Question

Kerosene flows through 3/4 standard type K drawn copper tube. The pressure drop measured at two points 50 m apart is 130 kPa. Determine the flow rate of the kerosene

Answer 1

Answer:

$Q=4.98\times 10^{-3}\ m^3/s$.

Explanation:

Given that

L= 50 m

Pressure drop = 130 KPa

copper tube is 3/4 standard type K drawn tube.

From standard chart ,the dimension of 3/4 standard type K copper tube given as

Outside diameter=22.22 mm

Inside diameter=18.92 mm

Dynamic viscosity for kerosene

$\mu =0.00164\ Pa.s$

We know that

$\Delta P=\dfrac{128\mu QL}{\pi d_i^4}$

Where Q is volume flow rate

L is length of tube

$d_i$ is inner diameter of tube

ΔP is pressure drop

μ is dynamic viscosity

Now by putting the values

$\Delta P=\dfrac{128\mu QL}{\pi d_i^4}$

$130\times 1000=\dfrac{128\times 0.00164\times 50Q}{\pi \times 0.01892^4}$

$Q=4.98\times 10^{-3}\ m^3/s$

So flow rate is $Q=4.98\times 10^{-3}\ m^3/s$.