Question

In a fully developed turbulent flow through a circular pipe, a head loss of h1 is observed. The diameter of the pipe is increased by 10% for the same flow rate and a head loss of h2 is noted. Assume friction factor for both the cases of pipe flow is the same. The ratio of \(\frac{h_2}{h_1}\) is closest to
1. 0.34
2. 0.62
3. 0.87
4. 1.00

Answer 1

Correct Answer - Option 2 : 0.62

Concept:

Head loss in the pipe due to friction is given as:

\(h_f = \frac {4f lV^2}{2gD}=\frac {8}{\pi ^2g}\frac {4flQ^2}{D^5} \)

where f = friction coefficient, 4f = friction factor, V = average velocity of flow, D = diameter of pipe, l = length of pipe

∴ head loss ∝ V2,  head loss ∝ Q2,  head loss ∝  l,  \(head~loss \propto \frac 1{D^5}\)

Calculation:

Given:

d₂ = 1.1d₁

From diameter proportional relation,

\(h_1 \propto \frac{1}{d_1^5}\)

\(h_2 \propto \frac{1}{d_2^5}\)

\(\Rightarrow h_2 \propto \frac{1}{(1.1d_1)^5}\)

\(\Rightarrow \frac{h_2}{h_1} = \frac{d_1^5}{(1.1d_1)^5}\)

\(\Rightarrow \frac{h_2}{h_1} = 0.621\)