Chapter 5 Principles of Convection¶

Exa 5.1¶

In [1]:
#Example 5.1
# water flow in a diffuser

#VARIABLE DECLARATION
Tw = 20 			# [degree celcius] water temperature
m_dot = 8 			# [kg/s] water flow rate
d1 = 0.03 			#[m] diameter at section 1
d2 = 0.07 			# [m] diameter at section 2

#CALCULATION
import math
A1 = math.pi*d1**(2)/4 		# [square meter] cross-sectional area at section 1
A2 = math.pi*d2**(2)/4 		# [square meter] cross-sectional area at section  2
gc = 1 				# [m/s**(2)] acceleration due to gravity
rho = 1000 			# [kg/cubic m] density of water at 20 degree celcius

# calculate the velocities from the mass-continuity relation
u1 = m_dot/(rho*A1) 		# [m/s]
u2 = m_dot/(rho*A2) 		# [m/s]
# the pressure difference is obtained by Bernoulli equation(5-7a)
p2_minus_p1 = rho*(u1**(2)-u2**(2))/(2*gc)	 # [Pa]

#RESULTS
print"the increase in static pressure between sections 1 and 2 is:",round(p2_minus_p1/1000,2)," kPa"

the increase in static pressure between sections 1 and 2 is: 61.88  kPa


Exa 5.2¶

In [2]:
#Example Number 5.2
# isentropic expansion of air

# Variable declaration

Ta = 300.0+273.0	 		# [K] air temperature
Pa = 0.7 				# [MPa] pressure of air
u2 = 300 				# [m/s] final velocity
gc = 1 					# [m/s^(2)] acceleration due to gravity
Y = 1.4 				# gama value for air
Cp = 1005 				# [J/kg degree celsius]
#the initial velocity is small and the process is adiabatic. in terms of 		temperature

#Calculation

T2 = Ta-u2**(2)/(2*gc*Cp)

#Result
print "The static temperature is:",round(T2,1),"K"

# we may calculate the pressure difference from the isentropic relation

p2 = Pa*((T2/Ta)**(Y/(Y-1)))

print "Static pressure is:",round(p2,1),"MPa"

# the velocity of sound at condition 2 is
a2 = (20.045*(T2**(0.5))) 		# [m/s]
#so that the mach no. is
M2 = u2/a2
print "Mach number is:",round(M2,3)

The static temperature is: 529.0 K
Static pressure is: 0.5 MPa
Mach number is: 0.651


Exa 5.4¶

In [3]:
#Example Number 5.4
#Calculate the heat transfereed in first 20 cm of the plate and the first 40 cm of the  plate

# Variable declaration

# total heat transfer over a certain length of the plate is desired, so we 		wish to calculate average heat transfer coefficients.
# for this purpose we use equations (5-44) and (5-45), evaluating the 		properties at the film temperature :
Tp = 60+273.15 				# [K] plate temperature
Ta = 27+273.15 				# [K] air temperature
Tf = (Tp+Ta)/2 				# [K]
u = 2 					# [m/s] air velocity

# from appendix A the properties are

v = 17.36*(10**(-6)) 			# [square meter/s] kinematic viscosity
x1 = 0.2 				# [m] distance from the leading edge of plate
x2 = 0.4 				# [m] distance from the leading edge of plate
k = 0.02749 				# [W/m K] heat transfer coefficient
Pr = 0.7 				# prandtl number
Cp = 1006 				# [J/kg K]

# at x = 0.2m

#Calculation

Re_x1 =(u*x1/v) 				# reynolds number
Nu_x1 = 0.332*(Re_x1**(0.5))*(Pr**(0.333)) 	# nusselt number
hx1 = Nu_x1*k/x1 				# [W/square meter K]

# the average value of the heat transfer coefficient is twice this value, or

h_bar1 = 2*hx1 					# [W/square meter K]

# the heat flow is

A1 = x1*1 					# [square meter] area for unit depth
q1 = h_bar1*A1*(Tp-Ta) 				# [W]

# at x = 0.4m

Re_x2 = u*x2/v 					# reynolds number
Nu_x2 = 0.332*Re_x2**(0.5)*Pr**(0.333) 		# nusselt number
hx2 = Nu_x2*k/x2 				# [W/square meter K]

# the average value of the heat transfer coefficient is twice this value, or

h_bar2 = 2*hx2 					# [W/square meter K]
# the heat flow is
A2 = x2*1 					# [square meter] area for unit depth
q2 = h_bar2*A2*(Tp-Ta) 				# [W]

