- #1

Matt Armstrong

## Homework Statement

A horizontal length of pipe starts out with an inner diameter (not radius!) of 2.60 cm, but then has a tapered middle part which narrows to a diameter of 1.60 cm. When water flows through the pipe at a certain rate, the gauge pressure is 34 kPa in the first (wider) section and 25 kPa in the second (narrower) section.

(a) What is the ratio of the flow speed in the narrow section to the flow speed in the wider section?

(b) What are the actual water flow speeds in the wider section and in the narrower section? (Hint: you will need to set up and solve Bernoulli’s equation. Bernoulli’s equation determines what flow rate is produced by this particular pair of inlet and outlet pressures.)

(c) What is the volume flow rate through the pipe?[/B]

## Homework Equations

r_1 = .013 m, r_2 = .008 m, P_1 = 34000 Pa, P_2 = 25000 Pa

A = pi*r^2

Q=Av

P + .5rho*v^2 = constant

P_1 + .5rho*(v_1)^2 = P_2 + .5rho*(v_2)^2[/B]

## The Attempt at a Solution

For a, since the flow rate is Q = v*a, I made a ratio between the velocities and the Areas and got a ratio of 2 (for narrow) to 5 (for wider). Part b is where I have trouble. I know that Bernoulli's principle can be written between any singular point or any two points along the pipe, and I tried to solve for v_1 and v_2 using the second equation, but that wouldn't be possible because there are two unknown variables. I know that in certain applications where a difference in height delta-y can be used in a formula for velocity, but since no height or reference frame is identified, I left those out of my principle equations as well as neglected to use them. I feel like if I can understand part b, then I can do part c on my own. What do I need to do?

Thanks for any information you can provide