Height and Pressure Relating the Two

[kurosaki_ichi]

Homework Statement

A cylindrical bucket, open at the top, is 24.0cm high and 15.0cm in diameter. A circular hole with a cross-sectional area 1.86cm^2 is cut in the center of the bottom of the bucket. Water flows into the bucket from a tube above it at the rate of 2.63×10−4 m^3/s


-Not sure how to begin please, anyone could provide me a hint, guidelines, a starting point thank you :)

 

[mgb_phys]

What is the question ?

 

[kurosaki_ichi]

Very soory the question was;

How high will the water in the bucket rise?

 

[kurosaki_ichi]

From what I know the units of the flow of water out has m^3/s i think i know this

Av=constant in our case Av=2.63E-4 m^3/s

 

[tiny-tim]

Welcome to PF!

A cylindrical bucket, open at the top, is 24.0cm high and 15.0cm in diameter. A circular hole with a cross-sectional area 1.86cm^2 is cut in the center of the bottom of the bucket. Water flows into the bucket from a tube above it at the rate of 2.63×10−4 m^3/s
…
-Not sure how to begin please, anyone could provide me a hint, guidelines, a starting point thank you :)

Hi kurosaki_ichi! Welcome to PF!

Hint: the water level will reach equilibrium when the flow out of the bottom equals the flow in …

so start with …

what is the pressure at the bottom when the height is h? ​

 

[mgb_phys]

You need to calculate the rate the water flows out of the bucket - which depends on the height of the water (ie the pressure) and the size of the hole.

Have you done Bernouilli's equation?

 

[kurosaki_ichi]

? I kinda have done berunillis equation, so do i use berunillis equation to cacuuluter pressure? or p=pa + rohgh?
help someone please, still lst lol

 

[tiny-tim]

? I kinda have done berunillis equation, so do i use berunillis equation to cacuuluter pressure? or p=pa + rohgh?
help someone please, still lst lol

(have a rho: ρ … and it's Bernoulli's equation )

Yes, they're the same thing in this case: p = p_a + ρgh

 

[kurosaki_ichi]

So if am right here am guessing i must:

pressure=Patomostpher density*g*h

so patm is 1.00 atm

density of water=1000kg/m^3

g=9.8m/s^2

height of water colum is the heigh of my cylinder? 24.0cm high

are those my given values?

 

[tiny-tim]

So if am right here am guessing i must:

pressure=Patomostpher density*g*h

so patm is 1.00 atm

density of water=1000kg/m^3

g=9.8m/s^2

height of water colum is the heigh of my cylinder? 24.0cm high

are those my given values?

Yup!

(but you won't need the value of atmospheric pressure, since it'll be the same on both sides of the equation )

 

[kurosaki_ichi]

THank you so much for ur help i will now try and attempt the problem furthiermore i am grateful i will ask if i need any more help :) much thankz

 

[mgb_phys]

Another way to calculate the rate of flow out is to use conservation of energy.
A particle of water will flow out of the bottom of the bucket with a kinetic energy equal to the potential energy if it dropped from the surface of the water
( - it's just another way of stating Bernouilli's law - but is simpler to understand)

 

[kurosaki_ichi]

I got my pressure to be 103.677 kPa now am not sure what to do? do i know use the continuty equation a1v1=a2v2?

 

[kurosaki_ichi]

how can I work with it using conservation of energy if am not given a entry speed or flowing out speed?

 

[mgb_phys]

You need to relate the water pressure at the hole to the flow speed, you can do this with Bernouilli's equation or as I suggested.

Imagine a drop of water (mass m) in the bucket at the surface, it has a potential energy =mgh, if all this energy goes into kinetic energy as it flows out (=mv^2) then you have the speed of the drop of water flowing out.

You don't need the input speed if it's a bucket - you assume the water stops when it is poured in (this is the same as saying there is no water pressure pushing down on the surface form the input flow).

Now you have the output flow speed, and the hole area you can easily work out a volume output flow rate.

 

[kurosaki_ichi]

So your telling me i should do it as Ea=Eb with potential energy at the top and kinetic energy at the bottom, can't i use the continuity equation i used it to solve for V2 i have pressure i just don't know how to relate evreything togethor i have my pressure solved also how does h play a part? is it mgh=1/2mv^2?

 

[mgb_phys]

Yes - all you need to find the flow rate (m^3/s) out of the bucket.
That depends on the area of the hole and the speed of the water coming out - you can get this from Bernouilli's equation and the pressure difference across the hole.

A simpler method (IMHO) is to image you were dropping small rocks from a distance 'h' above the bottom of the bucket through a hole in the bottom of the bucket. It's very easy to work out the speed they are going at the bottom of the bucket - right?

Now instead of rocks just picture a drop of water it starts at the surface of the water ( a distance 'h' ) from the bottom of the bucket and ends up going out of the hole with a certain speed - it's exactly the same as the rock.

 

[kurosaki_ichi]

serisouly confused :S...lost please help

 

[mgb_phys]

Ok - back to the start ;-)
You have a bucket of known volume with a known flow rate in.
To find how the water height changes with time you need to also know the flow rate out.

To get the flow rate out - you need to know the speed of the water flowing out.
The easiest way to do this is just conservation of energy.
This will give you an equation for the speed of the water out of the hole only depending on the water height

pe = ke = mgh = 1/2m v^2

 

[kurosaki_ichi]

mgh=1/2mv^2

okay mass cancels out

so i have g=9.8m/s^2

h=0.24 m

using the above equation i get 1.08m/s^2

so ineed the flow rate i multiply it by the area? the cross sectional area? correct

 

[mgb_phys]

mgh=1/2mv^2

okay mass cancels out

so i have g=9.8m/s^2

h=0.24 m

using the above equation i get 1.08m/s^2

so ineed the flow rate i multiply it by the area? the cross sectional area? correct

Exactly,
But you don't know that 'h=0.24' is the height of the water, that's just the top of the bucket.
You need to write an equation for the output rate of flow in terms of 'h'

The rate the water flows out varies with 'h' so since you know the rate it flows in - there will be some 'h' where the two are equal, that is the maximum height the water can reach.

 

[kurosaki_ichi]

I think i got the answer my final answer heigh of water colum was 10.2 cm

i solved for vb then i found the cross sectional area of the bucket, then applied burnilles equation and solved to get 10.2 cm for height hope that is correct.