Question

Water is flowing through a horizontal pipe of constant diameter and the flow is laminar. If the diameter of the pipe is increased by 50% keeping the volume flow rate constant, then the pressure drop in the pipe due to friction will decrease by

• A. 33%
• B. 50%
• C. 70%
• D. 80%

Answer 1

The correct option is D 80%

pressure drop in the pipe,

$\Delta p=\frac{32\mu \overline{u}L}{D^2}$
$Discharge:Q=AV=\frac{\pi }{4}{d}^{2}V$

$orV=\frac{4Q}{\pi d^2}$

$\therefore \Delta P=\frac{32\mu }{D^2}\times \frac{4Q}{\pi D^2}L$

$\Delta p=\frac{128\mu QL}{\pi D^4}$

$or\Delta p=\frac{C}{D^4}$

$whereC=\frac{128\mu QL}{\pi }$

$\Delta p_1=\frac{C}{D_1^4}=\frac{C}{D^4}where{D}_1=D$

$and\Delta p_2=\frac{C}{D_2^4}=\frac{C}{(1.5D{)}^4}where,D_2=1.5D$

% change in pressure drop.

$=\frac{\Delta p_1-\Delta p_2}{\Delta p_1}\times 100$

$=(1-\frac{\Delta p_2}{\Delta p_1})\times 100$

$=[1-\frac{C}{(1.5D{)}^4}\times \frac{D^4}{C}]\times 100$

$=(1-\frac{C}{(1.5{)}^4})\times 100=(1-\frac{1}{5.06})\times 100$

$=(1-0.1976)\times$ 100 = 80.24% $\approx 80$%