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:

$A_1=\frac{\pi d^2}{4}$

And the final area of the pipe is given by:

$A_2=\frac{\pi(d/2)^2}{4}\Rightarrow A_2=\frac{\pi d^2}{16}$

Now, we will have that generally:

$A_1v_1=A_2v_2$

So:

$\frac{\pi d^2}{4}\cdot1.7m/s=\frac{\pi d^2}{16}\cdot v_2\Rightarrow v_2=\frac{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:

$p_1+\frac{1}{2}\cdot\rho\cdot v^2_1=p_2+\frac{1}{2}\cdot\rho\cdot v^2_2$

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

$p_2=p_1+\frac{1}{2}\cdot\rho\cdot(v^2_1-v^2_2)\Rightarrow p_1=(110000)+\frac{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.