# Chapter 7 Natural Convection Systems¶

## Exa 7.1¶

In :
#Example Number 7.1
# constant heat flux from vertical plate

# Variable declaration

q_w = 800.0 			# [W/square meter] radiant energy flux
H = 3.5 			# [m] height of metal plate surface
W = 2 				# [m] width of metal plate
Ta = 30 			# [degree celsius] surrounding air temperature
# we treat this problem as one with constant heat flux on the surface since we 	do not know the surface temperature, we must make an estimate for determining 		Tf and the air properties.
h = 10 				# [W/square meter degree celsius]
dT = q_w/h 			# [degree celsius]
# then
Tf = (dT/2)+Ta 			# [degree celsius] approximately

#Calculation

# at Tf the properties of air are
v = 2.043*10**(-5) 		# [square meter/s]
k = 0.0295 			# [W/m degree celsius]
Pr = 0.7 			# prandtl number
Beta = 1.0/(Tf+273) 		# [K**(-1)]
# from equation (7-30), with
x = 3.5 			# [m]
g = 9.8 			# [square meter/s] acceleration due to gravity
Gr_x = (g*Beta*q_w*x**(4))/(k*v**(2))

# we may therefore use equation (7-32) to evaluate h_x

h_x = (k*0.17*(Gr_x*Pr)**(1.0/4.0))/x 	# [W/square meter degree celsius]

# in the turbulent heat transfer governed by equation (7-32), we note that
# Nu_x = h*x/k ~ (Gr_x)**(1/4) ~ x
# or h_x doest noy vary with x, and we may take this as the average value. the 	value of h

h = 5.41 			# [W/square meter degree celsius]
# is less than the approximate value we used to estimate Tf, recalculating dT, 	 we obtain
dT1 = q_w/h_x 			# [degree celsius]

# our new film temperature would be
Tf1 = Ta+dT1/2 			# [degree celsius]
# at Tf the properties of air are
v1 = 2.354*10**(-5) 		# [square meter/s]
k1 = 0.0320 			# [W/m degree celsius]
Pr1 = 0.695 			# prandtl number
Beta1 = 1/(Tf1+273) 		# [K**(-1)]

# then
Gr_x1 = (g*Beta1*q_w*x**(4))/(k1*v1**(2))
# and h_x is caalculated from
h_x1 = (k1*0.17*(Gr_x1*Pr1)**(1.0/4.0))/x 	# [W/square meter degree celsius]

# our new temperature difference is calculated as
dT2 = q_w/h_x1 			# [degree celsius]

# the average wall temperature is therefore
T_w_avg = dT2+Ta 		# [degree celsius]

#Result
print "The average wall temperature is",round(T_w_avg),"degree celsius"

The average wall temperature is 185.0 degree celsius


## Exa 7.2¶

In :
#Example Number 7.2
# heat transfer from isothermal vertical plate

# Variable declaration

H = 4.0 				# [m] height of vertical plate
Tp = 60.0 			# [degree celsius] plate temperature
Ta = 10.0 			# [degree celsius] atmospheric temperature
# we first determine the film temperature as
Tf = (Tp+Ta)/2 			# [degree celsius]
# the properties of interest are thus
v = 16.5*10**(-6) 		# [square meter/s]
k = 0.02685 			# [W/m degree celsius]
Pr = 0.7 			# prandtl number
Beta = 1/(Tf+273) 		# [K**(-1)]

#Calculation

g = 9.8 			# [square meter/s] acceleration due to gravity
# and
Gr_into_Pr = (g*Beta*(Tp-Ta)*H**(3)*Pr)/(v**(2))

# we then may use equation (7-29) to obtain
Nu_bar_root = (0.825+(0.387*(Gr_into_Pr)**(1.0/6.0))/(1+(0.492/Pr)**(9.0/16.0))**(8.0/27.0) )

Nu_bar = (Nu_bar_root)**(2)
# the heat transfer coefficient is
h_bar = Nu_bar*k/H 		# [W/square meter degree celsius]

# the heat transfer is
A = H*10 			# [square meter] for 10 m wide plate
q = h_bar*A*(Tp-Ta) 		# [W]

# as an alternative, we could employ the simpler relation
Nu = 0.1*(Gr_into_Pr)**(1/3)

#Result
print "Heat transfer if the plate is 10 m wide is",round(q),"W"

