Calc Volume Flow Rate of Water from Tap (45mm ht)

[coltsamuel96]

1. Homework Statement
Tap water flowing has an cross section area of 1.2cm^2 near the tap. after a height, h=45mm the cross section becomes 0.35cm^2...find the volume flow rate of water from the tap

Homework Equations

g=9.8m/s^2

The Attempt at a Solution

i think of using these two equations A0*V0=A*V and V0^2=V^2-2gh... but it doesn´t seem correct... any helps? thanks in advance

 

[dacruick]

Hi coltsamuel,

are you assuming that the speed at the instance it leaves the tap is zero?

 

[coltsamuel96]

sorry, it is volume flux rate and not volume flow rate...

 

[coltsamuel96]

no i guess it should have a speed

 

[dacruick]

You have 2 diameters and an acceleration, but you need an initial condition.

 

[gneill]

The rate of flow [V/T] past a point is the product of the velocity of the flow and the cross-sectional area of the flow. From two cross sections and their vertical separation you can pin down the two velocities because the change in velocity over a given distance due to gravitational acceleration depends upon the initial velocity (if an object is falling quickly it spends less time accelerating in the given distance).

So, what expressions can you write relating the velocities with the height change? How about the velocities and cross-sectional areas?

 

[coltsamuel96]

ok this is from a book, A0v0=Av(1st equation) v^2=v0^2+2gh(2nd equation) from 1st and 2nd eliminating v and solving for v0, we get vo=(2ghA^2/A0^2-A^2)^1/2...i don´t understand how we get this v0... any helps?

 

[gneill]

ok this is from a book, A0v0=Av(1st equation) v^2=v0^2+2gh(2nd equation) from 1st and 2nd eliminating v and solving for v0, we get vo=(2ghA^2/A0^2-A^2)^1/2...i don´t understand how we get this v0... any helps?

Are you asking where the equations come from, or to have the algebra detailed?

 

[coltsamuel96]

everything clear now... thank you