written 5.6 years ago by | modified 2.8 years ago by |
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
written 5.6 years ago by | modified 2.8 years ago by |
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
written 5.6 years ago by | modified 5.6 years ago by |
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.