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

AI Thread Summary
To determine the volume flow rate of water leaving the faucet, the gauge pressure of 102 kPa and the height of 10 m must be considered. The simplified Bernoulli equation is used to calculate the velocity of water, which is then multiplied by the cross-sectional area of the faucet (2.0 cm²) to find the volume flow rate. The initial attempt yielded an incorrect value of 8 cm³/m, indicating a misunderstanding in the conversion of velocity to volume flow rate. The correct approach involves ensuring that the velocity is in cm/s before applying it to the area for the final calculation. The solution requires careful attention to unit conversions and the application of the volume flow rate formula Q = Av.
kathmill
Messages
4
Reaction score
0

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?
 
Physics news on Phys.org
You solved for the speed v (in m/s). You need the volume flow rate, which is Av.
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top