Question

Calculate the pressure drop (ft of water) over a distance of 800 ft within a PVC pipe carrying 750 ft3/min of landfill gas at 115 ºF. The cross-sectional area of the pipe interior is 0.0884 ft2 (d = 4.026 in). Assume the pressure in the pipeline is close to atmospheric (13.6 lb/in2) and the molecular weight of the gas is 28 lb/mol. Assume the gas is incompressible and the viscosity at 115 º is 7.65×10^-6lb/ft·s. The pipe relative roughnessis 1.5x10-5. Use the Moody diagram

Answer 1

Ans) We know,

Pressure drop = Friction loss = f L V² / ( 2 g D)

where, f = Darcy friction factor

L = Length of pipe

V = Flow velocity

   g = Acceleration due to gravity

D = Pipe diameter

Also, Flow rate(Q) = A x V

=> V = Q / A

Q = 750 ft³ /min or 12.5 ft³/s

=> V = 12.5 / 0.0884 = 141.4 ft/s

To determine friction factor (f) , calculate Reynold number = V D / , where = dynamic viscosity

To calculate density() use relation, P = RT

where, R = Gas constant = 8.314 J /kg K

P = 13.6 psi or 93.76 kPa

T = Temperature = 115 F or 319.2 K

=> 93760 = x 8.314 x 319.2

=> = 35.3 kg/m³ or 0.0685 slugs/ft³

=> Re = 0.0685 x 141.4 x 0.333 / 7.65 x 10^-6

=> Re = 4.21 x 10⁵

Relative roughness = 0.000015

According to Moody diagram for Re = 4.21 x 10⁵ and relative roughness = 0.000015 , f = 0.0137

Putting values,

Pressure drop = 0.0137 (800) (141.4)² / ( 2 x 32.2 x 0.3355)

= 10142 ft

Pressure drop in psf = g H = 0.0685 x 32.2 x 10142 psf = 22371 psf

=> Pressure drop in ft of water = P / _w = 22371 psf / 62.4 = 358.5 ft