# Chapter 6 Empirical and Practical Relations for Forced Convection Heat Transfer¶

## Exa 6.1¶

In [2]:
#Example Number 6.1
# turbulent heat transfer in a tube
# Variable declaration

p = 2*101325 			# [Pa] pressure of air
Ta = 200+273.15			# [K] temperature of air
d = 0.0254 			# [m] diameter of tube
R = 287 			# [] gas constant
u = 10 				# [m/s] velocity of air
dT = 20 			# [deg C] temperature difference between wall and air

# we first calculate the reynolds number to determine if the flow is laminar 		or turbulent, and then select the appropriate empirical correlation to 			calculate the heat transfer

# the properties of air at a bulk temperature of 473 K are

#Calculaiton

rho = p/(R*Ta) 			# [kg/cubic meter] density of gas
mu = 2.57*10**(-5)		# [kg/m s] viscosity
k = 0.0386 			# [W/m degree celsius]
Cp = 1025 			# [J/kg K]
Pr = 0.681 			# prandtl no.
Re_d = rho*u*d/mu 		# reynolds number

print "Reynolds number is",round(Re_d)
print"so that the flow is turbulent"

# we therefore use equation (6-4a) to calculate the heat transfer coefficient

Nu_d = 0.023*Re_d**(0.8)*Pr**(0.4) 	# nusselt no.
h = Nu_d*k/d 				# [W/m**2 deg C] heat transfer coefficient

# the heat transfer per unit length is then

import math
q_by_L = h*math.pi*d*(dT) 		# [W/m]
L = 3 					# [m]
# we can now make an energy balance to calculate the increase in bulk 		temperature in a 3 m length of tube :
# q = m_dot*Cp*dT_b = L*(q_byL)
m_dot = rho*u*math.pi*d**(2)/4 		# [kg/s] gas flow rate
# so that we insert the numerical values in the energy balance to obtain
dT_b = L*q_by_L/(m_dot*Cp) 		# [degree celsius]

#Result

print"Heat transfer per unit length is",round(q_by_L,1)," W/m"
print"Bulk temperature increase over the length of 3 m on tube is",round(dT_b,2)," degree C"

Reynolds number is 14749.0
so that the flow is turbulent
Heat transfer per unit length is 103.5  W/m
Bulk temperature increase over the length of 3 m on tube is 40.04  degree C


## Exa 6.3¶

In [12]:
#Example Number 6.3
# heating of air in laminar tube flow for constant heat flux

#Variable declaration

import math
p = 101325 			# [Pa] pressure of air
Ta = 27 			# [degree celsius] temperature of air
d = 0.005 			# [m] diameter of tube
R = 287 			# [] gas constant
u = 3 				# [m/s] velocity of air
L = 0.1 			# [m] length of tube
Tb = 77 			# [degree celsius] exit bulk temperature
# we first must evaluate the flow regime and do so by taking properties at the 	average bulk temperature
Tb_bar = (Ta+Tb)/2 		# [degree celsius]
v = 18.22*10**(-6) 		# [square meter/s] kinematic viscosity
k = 0.02814 			# [W/m degree celsius]
Cp = 1006 			# [J/kg K]
Pr = 0.703 			# prandtl no.
Re_d = u*d/v 			# reynolds number
print "Reynolds number is",round(Re_d)
print"so that the flow is laminar\n"

#the tube length is short, so we expect a thermal entrance effect and shall 		consult figure(6-5)
# the inverse Graetz number is computed as
Gz_inverse = L/(Re_d*Pr*d)
# therefore, for qw = constant, we obtain the nusselt number at exit from 		figure (6-5) as
Nu = 4.7
# the total heat transfer is obtained in terms of the overall energy balance
# at entrance
rho = 1.1774 			 # [kg/cubic meter] density
# mass flow is
m_dot = rho*math.pi*d**(2)*u/4 	 # [kg/s]
q = m_dot*Cp*(Tb-Ta) 		 # [W]
# thus we may find the heat transfer without the actually determining wall 		temperatures or values of h. However, to determine Tw we must compute qw for 		insertion in equation(b). we have
qw = q/(math.pi*d*L) 		 # [W]
# now
Tw = Tb+(qw*d/(Nu*k)) 		 # [degree celsius]
# and the heat transfer coefficient is
h = qw/(Tw-Tb) 			 # [W/square meter degree celsius]
print "Total heat transfer is",round(q,2),"W"
print "\nExit wall temperature is",round(Tw)," degree celsius"
print "\nHeat transfer coefficient is",round(h,2)," W/sq meter degree C"