Heat transfer if the plate is 10 m wide is 9603.0 W


## Exa 7.3¶

In :
#Example Number 7.3
# heat transfer from horizontal tube in water

# Variable declaration

d = 0.02 			# [m] diameter of heater
Ts = 38 			# [degree celsius] surface temperature of heater
Tw = 27 			# [degree celsius] water temperature
# the film temperature is
Tf = (Ts+Tw)/2 			# [degree celsius]
# from appendix A the properties of water are
k = 0.630 			# [W/m degree celsius] thermal conductivity
# and the following term is particularly useful in obtaining the product GrPr 		product when it is multiplied by d**(3)*DT
# g*Beta*rho**(2)*Cp/(mu*k) = 2.48*10**(10) [1/m**(3) degree celsius]

K = 2.48*10**(10) 		# [1/m**(3) degree celsius]
Gr_into_Pr = K*(Ts-Tw)*d**(3)

# using table 7-1 (page number -328), we get

C = 0.53
m = 1/4
# so that

Nu = C*(Gr_into_Pr)**(1.0/4.0)
h = Nu*k/d 			# [W/sq m deg C] convection heat transfer coefficient
# the heat transfer is thus
import math

q_by_L = h*math.pi*d*(Ts-Tw) 	# [W/m]
print"free-convection heat loss per unit length of heater is",round(q_by_L),"W/m"

free-convection heat loss per unit length of heater is 443.0 W/m


## Exa 7.4¶

In [ ]:
#Example Number 7.4
# heat transfer from fine wire in air

# Variable declaration

d = 0.00002 			# [m] diameter of wire
L = 0.5 			# [m] length of wire whose temperature is maintained
Ts = 54.0 			# [degree celsius] surface temperature of wire
Pa = 101325.0 			# [Pa] pressure of air
Ta = 0 				# [degree celsius] temperature of air
# we first determine the film temperature as
Tf = (Ts+Ta)/2 			# [degree celsius]
# the properties of interest are thus
v = 15.69*10**(-6) 		# [square meter/s]
k = 0.02624 			# [W/m degree celsius]
Pr = 0.708 			# prandtl number
Beta = 1/(Tf+273) 		# [K^(-1)]
g = 9.8 			# [square meter/s] acceleration due to gravity
# and
Gr_into_Pr = (g*Beta*(Ts-Ta)*d**(3)*Pr)/(v**(2))
# from table 7-1 we find
C = 0.675
m = 0.058
# so that
Nu_bar = C*(Gr_into_Pr)**(m)
h_bar = Nu_bar*k/d 		# [W/square meter degree celsius]
# the heat required is
import math
A = math.pi*d*L 		# [square meter] surface area of wire
q = h_bar*A*(Ts-Ta) 		# [W]
print "Electric power  to maintain the the wire temperature if the length is 0.5 m is",round(q,3),"W"


## Exa 7.5¶

In :
#Example Number 7.5
# heated horizontal pipe in air

# Variable declaration

d = 0.3048 			# [m] diameter of pipe
Ts = 250.0 			# [degree celsius] surface temperature of pipe
Ta = 15.0 			# [degree celsius] temperature of air
# we first determine the Grashof-prandtl number product and then select the 	appropriate constants from table 7-1(page no.-328) for use with              	equation (7-25)
# the properties of air are evaluated at the film temperature:
Tf = (Ts+Ta)/2 			# [degree celsius]
# the properties of interest are thus
v = 26.54*10**(-6) 		# [square meter/s]
k = 0.03406 			# [W/m degree celsius]
Pr = 0.687 			# prandtl number
Beta = 1/(Tf+273) 		# [K**(-1)]
g = 9.8 			# [square meter/s] acceleration due to gravity
Gr_d_into_Pr = g*Beta*(Ts-Ta)*d**(3)*Pr/(v**(2))
# from table 7-1
C = 0.53
m = 1.0/4.0
Nu_d = C*(Gr_d_into_Pr)**(m)
h = Nu_d*k/d 			# [W/square meter degree celsius]
# the heat transfer per unit length is then calculated from
import math

q_by_L = h*math.pi*d*(Ts-Ta) 	# [W/m]
print "free-convection heat loss per unit length is",round(q_by_L/1000,1),"kW/m"

