Question

Water is flowing into a factory in a horizontal pipe with a radius of 0.0183 m at ground level. This pipe is then connected to another horizontal pipe with a radius of 0.0420 m on a floor of the factory that is 12.6 m higher. The connection is made with a vertical section of pipe and an expansion joint. Determine the volume flow rate that will keep the pressure in the two horizontal pipes the same.

Answer 1

`0.0168 m^3/s`

Step-by-step explanation:

We are given that

`r_1=0.0183 m`

`h_1=0`

`r_2=0.0420 m`

`h_2=12.6 m`

Let
`P_1=P_2=P`

By using Bernoulli theorem

`P+(1)/(2)\rho v^2_1+\rho gh_1=P+(1)/(2)\rho v^2_2+\rho gh_2`

`(1)/(2)\rho v^2_1+\rho gh_1=(1)/(2)\rho v^2_2+\rho gh_2`

`v^2_1+2gh_1=v^2_2+2gh_2`

`A_1v_1=A_2v_2`

`v_1=(A_2v_2)/(A_1)`

`((A_2)/(A_1))^2v^2_2+2g* 0=v^2_2+2* 9.8* 12.6`

`((\pi r^2_2)/(\pi r^2_1))^2v^2_2-v^2_2=246.96`

`v^2_2(((r^2_2)/(r^2_1))^2-1)=246.96`

`v^2_2=246.96(r^4_1)/(r^2_4-r^4_1)`

`v_2=\sqrt{246.96(r^4_1)/(r^4_2-r^4_1)}`

`v_2=\sqrt{246.96* ((0.0183)^4)/((0.042)^4-(0.0183)^4)}`

`v_2=3.038 m/s`

Volume flow rate =
`A_2v_2`

Volume flow rate =
`\pi r^2_2v_2=\pi (0.042)^2* 3.038=0.0168 m^3/s`