# Chapter 5 : free convection¶

## Example 5.1 Page No : 153¶

In :
# Variables
T1 = 65.                    			#C, furnace temp.
T2 = 25.                    			#C, ambient temp.
h = 1.5                     			#m, height of door
w = 1.                      			#m, width of door
Tf = (T1+T2)/2               			#c, average air film temp.

# Calculations
Pr = 0.695                 			#Prandtl no.
mu = 1.85*10**-5            			#m**2/s, vismath.cosity
beeta = 1/(Tf+273)         			#K**-1. coefficient of volumetric expension
k = 0.028                  			#W/m C, thermal conductivity
g = 9.8                    			#m/s**2, gravitational consmath.tant
Grl = g*beeta*(T1-T2)*h**3/(mu**2)        			#Grashof no.
Ral = Grl*Pr                            			#Rayleigh no.
#Nusslet no.
Nul = (0.825+(0.387*(Ral)**(1./6))/(1+(0.492/Pr)**(9./16))**(8./27))**2
hav = Nul*k/h              			#average heat transfer coefficient
Ad = h*w                   			#m**2, door area
dt = T1-T2                 			#temp. driving force
q = hav*Ad*dt              			#W,rate of heat loss

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

The rate of heat loss is 267 W


## Example 5.2 Page No : 154¶

In :
# Variables
T1 = 60.                    			#C, plate temp.
T2 = 25.                    			#C, ambient temp.
h = 1.
w = 1.                      			#m, width of door
q = 170.                    			#W, rate of heat transfer
Tf = (T1+T2)/2             			#c, average air film temp.
#Properties of air at Tf
Pr = 0.7                 			#Prandtl no.
mu = 1.85*10**-5            			#m**2/s, vismath.cosity
beeta = 1./(Tf+273)         			#K**-1. coefficient of volumetric expension
k = 0.028                  			#W/m C, thermal conductivity
g = 9.8                    			#m/s**2, gravitational consmath.tant

#Calculation
A = h*w                    			#m**2, plate area
P = 2*(h+w)                			#m,perimeter of plate
L = A/P                    			#m characteristic length
Grl = g*beeta*(T1-T2)*L**3/(mu**2)      			#Grashof no.
Ral = Grl*Pr                          			#Rayleigh no.
#Nusslet no.
Nul = 0.54*(Ral)**(1./4)                			#Nusslet no.
hav = Nul*k/L                        			#average heat transfer coefficient
Ts = q/(hav*A)+T2

# Results
print "the steady state temp. of the plate is %.1f C"%(Ts)

the steady state temp. of the plate is 61.6 C


## Example 5.3 Page No : 156¶

In :
import math
# Variables
d = 0.0254                  			#m, diameter of steel rod
l = 0.4                     			#m, length of rod
T1 = 80.                     			#C, initial temp.
T2 = 30.                     			#C, ambient temp.
T3 = 35.                     			#c, temp. after cooling
rho = 7800.                  			#kg/m**3 ,density of steel rod
cp = 0.473                  			#kj/kg C. specific heat

#Calculation
m = math.pi/4*d**2*l*rho         			#kg. mass of cylinder
A = math.pi*d*l                 			#m**2, area of cylinder
dt = T1-T2                  			#c, insmath.tanmath.taneous temp. difference
h = 1.32*(dt/d)**0.25        			#W/m**2 C, heat transfer coefficient

def f0(T):
return 1./(T**(5./4))

t = i/(3.306*A/(m*cp*10**3))

# Results
print "The required time for cooling is %.2f hr"%(t/3600.)

The required time for cooling is 2.30 hr


## Example 5.4 Page No : 157¶

In :
import math
# Variables
id_ = 78.*10**-3                    			#m, internal diameter
od = 89.*10**-3                    			#m, outer diameter
Pg = 15.                          			#kg/cm**2, gauge pressure
t = 2.*10**-2                      			#m, thickness of preformed mineral fibre
k = 0.05                         			#W/m C. thermal conductivity
Ta = 25.                          			#C, ambient air temp.
Pr = 0.705                       			#Prandtl no.
#assume
Ts = 50.                          			#C, skin temp.
l = 1.                            			#m, length
Ti = 200.5                        			#C, initial temp.
rs = od/2+t                      			#m, outer radius of insulation
ri = od/2                        			#m, inner radius of insulation

# Calculations
Q = 2*math.pi*l*k*(Ti-Ts)/(math.log(rs/ri))           			#W
#properties of air at taken at the mean film temp.
Tf = (Ta+Ts)/2                  			#C
mu = 1.76*10**-5                 			#m**2/s. vismath.cosity
beeta = (1/(Tf+273))            			#K**-1, coefficient of volumetric expansion
k1 = 0.027                      			#W/m C, thermal conductivity
ds = 2*rs                       			#m, outer dia. of insulated pipe
g = 9.8                         			#m/s**2, gravitational consmath.tant
Grd = g*beeta*(Ts-Ta)*ds**3/(mu**2)        			#Grashof no.
#from eq. 5.9
#Nusslet no.
hav = Nu*k1/ds                 			#W/ m**2 C, average heat transfer coefficient
Ts = (Q/(math.pi*ds*l*hav))+Ta     			#C, skin temp.
#revised calculation by assuming
Ts1 = 70.                      			#C, skin temp.
#Rate of heat transfer through insulation
Q1 = 2*math.pi*l*k*(Ti-Ts1)/(math.log(rs/ri))
Tf1 = (Ta+Ts1)/2                			#C, average aie mean film temp.
mu1 = 1.8*10**-5                  			#m**2/s. vismath.cosity
beeta1 = (1/(Tf1+273))            			#K**-1, coefficient of volumetric expansion
k1 = 0.0275                     			#W/m C, thermal conductivity
Pr1 = 0.703                     			#Prandtl no.
Grd1 = g*beeta1*(Ts1-Ta)*ds**3/(mu1**2)        			#Grashof no.
#from eq. 5.9
# average heat transfer coefficient, in 			#W/ m**2 C,
Ts2 = (Q1/(math.pi*ds*l*hav1))+Ta
#again assume skin temp. = 74
Ts2 = 74                      			#C, assumed skin temp.
Q3 = 2*math.pi*l*k*(Ti-Ts2)/(math.log(rs/ri))

