Question

Three pipes A, B and C have the following basic geometries :

Pipe	A	B	C
Diameter	D	D/2	2D
Length	L	L	4L

If these pipes are connected in series, by assuming the value of friction factor f to be same for all the three pipes and the equivalent pipe, this set of pipes in series is equivalent pipe, of length $L_e$ and diameter D and friction factor f with the equivalent length $L_e$ being equal to

• A. $5\frac{1}{8}L$
• B. $4\frac{1}{8}L$
• C. $26\frac{1}{8}L$
• D. $33\frac{1}{8}L$

Answer 1

The correct option is D $33\frac{1}{8}L$

The head loss in a circular pipe of diameter D, and length L, carrying a discharge Q is given by

$h_f=\frac{f\times L\times Q^2}{12.1\times D^5}$

When pipes are connected in series total head loss is given by

$h_f=h_{fA}+h_{fB}+h_{fC}$

$\Rightarrow \frac{f\times L_e\times Q^2}{12.1\times D^5}=\frac{f\times L\times Q^2}{12.1\times D^5}+\frac{f\times L\times Q^2}{12.1\times {\left(\frac{D}{2}\right)}^5}+\frac{f\times 4L\times Q^2}{12.1\times (2D{)}^5}$

$\Rightarrow \frac{L_e}{D^5}=\frac{L}{D^5}+\frac{32L}{D^5}+\frac{L}{8D^5}$

$\therefore L_e=L+32L+\frac{L}{8}$

$\Rightarrow L_e=33\frac{1}{8}L$