuD is nusselt number
print"If we use eq NuD=0.3+((0.62*Re**0.5*Pr**(1/3))/(1+(0.4/Pr**(2/3))**(1/4))*(1+(Re/282000)**(5/8))**(4/5)"
NuD=0.3+((0.62*Re**0.5*Pr**(1/3))/(1+(0.4/Pr)**(2/3))**(1/4))*(1+(Re/282000)**(5/8))**(4/5)
print"NuD=",NuD
#hbar is the average Heat transfer coefficient
print"The average Heat transfer coefficient in W/(m**2*K) is"
hbar=(k/(D*10**-3))*NuD
print"hbar=",hbar
#Heat transfer per unit length(qL) is given by pi*D*hbar*(Tw-Tinf)
print"Heat transfer per unit length in W/m is"
qL=math.pi*(D*10**-3)*hbar*(Tw-Tinf)
print"qL=",qL

Introduction to heat transfer by S.K.Som, Chapter 7, Example 7
Reynolds number is
Re= 141.176470588
Nusselt number is
NuD= 6.85819682626
The average Heat transfer coefficient in W/(m**2*K) is
hbar= 4629.28285773
Heat transfer per unit length in W/m is
qL= 14.5433210172
If we use eq NuD=0.3+((0.62*Re**0.5*Pr**(1/3))/(1+(0.4/Pr**(2/3))**(1/4))*(1+(Re/282000)**(5/8))**(4/5)
NuD= 7.66669771975
The average Heat transfer coefficient in W/(m**2*K) is
hbar= 5175.02096083
Heat transfer per unit length in W/m is
qL= 16.2578078327


## Ex7.8:pg-334¶

In :
import math

print"Introduction to heat transfer by S.K.Som, Chapter 7, Example 8"
#Mercury and a light oil flowing at Uinf=4mm/s in a smooth tube having diameter(D)=25mm at a bulk temprature of 80Â°C.
Uinf=4*10**-3;#in metre
D=25*10**-3;#in metre
#The pertinent properties of the fluid at that temprature are kinematic viscosity of mercury(nuHg=1*10**-7m**2/s),kinematic viscosity of oil(nuoil=6.5*10**-6m**2/s)
#Prandtl number of mercury(PrHg=0.019),Prandtl number of oil(Proil=85).
nuHg=1*10**-7;
nuoil=6.5*10**-6;
PrHg=0.019;
Proil=85;
#ReHg is Reynolds number for mercury
print"Reynolds number for mercury is"
ReHg=Uinf*D/nuHg
print"ReHg=",ReHg
#Reoil is Reynolds number for oil
print"Reynolds number for oil is"
Reoil=Uinf*D/nuoil
print"Reoil=",Reoil
#The hydrodynamic length are given by L=0.05*Re*D
#LeHg is the hydrodynamic entry length for mercury
print"The hydrodynamic entry length for mercury in m is"
LeHg=0.05*ReHg*D
print"LeHg=",LeHg
#Leoil the hydrodynamic entry length for oil
print"The hydrodynamic entry length for oil in m is"
Leoil=0.05*Reoil*D
print"Leoil=",Leoil
#The thermal entry length are given by L=0.05*Re*Pr*D
#LtHg is the thermal entry length for mercury
print"The thermal entry length for mercury in m is "
LtHg=0.05*ReHg*PrHg*D
print"LtHg=",LtHg
#Ltoil is the thermal entry length for oil
print"The thermal entry length for oil in m is"
Ltoil=0.05*Reoil*Proil*D
print"Ltoil=",Ltoil

Introduction to heat transfer by S.K.Som, Chapter 7, Example 8
Reynolds number for mercury is
ReHg= 1000.0
Reynolds number for oil is
Reoil= 15.3846153846
The hydrodynamic entry length for mercury in m is
LeHg= 1.25
The hydrodynamic entry length for oil in m is
Leoil= 0.0192307692308
The thermal entry length for mercury in m is
LtHg= 0.02375
The thermal entry length for oil in m is
Ltoil= 1.63461538462


