Answer :
Answer:
[tex]Q=4.98\times 10^{-3}\ m^3/s[/tex]
Explanation:
Given that
L= 50 m
Pressure drop = 130 KPa
For Copper tube is 3/4 standard type K drawn tube
Outside diameter=22.22 mm
Inside diameter=18.92 mm
Dynamic viscosity for kerosene
[tex]\mu =0.00164\ Pa.s[/tex]
Pressure difference given as
[tex]\Delta P=\dfrac{128\mu QL}{\pi d_i^4}[/tex]
Where
L is length of tube
μ is dynamic viscosity
Q is volume flow rate
d is inner diameter of tube
ΔP is pressure drop
Now by putting the values
[tex]\Delta P=\dfrac{128\mu QL}{\pi d_i^4}[/tex]
[tex]130\times 1000=\dfrac{128\times 0.00164\times 50\times Q}{\pi\times 0.0189^4}[/tex]
[tex]Q=4.98\times 10^{-3}\ m^3/s[/tex]
So flow rate is [tex]Q=4.98\times 10^{-3}\ m^3/s[/tex]