# Chapter 4 : Forced Convection¶

## Example 4.2 Page No : 112¶

In [4]:
# Variables
l = 2.                      			#m, length of flat surface
T1 = 150.                   			#C, surface temp.
p = 1.                      			#atm, pressure
T2 = 30.                    			#C, bulk air  temp.
V = 12.                     			#m/s, air velocity

#Calculation
Tf = (T1+T2)/2            			#C, mean air film temp.
mu = 2.131*10**-5            			#m**2/s, vismath.cosity
k = 0.031                   			#W/m C, thermal conductivity
rho = 0.962                 			#kg/m**3, density of air
cp = 1.01                   			#kj/kg C, specific heat of air
Pr = cp*10**3*mu/k                			#Prandtl no.
Remax = l*V*rho/mu          			#maximum Reynold no.
Re = 5.*10**5                 			#Reynold no. during transition to turbulent flow
L_ = (Re*mu)/(V*rho)        			#m,dismath.tance from the leading edge
#for laminar flow heat transfer coefficient h,
#h16.707*x**-(1/2)
#(a)
#h2 = 31.4*x**(-1/5)
#b
hav = 22.2
#c
Q = hav*l*p*(T1-T2)

# Results
print "The rate of heat loss is %.0f W"%(Q)

# rounding off error.

The rate of heat loss is 5328 W


## Example 4.3 Page No : 114¶

In [2]:
import math

# Variables
d = 7.24*10**-4                  			#m, diameter of wire
l = 1.                           			#m, length of wire
I = 8.3                         			#A, current in a wire
R = 2.625                       			#ohm/m, electrical resistance
V = 10.                          			#m/s, air velocity
Tb = 27.                         			#C, bulk air temp.
#the properties at bulk temp.
mu = 1.983*10**-5                			#m**2/s, vismath.cosity
k = 0.02624                     			#W/m C, thermal conductivity
rho = 1.1774                    			#kg/m**3, density of air
cp = 1.0057                     			#kj/kg C, specific heat of air

# Calculations and Results
Pr = cp*10**3*mu/k               			#Prandtl no.
Re = d*V*rho/mu                 			# Reynold no.
#from eq. 4.19, nusslet no.
Nu = 0.3+(0.62*Re**(1./2)*Pr**(1./3)/(1+(0.4/Pr)**(2./3))**(1./4))*(1+(Re/(2.82*10**5))**(5./8))**(4./5)
hav = Nu*k/d                    			#W/m**2 C, average heat transfer coefficient
Q = I**2*R                       			#W, rate of electrical heat generation
A = math.pi*d*l
dt = Q/(hav*A)                  			#C,temp. difference
T = dt+Tb                       			#C, steady state temp.
print "The steady state temprature is %.0f C"%(T)

Tm = (T+Tb)/2                   			#C, mean air film temp.
#the properties at Tm temp.
mu1 = 2.30*10**-5                  			#m**2/s, vismath.cosity
k1 = 0.0338                       			#W/m C, thermal conductivity
rho1 = 0.878                      			#kg/m**3, density of air
cp1 = 1.014                       			#kj/kg C, specific heat of air
Re1 = d*V*rho1/mu1                			# Reynold no.
Pr1 = (1.014*10**3*2.30*10**-5)/k1             			#Prandtl no.
#from eq. 4.19, nusslet no.
Nu1 = 0.3+(0.62*Re1**(1./2)*Pr1**(1./3)/(1+(0.4/Pr1)**(2./3))**(1./4))*(1+(Re1/(2.82*10**5))**(5./8))**(4./5)
hav1 = Nu1*k1/d                    			#W/m**2 C, average heat transfer coefficient
dt1 = Q/(hav1*A)                  			#C,temp. difference
T1 = dt1+Tb                       			#C, steady state temp.
print "The recalculated value  is almost equal to previous one."

The steady state temprature is 230 C
The recalculated value  is almost equal to previous one.


