# Chapter 15 : Heat transfer and fluid friction¶

## Example 15.1 page : 267¶

In [3]:
import math

# Variables
Tm = 70.;                                    # Average air temperature in degF
Tw = 60.;                                    # Pipe wall temperature in degF
thm = Tm-Tw;                                # Mean temperature difference in degF
# Thm is so small that the fluid properties may be based on 70 degF

v = 30.;                                     # Velocity in ft/sec
L = 1000.;                                   # Length of pipe
D = 3./12;                                   # Diameter in ft
y = 0.15;                                   # Specific weight in  lb/ft**3
p = 0.15/32.2;                              # Density in slug/ft**3
u = 0.00137;                                # Vismath.cosity in slug/ft/hr

# Calculations and Results
Nre = v*3600*D*p/u;                              # Reynolds number
f = 0.08/(Nre)**.25;                         # Nusselt number
delp = 2*f*L*p*(v**2)/D;                     # Pressure drop in lb/sq.in
print "The pressure drop is %d lb/sq.ft  "%(delp);

cp = 0.24*32.2;                             # Specific heat capacity in slug/degF
Cp = 0.24*0.15;                             # Heat capacity in Btu/ft**3-degF
k = 0.0148;                                 # Thermal conductivity in Btu/ft-hr-degF
Npr = u*cp/k;                               # Prandtls number
phi = math.sqrt(Npr)/(1+(750*math.sqrt(Npr)/Nre)+7.5*(Npr**0.25)/math.sqrt(Nre));
A = math.pi*L*D;                                # Area in ft**2
q = phi*f*Cp*A*v*thm*3600/(2*Npr);          # Heat loss in Btu/hr
print "Heat loss per hour of air is %.3f Btu/hr  "%(phi);
h = q/(A*thm);                             # Film coefficient
print "The film coefficient of heat transfer on the inner pipe wall is %.1f Btu/hr-ft**2-degF"%(h);

# note : rounding off error.

The pressure drop is 154 lb/sq.ft
Heat loss per hour of air is 0.821 Btu/hr
The film coefficient of heat transfer on the inner pipe wall is 10.3 Btu/hr-ft**2-degF