Question

Consider fully developed flow in a circular pipe with negligible entrance length effects. Assuming the mass flow rate, density and friction factor to be constant, if the length of the pipe is doubled and the diameter is halved, the head loss due to friction will increase by a factor of

• A. 4
• B. 16
• C. 32
• D. 64

Answer 1

The correct option is D 64

Method I:
Head loss due to friction,

$h_f=\frac{fL{V}^2}{2gd}$

Mass flow rate,
$\dot{m}=\rho AV=\rho \times \frac{\pi }{4}{d}^{2}V$
$orV=\frac{4\dot{m}}{\rho \pi d^2}$

$\therefore h_f=\frac{fL}{2gd}\times \frac{16{\dot{m}}^2}{{\rho }^2{\pi }^2{d}^4}$

$=\frac{8f{\dot{m}}^2}{{\rho }^2{\pi }^2}\times \frac{L}{d^5}$

$=C\times \frac{L}{d^5}whereC=\frac{8f{\dot{m}}^2}{{\rho }^2{\pi }^2}$

$h_{f_1}=C\times \frac{L}{d^5}$

$h_{f_2}=C\times \frac{2L}{(d/2{)}^5}=\frac{C\times 2L}{\frac{d^5}{2^5}}$

$=\frac{C\times 2\times 2^{5}L}{d^5}=\frac{C\times 64L}{d^5}$

$h_{f_2}=64h_{f_1}$

The head loss due to friction will increase by a factor of 64

Method II:
$\dot{m}=\rho Q$
(where m, $\rho$ and Q are constant)

Head loss due to friction,

$h_f=\frac{fL{V}^2}{2gD}=\frac{fL{Q}^2}{12D^5}$

$\frac{h_{f_2}}{h_{f_1}}=\frac{f\left(2L\right)Q^2}{12(D/2{)}^5}\times \frac{12D^2}{fL{Q}^2}$

=$2\times 2^5=64$

$h_{f_2}=64h_{f_1}$

So, head loss due to friction increases by a factor of 64