free-convection heat loss per unit length is 1.5 kW/m


## Exa 7.6¶

In :
#Example Number 7.6
# cube cooling in air

# Variable declaration

L = 0.2 				# [m] side length of cube
Ts = 60 				# [degree celsius] surface temperature of cube
Ta = 10 				# [degree celsius] air temperature
# this is an irregular solid so we use the information in the last entry of 	table 7-1(page no.-328) in the absence of a specific correlation for this    	geometry.
# the properties were evaluated as
v = 17.47*10**(-6)			 # [square meter/s]
k = 0.02685 				# [W/m degree celsius]
Pr = 0.70 				# prandtl number
Beta = 3.25*10**(-3) 			# [K**(-1)]
g = 9.8 				# [square meter/s] acceleration due to gravity
# the characteristic length is the distance a particle travels in the boundary 	layer, which is L/2 along the bottom plus L along the side plus L/2 on  the 	top or

#Calculation
Gr_into_Pr = (g*Beta*(Ts-Ta)*(2*L)**(3)*Pr)/(v**(2))
# from the last entry in table 7-1 we find
C = 0.52
n = 1.0/4.0
# so that
Nu = C*(Gr_into_Pr)**(n)
h_bar = Nu*k/(2*L) 			# [W/square meter degree celsius]
# the cube has six sides so the area is
A = 6*L**(2) 				# [square meter]
# the heat required is
q = h_bar*A*(Ts-Ta) 			# [W]

#Result
print "Heat transfer is",round(q,1),"W"

Heat transfer is 51.8 W


## Exa 7.7¶

In :
#Example Number 7.7
# calculation with simplified relations

#Variable declaration

# this example is calculation of heat transfer with simplified relations for 	example (7.5) so we use the data of example 7.5

d = 0.3048 			# [m] diameter of pipe
Ts = 250 			# [degree celsius] surface temperature of pipe
Ta = 15 			# [degree celsius] temperature of air
# we first determine the Grashof-prandtl number product and then select the 	appropriate constants from table 7-1 for use with equation (7-25)
# the properties of air are evaluated at the film temperature:

#Calculation

Tf = (Ts+Ta)/2 				# [degree celsius]
# the properties of interest are thus
v = 26.54*10**(-6)			# [square meter/s]
k = 0.03406 				# [W/m degree celsius]
Pr = 0.687 				# prandtl number
Beta = 1/(Tf+273) 			# [K**(-1)]
g = 9.8 				# [square meter/s] acceleration due to gravity
# in example (7.5) we found that a rather large pipe with a substantial 		temperature difference between the surface and air still had a GrPr product of  	1.57*10**(8)<10**(9), so laminar equation is selected from table 7-2(page 		no.-339). the heat transfer coefficient is given by
h = 1.32*((Ts-Ta)/d)**(1.0/4.0) 		# [W/square meter degree celsius]
# the heat transfer is then
import math

q_by_L = h*math.pi*d*(Ts-Ta) 		# [W/m]

#Result

print "heat transfer is",round(q_by_L/1000,2),"kW/m"

heat transfer is 1.57 kW/m


## Exa 7.8¶

In :
#Example Number 7.8
# heat transfer across vertical air gap

#Variable declaration

L = 0.5 			# [m] side length vertical square plate
d = 0.015 			# [m] distance between plates
p = 101325.0 			# [Pa] pressure of air
R = 287 			# [] universal gas constant
T1 = 100.0 			# [degree celsius] temperature of first plate
T2 = 40.0 			# [degree celsius] temperature of second plate
E = 0.2 			# emissivity of both surfaces
# the properties of air is evaluated at the mean temperature
Tf = (T1+T2)/2 			# [degree celsius]
rho = p/(R*(Tf+273)) 		# [Kg/m**(3)] density
k = 0.0295 			# [W/m degree celsius]
Pr = 0.70 			# prandtl number
Beta = 1/(Tf+273) 		# [K**(-1)]
mu = 2.043*10**(-5) 		# [Kg/m s] viscosity
g = 9.8 			# [square meter/s] acceleration due to gravity
# the Grashof-prandtl number product is now calculated as

