Volume Flow Rate: Solve for Water Leaving Faucet in cm^3/s

[kathmill]

Homework Statement

A pump at ground level creates a gauge pressure of 102 kPa in the water line supplying an apartment building. The water leaves the tank into a pipe at a negligible speed. It travels up 10 m through the building and exits through a faucet. The cross-sectional area of the faucet is 2.0 cm^2. What is the Volume Flow Rate of the water leaving the faucet in cm^3/s?

Homework Equations

Volume Flow Rate: Q=Av
Bernoulli equation simplified: v = square root of: (Gauge Pressure - density of water*g*height)/(1/2 density)

So put V into Q=Av and solve!?

The Attempt at a Solution

I found it to be: 8 cm^3/m

Where am I going wrong?

 

[Doc Al]

You solved for the speed v (in m/s). You need the volume flow rate, which is Av.

 

[thomate1]

Your approach is correct, but your answer is not in the correct units. The volume flow rate should be in cm^3/s, not cm^3/m. You can convert your answer by multiplying by the unit conversion factor (m/s to cm/s), which is 100 cm/m.

Also, when using the Bernoulli equation, be sure to use consistent units throughout the equation. In this case, the gauge pressure is given in kPa, so the density of water should also be in kPa (which can be converted from g/cm^3 using the unit conversion factor of 1000 g/kg). This will give you a more accurate answer.

Finally, be sure to include the cross-sectional area of the faucet in your calculation. Your final equation should look like Q = A*v, where Q is the volume flow rate, A is the cross-sectional area of the faucet, and v is the velocity of the water leaving the faucet.

So, the correct answer should be:

Q = (2.0 cm^2)*(100 cm/m)*(square root of (102 kPa - 1000 g/kg * 9.8 m/s^2 * 10 m)/(1/2 * 1000 g/kg))

= 200 cm^3/s

= 0.2 L/s

= 0.2*60*60 cm^3/hour

= 720 cm^3/hour

Therefore, the volume flow rate of the water leaving the faucet is 200 cm^3/s or 0.2 L/s or 720 cm^3/hour.