E4.15 Isothermal Gas Flow

Example 4.15: What is the pressure drop in a pipe when 120 scfm (standard cubic feet per minute) of air are flowing through a pipe that has a length of 105 ft. The inside diameter of the pipe is 1.38 in, and its roughness is 0.00015 ft. Air is flowing at 110 psia and 100 °F. Use the following information:

Molecular weight = 28.92; Viscosity of air = 1.28 ´ 10^-5 lb/(ft·s);

R = 1545 ft·lb_f/(lbmol·°R)

Solution: Flow rate is given in standard cubic feet per minute. Standard conditions are 1 atm and 0 °C. However, in the industrial setting, standard temperature is 60 °F and 1 lb mole of gas posses 379 ft³.

Calculate mass flow rate of air: m' = as 9.16 lb/min or 0.153 lb/s.

Calculate cross sectional area is 0.0104 ft².

For compressible fluids, Reynolds number is calculated by utilizing mass velocity, G. So mass velocity is

The Reynolds number is found to be

Relative roughness is 0.0013. We can find the friction factor to be 0.0057. We know that the initial pressure is 110 psi, which should be converted to consistent units (1.584 ´ 10⁴ lb_f/ft²). Plugging these numbers in the isothermal flow equation, exit pressure can be found by trial and error or by using Newton's method. Initial guess for P_b can be found to be

The value of P_b converges to 1.571 ´ 10⁴ lb_f/ft². The pressure drop can be calculated as 0.91 psi.