Bernoulli's Equation and a water pipe

[Zaros]

Homework Statement

The water supply of a building is fed through a main pipe 6.00cm in diameter. A 2.00cm diameter tap is located 2.00m above the main pipe. When the tap is turned on it fills a 25.0L container in 30.0s if no other taps are turned on at the same time.
a) What is the speed at which the water leaves the tap?
b) What is the gauge pressure in the 6cm main pipe?

Homework Equations

P1+0.5pv1^2 + pgy1 = P2 +0.5pv2^2 + pgy2
Pg = Ps - Pa

The Attempt at a Solution

I solved part (a) easily enough by dividing 25 by 30 to get 0.833 L/s
and I'm sure i have to use Bernoulli's equation to solve this but have been having problems understanding the equation and the concept behind it and how i would use this in the particular problem I'm faced with.

Thanks for helping
Zaros

 

[Zaros]

Is anyone able to help with this problem. I may be getting at it from the wrong direction so you might want to disregard some of the equations I'v used.

 

[zorro]

apply the equation of continuity for the first part.
Q(rate of flow) = 0.833 x 10^-3 m³/s
Q=A₂V₂
calculate V from here which is the answer.

Use Bernouillis equation for the second part

P1+0.5pV₁^2 = Pa +0.5pV₂^2 + pgh here h=height of the tap from the main pipe.

P1 will be your answer ( I got it as 2.3 x 10⁴ Pa)

 

[Zaros]

Great thanks for this you really helped. I was wondering how to use Bernoulli's equation but never thought that the second pressure would be that of the atmosphere.

Thanks.
Zaros

 

[Zaros]

i think i must be doing something wrong. Is V₁ = V₂ if not what should it be. Also if i try this i get 3011370600 Pa which is outrageously big. For part a i got a speed of 2.65 * 10³ m/s any ideas what i did wrong

 

[zorro]

You messed up with some units. V₂ is 2.65m/s
V₁ is not equal to V₂

Calculate V₁ from A₁V₁=A₂V₂