Reynolds number is 823.0
so that the flow is laminar

Total heat transfer is 3.49 W

Exit wall temperature is 161.0  degree celsius

Heat transfer coefficient is 26.45  W/sq meter degree C


## Exa 6.4¶

In [4]:
#Example Number 6.4
# heating of air with isothermal tube wall

# Variable declaration

p = 101325 			# [Pa] pressure of air
Ta = 27				# [degree celsius] temperature of air
d = 0.005 			# [m] diameter of tube
R = 287 			# [] gas constant
u = 3 				# [m/s] velocity of air
L = 0.1 			# [m] length of tube
Tb = 77 			# [degree celsius] exit bulk temperature

# we first must evaluate the flow regime and do so by taking properties at the 	average bulk temperature

Tb_bar = (Ta+Tb)/2 		# [degree celsius]
v = 18.22*10**(-6) 		# [square meter/s] kinematic viscosity
k = 0.02814 			# [W/m degree celsius]
Cp = 1006 			# [J/kg K]
Pr = 0.703 			# prandtl no.
Re_d = u*d/v 			# reynolds number
print "Reynolds number is",round(Re_d)
print "so that the flow is laminar"
# so that the flow is laminar
# now we determine Nu_d_bar for Tw = constant. for Gz_inverse = 0.0346 we read
Nu_d = 5.15
# we thus calculate the average heat transfer coefficient as

h_bar = Nu_d*k/d 		# [W/square meter degree celsius]
# we base the heat transfer on a mean bulk temperature of Tb_bar, so that
import math
Tw = 3.49/(h_bar*math.pi*d*L)+Tb_bar 	# [degree celsius]

print "Exit wall temperature is",round(Tw,2),"degree celsius"

Reynolds number is 823.0
so that the flow is laminar
Exit wall temperature is 128.66 degree celsius


## Exa 6.5¶

In [5]:
#Example Number 6.5
# heat transfer in a rough tube

# Variable declaration

Tw = 90 			# [degree celsius] temperature of tube wall
d = 0.02 			# [m] diameter of tube
u = 3 				# [m/s] velocity of air
Tw_i = 40 			# [degree celsius] entering water temperature
Tw_f = 60 			# [degree celsius] leaving water temperature
Cp = 4.174*10**3 		# [J/kg K]

#Calculation

Tb_bar = (Tw_i+Tw_f)/2 		# [degree celsius]

#we first calculate the heat transfer from q = m_dot*Cp*dTb

import math
q = 989*3*math.pi*0.01**(2)*4174*(Tw_f-Tw_i) 	# [W]

# for the rough tube condition, we may employ the Petukhov relation, equation 		  (6-7) The mean film temperaturee is

Tf = (Tw+Tb_bar)/2 		# [degree celsius]

# and the fluid properties are

rho = 978 			# [kg/cubic meter] density of gas
mu = 4.0*10**(-4) 		# [kg/m s] viscosity
k = 0.664 			# [W/m degree celsius]
Pr = 2.54 			# prandtl no.