## Example 4.4 Page No : 116¶

In [3]:
import math
# Variables
di = 0.04                       			#m, diameter of ice ball
V = 2.                           			#m/s, air velocity
T1 = 25.                         			#C, steam temp.
T2 = 0.
#the properties  of air
mu = 1.69*10**-5                 			#kg/ms, vismath.cosity
k = 0.026                       			#W/m C, thermal conductivity
rho = 1.248                     			#kg/m**3, density
cp = 1.005                      			#kj/kg C, specific heat
#propertice of ice
lamda = 334.                     			#kj/kg, heat of fusion
rhoice = 920.                    			#kg/m**3 density of ice

# Calculations and Results
Pr = cp*10**3*mu/k               			#Prandtl no.
Re = di*V*rho/mu                			# Reynold no.
#from eq. 4.19, nusslet no.
Nu = 2+(0.4*Re**0.5+0.06*Re**(2./3))*Pr**0.4
hav = Nu*k/di                          			#W/m**2 C, average heat transfer coefficient
Ai = math.pi*di**2                   			#initial area of sphere
Qi = Ai*hav*(T1-T2)                 			#W = J/s, initial rate of heat transfer
Ri = Qi/lamda                       			#initial  rate of melting of ice
print "initial  rate of melting of ice is %.4f g/s"%(Ri)

#(b)
#mass of ice ball 4/3*math.pi*r**3
#Rate of melting =  Rm =  -d/dt(m)
#Rate of heat input required  = -lamda*Rate of melting
#heat balance equation
# -lamda*(Rm) = h*4*math.pi*r**2*dt
#integrating and solving
rf = ((di/2)**3/2.)**(1./3)
#solving eq. 3
t1 = 1.355*10**-4/(8.136*10**-8)
print "The required time is is %.0f s"%(t1)

initial  rate of melting of ice is 0.0109 g/s
The required time is is 1665 s


## Example 4.5 Page No : 121¶

In [4]:
from scipy.integrate import quad
# Variables
Vo = 0.5                        			#m/s air velocity
T1 = 800.                        			#C, initial temp.
T2 = 550.                        			#C, final temp.
Tam = 500.                       			#C, air mean temp.
P = 1.2                         			#atm, pressure
#the properties of solid particles.
dp = 0.65*10**-3                 			#m, average particle diameter
cps = 0.196                     			#kcal/kg C, specific heat
rhos = 2550.                     			#kg/m**3, density
#Properties of air
mu = 3.6*10**-5                  			#kg/ms, vismath.cosity
k = 0.05                        			#kcal/hm C, thermal conductivity
rho = 0.545                     			#kg/m**3, density of air
cp = 0.263                      			#kcal/kg C, specific heat of air

#calculation
Pr = cp*mu*3600/k                    			#Prandtl no.
Redp = dp*Vo*rho/mu             			# Reynold no.
#from eq. 4.29(b) heat transfer coefficient
h = (k/dp)*(2+0.6*(Redp)**(1./2)*(Pr)**(1./3))
Tg = 500                        			#C, gas temp.
#from heat balance equation
# -(dTs/dt) = 6h/(dp*rhos*cps)*(Ts-Tg)

def f2(Ts):
return (1/(Ts-Tg))

t = (dp*rhos*cps/(6*h))* quad(f2,550,800)[0]

# Results
print "the required contact time is %.2f s"%(t*3600)

the required contact time is 1.43 s


## Example 4.6 Page No : 126¶

In [5]:
import math

# Variables
mo_ = 1000.           			#kg/h, cooling rate of oil
cpo = 2.05           			#kj/kg C, specific heat of oil
T1 = 70.              			#C, initial temp. of oil
T2 = 40.              			#C,  temp. of oil after cooling
cpw = 4.17           			#kj/kg C, specific heat of water
T3 = 42.              			#C, initial temp. of water
T4 = 28.              			#C,  temp. of oil after cooling
A = 3.                			#m**2, heat exchange area

