# Chapter 9 : Natural Convection¶

### Example 9.1¶

In :
import math
# Heat Loss from Hot Water Pipes

# Variables
l = 6	    		#Length[m]
d = 0.08			#diameter[m]
T_room = 20			#[degree Celcius]
Ts = 70		    	#Surface temperature of pipe[degree Celcius]
Tf = (Ts+T_room)/2			#Film temperature[degree Celcius]
#Properties of air at Tf
k = 0.02699			#[W/m.degree Celcius]
Pr = 0.7241			#Prandtl number
nu = 1.750*10**(-5)			#[m**2/s]
b = (1/(Tf+273))			#[K**-1]
g = 9.81		    	#Acc dur to gravity[m/s**2]
e = 1		    	#Emissivity

# Calculations and Results
Lc = d			#Characteristic length[m]
Ra_d = g*b*(Ts-T_room)*(d**3)*Pr/(nu**2);
print "The Rayleigh number is",Ra_d

Nu = ((0.6+((0.387*(Ra_d**(1./6)))/((1+((0.559/Pr)**(9./16)))**(8./27))))**2);
print " The natural convection Nusselt number is",Nu

h = k*Nu/d			#[W/m**2.degree Celcius]
As = math.pi*d*l			#[m**2]
Q = h*As*(Ts-T_room)			#[W]
print "The pipe loses heat to the air in the room at a rate of",round(Q),"W","by natural convection"

Q_rad = e*As*(5.76*10**(-8))*(((Ts+273)**4)-((T_room+273)**4))			#[W]
print "The radiation heat transfer is",Q_rad,"W"

The Rayleigh number is 0.0
The natural convection Nusselt number is 0.36
The pipe loses heat to the air in the room at a rate of 9.0 W by natural convection
The radiation heat transfer is 562.083528514 W


### Example 9.2¶

In :
import math
# Cooling of a Plate in different orientaions

# Variables
L = 0.6			#side of square plate[m]
T_surr = 30.			#[degree Celcius]
Tp = 90.			#Temp of plate[degree Celcius]
Tf = (Tp+T_surr)/2			#Film temperature[degree Celcius]
#Properties of air at Tf
k = 0.02808			#[W/m.degree Celcius]
Pr = 0.7202			#Prandtl number
nu = 1.896*10**(-5)			#Kinematic vismath.cosity[m**2/s]
b = 1./(Tf+273)			#[K**-1]
g = 9.81			#Acc due to gravity[m/s**2]

# Calculations and Results
Lc_a = L			#Characteristic length
Ra_1 = g*b*(Tp-T_surr)*(L**3)*Pr/(nu**2);
print "(a) The Rayleigh no is",Ra_1

Nu_a = ((0.825+(0.387*(Ra_1**(1./6)))/((1+((0.492/Pr)**(9./16)))**(8./27)))**2);
print "The natural convection Nusselt number is",Nu_a

h_a = k*Nu_a/L			#[W/m**2.degree Celcius]
As = L**2			    #[m**2]
Q_a = h_a*As*(Tp-T_surr)			#[W]
print "Heat loss to the surrounding is",Q_a,"W"

#Solution (b)
Lc_b = As/(4*L)			#[m]
Ra_2 = g*b*(Tp-T_surr)*(Lc_b**3)*Pr/(nu**2);
print "(b) The Rayleigh number is",Ra_2

Nu_b = 0.54*(Ra_2**(1/4));
print "The natural convection Nusselt number is",Nu_b

h_b = k*Nu_b/Lc_b			        #[W/m**2.degree Celcius]
Q_b = h_b*As*(Tp-T_surr)			#[W]
print "Heat Loss is",round(Q_b),"W"

#Solution (c)
Lc_c = Lc_b
Nu_c = (0.27*Ra_2**(1/4));
print "(c) Natural convection Nusselt number",Nu_c

h_c = k*Nu_c/Lc_c			        #[W/m**2.degree Celcius]
Q_c = h_c*As*(Tp-T_surr)			#[W]
print "Heat Loss is",Q_c,"W"

Q_rad = e*(5.67*10**(-8))*As*(((Tp+273)**4)-((T_surr+273)**4))			#[W]
print "Radiation heat loss is",round(Q_rad),"W"

(a) The Rayleigh no is 764902757.268
The natural convection Nusselt number is 113.344693409
Heat loss to the surrounding is 114.577883674 W
(b) The Rayleigh number is 11951605.5823
The natural convection Nusselt number is 0.54
Heat Loss is 2.0 W
(c) Natural convection Nusselt number 0.27
Heat Loss is 1.0917504 W
Radiation heat loss is 182.0 W


### Example 9.3¶

In :
import math
# Optimum Fin Spacing of a Heat Sink

# Variables
w = 0.12			#width[m]
l = 0.18			#length[m]
t = 0.001			#thickness[m]
H = 0.024			#height[m]
Ts = 80.	    		#Bast temperature[degree Celcius]
T_surr = 30.		#[degree Celcius]
Tf = (Ts+T_surr)/2			#[degree Celcius]
#Properties of air at film temperature
k = 0.02772			#[W/m.degree Celcius]
Pr = 0.7215			#Prandtl number
nu = 1.847*10**(-5)			#[m**2/s]
b = 1/(Tf+273)			#[K**-1]
g = 9.81			#[m/s**2]