# also

mu_b = 5.55*10**(-4) 		# [kg/m s] viscosity
mu_w = 2.81*10**(-4) 		# [kg/m s] viscosity

# the reynolds number is thus

Re_d = rho*u*d/mu

# consulting figure(6-14), we find the friction factor as

f_f = 0.0218

# because Tw>Tb, we take

n = 0.11

# and obtain

Nu_d=((f_f*Re_d*2.54)/((1.07+12.7*(f_f/8)**(0.5)*(2.54**(2.0/3.0)-1))*8))*(mu_b/mu_w)**(n)
h = Nu_d*k/d 			# [W/square meter degree celsius]

# the tube length is obtained from energy balance

L = q/(h*math.pi*d*(Tw-Tb_bar))  # [m]

print "The length of tube necessary to accomplish the heating is",round(L,2),"m"

The length of tube necessary to accomplish the heating is 1.4 m


## Exa 6.6¶

In [6]:
#Example Number 6.6
# turbulent heat transfer in a short tube

#Variable declaration

p = 101325			 # [Pa] pressure of air
Ta = 300 			 # [K] temperature of air
d = 0.02 			 # [m] diameter of tube
u = 40 				 # [m/s] velocity of air
L = 0.1				 # [m] length of tube
dT = 5.0 				 # [degree celsius] rise in temperature
#	 we first must evaluate the air properties at 300 K
v = 15.69*10**(-6) 		 # [square meter/s] kinematic viscosity
k = 0.02624 			 # [W/m degree celsius]
Cp = 1006.0			 # [J/kg K]
Pr = 0.70 			# prandtl no.
rho = 1.18 			# [kg/cubic meter]
Re_d = u*d/v 			# reynolds number

print "Reynolds number is",round(Re_d)
print"so the flow is turbulent"

# consulting figure (6-6) for this value of Re_d and L/d = 5 we find
Nu_x_by_Nu_inf = 1.15
# or the heat transfer coefficient is about 15 percent higher that it would be 	for thermally developed flow.
# we calculate heat-transfer for developed flow using
Nu_d = 0.023*Re_d**(0.8)*Pr**(0.4)
# and
h = k*Nu_d/d 			 # [W/square meter degree celsius]
# increasing this value by 15 percent
h = 1.15*h 			 # [W/square meter degree celsius]
# the mass flow is
import math
Ac = math.pi*d**(2)/4 		 # [square meter]
m_dot = rho*u*Ac 		 # [kg/s]
# so the total heat transfer is
A = math.pi*d*L			 # [square meter]
q_by_A = m_dot*Cp*dT/A		 # [W/square meter]
print "The constant heat flux that must be applied at the tube surface to result in an air temperature rise of 5 degree celsius is",q_by_A," W/square meter"
Tb_bar = (Ta+(Ta+dT))/2 	 # [K]
Tw_bar = Tb_bar+q_by_A/h 	 # [K]
print "average wall temperature is",round(Tw_bar),"K"

Reynolds number is 50988.0
so the flow is turbulent
The constant heat flux that must be applied at the tube surface to result in an air temperature rise of 5 degree celsius is 11870.8  W/square meter
average wall temperature is 370.0 K


## Exa 6.7¶

In [7]:
#Example Number 6.7
# airflow across isothermal cylinder

# Variable declaration

p = 101325 			# [Pa] pressure of air
Ta = 35+273.15 			# [K] temperature of air
d = 0.05 			# [m] diameter of tube
R = 287 			# [] gas constant
u = 50 				# [m/s] velocity of air
Tc = 150+273.15 		# [degree celsius] cylinder temperature

# we first find the reynolds number and then find the applicable constants 	from table(6-2) for use with equation (6-17)
# the properties of air are evaluated at the film temperature:

#Calculation

Tf = (Ta+Tc)/2 			# [K]
rho_f = p/(R*Tf) 		# [kg/cubic meter]
mu_f = 2.14*10**(-5) 		# [kg/m s]
k_f = 0.0312 			# [W/m degree celsius]
Pr_f = 0.695 			# prandtl number
Re_f = rho_f*u*d/mu_f 		# reynolds number

# from table (6-2) table

C = 0.0266
n = 0.805

# so from equation (6-17)

h = C*(Re_f)**(n)*(Pr_f)**(1.0/3.0)*k_f/d   # [W/sq m deg C] heat transfer coefficient

