# Chapter 8 : Internal Forced Convection¶

### Example 8.1¶

In :
import math
# Heating of water in a tube by Steam

# Variables
id_ = 0.025 			#Internal diameter[m]
Tin = 15.	    		#Initial temp[degree Celcius]
m_ = 0.3		    	#Flow rate[kg/s]
h = 800./1000			#avg heat transfer coefficient[W/m**2.degree Celcius]
Tf = 115.			#Final temp of water[degree Celcius]
Ts = 120.			#[degree Celcius]
Hs = 2203.			#Heat of condensation of steam at 120 degree Celcius[kJ/kg]
Tavg = (Tin+Tf)/2			#[degree Celcius]
Cp = 4187.			#Sp Heat of water at Tavg[J/kg.degree Celcius]

# Calculations and Results
Q_ = m_*Cp*(Tf-Tin)/1000			#[kW]
print "The rate of heat transfer is",Q_,"kW"

del_Tf = Ts-Tf			    #[degree Celcius]
del_Tin = Ts-Tin			#[degree Celcius]
ln_del_T = (del_Tf-del_Tin)/(math.log(del_Tf/del_Tin))			#[degree Celcius]
print "Logrithmic Mean temperature difference is",ln_del_T,"degree Celcius"

A = Q_/(h*ln_del_T)			#[m**2]
print "Heat Transfer surface area is",A,"m**2"

l = A/(math.pi*id_)			#[m]
print "Required tube length is",round(l),"m"

The rate of heat transfer is 125.61 kW
Logrithmic Mean temperature difference is 32.8458738753 degree Celcius
Heat Transfer surface area is 4.78028079253 m**2
Required tube length is 61.0 m


### Example 8.2¶

In :
import math
# Pressure Drop in a tube

# Variables
Tw = 5			#Temperature of water[degree Celcius]
#Properties of water at Tw
rho = 999.9			#[kg/m**3]
mu = 1.519*10**(-3)			#Vismath.cosity[kg/m.s]
d = 0.003			#diameter[m]
l = 10			#length[m]
v_avg = 0.9			#Average flow velocity[m/s]

# Calculations and Results
Re = (rho*v_avg*d)/mu;
print "The reynolds number is ",Re

f = 64/math.ceil(Re);
print "Friction factor is",f

del_P = f*l*rho*(v_avg**2)/(2*d)			#[N/m**2]
print "The Pressure drop is ",del_P/1000,"kPa"

V = v_avg*(math.pi*(d**2))/4			#[m**3/s]
print "Volumetric flow rate is",V,"m**3/s"

W_pump = V*del_P			#[W]
print "Mechanical Power Input of",W_pump,"W"
print ("is needed to overcome the frictional losses in the flow due to viscosity")

The reynolds number is  1777.3074391
Friction factor is 0.0359955005624
The Pressure drop is  48.5890663667 kPa
Volumetric flow rate is 6.36172512352e-06 m**3/s
Mechanical Power Input of 0.309110284233 W
is needed to overcome the frictional losses in the flow due to viscosity


### Example 8.3¶

In :
import math
# Flow of Oil in a Pipeline through a Lake

# Variables
Ts = 0			#Temp of lake[degree Celcius]
Ti = 20			#Temp of oil[degree Celcius]
d = 0.3			#Diameter[m]
l = 200			#length of pipe[m]

#At 20 degree Celcius
rho = 888.1			#[kg/m**3]
nu = 9.429*10**(-4)			#Kinematic vismath.cosity[m**2/s]
k = 0.145			#[W/m.degree Celcius]
Cp = 1880			#[J/kg.degree Celcius]
Pr = 10863			#Prandtl Number
v_avg = 2			#[m/s]

# Calculations and Results
Re = v_avg*d/nu;
print "The Reynolds number is",Re

Lt = 0.05*Re*Pr*d			#[m]
print "The thermal entry length is",Lt,"m"

Nu = 3.66+((0.065*(d/l)*Re*Pr)/(1+(0.04*(((d/l)*Re*Pr)**(2./3)))));
h = (k*Nu)/d            	        		#[W/m**2.degree Celcius]
As = math.pi*d*l		                  	#[m**2]
m_ = rho*math.pi*((d/2)**2)*v_avg			#[kg/s]
Te = Ts-((Ts-Ti)*math.exp((-h*As)/(m_*Cp)))			#[degree Celcius]
print "Exit temperature of oil is",Te,"degree Celcius"

#Solution(b):-
ln_del_T = (Ti-Te)/(math.log((Ts-Te)/(Ts-Ti)))			#[degree Celcius]
print "The logrithmic mean temperature difference is",ln_del_T,"degree Celcius"