#Result

print"The heat transfered in first case of the plate is",round(q1,2),"W"
print"and the heat transfered in second case of the plate is:",round(q2,1),"W"

The heat transfered in first case of the plate is 81.2 W
and the heat transfered in second case of the plate is: 114.8 W


Exa 5.5¶

In [4]:
#Example Number 5.5
# Calculate the av temperature difference along the plate & Temperature diff at the trailing edge

#Variable declaration

u = 5 					# [m/s] air velocity
l = 0.6 				# [m] plate length
Ta = 27+273.15 				# [K] temperature of airstream

# properties should be evaluated at the film temperature, but we do not know 	the plate temperature so for an initial calculation we take the properties at 	the free-stream conditions of

v = 15.69*10**(-6)	 		#[square meter/s] kinematic viscosity
k = 0.02624 				#[W/m deg celsius] heat transfer coefficient
Pr = 0.7 				# prandtl number
Re_l = l*u/v 				# reynolds number
P = 1000 				# [W] power of heater
qw = P/l**(2) 				# [W/square meter] heat flux per unit area

# from equation (5-50) the average temperature difference is

#Calculation

Tw_minus_Tinf_bar = qw*l/(0.6795*k*(Re_l)**(.5)*(Pr)**(0.333)) 	 # [degree celsius]

# now, we go back and evaluate properties at
Tf = (Tw_minus_Tinf_bar+Ta+Ta)/2 	# [degree celsius]

# and obtain

v1 = 28.22*10**(-6) 			# [square meter/s] kinematic viscosity
k1 = 0.035 				# [W/m deg celsius] heat transfer coefficient
Pr1 = 0.687 				# prandtl number
Re_l1 = l*u/v1 				# reynolds number
Tw_minus_Tinf_bar1 = qw*l/(0.6795*k1*(Re_l1)**(0.5)*(Pr1)**(0.333)) #[degree celsius]

# at the end of the plate(x = l = 0.6m) the temperature difference is obtained 	from equation (5-48) and (5-50) with the constant of 0.453

Tw_minus_Tinf_x_equal_l = Tw_minus_Tinf_bar1*0.6795/0.453 	# [degree celsius]

#Result

print "Average temperature difference along the plate is:",round(Tw_minus_Tinf_bar)," degree celsius"
print "Temperature difference at the trailing edge is:",round(Tw_minus_Tinf_x_equal_l,1),"degree celsius"

Average temperature difference along the plate is: 241.0  degree celsius
Temperature difference at the trailing edge is: 365.3 degree celsius


Exa 5.7¶

In [5]:
#Example Number 5.7
# Calculate the heat lost by the plate

# Variable declaration

u = 1.2 			# [m/s] oil velocity
l = 0.2 			# [m] plate length as well as width (square)
To = 20+273.15 			# [K] temperature of engine oil
Tu = 60+273.15 			# [K] uniform temperature of plate
# First we evaluate the film temperature
T = (To+Tu)/2 			# [K]
# and obtain the properties of engine oil are
rho = 876 			# [kg/cubic meter] density of oil
v = 0.00024 			# [square meter/s] kinematic viscosity
k = 0.144 			# [W/m degree celsius] heat transfer coefficient
Pr = 2870 			# prandtl number
# at the trailing edge of the plate the reynolds number is

#Calculation

Re = l*u/v 			# reynolds number

# because the prandtl no. is so large we will employ equation(5-51) for the 		solution.

# we see that hx varies with x in the same fashion as in equation(5-44) , i.e. 	hx is inversely proportional to the square root of x ,
# so that we get the same solution as in equation(5-45) for the average heat 		transfer coefficient.

# evaluating equation(5-51) at x = 0.2m gives

Nux = (0.3387*(Re**(1.0/2.0))*(Pr**(1.0/3.0)))/((1+(0.0468/Pr)**(2.0/3.0))**(1.0/4.0))

hx = Nux*k/l 			# [W/sq m degree celsius]  heat transfer coefficient

# the average value of the convection coefficient is

h = 2*hx 			# [W/square meter degree celsius]

# so that total heat transfer is

A = l**(2) 			# [square meter] area of the plate
q = h*A*(Tu-To) 		#[W]

print "Average value of the convection coefficient is",round(h,1),"W/sq meter degree celsius"
print " and the heat lost by the plate is",round(q,1),"W"

