Answer to Question #278759 in Astronomy | Astrophysics for Gingging

Question

2.A horizontal pipe is 2 cm in radius at one end and increases in size so that it is 5 cm in radius at

the other end. Water is pumped into the small end of the pipe at 10 m/s and a pressure of

300 kPa (kilopascal). Find the

a.) speed and

b.) pressure of the water at the pipe’s large end.

Answer 1

Explanations & Calculations

• Hello Gingging,

• This whole thing deal with the Bernoulli's law and the concept of volumetric flow rate.
• Since the water is incompressible, the volumetric flow rate ($\small Q = Av$ ) is constant throughout the pipe.

a)

• Applying volumetric flow rate to the input and output ends,

$\qquad\qquad \begin{aligned} \small Q=A_1v_1&=\small A_2v_2\\ \small v_2&=\small \frac{A_1}{A_2}.v_1\\ &=\small \frac{\pi r_1^2}{\pi r_2^2}.v_1\\ &=\small \Big(\frac{r_1}{r_2}\Big)^2.v_1 \end{aligned}$

• Here $\small r_1\,\&\,v_1$ refer to the input side and the out put side follows the notation accordingly.

b)

• For this part, you can use the Bernoulli's equation — $\small P+\frac{1}{2}\rho v^2+\rho g h=\text{constant}$ — for the input and out put conditions.
• Since the pipe is level throughout, any graduation is neglected, so that $\small h = 0$ .
• By now you have found the speed at the output ($\small v_2$ ) which is needed for this step.

$\qquad\qquad \begin{aligned} \small P_1+\frac{1}{2}\rho v_1^2 &=\small P_2+\frac{1}{2}\rho v_2^2\\ \small P_2 &=\small P_1+\frac{1}{2}\rho(v_1^2-v_2^2) \end{aligned}$

• Now you can give it a try substituting the values accordingly and getting the answers.
• use the units correctly, when the lengths are input in meters, the speeds are obtained in meters per second.
• Let me know in comments if you find any difficulty.