Question

Water flows at the rate of 200 I/s upwards through a tapered vertical pipe. The diameter at Marks(3) CLO5) the bottom is 240 mm and at the top 200 mm and the length is 5m. The pressure at the bottom is 8 bar, and the pressure at the topside is 7.3 bar. Determine the head loss through the pipe. Express it as a function of exit velocity head.

Answer 1

Answer: 5.35m

Step-by-step explanation:

By using energy equation:

`(P_1)/(\gamma)+z_1+(v_1^(2) )/(2g)  =(P_2)/(\gamma)+z_2+(v_2^(2) )/(2g)+h_(L)`

`\gamma=specific weight`

`v_(1) =(Q_1)/(A_1) =(0.2)/((\pi )/(4) * 0.24^2) =4.42 m/s\\v_(2) =(Q_2)/(A_2) =(0.2)/((\pi )/(4) * 0.2^2) =6.37 m/s`

`h_(L)=(P_1-P_2)/(\gamma)+z_1-z_2+(v_1^(2)-v_2^(2) )/(2g)`

`h_(L)=((8-7.3)* 100 )/(9.81)  +0+5+(4.42^2-6.37^2)/(2* 9.81)`

`h_L=7.135+3.927\\h_L=11.062m`

exit velocity head =
`(v_2^(2) )/(2g)`=2.068m

head loss as a function of exit velocity head is=
`(11.062)/(2.068)`

`h_L=K* V_e`

head loss as a function of exit velocity head =5.35m