Question

A horizontal pipe contains water at a pressure of 110 kPa flowing with a speed of 1.7m/s. The pipe narrows to one-half its original diameter. A. What is the speed of the water when the pipe narrows?B. What is the pressure of the water when the pipe narrows?

Answer 1

We will have the following:

We are given:

`\begin{gathered} P=110kPa\Rightarrow P=110000Pa \\ v=1.7m/s \end{gathered}`

Then, We will have that the area of the pipe is given by:

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

And the final area of the pipe is given by:

`A_2=(\pi(d/2)^2)/(4)\Rightarrow A_2=(\pi d^2)/(16)`

Now, we will have that generally:

`A_1v_1=A_2v_2`

So:

`(\pi d^2)/(4)\cdot1.7m/s=(\pi d^2)/(16)\cdot v_2\Rightarrow v_2=(1)/(4)(16)(1.7m/s)`
`\Rightarrow v_2=6.8m/s`

A. The speed of the water when the pipe narrows is then 6.8 m/s.

For the pressure we determine it using Bernuolli's equation of constant height to calculate it, that is:

`p_1+(1)/(2)\cdot\rho\cdot v^2_1=p_2+(1)/(2)\cdot\rho\cdot v^2_2`

Here rho is the density of water, that we know is 1000kg/m^3; then:

`p_2=p_1+(1)/(2)\cdot\rho\cdot(v^2_1-v^2_2)\Rightarrow p_1=(110000)+(1)/(2)(1000)\cdot(1.7^2-6.8^2)`
`\Rightarrow p_1=88325Pa\Rightarrow p_1=88.325\text{kPa}`

B. The pressure of the water when the pipe narrowsis then 88.325 kPa.