Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

What we resources 2. Not all engineers become designers: some work for regulator

ID: 1713335 • Letter: W

Question

What we resources 2. Not all engineers become designers: some work for regulatory agencies and review permit applications involving proposed designs. In this capacity, the reviewer often must verify the calculations by the design engineer. For this problem you are the reviewer and have been given the following solution for a piping system involving 4 parallel pipes. It is suspected the flow for one of the parallel pipes was caleulated incorreetly and therefore the estimate of the total system flow is wrong (a) State the two governing constraints, or properties, for this type of pipe flow problem. Ans: 1 (b) For which pipe was the flow computed incorrectly? c) What is the correet flow for that pipe? d) What is the correct flow for the total system? Ans: Pipe Number Ans: Corect Flow Ans: System Flow The following table gives data for the parallel pipe system. Flow rates are rounded to the nearest integer value. Pipe Hazen-Willias Q No 660 8.2 591 506 449 10

Explanation / Answer

(a) For parallel pipe system,

1) head loss is the same in each pipe

(2) total flow rate is the sum of the flow rates in individual pipes

(b) Head loss due to friction, H = KpL v2/2g = KpL (Q/A)2/2g

Since, H, A and g are constants, so,

KpLQ2 = constant

For pipe 1, KpLQ2 = 4 x 8.2 x 6602 = 14287680

For pipe 2, KpLQ2 = 6 x 10 x 5912 = 20956860

For pipe 3, KpLQ2 = 8 x 10 x 5062 = 20482880

For pipe 4, KpLQ2 = 10 x 10 x 4492 = 20160100

The above value for pipe 1 is very much different from the other pipes. So, it can be inferred that the flow is computed incorrectly for Pipe 1. The above values for pipe 2, 3 and 4 vary slightly because the flow rates are rounded to the nearest integer value.

(c) For pipe 1, KpLQ2 = avg ( KpLQ2 for pipe 2,  KpLQ2 for pipe 3 and KpLQ2 for pipe 4)

=> 4 x 8.2 x Q2 = 20533280

=> Approx value of Q = (20533280 / (4 x 8.2))2 = 791.21 gpm

(d) Flow for total system = Q1 + Q2 + Q3 + Q4 = 791.21 + 591 + 506 + 449 = 2337.21 gpm

Hire Me For All Your Tutoring Needs
Integrity-first tutoring: clear explanations, guidance, and feedback.
Drop an Email at
drjack9650@gmail.com
Chat Now And Get Quote