Q = h*As*ln_del_T			#[W]
print "The rate of heat loss from the oil are",Q,"W"

#Solution(c)
f = 64/Re			                    #Friction factor is
del_P = l*rho*(v_avg**2)/(2*d)			#[N/m**2]
print (del_P);

W_pump = m_*del_P/rho			#[kW]
print "We need a ",W_pump/1000,"pump just to overcome the friction in the pipe as the oil flows"

The Reynolds number is 636.334712059
The thermal entry length is 103687.559656 m
Exit temperature of oil is 19.7139328074 degree Celcius
The logrithmic mean temperature difference is -19.856622966 degree Celcius
The rate of heat loss from the oil are -67522.7559473 W
1184133.33333
We need a  167.402906139 pump just to overcome the friction in the pipe as the oil flows


### Example 8.4¶

In :
import math
# Pressure Drop in a Water tube

# Variables
Tw = 15			#temp of water while entering[degree Celcius]
rho = 999.1			#[kg/m**3]
mu = 1.138*10**(-3)			#Vismath.cosity[kg/m.s]
id_ = 0.05			#Internal diameter[m]
V = 5.5*10**(-3)			#Flow rate[m**3/s]
l = 60			#length of tube[m]
e = 0.002*10**(-3)			#[m]
#Solution:-
v = V/(math.pi*(id_**2)*(1./4))			#Mean Velocity[m/s]
Re = rho*v*id_/mu;
print "Reynolds Number is",Re

#Flow is turbulent
r = e/id_			#Relative roughness of the tube
def fric(fac):
Func = (1/(fac**(1./2)))+(2*math.log((0.00004/3.7)+(2.51/(122900*fac**(1./2)))));

#print "Friction Factor is",xs
#del_P = xs*l*rho*(v**2)/(2*id_)			#[kPa]
#print "The pressure drop is",del_P,"Pa"
W_pump = V*del_P			            #[W]
print "The required poer input tp overcome the frictional losses in the tube is",W_pump,"W"

 Reynolds Number is 122961.598599
The required poer input tp overcome the frictional losses in the tube is 6512.73333333 W


### Example 8.5¶

In :
import math
# Heating of water by Resismath.tance Heaters in a tube

# Variables
Ti = 15 		        	#Initial Temp[degree Celcius]
Tf = 65	        		    #Final Temp[degree Celcius]
d = 0.03	     	    	#Internal diameter[m]
l = 5   		        	#length[m]
V = 10.*10**(-3)			#flow rate of water[m**3/s]
Tavg = (Ti+Tf)/2			#[degree Celcius]
#Properties of water at Tavg
rho = 992.1		        	#[kg/m**3]
Cp = 4170	        		#[J/kg.degree Celcius]
k = 0.631       			#[W/m.degree Celcius]
nu = 0.658*10**(-6)			#[m**2/s]
Pr = 4.32       			#Prandtl Number

# Calculations and Results
Ac = math.pi*(d**2)*(1./4)		#[m**2]
As = math.pi*d*l		    	#[m**2]
m_ = rho*V*(1./60)	    		#[kg/s]
Q_ = m_*Cp*(Tf-Ti)/1000			#[kW]
print "The power rating of the heater is",Q_,"kW"

qs = Q_/As      			#[kW/m**2]
print "Heat flux is",qs,"kW/m**2"

v_avg = V/(Ac*60)			#[m/s]
Re = v_avg*d/nu		    	#[Reynolds Number]
Lt = 10*d		    	    #Entry length [m]
Nu = 0.023*(Re**(0.8))*(Pr**(0.4));
print "The nussel number is",Nu

h = k*Nu/d	        		#[W/m**2]
Ts = Tf+(qs*1000/h)			#[degree Celcius]
print "The surface temperature of the pipe at the exit becomes",round(Ts),"degree Celcius"

The power rating of the heater is 34.475475 kW
Heat flux is 73.1592301559 kW/m**2
The nussel number is 69.3506779058
The surface temperature of the pipe at the exit becomes 115.0 degree Celcius


### Example 8.6¶

In :
import math
#  Heat Loss from the ducts of a Heating System

# Variables
Ti = 80			#Inlet temp[degree Celcius]
A = 0.2*0.2			#Area of cross section[m**2]
l = 8			#Length of tube[m]
V = 0.15			#[m**3/s]
Td = 60			#Temperature of duct[degree Celcius]
#Properties of air at inlet conditions
rho = 0.9994			#[kg/m**3]
Cp = 1008			#[J/kg.degree Celcius]
k = 0.02953			#[W/m.degree Celcius]
nu = 2.097*10**(-5)			#[m**2/s]
Pr = 0.7154			#Prandtl number

