# Chapter 11 : Boundary layer heat transfer¶

## Example 11.1 Page No : 478¶

In :
import math
#Variable
v  = 1. 			#m/s
#temprature
T = 25.            			# degree celcius
#length of plate,l = 1m
l = 1.             			#m
#width of plate,w = 0.5m
w = 0.5 			#m
#angle of incidence,theta = 0 degree
theta = 0.         			#degree

#Calculation
#for water at 25 degree celcius ,momentum diffusivity,
MD = 8.63*(10**-7)  			# m**2/s
#local Reynold no.
x = 0.5 			#m
Re = x*v/MD
#from Eq. 11.39,the boundary layer thickness is
t = 5*x/(Re**0.5)

#Results
print  "i) Boundary layer thickness is %.4f  m"%(t)

#local drag coefficient
#CD = local drag force per unit area (F)/kinetic energy per unit volume(KE)
#F = 0.332*rho*v**2*Re**0.5 and KE =  0.5*rho*v**2
CD = 0.332*v**2*(Re**-0.5)/(0.5)*v**2

print "Local drag coefficient is %.2e "%(CD)

#From eq 11.44, the drag force acting on one side of the plate is
#kinetic viscocity
mu = 8.6*(10**-4)
fd = 0.664*mu*v*(l*v/MD)**0.5*w
#the total force acting on both sides of the plate

tfd = 2*fd
print "total drag force is %.3f  N "%(tfd)

#shear stress at any point in the boundary layer
#at a point in the boundary layer,
x = 0.5 			#m
y = t/2
# n = blasius dimensionless variable
n = y/(MD*x/v)**0.5
#From table 11.1, at n = 2.5,f"(n) = 0.218
#shear stress =  tau
fn = 0.218 			#f"(n) = fn
tau = (mu*v*(v/(MD*x))**0.5)*fn
print "Shear stress is %.3f N/m**2"%(tau)

i) Boundary layer thickness is 0.0033  m
Local drag coefficient is 8.72e-04
total drag force is 0.615  N
Shear stress is 0.285 N/m**2


## Example 11.2 Page No : 488¶

In :
#Variable
Ts = 200.           			# C,temp. of air
Ta = 30.            			#C, temp .of surface
Va = 8.             			#m/s, velocity of air
d = 0.75           			#m, dismath.tant from leading edge

#Calculation and Results
Tm = (Ts+Ta)/2     			#C, Mean temp. of boundary layer
mu = 2.5*10**-5    			#m**2/s, vismath.cosity
P = 0.69           			#prndatl no.
k = 0.036          			#W/m c, thermal conductivity
Re = d*Va/mu       			#reynold no.
t = 5*d/(Re**0.5*P**(1./3))           			#m, thermal boundary layer thickness
print "Thermal boundary layer thickness is %.1f mm "%(t*10**3)

N = (0.332*Re**(0.5)*P**(1./3))   			#Nusslet no.
h = k*N/d                                     			#heat transfer coefficent
print "heat transfer coeff is %.1f W/m**2 C"%(h)

Thermal boundary layer thickness is 8.7 mm
heat transfer coeff is 6.9 W/m**2 C


## Example 11.3 Page No : 489¶

In :
# Variables
#Free strean velocity (v1) and temp.(t1) on side 1
v1 = 6. 			#m/s
t1 = 150. 			#degree celcius
#same on  side 2
v2 = 3. 			#m/s
t2 = 50. 			#degree celcius
#dismath.tant
x = 0.7 			#m
#The plate temp. is assumed to be equal to the mean of the bulk air temp on the two sides of the plates
T = 100. 			#degree celcius

# Calculations
#Side 1
#mean air temp.
tm1 = (T+t1)/2
#From thermophysical properties:kinetic vismath.cosity (kv),Prandtl no.(P), thermal conductivity (k)
kv1 = 2.6*10**-5 			#m**2/s
P1 = 0.69
k1 = 0.0336 			#W/m degree celcius
#Reynold no.
Re1 = x*v1/kv1
#Nusslet no(N1)
a = 1/3.
N1 = 0.332*(Re1)**0.5*P1**a
h1 = k1*N1/x
#Side 2 of the plate
tm2 = (T+t2)/2
#Similarly
kv2 = 2.076*(10)**-5 			#m**2/s
P2 = 0.70
k2 = 0.03 			#W/m degree celcius
Re2 = x*v2/kv2
N2 = 0.332*(Re2)**0.5*P2**a
h2 = k2*N2/x
#overall heat transfer coeff.
U = h1*h2/(h1+h2)
#The local rate of heat exchange
RH = U*(t1-t2)

# Results
print "Local rate of heat exchange is  %.0f W/m2"%(RH)
#the plate temp is given by
TP = t2+(t1-t2)*U/h2
print "Plate temperature is :%.0f Celsius "%(TP)

Local rate of heat exchange is  235 W/m2
Plate temperature is :108 Celsius


## Example 11.4 Page No : 490¶

In :
import math
# Variables
T1 = 120.                             			#C, initial temp.
T2 = 25.                              			#C, Final temp.
Tm = (T1+T2)/2                       			#C, mean temp.
rho = 8880.                           			#kg/m**3, density of plate
#Properties of air at mean temp.
mu = 2.07*10**-5                      			#m**2/s, vismath.cosity
Pr = 0.7                             			#Prandtl no.
k = 0.03                             			#W/m C, thermal conductivity
l = 0.4                              			#m, length of plate
w = 0.3                              			#m, width of plate
d = 0.0254                           			#m, thickness of plate
Vinf = 1.                             			#m/s, air velocity
Re = l*Vinf/mu                       			#REynold no.

#from eq. 11.90 (b)
Nu = 0.664*(Re)**(1./2)*(Pr)**(1./3)     			#average Nusslet no.
#Nu = l*h/k
h = Nu*k/l                           			#W/m**2 C, heat transfer coefficient
#Rate of change of temp. is given by
A = 2*l*w                            			#m**2. area of plate
t = 1*3600.                           			#s, time
cp = 0.385*10.**3                      			#j/kg K, specific heat
m = l*w*d*rho                        			#kg, mass of plate

#-d/dt(m*cp8dt) = A*hv*(T1-T2)
#appling the boundary condition
T = (T1-T2)*math.exp(-A*h*t/(m*cp))+T2
print "The  temprature of plate after 1 hour is %.0f C"%(T)

The  temprature of plate after 1 hour is 82 C


## Example 11.5 Page No : 508¶

In :
import math
# Variables
#Reynold no (Re),friction factor(f),Prandlt no. (P)
Re = 7.44*(10**4)
f = 0.00485
P = 5.12
x = P-1            			#assume

# Calculations
#according to Von Karmen anamath.logy
N = ((f/2)*Re*P)/(1+(5*math.sqrt(f/2))*(x+math.log(1+(5./6)*x)))

# Results
print "Nusslet no is: %.0f "%(N)

Nusslet no is: 388