Average value of the convection coefficient is 219.1 W/sq meter degree celsius
and the heat lost by the plate is 350.6 W


Exa 5.8¶

In [6]:
#Example Number 5.8
# Compute the drag force on the first 40 cm of the plate

# Variable declaration

# data is used from example 5.4
# we use equation (5-56) to compute the friction coefficient and then 		calculate the drag force .
# an average friction coefficient is desired, so st_bar*pr**(2/3) = Cf_bar/2

p = 101325 			# [Pa] pressure of air
x = 0.4				#[m] drag force is computed on first 0.4 m of the 			                  plate
R = 287 			# []
Tf = 316.5 			# [K]
u = 2 				# [m/s] air velocity
Cp = 1006 			# [J/kg K]
Pr = 0.7 			# prandtl no.
rho = p/(R*Tf) 			# [kg/cubic meter] density at 316.5 K
h_bar = 8.698 			# [W/square meter K]  heat transfer coefficient

#Calculation
# for the 0.4m length

st_bar = h_bar/(rho*Cp*u)

# then from equation (5-56)

Cf_bar = st_bar*Pr**(2.0/3.0)*2

# the average shear stress at the wall is computed from equation(5-52)

tau_w_bar = Cf_bar*rho*u**(2)/2 	# [N/square meter]
A = x*1 				# [square meter] area per unit length

# the drag force is the product of this shear stress and the area,

D = tau_w_bar*A 			# [N]

#Result

print "Drag force exerted on the first 0.4 m of the plate is",round(D*1000,2),"mN"

Drag force exerted on the first 0.4 m of the plate is 5.45 mN


Exa 5.9¶

In [7]:
#Example Number 5.9
# Calculate the heat transfer from the plate

# Variable declaration

p = 101320.0 				# [Pa] pressure of air
R = 287.0 				# []
Ta = 20+273 				# [K] temperature of air
u = 35 					# [m/s] air velocity
L = 0.75 				# [m] length of plate
Tp = 60+273 				# [K] plate temperature

# we evaluate properties at the film temperature

#Calculations

Tf = (Ta+Tp)/2 				# [K]

rho = (p/(R*Tf)) 			# [kg/cubic meter]

mu = 1.906*(10**(-5)) 			# [kg/m s] viscosity
k = 0.02723 				# [W/m degree celsius]
Cp = 1007 				# [J/kg K]
Pr = 0.7 				# prandtl no.

# the reynolds number is

Rel = (rho*u*L)/mu
Rel=round(Rel)

# and the boundary layer is turbulent because the reynolds number is 			greater than 5*10**(5).
# therefore, we use equation(5-85) to calculate the average heat 			transfer over the plate:

Nul_bar = (Pr**(1.0/3.0))*(0.037*(Rel**(0.8))-871)

print "Nul_bar is",round(Nul_bar)
A = L*1 				# [square meter] area of plate per unit depth
h_bar = Nul_bar*k/L  			# [W/square meter degree celsius]
q = h_bar*A*(Tp-Ta) 			# [W] heat transfer from plate

#Result

print "Heat transfer from plate is",round(q),"W"

Nul_bar is 2175.0
Heat transfer from plate is 2369.0 W


Exa 5.10¶

In [8]:
#Example Number 5.10
# Calculate turbulent-boundary-layer thickness at the end of plate

# Variable declaration

# we have to use the data from example 5.8 and 5.9
Rel = 1.553*10**6 					# from previous example
L = 0.75 						# [m] length of plate
# it is a simple matter to insert this value in equations(5-91) and 		(5-95) 	along with
x = L 							# [m]
# turbulent-boundary-layer thickness are
# part a.  from the leading edge of the plate

#Calculation

del_a = x*0.381*Rel**(-0.2) 				# [m]
# part b   from the transition point at Recrit = 5*10**(5)

del_b = x*0.381*Rel**(-0.2)-10256*Rel**(-1) 		# [m]

#Result

print "Turbulent-boundary-layer thickness at the end of the plate from the leading edge of the plate is",round(del_a*1000,1),"mm"
print "Turbulent-boundary-layer thickness at the end of the plate from the transition point at Re_crit = 5*10**(5) is",round(del_b*1000,1)," mm"

Turbulent-boundary-layer thickness at the end of the plate from the leading edge of the plate is 16.5 mm
Turbulent-boundary-layer thickness at the end of the plate from the transition point at Re_crit = 5*10**(5) is 9.9  mm