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ISRO IPRC Technical Assistant Mechanical held on 28/08/2016

Option 2 : 2 times

ST 1: Building Material and Concrete Technology

19807

20 Questions
20 Marks
12 Mins

**Concept:**

**The pressure drop head in a laminar flow through the circular pipe is given by:**

\(\frac{{{P_1} - {P_2}}}{{ρ g}} = \frac{{32\;μ UL}}{{ρ g{D^2}}}\)

\(Δ P = \frac{{32\;μ UL}}{{{D^2}}}\)

where, ΔP = Pressure drop, μ = viscosity of the fluid, U = Mean velocity of flow, L = Length of pipe, D = diameter of the pipe, ρ = Density of the fluid

**we know Q = U × A**

where Q = Discharge, A = Cross-sectional area of pipe

∴ \(Δ P= \frac{{32\;μ QL}}{{A {D^2}}}\)

If we keep all the other parameters to be constant then:

**Pressure drop (ΔP) ∝ Discharge (Q)**

**Calculation:**

**Given:**

Q_{2} = (2 × Q_{1})

where Q_{1} = Original discharge, Q_{2} = final discharge

Let ΔP_{1} = Original pressure drop ΔP_{2} = Final pressure drop

then we know,** **ΔP ∝ Q

∴ \(\frac{Δ P_2}{Δ P_1}=\frac{Q_2}{Q_1}\)

\(\frac{Δ P_2}{Δ P_1}=\frac{2 × Q_1}{Q_1}\)

**ΔP _{2} = 2 × ΔP_{1}**

**Hence, If the flow rate of water in a pipeline is doubled, then the pressure drop will be doubled the original value.**