# Results
print "the rate of heat loss by free convection per meter length of pipe. is %.0f W"%(Q3)

the rate of heat loss by free convection per meter length of pipe. is 107 W


## Example 5.5 Page No : 159¶

In :
from scipy.optimize import fsolve
import math

# Variables
Ts = 65.                       			#C, skin temp.
To = 30.                       			#C, ambient temp.
Tw = 460.                      			#C, wall temp.
Tf = (Ts+To)/2                  			#C,mean air film temp.
beeta = (1./(Tf+273))                			#K**-1, coefficient of volumetric expansion
g = 9.8                         			#m/s**2, gravitational consmath.tant
mu = 1.84*10**-5               			#m**2/s, vismath.cosity
L = 10.5                      			#m, height of converter
di = 4.                        			#m,diameter of converter
Pr = 0.705                    			#Prandtl no.
k = 0.0241                    			#kcal/h m C, thermal conductivity

#Calculation
Grl = g*beeta*(Ts-To)*L**3/(mu**2)        			#Grashof no.
x = di/L                      			#assume di/l = x
y = 35/(Grl)**(1./4)            			#assume 35/(Grl)**(3/4) =  y
#for a verticla flat plate,  from eq. 5.3
Ral = Grl*Pr                  			#Rayleigh no.
#nusslet no.
Nu = (0.825+(0.387*(Ral)**(1./6))/(1+(0.496/Pr)**(9./16))**(8./27))**2
hav = Nu*k/L                 			#kcal/h m**2 C, average heat transfer coefficient
#w = poly(0,"w")
#Dav = (4+(4+2*w))/2                 			#average diameter
#Aav = math.pi*Dav*L                     			#average heat transfer area
#Qi = math.pi*Dav*L*0.0602*(Tw-Ts)/w     			#Rate of heat transfer through insulation
#rate of heat transfer from the outer surface of the insulation by free convection
#Qc = hav*math.pi*Dav*L*(Ts-To)
#Qi = Qc
def f(w):
return math.pi*(4+w)*L*0.0602*(Tw-Ts)/w-hav*math.pi*(4+2*w)*L*(Ts-To)
w = fsolve(f,0.1)

# Results
print "The required insulation thickness is %.3f m"%(w)

The required insulation thickness is 0.188 m


## Example 5.6 Page No : 162¶

In :
# Variables
L = 1.6                  			#m,height of enclosure
w = 0.04                 			#m, width of enclosure
b = 0.8                  			#m, breath
T1 = 22.                  			#C,surface temp.
T2 = 30.                  			#C, wall temp.
Tm = (T1+T2)/2           			#C, Mean air temp.
Pr = 0.7                 			#Prandtl no.

# Calculations
#fpr air at 26 C
beeta = 1./(Tm+273)       			#K**-1. coefficient of volumetric expension
mu = 1.684*10**-5         			#m**2/s,  vismath.cosity
k = 0.026                			#W/m C, thermal conductivity
alpha = 2.21*10**-5       			#m**2/s, thermal diffusity
g = 9.8                  			#m/s**2, gravitational consmath.tant
Raw = g*beeta*(T2-T1)*w**3/(mu*alpha)        			#Rayleigh no.
Nuw = 0.42*(Raw)**0.25*Pr**0.012*(L/w)**-0.3   			#Nusslet no.
h = Nuw*k/w                                 			#kcal/h m**2 C, heat transfer coefficient
q = h*(T2-T1)*(L*b)                         			#W,the rate of heat transfer

# Results
print "the rate of heat  transfer is %.1f W"%(q)

the rate of heat  transfer is 13.4 W


## Example 5.7 Page No : 163¶

In :
import math
# Variables
Ts = 60.              			#C, surface temp
To = 30.              			#C, bulk temp.
d = 0.06             			#m, diameter  of pipe
l = 1.                			#m, length
Tm = (Ts+To)/2
#for air at Tm
rho = 1.105                     			#kg/m**3, density
cp = 0.24                       			#kcal/kg C. specific heat
mu = 1.95*10**-5                 			#kg/m s. vismath.cosity
P = 0.7                         			#Prandtl no.
kv = 1.85*10**-5                     			#m**2/s, kinetic vismath.cosity
k = 0.0241                      			#kcal/f m C, thermal conductivity
beeta = (1./(Tm+273))                  			#K**-1. coefficient of volumetric expension
V = 0.3                         			#m/s, velocity
g = 9.8                         			#m/s**2, gravitational consmath.tant

#Calculation
#from eq. 5.9
#calculation of forced convection nusslet no.
#from eq. 4.19
Re = d*V/(kv)
Nuforced = 0.3+(0.62*Re**(1./2)*P**(1./3)/(1+(0.4/P)**(2./3))**(1./4))*(1.+(Re/(2.82*10**5))**(5./8))**(4./5)
Nu = (Nuforced**3+Nufree**3)**(1./3)       			#nusslet no. for mixed convection
#Nu = h*d/k
h = Nu*k/d                     			#kcal/h m**2 C, heat transfer corfficient
q = h*math.pi*d*l*(Ts-To)

# Results
print "the rate of heat loss per meter length is %.1f kcal/h"%(q)

the rate of heat loss per meter length is 39.7 kcal/h