Problem 4.10.2

A piping arrangement is showing below. A pipe section AB is 2500 foot long. A branch, AC, is taken out and has a length of 1000 feet. The pipe is connected back to main pipe via a branch CB, 2000 feet in length. All pipes are at same elevation. Pressure drop for main branch, AB, is 40 psi. What are the flow rates through the pipe branches? Viscosity of fluid is 1 cP.

Solution:

Applying Bernoulli equation between points A and B: P_A over rho is equal to P_B over rho plus h_f1.

h_f1 can be found to be 92.31 foot pound-force per pound-mass. Now we can find out the velocity through the pipe that offers this frictional loss.

The term Re₁ sqrt f₁ is equal to rho over mu sqrt (d₁³ g_c h_f1 over 2 L₁ ) or 9.255 time 10³. Friction factor, f₁, can be read as 0.005.

From Figure 5-11, Reynolds number can be read directly as 1.307 times 10⁵. This corresponds to velocity, V₁, of 5.507 feet per second. Volumetric flow rate, q₁, is 0.283 cubic foot per second.

Assuming that pipes 2 and 3 have same friction factor and noting the same volumetric flow rate passes through both pipes,

Re₂ sqrt f₂ is equal to rho d₁ over mu sqrt (g_c h_f1 over 2 L₂ plus 2 L₃ (d₂/d₃)⁴).

Re₂ sqrt f₂ is equal to 2912. So friction factor, f₂, is equal to 0.00589. This gives a Reynolds number is 3.795 times 10⁴. This corresponds to a velocity, V₂, of 1.218 feet per second.

Velocity in second segment, V₃, can be calculated as 4.621 feet per second. Reynolds number, Re₃, can be found as 7.392 times 10⁴. Friction factor, f₃, can be read as 0.00562.

Now these revised values of f₂ and f₃ can used and process repeated. Final converged values are:

V₂ is 1.248 feet per second. V₃ is 4.733 feet per second.