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Oil of specific gravity 0.82 is pumped through a horizontal pipeline 15mm in diameter and 3Km long at the rate of 0.015m /s.

The pump has an efficiency of 68% requires 7.5 KW to pump the oil. (i) What is the dynamic viscosity of oil? (ii) is the flow laminar?


Subject: Fluid Mechanics 2

Difficulty: Hard

Marks: 10M

2 Answers
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Given : Specific gravity = 0.82 of oil.

$\therefore \text{density of oil } \rho = 0.82 \times 1000 = 820 kg/m^3$

Dia. of pipe D = 0.015m (D = 15mm)

Length L = 3 km = 3000 m.

discharge Q = 0.015 $m^3/s$

Power = 7.5 kw $ = 7.5 \times 10^3 w$ and $ \eta = 68%$

Now, Q = velocity x Area.

$0.015 = ū \times \frac{\pi}{4} \times D^2$ (D = 0.015m)

$\therefore ū = 84.88 m/s$

Now, Power $p = \eta \times w \times hf$ (watts) ----------- (1)

w = weight of oil flowing per sec = $\rho g \times Q$ ----- (2)

$h_f = \frac{32 \mu ū L}{\rho g D^2}$

$p = \eta \times (\rho g \times Q) \times \frac{32 \mu ū L}{\rho g D^2}$

$\therefore 7.5 \times 10^3 = 0.68 \times 0.015 \times 32 \times \mu \times 84.88 \times 3000$

Dynamic viscosity is, $(\mu)$

$\therefore M = 0.090 NS/m^2$

Now, Reynold Number is Re = $\frac{ū \times D}{v}$

$R_e = \frac{ū \times D}{\mu / \rho } = \frac{84.88 \times 0.015}{0.090/820}$

$R_e = 11600.267$

$\therefore$ For Laminar flow, Re value must lies below 2000.

Hence flow is not Laminar flow.

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