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A lubricating oil of viscosity 1 poise and specific gravity 0.9 is pumped through a 30mm diameter pipe. If the pressure drop per meter length of pipe is $20 kN/m^2$ determine:

### iv) The power required per 50 m length of the pipe to maintain the flow.

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Given,

$μ = 1.0\ poise = 0.1\ Ns/m^2$

$s = 0.9\ \implies ρ = 900\ kg/m^3$

$L = 50m$

$D = 300mm \implies 0.03m$

$R = D/2 = 0.015m$

$Pressure\ Drop\ = 20KN/m^2$

(i) Mass flow rate (m)

Q = ūA

$ū = \frac{(P_1-P_2)D^2}{32ūL}$

$(P_1-P_2) = \frac{32ūμL}{D^2} = \frac{32\times0.1\timesū\times50}{0.03m^2}$

$\implies 20,000 = 5333.33ū$

$ū = 3.75m/s$

$A = \frac{π}{4}D^2 \implies \frac{π}{4} \times 0.03^2 \implies A = 0.00070685m^2$

$Q = 3.75 \times 0.00070685 = 0.002650m^3/s$

$m = Q\timesρ = 0.002650 \times 900 = 2.385 kg/sec = 143.1kg/min$

Ans.

(ii) Shear stress at pipe wall (τ)

$τ = \frac{-∂P }{∂x} \times \frac{R}{2}$

$\frac{-∂P }{∂x} = \frac{-(P_2-P_1)}{x_2-x_1} = \frac{20,000}{50} = 400N/m^2$

$τ = 400 \times \frac{0.015}{2} = 3N/m^2$

Ans.

(iii) Reynolds (R)

$R = \frac {ρvD}{μ} = \frac{900\times3.75\times0.03}{0.1} = 1012.5$

Ans.

(iv) Power required per 50m length (P)

$P = W \times hf$

$W = Q \times ρg \implies 0.002650 \times 900 \times 9.81 = 23.39 kg-m/s^2$

$hf = \frac{32μūL}{ρgD^2} \implies \frac{32 \times 0.1 \times 3.75 \times 50}{900 \times 9.81 \times (0.03)^{2}}$

$hf = 75.50 m$

$P = 23.3968 \times 75.50 = 1766.66 N$