Question

A liquid of density 1250 kg/m3 flows steadily through a pipe of varying diameter and height. At Location 1 along the pipe, the flow speed is 9.93 m/s and the pipe diameter d1 is 11.1 cm. At Location 2, the pipe diameter d2 is 16.7 cm. At Location 1, the pipe is Δy=8.89 m higher than it is at Location 2. Ignoring viscosity, calculate the difference ΔP between the fluid pressure at Location 2 and the fluid pressure at Location 1.

Answer 1

The pressure difference is
`\Delta P = 1.46 *10^(5)\ Pa`

Step-by-step explanation:

From the question we are told that

The density is
`\rho = 1250 \ kg/m^3`

The speed at location 1 is
`v_1 = 9.93 \ m/s`

The diameter at location 1 is
`d_1 = 11.1\ cm = 0.111 \ m`

The diameter at location 2 is
`d_1 = 16.7\ cm = 0.167 \ m`

The height at location 1 is
`h_1 = 8.89 \ m`

The height at location 2 is
`h_2 = 1 \ m`

Generally the cross- sectional area at location 1 is mathematically represented as

`A_1 = \pi * (d^2)/(4)`

=>
`A_1 = 3.142 * ( 0.111^2)/(4)`

=>
`A_1 = 0.0097 \ m^2`

Generally the cross- sectional area at location 2 is mathematically represented as

`A_2 = \pi * (d_1^2)/(4)`

=>
`A_2= 3.142 * ( 0.167^2)/(4)`

=>
`A_2 =0.0219 \ m^2`

From continuity formula

`v_1 * A_1 = v_2 * A_2`

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

=>
`v_2 = (0.0097 * 9.93)/(0.0219 )`

=>
`v_2 = 4.398 \ m/s`

Generally according to Bernoulli's theorem

`P_1 + \rho * g * h_1 + (1)/(2) \rho * v_1^2 = P_2 + \rho * g * h_2 + (1)/(2) \rho * v_2^2`

=>
`P_2 - P_1 = (1)/(2) \rho (v_1 ^2 - v_2^2 ) + \rho* g (h_1 - h_2)`

=>
`\Delta P = (1)/(2)* 1250* (9.93 ^2 - 4.398^2 ) + 1250* 9.8 (8.89- 1)`

=>
`\Delta P = 1.46 *10^(5)\ Pa`