# Calculations and Results
Ra_l = g*b*(Ts-T_surr)*(l**3)*Pr/(nu**2);
print "The Rayleigh number is",Ra_l

S_opt = 2.714*l/(Ra_l**(0.25))			#[m]
print "The optimum spacing is",S_opt*100,"mm"

n = w/(S_opt+t);
print "The no of for this optimum fin spacing are",round(n)

Nu_opt = 1.307      		    	#Optimum Nusselt number
h = Nu_opt*k/S_opt	        		#[W/m**2.degree Celcius]
Q = h*2*round(n)*l*H*(Ts-T_surr)			#[W]
print "The rate of natural convection heat transfer",Q,"W"

The Rayleigh number is 18445278.9656
The optimum spacing is 0.745437165061 mm
The no of for this optimum fin spacing are 14.0
The rate of natural convection heat transfer 29.3947353567 W


### Example 9.4¶

In :
import math
# Heat Loss through a Double Pane Window

# Variables
H = 0.8 			#Height[m]
L = 0.02			#Air gap[m]
w = 2.	    		#Width[m]
T1 = 12.
T2 = 2.		    	#Glass Surface temperatures across the air gap
Tavg = (T1+T2)/2	#[degree Celcius]
k = 0.02416			#[W/m.degree Celcius]
Pr = 0.7344			#Prandtl Number
nu = 1.4*10**(-5)	#Kinematic Vismath.cosity[m**2/s]
g = 9.81			#[m/s**2]

# Calculations and Results
Lc = L			#Characteristic length
b = 1/(Tavg+273)			#[K**-1]
Ra_L = g*b*(T1-T2)*Pr*(Lc**3)/(nu**2);
print "The Rayleigh Number is",Ra_L

Nu = 0.42*(Ra_L**(1./4))*(Pr**(0.012))*((H/L)**(-0.3));
print "The Nusselt Number is",Nu

As = H*w			#[m**2]
h = k*Nu/L			#[W/m**2.degree Celcius]
Q = h*As*(T1-T2);
print "Rate at which Heat is Lost through the window is",Q,"W"

The Rayleigh Number is 10502.1341108
The Nusselt Number is 1.40068950621
Rate at which Heat is Lost through the window is 27.0725267761 W


### Example 9.5¶

In :
import math
# Heat Transfer through a Spherical Enclosure

# Variables
Di = 0.2			#Inner Diameter[m]
Do = 0.3			#Outer Diameter[m]
Ti = 320
To = 280			#The surface temperatures of two spheres enclomath.sing the air[K]
Tavg = (Ti+To)/2.			#[K]
#Properties at Tavg
k = 0.02566			#[W/m.K]
Pr = 0.7290			#Prandtl Number
nu = 1.58*10**(-5)			#[m**2/s]
b = (1/Tavg);
g = 9.81			#[m/s**2]

# Calculations and Results
Lc = (Do-Di)/2			#Characteristic length[m]
Ra_L = g*b*(Ti-To)*(Lc**3)*Pr/(nu**2);
print "The Rayleigh Number is",Ra_L

Fsph = Lc/(((Di*Do)**4)*((((Di**(-7./5))+(Do**(-7./5))))**5));
keff = 0.74*k*((Pr/(0.861+Pr))**(1./4))*((Fsph*Ra_L)**(1./4))			#[W/m.K]
print (Fsph,keff)

Q = keff*(math.pi*Di*Do/Lc)*(Ti-To)			#[W]
print "The rate of heat transfer between the spheres is",Q,"W"

The Rayleigh Number is 477453.533088
(0.0052291424278198, 0.11044889324463623)
The rate of heat transfer between the spheres is 16.6553007175 W


### Example 9.6¶

In :
# Heating Water in a Tube by Solar Enegy
import math

# Variables
Ts = 40.        		#Glass Temp[degree Celcius]
T_surr = 20.			#Surrounding temperature[degree Celcius]
Tavg = (Ts+T_surr)/2			#[degree Celcius]
Do = 0.1		    	#[m]
Di = 0.05			    #[m]
L = 1			        #[m]
#Properties of glass at Tavg
k = 0.02588 			#[W/m.degree Celcius]
Pr = 0.7282	    		#Prandtl Number
nu = 1.608*10**(-5)			#[m**2/s]
b = 1/(Tavg+273)			#[K**-1]

Q = 30			#Rate pof absorpto\ion of solar radiation[W]
g = 9.81			#[m/s**2]

# Calculations and Results
Ao = math.pi*Do*L			#Heat transfer surface area of the glass cover[m**2]
Ra_D = g*b*(Ts-T_surr)*(Do**3)*Pr/(nu**2);
print "The Rayleigh Number is",Ra_D

