A horizontal 20-mm diameter tube is to carry a laminar flow of SAE 30 oil. What pressure drop. (p1-p2) is needed to produce a flow rate of 72 liters/hr?

Question

asked Dec 9, 2020 186k views

1 Answer

Sunil B N · Answer 1 · 2020-12-14T17:26:48+0000

Answer:

Step-by-step explanation:

Given that diameter of tube= 20 mm

flow rate = 72 lt/hr

The properties of SAE 30 oil at 30°C

$\mu =155.31 mPa$

$\rho =872.5 (kg)/(m^3)$

We know that volume flow rate Q= AV

$2* 10^(-5)=(\pi)/(4)*{0.02^2}V$

V= 0.064 m/s

Now lets find Reynolds number to check the type of flow

$Re=(\rho VD)/(\mu )$

Re=(872.5* 0.064*0.020)/(0.15531 )

Re=7.18 So we can say that this is laminar flow and we know that pressure drop for laminar flow is given as

$P_1-P_2=(32\mu VL)/(d^2)$

Now by putting the values the pressure drop for per unit length

P_1-P_2=(32* 0.15531* 0.064 )/(0.02^2)