Gr_into_Pr = (g*rho**(2)*Beta*(T1-T2)*(d)**(3)*Pr)/(mu**(2))

# we may now use eq(7-64) to calculate the effective thermal 		conductivity, with
L = 0.5 			# [m]
deli=0.015			# [m]
# and the constants taken from table 7-3:

#Calculation

Ke_by_K = 0.197*(Gr_into_Pr)**(1.0/4.0)*(L/deli)**(-1.0/9.0)
# the heat transfer may now be calculated with eq(7-54). the area is
A = L**(2) 			# [square meter]
q = Ke_by_K*k*A*(T1-T2)/deli 	# [W]
# the radiation flux is calculated with equation(7-67), taking
T1 = 373 			# [K]
T2 = 313 			# [K]
E1 = E
E2 = E
sigma = 5.669*10**(-8) 			# [W/square meter K**(4)]
q_A = sigma*(T1**(4)-T2**(4))/((1/E1)+(1/E2)-1) 	# [W/square meter]

#Result
print "Free-convection heat transfer across the air space is",round(q,2),"W"

Free-convection heat transfer across the air space is 39.64 W
Radiation heat transfer across the air space is 15.37 W


## Exa 7.9¶

In :
#Example Number 7.9
# heat transfer across horizontal air gap

# Variable declaration

a = 0.2 			# [m] side length of plate
d = 0.01 			# [m] seperation between two plates
p = 101325.0 			# [Pa] pressure of air
R = 287 			# [] universal gas constant
T1 = 100.0 			# [degree celsius] temperature of first plate
T2 = 40.0 			# [degree celsius] temperature of second plate
# the properties are the same as given in example(7.8)
Tf = (T1+T2)/2 			# [degree celsius]
rho = p/(R*(Tf+273)) 		# [Kg/m**(3)] density
k = 0.0295 			# [W/m degree celsius]
Pr = 0.70 			# prandtl number
Beta = 1/(Tf+273) 		# [K**(-1)]
mu = 2.043*10**(-5) 		# [Kg/m s] viscosity
g = 9.8 			# [sq m/s] acceleration due to gravity
# the GrPr product is evaluated on the basis of the separating distance, so we 	  have

#Calculation

Gr_into_Pr = (g*rho**(2)*Beta*(T1-T2)*(d)**(3)*Pr)/(mu**(2))
# consulting table 7-3(page no.-344) we find
C = 0.059
n = 0.4
m = 0
Ke_by_K = C*(Gr_into_Pr)**(n)*(a/d)**(m)
A = a**(2) 			# [square meter] area of plate
q = Ke_by_K*k*A*(T1-T2)/d 	# [W]

#Result

print "Heat transfer across the air space is",round(q,2),"W"

Heat transfer across the air space is 10.34 W


## Exa 7.10¶

In :
#Example Number 7.10
# heat transfer across water layer

# Variable declaration

L = 0.5 			# [m] length of square plate
d = 0.01 			# [m] seperation between square plates
T1 = 100			# [degree F] temperature of lower plate
T2 = 80 			# [degree F] temperature of upper plate

dT=5.0*(T1-T2)/9.0			#in Degree C
# we evaluate properties at mean temperature of 90 deg F and 				obtain, for water
k = 0.623 			# [W/m degree celsus]
# and the following term is particularly useful in obtaining the product GrPr
# g*Beta*rho**(2)*Cp/(mu*k) = 2.48*10**(10) [1/m**(3) degree celsius]
# the Grashof-prandtl number product is now evaluated using the plate spacing 		of 0.01 m as the characterstic dimension

#Calculation

K = 2.48*10**(10.0) 		# [1/m**(3) degree celsius]

Gr_into_Pr = K*(T1-T2)*(5.0/9.0)*d**(3.0)

# now, using equation 7-64 and consulting table 7-3(page no.-344) we obtain
C = 0.13
n = 0.3
m = 0.0
# therefore, equation (7-64) becomes
Ke_by_K = C*Gr_into_Pr**(n)

# the effectve thermal conductivity is thus
ke = k*Ke_by_K 			# [W/m degree celsius]

print ke
# and the heat transfer is
A = L**(2.0) 			# [square meter] area of plate
q = ke*A*(dT)/d 	# [W]

#Result
print "heat lost by the lower plate is ",round(q),"W"