## Ex7.9:pg-336¶

In :
import math

print"Introduction to heat transfer by S.K.Som, Chapter 7, Example 9"
#Air at one atmospheric pressure and temprature(Tbi=75Â°C) enters a tube of internal diameter(D)=4.0mm with average velocity(U)=2m/s
Tbi=75;
D=4*10**-3;#in metre
U=2;
#The tube length is L=1.0m and a constant heat flux is imposed by the tube surface on the air over the entire length.
L=1;
#An exit bulk mean temprature(Tbo)=125Â°C is required.
Tbo=125;
#The properties of air 100Â°C are density(rho=0.95kg/m**3),Prandtl number(Pr=0.70),conductivity(k=0.03W/(m*K)),viscosity(mu=2.18*10**-5kg/(m*s)),specific heat(cp=1.01kJ/(kg/K))
rho=0.95;
Pr=0.70;
k=0.03;
mu=2.18*10**-5;
cp=1.01*10**3;
#Re is reynolds number
print"Reynold number is"
Re=rho*U*D/mu
print"Re=",Re
#Leh is the hydrodynamic entrance length
print"Therefore the flow is laminar.The hydrodynamic entrance length in m is"
Leh=0.05*Re*D
print"Leh=",Leh
#Let is the thermal entrance length
print"The thermal entrance length in m is"
Let=0.05*Re*Pr*D
print"Let=",Let
#The length of tube is given as 1m.A reasonable approach is to consider the flow to be fully developed for both velocity and tempratures over the entire profile lengths.
#For a fully developed flow with constant surface heat flux,Nusselt number is Nu=4.36
Nu=4.36;
#h is the heat transfer coefficient
print"The heat transfer coefficient in W/(m**2*K) is "
h=Nu*(k/D)
print"h=",h
#Here h=hL Since the heat transfer coefficient is constant over the entire length of tube.
#hL is the local heat transfer coefficient
hL=h;
#from an energy balance qw*pi*D*L=mdot*cp*(Tbo-Tbi)
#mdot is mass flow rate
print"The mass flow rate of air in kg/s is"
mdot=rho*(math.pi/4)*D**2*U
print"mdot=",mdot
#qw is the constant surface heat flux
print"Therefore the constant surface heat flux qw in W/m**2 is"
qw=(mdot*cp*(Tbo-Tbi))/(math.pi*D*L)
print"qw=",qw
#Let Twe be the surface temprature at the exit plane.Then we can write hL*(Twe-Tbo)=qw
print"The tube surface temprature at the exit plane in Â°C is "
Twe=Tbo+(qw/hL)
print"Twe=",Twe

Introduction to heat transfer by S.K.Som, Chapter 7, Example 9
Reynold number is
Re= 348.623853211
Therefore the flow is laminar.The hydrodynamic entrance length in m is
Leh= 0.0697247706422
The thermal entrance length in m is
Let= 0.0488073394495
The heat transfer coefficient in W/(m**2*K) is
h= 32.7
The mass flow rate of air in kg/s is
mdot= 2.38761041673e-05
Therefore the constant surface heat flux qw in W/m**2 is
qw= 95.95
The tube surface temprature at the exit plane in Â°C is
Twe= 127.934250765


