Water Bottle: relationship between Volume and Time

[srsthsr]

To all who have given up their time to help. Could you please have a look at my experiment, I am really stuck with this and don't know how to go on with it. I am unable to find a suitable equation to prove it mathematically as there is a varying change in both time taken and the pressure inside the bottle, could anyone please give me a hint so I can get on with it

Homework Statement

To investigate the relationship between the initial volume and the time it takes for the water to leak out of a hole beneath the bottle.
Variables:
Dependent Variable: Time it takes for all the water to flow out of the bottle
Independent Variable: Initial Volume of Water
Controlled Variable:Bottle and Gap size
Vol(ml)Time(s)
800 30.38
700 27.78
600 24.98
500 21.54
400 18.57
300 14.52
200 11.69
100 6.69
50 3.78

Experiment:(brief)
A certain volume of water is poured into a bottle with a blocked hole in the bottom.
unblock the hole and start the timer.
Measure the time it takes for the water to stop drippiing.

Homework Equations

p = ρgh?

Thank you so much

 

[tiny-tim]

welcome to pf!

hi srsthsr! welcome to pf!

(have a rho: ρ )

if A is the cross-section area of the bottle, and if V and h are the height and the volume of the water, and B is the cross-section area of the hole,

then V = Ah …

so what is dV/dt? ​

 

[srsthsr]

hi srsthsr! welcome to pf!

(have a rho: ρ )

if A is the cross-section area of the bottle, and if V and h are the height and the volume of the water, and B is the cross-section area of the hole,

then V = Ah …

so what is dV/dt? ​

Thank you so much for the reply.
However, I think I am missing something out.
if V=Ah then dV/dt = Adh/dt + hdA/dt right? how does it relate to B which is the cross section area of the hole? I am missing something out

I found out that the hight at Vol 800ml is 14.45 cm, the radius of the bottle is 4.77465, area cross section A is 225, and Area cross section B is 3.519...

 

[tiny-tim]

hi srsthsr!

if V=Ah then dV/dt = Adh/dt + hdA/dt right?
…
I found out that the hight at Vol 800ml is 14.45 cm, the radius of the bottle is 4.77465, area cross section A is 225, and Area cross section B is 3.519...

then dA/dt = 0, so it's just dV/dt = Adh/dt

how does it relate to B which is the cross section area of the hole?

it doesn't …

dV/dt = Adh/dt is just geometry

now you need a physics equation relating dV/dt to h and B

(in english: how does the loss of water depend on the depth and on B?)

 

[srsthsr]

hi srsthsr!


then dA/dt = 0, so it's just dV/dt = Adh/dt


it doesn't …

dV/dt = Adh/dt is just geometry

now you need a physics equation relating dV/dt to h and B

(in english: how does the loss of water depend on the depth and on B?)

I am sooooooo sorry for the late reply
I still however don't get it, i mean the depth? and the size of the hole? I am really confused i can't seem to find out the equation anywhere! i don't even know where to start looking =[...

Thank you so much for helping me out!

 

[tiny-tim]

well, say it in English …

how does the amount of water leaking out depend on the depth (the height) of water and on the size of the hole?

 

[srsthsr]

okay... in English... the bigger the hole the more amount of water leaks out in a certain amount of time. The greater the depth, the more pressure, also the more amount of water leaking out...
I get it, but i can't find an equation to relate all of them together and prove that the numbers that i got from the experiment actually is correct (to an extent) =[

 

[tiny-tim]

hi srsthsr!

The greater the depth, the more pressure, also the more amount of water leaking out...

right!

so you need two equations, one to tell you the https://www.physicsforums.com/library.php?do=view_item&itemid=80" , and one to tell how the pressure affects the speed

i] pressure = ρgh (ρ is density)

ii] https://www.physicsforums.com/library.php?do=view_item&itemid=115"

… the bigger the hole the more amount of water leaks out in a certain amount of time.

this is just geometry … if you know the speed, then just write out the definition of flow …

iii] flow = volume per time = area times distance per time = area times speed

 

[srsthsr]

Thank you for the rapid response, the problem is though, that the speed is not constant and varies as pressure and volume changes doesn't it?... how would I find the average speed?