# Calculations and Results
Dh = 4*A/(4*0.2)			#Hydraulic Diameter[m]
v_avg = V/A			#[m/s]
Re = v_avg*Dh/nu;
print "Reynolds number is",Re

Lt = 10*Dh			#Entry length
Nu = 0.023*(Re**(0.8))*(Pr**(0.3));
h = Nu*k/Dh			#[W/m**2.degree Celcius]
As = 4*0.2*l			#[m**2]
m_ = rho*V			#[kg/s]
Te = Td-((Td-Ti)*math.exp((-h*As)/(m_*Cp)))			#[degree Celcius]
print "The exit temperature of air is",Te,"degree Celcius"

ln_delT = (Ti-Te)/(math.log((Td-Te)/(Td-Ti)))			#[degree Celcius]
Q = h*As*ln_delT			#[W]
print "The math.logrithmic mean temperature difference and the rate of heat loss from the air are" \
,ln_delT,"degree Celcius","and",round(Q),"W","respectively"

Reynolds number is 35765.379113
The exit temperature of air is 71.2940188449 degree Celcius
The math.logrithmic mean temperature difference and the rate of heat loss from the air are -15.2346559432 degree Celcius and -1316.0 W respectively


### Example 8.7¶

In :
import math
# Non-isothermal fully developed Friction in the Transition Region

# Variables
q = 8			#Wall heat flux[kW/m**2]
xm = 0.34			#Mass fraction
d = 0.0158			#Inside diameter[m]
V = 1.32*10**(-4)			#Flow rate[m**3/s]
Pr = 11.6			#Prandtl Number
nu = 1.39*10**(-6)			#[m**2/s]
p = 1.14			#(mu_b/mu_s)i.e. ratio of vismath.cosities of two subsmath.tances
Gr = 60800			#Grashof number

# Calculations and Results
Ac = math.pi*(d**2)*(1./4)			#[m**2]
Re = (V/Ac)*d/nu;
print "Reynolds number is",Re

#For bell mouth inlet shape
Cf1 = ((1+((round(Re)/5340)**(-0.099)))**(-6.32))*(p**(-2.58-0.42*(60.800**(-0.41))*(11.6**0.265)));
print "For bell mouth inlet friction coefficient is",Cf1

#For square edged inlet Case
Cf2 = (0.0791/(Re**(0.25)))*(p**(-0.25));
print "For square edged inlet case coefficient of friction is",Cf2

Reynolds number is 7652.65549153
For bell mouth inlet friction coefficient is 0.00978680626691
For square edged inlet case coefficient of friction is 0.00818459567862


### Example 8.8¶

In :
import math
# Heat transfer in the Transition Region

# Variables
xm = 0.6			#mass fraction of glycol
V = 2.6*10**(-4)			#Flow rate[m**3/s]
d = 0.0158			#inside diameter[m]
Gr = 51770			#grashof number
Pr = 29.2			#Prandtl number
nu = 3.12*10**(-6)			#[m**2/s]
p = 1.77			#mu_t/mu_s
q = 90			#A particular loctaion x with x/d = q

# Calculations and Results
Ac = math.pi*(d**2)/4;
Re = (V/Ac)*d/nu;
print "Reynolds Number is",Re

#Value of Re lies in transition Region
Nu_lam = 1.24*(((Re*Pr/q)+(0.025*((Gr*Pr)**(0.75))))**(1./3))*(p**(0.14));
Nu_tur = 0.023*(Re**(0.8))*(Pr**0.385)*(q**(-0.0054))*(p**(0.14));
#(a)
Nu_tran_a = Nu_lam+((math.exp((1766-Re)/276)+(Nu_tur**(-0.955)))**(-0.955));
print "(a) Nusselt number for re-entrant inlet is",Nu_tran_a

Nu_tran_b = Nu_lam+((math.exp((2617-Re)/207)+(Nu_tur**(-0.950)))**(-0.950));
print "(b) Nusselt number for square edged inlet is",Nu_tran_b

Nu_tran_c = Nu_lam+((math.exp((6628-Re)/237)+(Nu_tur**(-0.980)))**(-0.980));
print "(c) Nusselt number for bell mouth inlet is",Nu_tran_c

Reynolds Number is 6715.3984427
(a) Nusselt number for re-entrant inlet is 88.259512708
(b) Nusselt number for square edged inlet is 85.3094482545
(c) Nusselt number for bell mouth inlet is 21.3233283966