# the heat transfer per unit length is
import math

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

#Result

print "Heat loss per unit length of cylinder is",round(q_by_L),"W/m"

Heat loss per unit length of cylinder is 3100.0 W/m


## Exa 6.8¶

In [8]:
#Example Number 6.8
# heat transfer from electrically heated

# Variable declaration

p = 101325 			# [Pa] pressure of air
Tw = 25+273.15 			# [K] temperature of air
d = 3.94*10**(-5) 		# [m] diameter of wire
R = 287 			# [] gas constant
u = 50 				# [m/s] velocity of air perpendicular to the air
Tr = 50+273.15 			# [degree celsius] rise in surface temperature
# we first obtain the properties at the film temperature :

#Calculation

Tf = (Tw+Tr)/2 			# [K]
v_f = 16.7*10**(-6) 		# [square meter/s]
k = 0.02704 			# [W/m degree celsius]
Pr_f = 0.706 			# prandtl number
Re_d = u*d/v_f 			# reynolds number
# the Peclet number is
Pe = Re_d*Pr_f
# and we find that equations (6-17),(6-21), or (6-19) apply.
# let us make the calculation with both the simplest expression, (6-17),and 		the most complex,(6-21), and compare results.
# using equation (6-17) with
C = 0.683
n = 0.466
# we have
Nu_d = 0.683*Re_d**(n)*Pr_f**(1/3)
# the value of heat transfer coefficient is
h = Nu_d*k/d 			 # [W/square meter degree celsius]
# the heat transfer per unit length is then
import math
q_by_L = math.pi*d*h*(Tr-Tw)	 # [W/m]
# using equation (6-21), we calculate the nusselt no as
Nu_d1=0.3+((0.62*Re_d**(1.0/2.0)*Pr_f**(1.0/3.0))/((1+(0.4/Pr_f)**(2.0/3.0))**(1.0/4.0))*((1+(Re_d/282000)**(5.0/8.0))**(4.0/5.0)))
h1 = Nu_d1*k/d 			 # [W/square meter degree celsius]
# and
q_by_L1 = h1*math.pi*d*(Tr-Tw) 	  # [W/m]

#Result

print "Heat lost per unit length by the wire is",round(q_by_L1,2),"W/m"

Heat lost per unit length by the wire is 11.88 W/m


## Exa 6.9¶

In [9]:
#Example Number 6.9
# heat transfer from sphere

# Variable declaration

p = 101325 			# [Pa] pressure of air
Ta = 27+273.15 			# [K] temperature of air
d = 0.012 			# [m] diameter of sphere
u = 4 				# [m/s] velocity of air
Ts = 77+273.15 			# [degree celsius] surface temperature of sphere
# consulting equation (6-30) we find that the reynolds number is evaluated at 		the free-stream temperature.
# we therefore need the following properties at Ta = 300.15K
v = 15.69*10**(-6) 		# [square meter/s]
k = 0.02624 			# [W/m degree celsius]
Pr = 0.708 			# prandtl number
mu_inf = 1.8462*10**(-5) 	# [kg/m s]
# at Ts = 350K
mu_w = 2.075*10**(-5) 		# [kg/m s]
Re_d = u*d/v 			# reynolds number

#Calculation
# from equation (6-30),
Nu_bar=2+((0.4)*(Re_d)**(1.0/2.0)+0.06*(Re_d)**(2.0/3.0))*(Pr**(0.4))*((mu_inf/mu_w)**(1.0/4.0))

# and

h_bar = Nu_bar*k/d 		# [W/sq m degree celsius] heat transfer coefficient

# the heat transfer is then

import math

A = (4*math.pi*d**(2))/4 	# [square meter] area of sphere

q = h_bar*A*(Ts-Ta) 		# [W]

