Subject: Fluid Mechanics 2

Topic: Laminar flow

Difficulty: High

Oil of specific gravity 0.9 is pumped through a horizontal pipeline 200 mm in diameter and 3km long at the rate of $0.015 m^3/s.$ the pump has an efficiency of 68 % and requires 7.5 kW to pump the oil. What is the dynamic viscosity of oil and whether the flow is laminar or not?

Given :

Specific gravity $= 0.9$ of oil

Density $= 900 $ kg/m³

d $ = 200 $ mm $ = 0.2$ m

L $= 3 $ Km $= 3000$ m

Q $= 0.015$ m³/s

Power (P) $= 7.5$ KW

$ \begin{array}{I} \eta=68 \% \end{array} $

Solution :

Let

$\begin{aligned} Q &=\text {Velocity} \times \text {Area} \\ Q &=\text { VA } \end{aligned}$

$$ \begin {aligned} V &=\frac{Q}{A}=\frac{0.015}{\frac{\pi}{4} \times(0.2)^{2}}=0.4775 \mathrm{~m} / \mathrm{s} . \end{aligned} $$

$$ \begin{aligned} P &=\eta \times w \times h_{f} \\ P &=n \times(\rho g Q) \times \frac{32 \mu V L}{\rho g d^{2}} \\ \mu &=\frac{P \times d^{2} \times Q}{\eta \times Q \times 32 \mathrm{VL}} \\ &=\frac{7.5 \times (10.2)^{2} \times 10^{3}}{0.68 \times 0.4775 \times 32 \times 0.015 \times 3000} \\ &=0.6416 \mathrm{N s} / \mathrm{m}^{2}\\ Now\\ R_{e} &=\frac{\rho V d}{\mu} \\ R_{e} &=\frac{900 \times 0.4775 \times 0.2}{0.6416} \\ &={133.96} \end{aligned} $$

Hence flow is Laminar flow.