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Consider water flow in a piping network. The pressure, velocity, and elevation at a specified point (point 1) of the flow are 300 kPa, 2.4 m/s, and 5 m. The velocity and elevation at point 2 are 1.9 m/s and 18 m. Take the correction factor to be 1.

If the irreversible head loss between point 1 and point 2 of the pipe is 2 m, determine the pressure of water at point 2 by using appropriate software.


a.131 kPa

b.143 kPa

c.154 kPa

Respuesta :

Answer:

pressure of water at point 2 is 151.1 kPa

Explanation:

given data

pressure p1 = 300 kPa = 300 × 10³ Pa

velocity v1 =  2.4 m/s

elevation h1 =  5 m

velocity v2 = 1.9 m/s

elevation h2 = 18 m

correction factor α = 1

head loss h(L)= 2 m

here g = 9.81 m/s

r = 10 × 10³ N/m³

solution

we apply here bernoulli's equation that is

[tex]\frac{P1}{r}[/tex] + [tex]\frac{\alpha v1^2}{2g}[/tex] + h1  =   [tex]\frac{P2}{r}[/tex] + [tex]\frac{\alpha v2^2}{2g}[/tex] + h2 + h(L) ........................1

put here value and we will get P2

[tex]\frac{300*10^3}{10*10^3}[/tex] + [tex]\frac{1*2.4^2}{2*9.81}[/tex]  + 5 = [tex]\frac{P2}{10*10^3}[/tex] + [tex]\frac{1*1.9^2}{2*9.81}[/tex] + 18 + 2  

solve it and we get  

P2 = 151.1 kPa

so pressure of water at point 2 is 151.1 kPa

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