 

[tiny-tim]

Thank you for the rapid response, the problem is though, that the speed is not constant and varies as pressure and volume changes doesn't it?... how would I find the average speed?

i don't understand … why do you want to find the average speed?

all the question asked for was …

To investigate the relationship between the initial volume and the time it takes for the water to leak out of a hole beneath the bottle.

 

[srsthsr]

OK, I will give it a try, I think I've found out what you meant =P thank you! will reply soon =]

 

[srsthsr]

Okay I tried out your equation, but the constant from Bernoulli's equation keeps changing.
Could you please tell me what I did wrong?
1) Flow rate of 0.8L
(0.8e-3)/30.38=2.63e-5(m^3/s)
2)speedxarea(of hole)=volume/time
so speed=2.63e-5/0.0056x2x3.1416=7.484e-4
3) P=pgh=1x9.81x0.1445=1.417545
4)using Bernoulli's equation along any streamline of a steady incompressible non-viscous flow: constant=2.8351
5)Flow=Vol/t=speed x area
6)so speed=V/at
7)rearranging equation, 2(pgh)+(0.5(V/at)^2) should equal to 2.8351, but it doesn't for other values! i.e. Vol=0.7L and time=27.78, =[ please help

 

[tiny-tim]

hi srsthsr!

no, your v in https://www.physicsforums.com/library.php?do=view_item&itemid=115" has to be the instantaneous velocity at time t …

your data …

Dependent Variable: Time it takes for all the water to flow out of the bottle
Independent Variable: Initial Volume of Water

… are giving you initial volume over total time,

which gives you an average velocity instead

 

[srsthsr]

then what should i do? I am so confused ;(

 

[tiny-tim]

you need to go back to your data and find dV/dt as a function of the volume V …

for example, you could draw a graph and measure the slope

 

[srsthsr]

OKAY, will do that, i have a plotted graph already, however, isn't the relationship parabolic? or is it a straight line?
once i get dV/dT which is A(dh/dt) then do i divide the answer by A and then that would give me dh/dt or in english, rate of change of height per time... which is the instantaneous speed?

I hope i am not being too much of a trouble for you =[

Be back soon, looking forward to your reply =]

 

[tiny-tim]

no, your v in https://www.physicsforums.com/library.php?do=view_item&itemid=115" needs to be the speed at the hole, which will be dh/dt times … ?

 

[srsthsr]

the area of the hole!

 

[tiny-tim]

no that can't be right …

look at the dimensions …

dh/dt is a speed (L/T), so (dh/dt)B would be speed times area (L³/T) …

so you need an extra … ?

 

[srsthsr]

AHHHH, I am going crazy =/... ok L^3/T is the rate of flow... but we want velocity... so do i times it by 1/h? i don't know... and what does the gradient represent, i understand that its Adh/dt so what does Adh/dt mean? =[ this is confusing me so much i don't know what to do!

 

[tiny-tim]

hi srsthsr!

it's really very logical …

dV/dt = Adh/dt

ie flow rate = bottle area times bottle speed

similarly

flow rate = hole area times hole speed

ie dV/dt = Bv

so v = … ?

(btw, this is all geometry, not physics!)

(and next you put this v into Bernoulli's equation)

 

[srsthsr]

so r u saying that (dV/dt)/B is v? and i just put that into the equation and i get the constant?

also, is the P in Bernoulli's equation the same as pgh?

 

[tiny-tim]

is the P in Bernoulli's equation the same as pgh?

yes

so r u saying that (dV/dt)/B is v? and i just put that into the equation and i get the constant?

Bernoulli will give you an equation in h, so you'll need v as a function of h, not of V, won't you?