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UPRVUNL JE ME 2016 Official Paper Shift 1

Option 4 : 1.91 m

__ Concept__:

For pipe:

The loss of pressure head in case of laminar flow is given by:

\({h_{L}} = \frac{{32{μ\bar{U}L}}}{{ ρ gd^2}}\)

where U̅ = average velocity, L = length of pipe, ρ = density of fluid flowing in the pipe, d = diameter of the pipe and μ = dynamic viscosity of the pipe.

**Specific gravity**:

Ratio of fluid density with the density of water i.e.

\(S.G=\frac{ρ}{ρ_w}\)

__ Calculation__:

__ Given__:

d = 20 cm = 0.2 m, L = 70 m, S.G = 0.95, μ = 0.23 Ns/m2, U̅ = 1.38 m/s.

\(R=\frac{\rho \bar{U} D}{\mu}=\frac{\rho _W\times S.G \times \bar {U}\times D}{\mu}\)

\(R=\frac{1000\times 0.95 \times 1.38\times 0.2}{0.23}=1140\)

\(S.G=\frac{ρ}{ρ_w}\)

**ρ = ρ _{w} × S.G**

\({h_{L}} = \frac{{32{μ \bar{U}L}}}{{ ρ gd^2}}=\frac{{32{μ \bar{U}L}}}{{ ρ_w \times S.G\times gd^2}}\)

\({h_{L}}=\frac{{32{\times0.23 \times 1.38 \times 70}}}{{ 1000 \times 0.95\times 9.81 \times 0.2^2}}=1.907 \approx1.91\;m\)