# Calculation and Results
mw_ = mo_*cpo*(T1-T2)/(cpw*(T3-T4))
print "the required rate of flow of water is %.0f kg/h "%(mw_)
Q = mo_*cpo*(T1-T2)/3600       			#kw, heat duty
dt1 = T1-T3                  			#C, hot end temp. difference
dt2 = T2-T4                  			#C, cold end temp. difference
LMTD = (dt1-dt2)/(math.log(dt1/dt2))  			#math.log mean temp. difference
dtm = LMTD
U = Q*10**3/(A*dtm)
print "the overall heat transfer coefficient is %.0f W/m**2 C"%(round(U,-1))

the required rate of flow of water is 1053 kg/h
the overall heat transfer coefficient is 300 W/m**2 C


## Example 4.7 Page No : 126¶

In [6]:
from scipy.optimize import fsolve
import math

# Variables
Q = 38700.             			#kcal/h, heat duty
W = 2000.              			#kg/h gas flow rate
cp = 0.239            			#kcal/kg C, specific heat of nitrogen
A = 10.                			#m**2 ,heat exchanger area
U = 70.                			#kcal/hm**2 C, overall heat transfer coefficient
n = 0.63              			#fin efficiency

#Calculation
dt = Q/(W*cp)         			#C, temp. difference
#To-Ti = dt.........................(i)
dtm = Q/(U*A*n)
#(To-Ti)/(math.log((160-Ti)/(160-To))) = 87.8........(2)
#solving  1  and 2
def f(To):
return (To-(To-dt))/(math.log((160-(To-dt))/(160-To)))-87.8

To = fsolve(f,100)
Ti = To-dt

# Results
print "The inlet temprature is  Ti = %.0f C"%(Ti)
print "The outlet temprature is To = %.0f C"%(To)

# note : answers are slightly different because of fsolve function of python.

The inlet temprature is  Ti = 26 C
The outlet temprature is To = 107 C


## Example 4.8 Page No : 127¶

In [7]:
import math

# Variables
V = 1.8                  			#m/s, velocity of hot water
T1 = 110.                 			#C, initial temp.
l = 15.                   			#m, length  of pipe
t = 0.02                 			#m, thickness of insulation
kc = 0.12                			#W/mC,thermal conductivity of insulating layer
ho = 10.                  			#Wm**2 C, outside film coefficient
T2 = 20.                  			#C, ambient temp.
#the properties of water at 110 C
mu = 2.55*10**-4          			#m**2/s, vismath.cosity
k = 0.685                			#W/m C, thermal conductivity
rho = 950.                			#kg/m**3, density of air
cp = 4.23                			#kj/kg C, specific heat of air
di = 0.035               			#m, actual internal dia. of pipe
ri = di/2.                			#m,internal radius
t1 = 0.0036              			#m, actual thickness of 1-1/4 schedule 40 pipe
ro = ri+t1               			#m, outer radius of pipe
r_ = ro+t                			#m, outer radius of insulation
kw = 43.                  			#W/mC, thermal conductivity of steel

#calculation
Pr = cp*10**3*mu/k        			#Prandtl no.
Re = di*V*rho/mu         			# Reynold no.
#from eq. 4.9,  Nusslet no.
Nu = 0.023*(Re)**0.88*Pr**0.3
hi = Nu*k/di             			#W/m**2 C, average heat transfer coefficient
#the overall coefficient inside area basis Ui
Ui = 1./(1/hi+(ri*math.log(ro/ri))/kw+(ri*math.log(r_/ro))/kc+ri/(r_*ho))
Ai = math.pi*di*l           			#m**2, inside area basis
W = math.pi*ri**2*V*rho      			#kg/s, water flow rate
#from the relation b/w LMTD and rate of heat loss

def f(To):
return (W*cp*10**3)/(Ui*Ai)*(T1-To)-((T1-To)/math.log((T1-T2)/(To-T2)))
To = fsolve(f,100)

