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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%
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B
50%
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C
70%
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D
80%
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Solution

The correct option is D 80%pressure drop in the pipe, Δp=32μ¯uLD2 Discharge:Q = AV = π4d2V or V=4Qπd2 ∴ΔP=32μD2×4QπD2L Δp=128μQLπD4 or Δp=CD4 where C=128μQLπ Δp1=CD41=CD4 where D1=D and Δp2=CD42=C(1.5D)4 where,D2=1.5D % change in pressure drop. =Δp1−Δp2Δp1×100 =(1−Δp2Δp1)×100 =[1−C(1.5D)4×D4C]×100 =(1−C(1.5)4)×100=(1−15.06)×100 =(1−0.1976)× 100 = 80.24% ≈80%

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