3.47131600256
heat lost by the lower plate is  964.0 W


## Exa 7.11¶

In :
#Example Number 7.11
# reduction of convection in ar gap

# Variable declaration

Tm = 300.0 			# [K] mean temperature of air
dT = 20.0 			# [degree celsius] temperature difference
R = 287 			# [] universal gas constant
g = 9.8 			# [m/s**(2)] acceleration due to gravity
p_atm = 101325.0 			# [Pa] atmospheric pressure
# consulting table 7-13,we find that for gases, a value Grdeli_into_Pr<2000 is 	necessary to reduce the system to one of pure 	conduction.
# at 300 K the properties of air are
k = 0.02624 			# [W/m degree celsius]
Pr = 0.7 			# prandtl no.
mu = 1.846*10**(-5) 		# [Kg/m s]
Beta = 1.0/300.0

# we have
Grdel_into_Pr = 2000.0

# Part A for spacing of 1cm
import math
deli = 0.01 			# [m] spacing between plate
p = math.sqrt((Grdel_into_Pr*((R*Tm)**(2))*mu**(2))/(g*Beta*dT*deli**(3)*Pr)) 	# [Pa]
p=math.sqrt(7773/deli**3)

# or vacuum
vacuum = p_atm-p 		# [Pa]

# Part B for spacing of 2cm

deli1 = 0.02 			# [m] spacing between plate
p1 = math.sqrt(Grdel_into_Pr*(R*Tm)**(2)*mu**(2)/(g*Beta*dT*deli1**(3)*Pr)) # [Pa]
p1=math.sqrt(7773/deli1**3)
# or vacuum
vacuum1 = p_atm-p1 		# [Pa]

#Result
print "Vacuum necessary for glass spacings of 1 cm is",round(vacuum),"Pa"
print "Vacuum necessary for glass spacings of 2 cm is",round(vacuum1),"Pa"

Vacuum necessary for glass spacings of 1 cm is 13160.0 Pa
Vacuum necessary for glass spacings of 2 cm is 70154.0 Pa


## Exa 7.13¶

In :
#Example Number 7.13
# combined free and forced convection with air

# Variable declaration

p = 101325.0 			# [Pa] pressure of air
Ta = 27.0 			# [degree celsius] temperature of air
d = 0.025 			# [m] diameter of tube
u = 0.3 			# [m/s] velocity of air
Tw = 140.0 			# [degree celcius] temperature of tube wall
L = 0.4 			# [m] length of tube
R = 287 			# [] universal gas constant
# the properties of air are evaluated at the film temperature:
Tf = (Tw+Ta)/2 			# [degree celcius]
# the properties of interest are thus
kf = 0.0305 			# [W/m degree celcius]
Pr = 0.695 			# prandtl number
Beta = 1/(Tf+273) 		# [K**(-1)]
g = 9.8 			# [square meter/s] acceleration due to gravity
mu_f = 2.102*10**(-5) 		# [Kg/m s]
mu_w = 2.337*10**(-5) 		# [Kg/m s]

#Calculations

rho_f = p/(R*(Tf+273)) 		# [Kg/cubic meter]
# let us take the bulk temperature as 27 degree celsius for evaluating mu_b 	then
mu_b = 1.8462*10**(-5) 		# [Kg/m s]
# the significant parameters are calculated as
Re_f = rho_f*u*d/mu_f
Gr = rho_f**(2)*g*Beta*(Tw-Ta)*d**(3)/mu_f**(2)
Z = Gr*Pr*d/L 			# constant
# according to figure(7-14)(page no.-354), the mixed convection flow regime is 	encountered. thus we must use equation(7-77).
# The graetz number is calculated as
Gz = Re_f*Pr*d/L
# and the numerical calculation for equation(7-77) becomes
Nu = 1.75*(mu_b/mu_w)**(0.14)*(Gz+0.012*(Gz*Gr**(1.0/3.0))**(4.0/3.0))**(1.0/3.0)
# the average heat transfer coefficient is calculated as
h_bar = Nu*kf/d 		# [W/square meter degree celsius]

#Result

print "heat transfer coefficient is",round(h_bar,2),"W/square meter degree celsius"

heat transfer coefficient is 9.4 W/square meter degree celsius