Nu = ((0.6+((0.387*(Ra_D**(1/6)))/((1+((0.550/Pr)**(9/16)))**(8/27))))**2);
print "The Nusselt number is",Nu

ho = k*Nu/Do			#[W/m**2.degree Celcius]
Qo = ho*Ao*(Ts-T_surr)			#[W]
print "The rate of natural convection heat transfer from the glass cover to the ambient air is",Qo,"W"

#Value of Qo is less than 30W so assuming a higher temp of glass cover
T_surr1 = 41    	    		#[degree Celcius]
Ts1 = 90    	    	    	#[degree Celcius]
Tavg1 = (T_surr1+Ts1)/2			#[degree Celcius]
b1 = 1./(Tavg1+273)			#[K**-1]
Lc = (Do-Di)/2			#Characteristic length [m]
Ra_L1 = g*b1*(Ts1-T_surr1)*(Lc**3)*Pr/(nu**2);
print "The Rayleigh number on assuming higher temperatures",Ra_L1

Fcyl = ((math.log(Do/Di))**4)/((Lc**3)*(((Di**(-3./5))+(Do**(-3./5)))**5));
keff = 0.386*k*((Pr/(0.861+Pr))**(1./4))*((Fcyl*Ra_L1)**(1./4))			#[W/m.degree Celcius]
Q1 = 2*math.pi*keff*(Ts1-T_surr1)/(math.log(Do/Di))         			#[W]
print "The rate of heat transfer between the cylinders is",Q1,"W"

#Obtained value of Q1 is more than 30 W, so umath.sing hit and trial aand suuming more values we get the tube temperature to be 82 degree Celcius,
print ("Therefore tube will reach an equilibrium temperature of 82 degree Celcius when the pump fails")

The Rayleigh Number is 1823622.57723
The Nusselt number is 0.974169
The rate of natural convection heat transfer from the glass cover to the ambient air is 1.58408486914 W
The Rayleigh number on assuming higher temperatures 62581.6484932
The rate of heat transfer between the cylinders is 35.7293537245 W
Therefore tube will reach an equilibrium temperature of 82 degree Celcius when the pump fails


### Example 9.7¶

In :
import math
# U factor for Center of glass Section of Windows

# Variables
e = 0.84			#Emissivity
#For winter season
hi = 8.29			#[W/m**2.degree Celcius]
ho = 34.0			#[W/m**2.degree Celcius]

# Calculations
e_eff = 1./((1/e)+(1/e)-1)			#Effective emissivity of air space
#the effective emissivity and an average air space temperature of 0 degree Celcius read
h_space = 7.2           			#[W/m**2.degree Celcius]
U_center = 1/((1/hi)+(1/ho)+(1/h_space))			#[W/m**s.degree Celcius]

# Results
print "The center of glass U-factor value is",U_center,"W/m**2.degree Celcius"

The center of glass U-factor value is 3.46107089988 W/m**2.degree Celcius


### Example 9.8¶

In :
import math
# Heat Loss through Aluminium Framed Windows

# Variables
H = 1.2			#Height[m]
w = 1.8			#Width[m]
Ti = 22			#Inside temp[degree Celcius]
To = -10		#Outside temp[degree Celcius]
U_a = 6.63
U_b = 3.51
U_c = 1.92
hi = 8.3		#[W/m**.degree Celcius]

# Calculations and Results
A_win = h*w			                #[m**2]
Q_win_a = U_a*A_win*(Ti-To)			#[W]
T_glass_a = Ti-(Q_win_a/(hi*A_win))			#[degree Celcius]
print "(a) The Inner surface temperature of the window glass is",T_glass_a,"degree Celcius"

Q_win_b = U_b*A_win*(Ti-To)			#[W]
T_glass_b = Ti-(Q_win_b/(hi*A_win))			#[degree Celcius
print "(b) The Inner surface temperature of the window glass is",T_glass_b,"degree Celcius"

Q_win_c = U_c*A_win*(Ti-To)			#[W]
T_glass_c = Ti-(Q_win_c/(hi*A_win))			#[degree Celcius]
print "(c) The Inner surface temperature of the window glass is",T_glass_c,"degree Celcius"

(a) The Inner surface temperature of the window glass is -3.56144578313 degree Celcius
(b) The Inner surface temperature of the window glass is 8.46746987952 degree Celcius
(c) The Inner surface temperature of the window glass is 14.5975903614 degree Celcius


### Example 9.9¶

In :
import math
# U-Factor of a Double-Door Window

# Variables
A_win = 1.8*2.0			        #[m**2]
A_glazing = 2*1.72*0.94			#[m**2]
U_c = 3.24
U_e = 3.71
U_f = 2.8			#U factors for the center edge and frame sections respectively [W/m**2.degree Celcius]

# Calculations
A_frame = A_win-A_glazing		            	#[m**2]
A_center = 2*(1.72-0.13)*(0.94-0.13)			#[m**2]
A_edge = A_glazing-A_center			            #[m**2]
U_win = ((U_c*A_center)+(U_e*A_edge)+(U_f*A_frame))/A_win			#[W/m**2.degree Celcius]

# Results
print "The overall U factor of the entire window is",U_win,"W/m**2.degree Celcius"

The overall U factor of the entire window is 3.28109722222 W/m**2.degree Celcius