# Results
print "The outlet eater temp. is %.1f C"%(To)

The outlet eater temp. is 109.8 C


## Example 4.9 Page No : 129¶

In [9]:
import math

# Variables
T1 = 28.              			#C, inlet temp.
T2 = 250.             			#C,bulk temp.
V = 10.               			#m/s, gas velocity
l = 20.               			#m, length of pipe
mw = 1.*3600          			#kg/h, water flow rate
di = 4.1*10**-2       			#m, inlet diameter
Tm = (T1+T2)/2       			#C, mean temp.
ro = 0.0484          			#m, outside radius
#properties of water
mu = 8.6*10**-4       			#kg/ms, vismath.cosity
kw = 0.528           			#kcal/h m C, thermal conductivity
kw_ = 0.528*1.162    			#W/ m C, thermal conductivity
rho = 996.            			#kg/m**3, density of air
cp = 1*4.18          			#kj/kg C, specific heat of air
cp_ = 1.              			#kcal/kg C
#properties of flue gas
mu1 = 2.33*10**-5     			#kg/ms, vismath.cosity
ka = 0.0292          			#kcal/h m C, thermal conductivity
rho1 = 0.891         			#kg/m**3, density of air
cp1 = 0.243          			#kcal/kg C, specific heat of air
Pr = 0.69

#calculation
A = math.pi/4*di**2       			#m**2, cross section of pipe
Vw = 1/(rho*A)       			#m/s, velocity of warer
Re = di*Vw*rho/mu    			# Reynold no.
Pr1 = cp*10**3*mu/kw_   			#Prandtl no. for water
Nu = 0.023*Re**0.8*Pr1**0.4               			#Nusslet no.
#water side heat transfer coefficient  hi
hi = 206*kw/di
#gas side heat transfer coefficient    ho
a = 41              			#mm, i.d. schedule
Tw = 3.7            			#mm, wall thickness
do = a+2*Tw         			#mm, outer diameter of pipe
Re1 = do*10**-3*V*rho1/mu1    			# Reynold no
#from eq. 4.19, nusslet no.
Nu1 = 0.3+(0.62*Re1**(1./2)*Pr**(1./3)/(1+(0.4/Pr)**(2./3))**(1/4.))*(1+(Re1/(2.82*10**5))**(5./8))**(4/5.)
ho = (Nu1*ka/do)*10**3      			#kcal/h m**2 C
Uo = 1/(ro/(di/2*hi)+1/ho)   			#kcal/h m**2 C, overall heat transfer coefficient

#Heat balance
A1 = math.pi*ro*l         			#m62, outside area of pipe
#from the formula of LMTD
def f(T2_):
return mw*cp_*(T2_-T1)-Uo*A1*((T2_-T1)/math.log((T2-T1)/(T2-T2_)))
T2_ = fsolve(f,1)

# Results
print "The exit water temp is %.0f C"%(T2_)

The exit water temp is 36 C


## Example 4.10 Page No : 131¶

In [10]:
import math
# Variables
dti = 0.0212        			#m inner tube
dto = 0.0254        			#cm, outer tube
dpi = 0.035         			#cm, outer pipe
mo_ = 500.           			#kh/h, cooling rate of oil
To2 = 110.           			#C, initial temo. of oil
To1 = 70.            			#C, temp. after cooling of oil
Tw2 = 40.            			#C, inlet temp. of water
Tw1 = 29.            			#C, outlet temp. of water
#properties of oil
cpo = 0.478         			#kcal/kg C
ko = 0.12           			#kcal/h m C, thermal conductivity
rho = 850.           			#kg/m**3, density of oil
#properties of water
kw = 0.542          			#kcal/h m C, thermal conductivity
kw_ = (kw*1.162)    			#kj/kg C
muw = 7.1*10**-4     			#kg/ms, vismath.cosity of water
cpw = 1.             			#kcal/kg C
cpw_ = cpw*4.17     			#kcal/kg C
rhow = 1000.         			#kg/m**3, density

