Respuesta :
Explanation:
The given data is as follows.
Diameter = 0.1 m, [tex]P_{1}[/tex] = 1000 kPa
[tex]P_{2}[/tex] = 500 kPa
Change in pressure [tex]\Delta P[/tex] = 1000 kPa - 500 kPa = 500 kPa
Since, 1000 Pa = 1 kPa. So, 500 kPa will be equal to [tex]500 \times 10^{3}[/tex].
Q = 5 [tex]m^{3}/min[/tex] = [tex]\frac{5}{60} m^{3}/sec[/tex] = 0.0833 [tex]m^{3}/sec[/tex]
It is known that Q = [tex]A \times V[/tex]
where, A = cross sectional area
V = speed of the fluid in that section
Hence, calculate V as follows.
V = [tex]\frac{Q}{A}[/tex]
= [tex]\frac{Q \times 4}{\pi \times d^{2}}[/tex]
= [tex]\frac{0.0833 \times 4}{3.14 \times (0.1)^{2}}[/tex]
= 10.61 m/sec
Also it is known that Reynold's number is as follows.
Re = [tex]\frac{\rho \times V \times d}{\mu}[/tex]
= [tex]\frac{1000 \times 10.61 \times 0.1}{10^{-3}}[/tex]
= 1061032.954
As, it is given that the flow is turbulent so we cannot use the Hagen-Poiseuille equation as follows. Therefore, by using Blasius equation for turbulent flow as follows.
[tex]\Delta P = \frac{0.241 \times \rho^{0.75} \times \mu^{0.25} \times L}{D^{4.75}} \times Q^{1.75}[/tex]
[tex]500 \times 10^{3} = \frac{0.241 \times (1000)^{0.75} \times (10^{-3})^{0.25} \times L}{(0.1)^{4.75}} \times (0.0833)^{1.75}[/tex]
[tex]500 \times 10^{3} = 5535.36 \times L[/tex]
L = [tex]\frac{500 \times 10^{3}}{5535.36}[/tex]
= 90.328 m
Thus, we can conclude that 90.328 m length galvanized iron line should be used to reach the desired outlet pressure.
