Example 6

Example 6 – Determining the diameter of an air duct

Heated air at 1 atm and 35°C is to be transported in a 150-m-long circular plastic duct at a rate of 0.35 m^3/s ( Fig, 8-33 ). If the head loss in the pipe is not exceed 20 m, determine the minimum diameter of the duct.

SOLUTION :

The flow rate  and the head loss in an air duct are given. The diameter of the duct is to be determined.

Assumptions :

|1. The flow is steady and incompressible.

2. The entrance effects are neglible, and thus the flow is fully developed,

3. The  duct involve no components such as bends, valves, and connectors.

4. Air is an ideal gas.

5. The duct is smooth since it is made of plastic 6.The flow is turbulent ( to be verified )

Properties :

The density, dynamic viscosity and kinematic viscosity of air at 35°C are ρ = 1.145 kg/m^3 , μ = 1.895 ×〖10〗^(-5) kg/m.s, and v = 1.655 × 〖10〗^(-5) m^2/s

Analysis : 

The average velocity, the Reynolds number,the friction factor, and the head loss relations are expressed as ( D is in m, V is in m/s, and Re and  f are dimensionless )

V= ὐ/A_c  = ὐ/(〖πD〗^2/4) = (〖0.35m〗^3/s)/(〖πD〗^2/4)

Re = VD/v = VD/(1.655 ×(〖10〗^(-5) m^2)/s)

1/√f = -2.0 log ((ε/D)/3.7 + 2.51/(Re√f)) = - 2.0 log (2.51/(Re√f))

h_L=f L/D  V^2/2g →20m=f 150m/D  V^2/2(9.81 m/s^2 ) 

D= 0.267 m      f = 0.0180     V = 6.24 m/s  and Re = 100,800

The diameter of the duct should be >26.7 cm if head loss is not to exceed 20m.

Re > 4000 , the turbulent flow assumption is verified.

The diameter  can also be determined directly from the third Swamee-Jain formula

D = 0.66 [ ε^(1.25 )( (LV^2)/(gh_L )) + vὐ^9.4 (L/(gh_L ) )^5.2 ]^0.04

    = 0.66 [ 0 + (1.655× 〖10〗^(-5 ) m^2/s)(0.35 m^3/s)^9.4(150m/((9.81 m/s^2 )(20m)) )^5.2 ]^0.04

    =  0.271 m