The pressure drop in a section of a pipe whose relative roug

The pressure drop in a section of a pipe whose relative roughness (k/D) is 0.05 is determined from tests with water. A pressure drop of 2 atm is obtained at a water(rho=1000kg/m^3) flow rate of 15kg/s. What will be the pressure drop when liquid oxygen (rho=1240kg/m^3) flows through the pipe at the rate of 20kg/s. Here both of the flows of water and liquid oxygen are assumed to be fully turbulent (Re g 10^5).

Solution

The pressure drop in the pipe section is given by,

p = (l / d_h) ( v² / 2)  

where p - pressure loss (N/m²), - Darcy-Weisbach friction coefficient, l - length of duct or pipe (m), v - velocity (m/s), d_h - hydraulic diameter (m), - density (kg/m³)

The mass flow rate, dot m_water = rho * A * v

Solve for d from the above eqn,

rho * 3.14 * d^2 * v = 15

d = sqrt (4 / rho * v) = 2 / sqrt rho * v

For water d = 2 / sqrt 1000 * v = 0.063 / v

The Reynolds number for the flow is assumed to be Re 10^5

Re = u L / where is the kinematic viscosity 0.326 * 10^-6 m2 / s

100000 = u L / 0.326 * 10^-6

L = 0.1 / u where u = velocity (v)

The friction coefficient lambda is found using,

1 / ^1/2 = -2 log [ 2.51 / (Re ^1/2) + (k / d_h) / 3.72 ]

Rplacing with the values for Re and relative roughness we get,

1 / ^1/2 = -2 log [ 2.51 / (10^5 ^1/2) + 0.05 / 3.72 ]

1 / ^1/2 = - 2 log [ 0.0000251/ ^1/2 + 0.0134]

1 / ^1/2 = -9.2 + 1.87 + 2 log ^1/2

1/ x = -7.32 + 2 log x

2 log x - 1/x = 7.32

x = 0.282 and = 0.08

Now, the velocity of the flow is found using the relation eqn 1,

p = (l / d_h) ( v² / 2)

2 = 0.08 * (0.1 / 0.063) * (1000 * v^2 / 2)

v = 0.177 m/s

So, the length and diameter of the section is found as,

L = 0.1 / v = 0.56 m or 56 cm

D = 0.063 / v = 0.355 m or 35.5 cm

Now the pressure drop in the pipe due to air flowing at the same velocity v = 0.5 m/s through L = 56 cm and dia = 35.5 cm using the eqn 1

p = (l / d_h) ( v² / 2)

p = 0.08 * 0.56 / 0.355 * 1.22 * 0.177^2 /2

p = 0.002 atm or 2 * 10^-3 N /m2