Question

Water is pumped through a 60 m pipe of 0.3 m diameter from a lower reservoir to a higher reservoir whose surface is 10 m above the lower one. The sum of minor loss coefficients is KL = 14.5. When the pump adds 40 kW to the water the flowrate is 0.20 m3 /s. Determine the pipe roughnes

Answer 1

`\epsilon = 0.028*0.3 = 0.0084`

Step-by-step explanation:

`(P_1)/(\rho) + (v_1^2)/(2g) +z_1 +h_p - h_l =(P_2)/(\rho) + (v_2^2)/(2g) +z_2`

where
`P_1 = P_2 = 0`

V1 AND V2 =0

Z1 =0

`h_P = (w_p)/(\rho Q)`

`=(40)/(9.8*10^3*0.2) = 20.4 m`

`20.4 - (f [(l)/(d)] +kl) (v_1^2)/(2g) = 10`

we know thaT
`V  =(Q)/(A)`

`V = (0.2)/(\pi (0.3^2)/(4)) =2.82 m/sec`

`20.4 - (f (60)/(0.3) +14.5) (2.82^2)/(2*9.81) = 10`

f = 0.0560

`Re =(\rho v D)/(\mu)`

`Re =(10^2*2.82*0.3)/(1.12*10^(-3)) =7.53*10^5`

fro Re = 7.53*10^5 and f = 0.0560

`\epsilon = 0.028*0.3 = 0.0084`