## Ex7.10:pg-338¶

In :
import math

print"Introduction to heat transfer by S.K.Som, Chapter 7, Example 10"
#Air at one atmospheric pressure and temprature(Tbi=75Â°C) enters a tube of internal diameter(D)=4.0mm with average velocity(U)=2m/s
Tbi=75;
D=4*10**-3;
U=2;
#The heated tube length is L=0.04m and a constant heat flux is imposed by the tube surface on the air over the entire length.
L=0.04;
#An exit bulk mean temprature(Tbo)=125Â°C is required.
Tbo=125;
#The properties of air 100Â°C are density(rho=0.95kg/m**3),Prandtl number(Pr=0.70),conductivity(k=0.03W/(m*K)),viscosity(mu=2.18*10**-5kg/(m*s)),specific heat(cp=1.01kJ/(kg/K))
rho=0.95;
Pr=0.70;
k=0.03;
mu=2.18*10**-5;
cp=1.01*10**3;
#Re is the reynolds number
print"Reynold number is"
Re=rho*U*D/mu
print"Re=",Re
#Leh is the hydrodynamic entrance length
print"Therefore the flow is laminar.The hydrodynamic entrance length in m is"
Leh=0.05*Re*D
print"Leh=",Leh
#Let is thermal entrance length
print"The thermal entrance length in m is"
Let=0.05*Re*Pr*D
print"Let=",Let
print"The thermal entrance length is greater than the tube length Therefore the flow is hydrodynamically developed but not thermally developed"
#We calculate the inverse graetz number at x=L=0.04m
x=0.04;
#Gr_1 is inverse of graetz number
print"The inverse of graetz number Gr_1 is"
Gr_1=(x/D)*(1/(Re*Pr))
print"Gr_1=",Gr_1
#For constant surface heat flux nusselt number is Nu=4.7 and Graetz number is Gr=4.1*10**-2
Nu=4.7;
Gr=4.1*10**-2;
#hL is the local heat transfer coefficient
print"Therefore the local heat transfer coefficient in W/(m**2*K) is"
hL=Nu*(k/D)
print"hL=",hL
#from an energy balance qw*pi*D*L=mdot*cp*(Tbo-Tbi)
#mdot is the mass flow rate
print"The mass flow rate of air in kg/s is"
mdot=rho*(math.pi/4)*D**2*U
print"mdot=",mdot
#qw is the surface heat flux
print"Therefore surafce heat flux qw in W/m**2 is"
qw=(mdot*cp*(Tbo-Tbi))/(math.pi*D*L)
print"qw=",qw
#Let Twe be the surface temprature at the exit plane.Then we can write hL*(Twe-Tbo)=qw
print"The tube surface temprature at the exit plane in Â°C is "
Twe=Tbo+(qw/hL)
print"Twe=",Twe

Introduction to heat transfer by S.K.Som, Chapter 7, Example 10
Reynold number is
Re= 348.623853211
Therefore the flow is laminar.The hydrodynamic entrance length in m is
Leh= 0.0697247706422
The thermal entrance length in m is
Let= 0.0488073394495
The thermal entrance length is greater than the tube length Therefore the flow is hydrodynamically developed but not thermally developed
The inverse of graetz number Gr_1 is
Gr_1= 0.040977443609
Therefore the local heat transfer coefficient in W/(m**2*K) is
hL= 35.25
The mass flow rate of air in kg/s is
mdot= 2.38761041673e-05
Therefore surafce heat flux qw in W/m**2 is
qw= 2398.75
The tube surface temprature at the exit plane in Â°C is
Twe= 193.04964539


