Question

A 6cm diameter horizontal pipe gradually narrows to 4cm.

Whenwater flows throught this pipe at a certain rate, the
guagepressure in these two sections is 32.0kPa and 24kPa
respectively.What is the volume rate of flow?

Answer 1

Answer:
`Q=0.5612 m^3/s`

Step-by-step explanation:

Given

diameter of pipe(
`d_1`)=6 cm

diameter of pipe(
`d_2`)=4 cm

`P_1=32 kPa`

`P_2=24 kPa`

`A_1=(\pi )/(4)6^2=9\pi cm^2`

`A_2=(\pi )/(4)4^2=4\pi cm^2`

`v_1=(Q)/(A_1)`

Applying bernoulli's equation

`(P_1)/(\rho g)+(v^2_1)/(2g)+z_1=(P_2)/(\rho g)+(v^2_2)/(2g)+z_2`

`(P_1)/(\rho g)+((Q^2)/(A_1^2))/(2g)+z_1=(P_2)/(\rho g)+((Q^2)/(A_2^2))/(2g)+z_2`

since
`z_1=z_2`

`(32* 10^3)/(10^3* 9.81)+(Q^2)/(2A_1^2g)=(24* 10^3)/(10^3* 9.81)+(Q^2)/(2A_2^2g)`

`Q^2=(8* 2* 81\pi ^2* 16\pi ^2* 10^(-4))/(65\pi ^2)`

`Q^2=3149.3722* 10^(-4)`

`Q=\sqrt{3149.3722* 10^(-4)}`

`Q=0.5612 m^3/s`