# for comparison purposes let us also calculate the heat-transfer coefficient 		 using equation(6-25). the film temperature is
Tf = (Ta+Ts)/2 			# [K]
v_f = 18.23*10**(-6) 		# [square meter/s]
k_f = 0.02814 			# [W/m degree celsius]
# reynolds number is
Re_d1 = u*d/v_f
# from equation (6-25)
Nu_f = 0.37*(u*d/v_f)**(0.6)
# and h_bar is calculated as

h_bar = Nu_f*k_f/d 		# [W/sq m degree celsius]

#Result

print "Heat lost by the sphere is",round(q,3),"W"

Heat lost by the sphere is 1.554 W


## Exa 6.11¶

In [10]:
#Example Number 6.11
# alternate calculation method

# Variable declaration

# data for this example is taken from previous example (6-10)
# properties for use in equation (6-34) are evaluated at free-atream 	conditions of 10 degree celsius
v = 14.2*10**(-6)			 # [square meter/s]
k = 0.0249 				 # [W/m degree celsius]
Pr = 0.712 				 # prandtl number
Pr_w = 0.70 				# prandtl number
u = 7 					# [m/s] velocity of air
Sp = 0.0381 				# [m] spacing between normal and parallel 						direction to the flow
Sn = 0.0381 				# spacing between normal and parallel 					 direction to the flow
d = 0.0254 				# [m] diameter of tube
#maximum velocity is
u_max = u*(Sn/(Sn-d)) 			# [m/s]
# the reynolds number is
Re_d_max = u_max*d/v
# so that the constants for equation (6-34) are
C = 0.27
n = 0.63
# inserting values we obtain
h = C*Re_d_max**(n)*(Pr/Pr_w)**(1/4)*k/d 	# [W/sq m degree C] heat transfer 							coefficient
# multiplying by 0.92 from table 6-7 (page no.-300) to correct for only five 	tube rows gives
h = 0.92*h 					# [W/square meter degree celsius]
print "Heat transfer coefficient is",round(h,1),"W/square meter degree celsius"

h_in = 163.46432 			# [W/sq m deg C] from previous example

print "Heat transfer coefficient for previous problem is",round(h_in,1),"W/sq meter degree C"
P = (h-h_in)*100/h_in
print "Percentage increase in value of h is",round(P)

Heat transfer coefficient is 185.6 W/square meter degree celsius
Heat transfer coefficient for previous problem is 163.5 W/sq meter degree C
Percentage increase in value of h is 14.0


## Exa 6.12¶

In [11]:
#Example Number 6.12
# heating of liquid bismuth in tube

# variable declaration

m_dot = 4.5 			# [Kg/s] flow rate of bismuth
d = 0.05 			# [m] diameter of steel tube

Ti = 415 			# [degree celsius] initial temperature of bismuth
Tf = 440 			# [degree celsius] final temperature of bismuth
# because a constant heat flux is maintained, we may use equation 6-47 to 		calculate the heat transfer coefficient.
# the properties of bismuth are evaluated at the average bulk temperature of
#Calculation

Ta = (Ti+Tf)/2 			# [degree celsius]
mu = 1.34*10**(-3) 		# [Kg/m s] viscosity
Cp = 149 			# [J/Kg degree celsius] heat
k = 15.6 			# [W/m degree celsius]
Pr = 0.013 			# prandtl number
# the total transfer is calculated from
q = m_dot*Cp*(Tf-Ti) 		# [W]
# we calculate reynolds and peclet number as

import math
G = m_dot/(math.pi*d**(2)/4)
Re_d = d*G/mu
Pe = Re_d*Pr
# the heat transfer coefficient is calculated from equation 6-47
Nu_d = 4.82+0.0185*Pe**(0.827)
h = Nu_d*k/d 				# [W/square meter degree celsius]
# the total required surface area of the tube may now be computed from q=h*A*DT
# where we use the temperature difference of
DT = 20 			# [degree celsius]
A = q/(h*DT) 			# [square meter]
# the area in turn can be expressed in terms of tube length
L = A/(math.pi*d) 		# [m]

#Result
print"Length of tube required to effect the heat transfer is",round(L,2),"m"

Length of tube required to effect the heat transfer is 1.56 m