The following questions are from a simulation using this link: http://phet.color

Question

The following questions are from a simulation using this link: http://phet.colorado.edu/en/simulation/fluid-pressure-and-flow

1.How does flow rate (volume per second) change as a pipe becomes smaller (as in #3)?

2.How does the flow rate (volume per second) change as a pipe rises (as in #2)?

Explain your answers to the two questions above.

3.If one trillion water molecules are added to one end of a filled garden hose, what happens at the other end of the garden hose?

4.As fluid density increases, the fluid speed (increases / decreases / stays the same) and the pressure (increases / decreases / stays the same). Explain your answers.

5.Imagine two identical houses, one at the top of a hill and one at the bottom of a valley. In which one would you rather take a shower?

Why?

6.Shower pipes in newer houses have smaller diameters than those in older houses. Why do you suppose this is?

7.Using the formulas on the front page, calculate this: water flows horizontally through a garden hose with an inner diameter of .012 m at a speed of 7.8 m/s. It exits out a small nozzle with a diameter of only 0.0085 m. How fast it is travelling out of the nozzle? ____m/s

Explanation / Answer

1. Flow rate is independent of pipe diameter. (Q = V*A = constant)

2. Flow rate remains the same if the pipe rises. (P + 1/2*rho*V^2 + rho*g*z = constant)

3. One trillion water molcules will come out from other end of the hose. (Mass conservation)

4. Speed and Pressure remain the same if density is changed. They are independent variables.

5. In the bottom of the valley because pressure will be higher there. (P + 1/2*rho*V^2 + rho*g*z = constant. As z increases, P decreases.)

6. Reducing the diameter will increase the increase the velocity. (Q = VA = constant)

7.

V*A = constant

V*pi/4*d^2 = constant

V*d^2 = constant

V1*d1^2 = V2*d2^2

V2 = V1*(d1/d2)^2

V2 = 7.8*(0.012 / 0.0085)^2

V2 = 15.546 m/s

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