Answer to Question #106558 in Physics for simran

Question

Suppose a viscous oil, whose flow is in the laminar regime is to be pumped through a 10 cm
diameter horizontal pipe over a distance of 15km at a rate of s/ 10 m
−3 3
. Viscosity of the oil is
03.0 poise. What is the required pressure drop to maintain such a flow?

Answer 1

Suppose a viscous oil, whose flow is in the laminar regime is to be pumped through a 10 cm diameter horizontal pipe over a distance of 15km at a rate of $10^{-3}\ m^3/s$ . Viscosity of the oil is 3.0 poise. What is the required pressure drop to maintain such a flow? 

Solution

For a viscous oil at 20°C take $\rho=100\ kg/m^3 .$ Given that $\mu=3 \ poise=0.3\ \dfrac{N\cdot s}{m^2},$

$D=10 \ cm=0.1\ m, L=15\ km=15000\ m, Q=10^{-3}\ m^3/s.$

Calculate the pipe velocity and Reynolds number:

$$V={Q \over A}={Q \over (\pi D^2)/4}={10^{-3}\ m^3/s \over (\pi (0.1\ m)^2)/4}\approx0.1273\ m/s$$

$$Re={\rho VD \over \mu}={1000\dfrac{kg}{m^3}\cdot0.1273\dfrac{m}{s}\cdot0.1\ m\over 0.3\dfrac{N\cdot s}{m^2}}=42.4<2100$$

The flow is laminar flow. Find the required pressure drop to maintain such a flow

$$\Delta p=p_2-p_1={128\mu LQ \over \pi D^4}$$

$$\Delta p={128\cdot0.3\ \dfrac{N\cdot s}{m^2}\cdot15000\ m\cdot10^{-3}\ \dfrac{m^3}{s} \over \pi (0.1\ m)^4 }=$$

$$=183346.5 \text{Pa}=1.833465\times10^5 \text{Pa}=0.183\ \text{MPA}$$