Practice Question: Flow to a Parallel Pipe System Based on Head Loss

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Solution

Answer: A) 2 cfs

Head loss for a parallel system is equal for each pipe:
$$ h_{f-total}=h_{fA}=h_{fB} $$
Head loss per the Darcy-Weisbach equation:
$$ h_f=f\ast\frac{L}{D}\ast\frac{v^2}{2g} $$
Calculate velocity in Pipe A:
$$ 5\ ft=0.018\ast\frac{100\ ft}{\frac{4}{12}\ ft}\ast\frac{v_A^2}{2\left(32.2\frac{ft}{s}\right)} $$
$$ v_A=7.72\frac{ft}{s} $$
Calculate velocity in Pipe B:
$$ 5\ ft=0.018\ast\frac{200\ ft}{\frac{6}{12}\ ft}\ast\frac{v_B^2}{2\left(32.2\frac{ft}{s}\right)} $$
$$ v_B=6.69\frac{ft}{s} $$
Total input flow is equal to flow through Pipes A and B:
$$ Q_{total}=Q_A+Q_B $$
Calculate total input flow:
$$ Q_{total}=A_Av_A+A_Bv_B $$
$$ Q_{total}=\frac{\pi\ast\left(\frac{4}{12}ft\right)^2}{4}\ast\left(7.72\frac{ft}{s}\right)+\frac{\pi\ast\left(\frac{6}{12}ft\right)^2}{4}\ast\left(6.69\frac{ft}{s}\right)\approx2\ cfs $$
 

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