## Ex7.11:pg-339¶

In :
import math

print"Introduction to heat transfer by S.K.Som, Chapter 7, Example 11"
#Liquid sulphur di oxide in a saturated state flows inside a L=5m long tube and D=25mm internal diameter with a mass flow rate(mdot) of 0.15 kg/s.
#The tube is heated at a constant surface temprature(Tw) of -10Â°C and the inlet fluid temprature is Tbi=-40Â°C
Tw=-10;
Tbi=-40;
mdot=0.15;
D=0.025;#in metre
L=5;
#The properties to be used shoud be estimated at a temprature which is arithmetic mean of Tbi and Tbo.
#Since (outlet fluid temprature Tbo) is not known a priori,the solution has to be based on an iterative method starting with a guess value of Tb1=(Tbi+Tbo)/2
#Here we denote bulk mean temprature as Tb.The superscript refers to the no. of trials
#For first trial,guess Tbo1=-20Â°C;so Tb1=-30Â°C
#We have the property values as follows at a temprature of -30Â°C.
rhob1=1520.64;#density in kg/m**3
nub1=0.371*10**-6;#kinematic viscosity in m**2/s
kb1=0.23;#conductivity in W/(m*Â°C)
Prb1=3.31;#Prandtl number
mub1=nub1*rhob1;#viscosity in kg/(m*s)
cpb1=1361.6;#specific heat in J/(kg*K)
#muw=nuw*rhow at Tw=10Â°C
nuw=0.288*10**-6;#kinematic viscosity at Tw in m**2/s
rhow=1463.61;#density at Tw in kg/m**3
muw=nuw*rhow;#viscosity at Tw in kg/(m*s)
#The reynolds number is found as Re1=(4*mdot)/(math.pi*D*mub1)
print"Reynold number is"
Re1=(4*mdot)/(math.pi*D*mub1)
print"Re1=",Re1
#Hence the flow is turbulent
#Now using equation, nusselt number is,Nubar1=0.027*(Re1)**0.8*Prb1**(1/3)*(mub1/muw)**0.14
print"Nusselt number is"
Nubar1=0.027*(Re1)**0.8*Prb1**(1/3)*(mub1/muw)**0.14
print"Nubar1=",Nubar1
#The heat transfer transfer coefficient hbar1=(kb1/D)*Nubar1
print"The heat transfer transfer coefficient in W/(m**2*Â°C) "
hbar1=(kb1/D)*Nubar1
print"hbar1=",hbar1
#The outlet fluid temprature can be found by making use of eqn Tbo2=Tw-(Tw-Tbi)*math.e((-math.pi*D*L*hbar1)/(mdot*cpb1))
print"Outlet fluid temprature in first iteration is Tbo2 in Â°C is"
Tbo2=Tw-(Tw-Tbi)*math.e**((-math.pi*D*L*hbar1)/(mdot*cpb1))
#Tb2 is the bulk mean temprature.
print"Tb2 in Â°C is"
Tb2=(Tbi+Tbo2)/2
print"Tb2=",Tb2
#Since the value differs from the assumed value of Tb1=-30Â°C,WE require furtheriteration,Therfore we start second trial with Tb2=-28.36Â°C
#We have the property value at a temprature of -28.36Â°C as follows
rhob2=1514;#density in kg/m**3
nub2=0.362*10**-6;#kinematic viscosity in m**2/s
kb2=0.229;#conductivity in W/(m*Â°C)
Prb2=3.23;#Prandtl number
mub2=nub2*rhob2;#viscosity in kg/(m*s)
cpb2=1362;#specific heat in J/(kg*K)
#muw=nuw*rhow at Tw=10Â°C
nuw=0.288*10**-6;#viscosity at Tw in m**2/s
rhow=1463.61;#density at Tw in kg/m**3
muw=nuw*rhow;#kinematic viscosity at Tw in kg/(m*s)
#The reynolds number is found as Re2=(4*mdot)/(math.pi*D*mub2)
print"Reynold number is"
Re2=(4*mdot)/(math.pi*D*mub2)
print"Re2=",Re2
#Now using equation, nusselt number is,Nubar2=0.027*(Re2)**0.8*Prb2**(1/3.0)*(mub2/muw)**0.14
print"Nusselt number is"
Nubar2=0.027*(Re2)**0.8*Prb2**(1/3.0)*(mub2/muw)**0.14
#The heat transfer transfer coefficient hbar2=(kb2/D)*Nubar2
print"The heat transfer transfer coefficient in W/(m**2*Â°C) "
hbar2=(kb2/D)*Nubar2
print"hbar2=",hbar2
#The outlet fluid temprature can be found by making use of eqn Tbo3=Tw-(Tw-Tbi)*math.e((-math.pi*D*L*hbar2)/(mdot*cpb2))
print"Outlet fluid temprature in second iteration is Tbo3 in Â°C is"
Tbo3=Tw-(Tw-Tbi)*math.e**((-math.pi*D*L*hbar2)/(mdot*cpb2))
print"Tbo3=",Tbo3
#Tb3 is the bulk mean temprature.
print"Tb3 in Â°C is"
Tb3=(Tbi+Tbo3)/2
#We see that difference between Tbo2 and Tbo3 and that between Tb2 and Tb3 is marginal.Therfore we can stop iteration and present the result as Tbo=-16.67Â°C
print"The Exit fluid temprature after second iteration is obtained as Tbo=-16.67Â°C"
print"Tb3=",Tb3

 Introduction to heat transfer by S.K.Som, Chapter 7, Example 11
Reynold number is
Re1= 13541.3214942
Nusselt number is
Nubar1= 56.808608087
The heat transfer transfer coefficient in W/(m**2*Â°C)
hbar1= 522.6391944
Outlet fluid temprature in first iteration is Tbo2 in Â°C is
Tb2 in Â°C is
Tb2= -30.4912413164
Reynold number is
Re2= 13938.8493187
Nusselt number is
The heat transfer transfer coefficient in W/(m**2*Â°C)
hbar2= 784.03829067
Outlet fluid temprature in second iteration is Tbo3 in Â°C is
Tbo3= -16.646852652
Tb3 in Â°C is
The Exit fluid temprature after second iteration is obtained as Tbo=-16.67Â°C
Tb3= -28.323426326