#calculation
HL = mo_*cpo*(To2-To1)      			#kcal/h, heat load of exchanger
mw_ = HL/(cpw*(Tw2-Tw1))    			#kg/h water flow rate
mw_1 = mw_/(3600*10**3)      			#m**3/s water flow rate
A1 = (math.pi/4)*(dti)**2        			#m**2, flow area of tube
Vw = mw_1/A1                			#m/s water velocity
Rew = dti*Vw*rhow/muw       			#Reynold no.
Prw = cpw_*10**3*muw/kw_      			#Prandtl no.
Nuw = 0.023*Rew**0.8*Prw**0.4 			#nusslet no.
#water side heat transfer coefficient   hi
hi = Nuw*kw/dti

#oil side heat transfer coefficient
A2 = math.pi/4*(dpi**2-dto**2)    			#m**2, flow area of annulus
Vo = mo_/(3600*rho*A2)      			#m/s velocity of oil
de = (dpi**2-dto**2)/dto      			#m, equivalent dia of annulus
Tmo = (To2+To1)/2           			#C,mean oil temp.
muoil = math.exp((5550./(Tmo+273))-19)  			#kg/ms, vismath.cosity of oil
Reo = de*Vo*rho/muoil
Pro = cpo*muoil*3600/ko     			#prandtl no. for oil

#assume (1st approximation)
Nuo = 3.66
ho = Nuo*ko/de             			#kcal/h m**2 c
L = 1                      			#assume length of tube
Ai = math.pi*dti*L
Ao = math.pi*dto*L
#overall heat transfer coefficient 1st approximation
Uo = 1/(1/ho+Ao/(Ai*hi))
LMTD = ((To2-Tw2)-(To1-Tw1))/(math.log((To2-Tw2)/(To1-Tw1)))
Ao1 = HL/(Uo*LMTD)         			                #m**2, heat transfer area
Lt = Ao1/(math.pi*dto)          			        #m, tube length
#from eq.  4.8
Nuo1 = 1.86*(Reo*Pro/(Lt/de))**(1./3)   			#Nusslet no.
ho1 = Nuo1*ko/de
Tmw = (Tw1+Tw2)/2         			#C, mean water temp.
#balancing heat transfer rate of oil and water

#average wall temp. Twall
Twall = ((hi*dti*(-Tmw))-(ho1*dto*Tmo))/(-65.71216)
#vismath.cosity of oil at this temp.
muwall = math.exp((5550/(Twall+273))-19)  			#kg/ms, vismath.cosity of oil
#Nusslet no.
Nuo2 = 1.86*(Reo*Pro/(Lt/de))**(1./3)*(muoil/muwall)**0.14
ho2 = Nuo2*ko/de
Uo2 = 1/((1/ho2)+(Ao/(Ai*hi)))
Ao2 = HL/(Uo2*LMTD)
Lt_ = Ao2/(math.pi*dto)

# Results
print "The tube length is %d m"%(Lt_)

# rounding off error.

The tube length is 123 m


## Example 4.11 Page No : 135¶

In [11]:
import math

# Variables
Ti = 260.                 			#C, initial  temp.
Ts = 70.                  			#C, skin temp.
St = 0.15                			#m,space between tubes in equilateral triangular arrangement
Sd = St                  			#space between tubes
mu = 4.43*10**-5                			#m**2/s, momentum diffusity
k = 0.0375                     			#W/m C, thermal conductivity
rho = 0.73                     			#kg/m**3, density of air
cp = 0.248                     			#kj/kg C, specific heat of air
V = 16.                         			#m/s, velociity
d = 0.06                       			#m, outside diameter of tube
Nt = 15.                        			#no. of tubes in transverse row
Nl = 14.                        			#no. of tubes in longitudinal row
N = Nl*Nt                      			#total no. of tubes
L = 1.                          			#m, length
#Calculation
Sl = (math.sqrt(3)/2)*St
Pr = cp*mu*3600*rho/k          			#Prandtl no. of bulk air
Pr = 0.62
Prw = 0.70                     			#Prandtl no. of air at wall temp. 70 C
#from eq. 4.25
Vmax = (St/(St-d))*V
#from eq. 4.26
Vmax1 = (St/(2*(St-d)))*V
Redmax = d*Vmax/mu
p = St/Sl                     			#pitch ratio
#from table 4.3
m = 0.6
C = 0.35*(St/Sl)**0.2
h = round((k/d)*C*(36163)**m*(Pr)**(0.36)*(Pr/Prw)**(0.25))
#from eq. 4.28
dt = round(190*math.exp(-math.pi*d*N*h/(rho*V*3600*Nt*St*cp)))
LMTD = ((Ti-Ts)-(dt))/math.log((Ti-Ts)/dt)
A = round(math.pi*d*L*N,1)                			#m**2, heat transfer area
Q = h*A*LMTD

