Question

A horizontal pipe contains water at a pressure of 110 kPa flowing with a speed of 1.4 m/s. When the pipe narrows to one half its original diameter, what is (a) the speed and (b) the pressure of the water?

Answer 1

a

`v_2 = 5.6 \ m/s`

b

`P_2 = 80600 \ Pa`

Step-by-step explanation:

From the question we are told that

The pressure of the water in the pipe is
`P_1= 110 \ kPa = 110 *10^(3 ) \ Pa`

The speed of the water is
`v_1 = 1.4 \ m/s`

The original area of the pipe is
`A_1 = \pi (d^2 )/(4)`

The new area of the pipe is
`A_2 = \pi * ([(d)/(2) ]^2)/(4) = \pi * ((d^2)/(4) )/(4) = \pi (d^2)/(16)`

Generally the continuity equation is mathematically represented as

`A_1 * v_1 = A_2 * v_2`

Here
`v_2` is the new velocity

So

`\pi * (d^2)/(4) * 1.4 = \pi * (d^2)/(16) * v_2`

=>
`(d^2)/(4) * 1.4 = (d^2)/(16) * v_2`

=>
`d^2 * 1.4 = (d^2)/(4) * v_2`

=>
`1.4 = 0.25 * v_2`

=>
`v_2 = 5.6 \ m/s`

Generally given that the height of the original pipe and the narrower pipe are the same , then we will b making use of the Bernoulli's equation for constant height to calculate the pressure

This is mathematically represented as

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

Here
`\rho` is the density of water with value
`\rho = 1000 \ kg /m^3`

`P_2 = P_1 + (1)/(2) * \rho [ v_1^2 - v_2^2 ]`

=>
`P_2 = 110 *10^(3) + (1)/(2) * 1000 * [ 1.4 ^2 - 5.6 ^2 ]`

=>
`P_2 = 80600 \ Pa`