# Chapter 6 : Fundamentals of convective heat transfer¶

## Example 6.2 Page No : 241¶

In [1]:
import math

# Variables
L = 1;			#Length of the palte in m
W = 1;			#Width of the plate in m
v = 2.5;			#Velocity of air in m/s
Re = (5*10**5);			#Reynolds number at the transition from laminar to turbulant
p = (0.85*10**-5);			#Dynamic vismath.cosity in N.s/m**2
r = 1.12;			#Density in kg/m**3

# Calculations
x = (p*Re)/(r*v);			#Calculated length in m

# Results
print 'The actual length of the plate is %i m, which is less than %3.2f m'%(L,x)

The actual length of the plate is 1 m, which is less than 1.52 m


## Example 6.6 Page No : 247¶

In [2]:
# Variables
p = 0.8;			#Dynamic viscosity in N.s/m**2
k = 0.15;			#Thermal conductivity in W/m.K
Tb = 10;			#Temperature of bearing in degree C
Ts = 30;			#Temperature of the shaft in degree C
C = 0.002;			#Clearance between bearig and shaft in m
U = 6;			#Velocity in m/s

# Calculations
qb = (((-p*U**2)/(2*C))-((k/C)*(Ts-Tb)))/1000;			#Surface heat flux at the bearing in kW/m**2
qs = (((p*U**2)/(2*C))-((k/C)*(Ts-Tb)))/1000;			#Surface heat flux at the shaft in kW/m**2
Tmax = Tb+(((p*U**2)/(2*k))*(0.604-0.604**2))+((Ts-Tb)*0.604);			#Maximum temperature in degree C occurs when ymax = 0.604L

# Results
print 'Maximum temperature rise is %3.3f degree C  \n \
Heat fux to the bearing is %3.1f kW/m**2  \n \
Heat fux to the shaft is %3.1f kW/m**2'%(Tmax,qb,qs)

Maximum temperature rise is 45.042 degree C
Heat fux to the bearing is -8.7 kW/m**2
Heat fux to the shaft is 5.7 kW/m**2


## Example 6.7 Page No : 257¶

In [3]:
# Variables
D = 0.02;			#I.D of the tube in m
Q = 1.5;			#Flow rate in litres per minute
k = (1*10**-6);			#kinematic vismath.cosity in m**2/s

# Calculations
um = ((Q/60)*10**-3)/(3.14*(D**2/4));			#Average velocity in m/s
Re = (um*D)/k;			#Reynolds number
x = 0.05*D*Re;			#Entry length in m

# Results
print 'Re which is %3.0f less than 2300, the flow is laminar.  \n \
Entry length is %3.3f m'%(Re,x)

Re which is 1592 less than 2300, the flow is laminar.
Entry length is 1.592 m


## Example 6.8 Page No : 257¶

In [5]:
# Variables
L = 3000;			#Distance transported in m
D = 0.02;			#I.D of the tube in m
Q = 1.5;			#Flow rate in litres per minute
k = (1*10**-6);			#kinematic vismath.cosity in m**2/s
pw = 1000;			#Density of water in kg/m**3

# Calculations
um = ((Q/60)*10**-3)/(3.14*(D**2/4));			#Average velocity in m/s
Re = (um*D)/k;			#Reynolds number
x = 0.05*D*Re;			#Entry length in m
hL = ((64./Re)*L*um**2)/(2*D*9.81)			#Head loss in m
P = (pw*9.81*(3.14/4)*D**2*um*hL);			#Power required to maintain this flow rate in W

# Results
print 'Head loss is %3.2f m  \n \
Power required to maintain this flow rate is %3.4f W'%(hL,P)

Head loss is 1.95 m
Power required to maintain this flow rate is 0.4777 W


## Example 6.9 Page No : 258¶

In [4]:
# Variables
L = 100;			#Length of rectangular duct in m
A = [0.02,0.025];			#Area of duct in m**2
Tw = 40;			#Temperature of water in degree C
v = 0.5;			#Velocity of flow in m/s
k = (0.66*10**-6);			#kinematic viscosity in m**2/s
p = 995;			#Density of water in kg/m**3

# Calculations
P = 2*(A[0]+A[1]);			#Perimeter of the duct in m
Dh = (4*(A[0]*A[1]))/P			#Hydraulic diameter of the duct in m
Re = (v*Dh)/k;			#Reynolds number
f = 0.316*Re**(-0.25);			#Friction factor
hL = (f*L*v**2)/(2*Dh*9.81);			#Head loss in m
P = (hL*9.81*p)/10**4;			#Pressure drop in smooth rectangular duct in 10**4 N/m**2

# Results
print 'Pressure drop in smooth rectangular duct is %3.4f*10**4 N/m**2'%(P)

Pressure drop in smooth rectangular duct is 1.5527*10**4 N/m**2