# Results
print " the rate of heat transfer to water.is %.2e kcal/h"%(Q)

# Note : Value of LMTD is wrong in book please check.

 the rate of heat transfer to water.is 6.93e+05 kcal/h


## Example 4.12 Page No : 140¶

In [12]:
import math

# Variables
W = 0.057                      			#m**3/min/tube, flow rate of water
W_ = W*16.66                   			#kg/s. water flow rate
di = 0.0212                    			#m,inside diameter
Ti = 32.                        			#C, inlet water temp.
Tw = 80.                        			#C, wall temp.
L = 3.                          			#m, length of pip

#Calculation
V = (W/60)*(1/((math.pi/4)*di**2))   			#m/s, water velocity
#the properties of water at mean liquid temp..
mu = 7.65*10**-4                 			#m**2/s, vismath.cosity
k = 0.623                       			#W/m C, thermal conductivity
rho = 995.                       			#kg/m**3, density of air
cp = 4.17                       			#kj/kg C, specific heat of air

Pr = cp*10**3*mu/k               			#Prandtl no.
Re = di*V*rho/mu                			# Reynold no.
#from eq. 4.19, nusslet no.
#from dittus boelter eq.
Nu = 0.023*Re**0.8*Pr**0.4        			#Prandtl no.
f = 0.0014+0.125*Re**-0.32       			#friction factor
#Reynold anamath.logy
St = f/2                       			#Smath.tanton no.
Nu1 = Re*Pr*St
#Prandtl anamath.logy
St1 = (f/2)/(1+5*(Pr-1)*math.sqrt(f/2))
Nu2 = St1*Re*Pr
#colburn analogy
Nu3 = Re*Pr**(1./3)*(f/2)
h = Nu3*k/(di)               			#W/m**2 C av heat transfer coefficient
#Q = W_*cp*10**3*(To-Ti) = h*A*LMTD
A = math.pi*di*L       			#m**2
def f(To):
return W_*cp*10**3*(To-Ti)-h*A*((To-Ti)/math.log((Tw-Ti)/(Tw-To)))
To = fsolve(f,1)
#Revised calculation
Tm = (Ti+To)/2               			#C, mean liquid temp.
#the properties of water at new mean liquid temp..
mu1 = 6.2*10**-4                 			#m**2/s, vismath.cosity
k1 = 0.623                       			#W/m C, thermal conductivity
rho1 = 991.                       			#kg/m**3, density of air
cp1 = 4.17                       			#kj/kg C, specific heat of air

Pr1 = cp1*10**3*mu1/k1               			#Prandtl no.
Re1 = di*V*rho1/mu1                			# Reynold no.
#from dittus boelter eq.
f1 = 0.0014+0.125*Re1**(-0.32)       			#friction factor
#colburn anamath.logy
Nu4 = Re1*Pr1**(1./3)*(f1/2)
h1 = Nu4*k1/(di)               			#W/m**2 C av heat transfer coefficient
def f(To_):
return W_*cp*10**3*(To_-Ti)-h1*A*((To_-Ti)/math.log((Tw-Ti)/(Tw-To_)))
To_ = fsolve(f,1)

print "Outlet temp. of water for one pass through the tubes is %.0f C"%(To_)

Outlet temp. of water for one pass through the tubes is 51 C