Question

A 7.5-cm-diameter horizontal pipe gradually narrows to 4.5 cm . When water flows through this pipe at a certain rate, the gauge pressure in these two sections is 31.0 kPa and 25.0 kPa , respectively. What is the volume rate of flow?

Answer 1

Step-by-step explanation:

Given

Diameter at section 1 is
`d_1=7.5\ cm`

diameter at section 2 is
`d_2=4.5\ cm`

Pressure
`P_1=31\ kPa`

`P_2=25\ kPa`

`A_1=(\pi d_1^2)/(4)=44.184\ cm^2`

`A_2=(\pi d_2^2)/(4)=15.906\ cm^2`

Applying Bernoulli's Equation we get

`P_1+\rho v_1^2+\rho gh_1=P_2+\rho v_2^2+\rho gh_2`

since there is no change in height therefore
`h_1=h_2` from continuity equation we can write as

`A_1v_1=A_2v_2`

`v_1=(A_2)/(A_1)v_2`

`P_1-P_2=(\rho )/(2)\left [ v_2^2-v_1^2\right ]`

`P_1-P_2=(\rho )/(2)\left [ v_2^2-((A_2)/(A_1))^2v_1^2\right ]`

`(A_2)/(A_1)=0.359`

`(31-25)* 10^3=(10^3)/(2)\left [ v_2^2-(0.3599)^2v_2^2\right ]`

`6* 2=v_2^2=\left [ 1-0.359^2\right ]`

`v_2=3.713\ m/s`

`Q=A_2v_2`

`Q=15.906* 10^(-4)* 3.173`

`Q=59* 10^(-4)\ m^3/s`