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Torricelli's Law With Water Drawn In

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  1. Jan 29, 2015 #1
    1. The problem statement, all variables and given/known data
    An open cylinder of height 5ft and cross sectional area of 1 ft2 is initially empty. There is a small hole at the bottom of the cylinder with an area of 0.005 ft2. Water is drawn into the tank at a rate of 4.8ft3/min. At the same time water is discharged out of the cylinder through a small hole at the bottom of the tank. The outlet velocity may be estimated from Torricelli’s theorem. Estimate the time required for the water level in the tank to reach one foot depth.

    2. Relevant equations
    dVolume/dTime = -a(2gy)^(1/2)

    3. The attempt at a solution
    I tried to solve this by Torricelli's Law and setting up a model that :
    Volume of water remain in tank = Volume water in-volume water out
    But I found that time need for water level in tank = 1 foot is only 0.21 second
    I think the model is wrong, please help me with the model.
     
  2. jcsd
  3. Jan 29, 2015 #2

    Nathanael

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    Homework Helper

    You'll need to show your work in more detail than that.

    Of course, if there was no hole, it would take less time to fill. So how long would it take if there was no hole?
    (You can use that time to check if your answer is reasonable.)
     
  4. Jan 29, 2015 #3
    Thank you for your hint.
    I calculate the time to reach height of 1 foot is 12.5 second if there is no hole, then if there is a hole, it will takes longer time to reach that height. It proof my answer 0.21 second is wrong.

    here is my detailed calculation :
    cross sectional area of 1 ft2 = B
    area of hole 0.005 ft2 = A
    velocity in = 4.8 ft^3/min
    velocity out = squareroot of 2gh

    model :
    Volume of water remain in tank = Volume water in-volume water out

    equation :
    B (delta h) = 4.8(delta t) - A (squareroot of 2gh) (delta t)
    B (delta h) = [4.8- A (squareroot of 2gh)] (delta t)
    (1) (delta h) = [4.8 - 0.04 {squareroot (h)}] (delta t)
    dh = [4.8 - 0.04 {squareroot (h)}] dt
    Integration of both sides :
    h = [4.8 - 0.04 {squareroot (h)}] t

    To reach h = 1 :
    1 = [4.8 - 0.04 {squareroot(1)}] t
    1 = 4.76 t
    t = 0.21 second
     
  5. Jan 29, 2015 #4

    Nathanael

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    Homework Helper

    I think I see three mistakes. A big one is that you didn't convert 4.8 ft3/minute into ft3/second! That's why your answer was off by so much. Also you seem to be saying [itex]A\sqrt{2g}=0.04[/itex] which is not correct.

    But there is also an important mistake with your integration; you integrated h with respect to dt as if h were a constant, but h is not constant over time so you're not allowed to do integrate like you did. The correct integral is not as nice :(
     
  6. Jan 29, 2015 #5
    Thank you for your inspiration again :) , yup i made mistakes and tried to redo it again.
    I recalculate as :
    cross sectional area of 1 ft2 = B
    area of hole 0.005 ft2 = A
    g = 32.17 ft/sec2
    velocity in = 4.8 ft3/min = 0.08 ft3/sec
    velocity out = α√2gh ; α = 0.6 (source : J.C Borda, based on Edwin Kreyzig's Book) = (0.6) √(2)(32.17)h = 4.812 √h

    Δh = (ϑin Δt)/B - (ϑout Δt A)/B
    Δh = {0.08 - (0.005) (4.812 √h) } Δt / B
    Δh = {0.08 - (0.005) (4.812 √h) } Δt / 1
    Δh = {0.08 - (0.005) (4.812 √h) } Δt


    After this, what should I do next to get the equation of depth of the water in the tank at any time ?
     
  7. Jan 30, 2015 #6

    Nathanael

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    Firstly, sorry for saying your 0.04 was wrong; I'm used to using units of meters (where g is 9.8) so I got that mixed up.

    You have the equation [itex]dh=\big(0.08-A\sqrt{2gh}\big)dt[/itex]
    You can integrate both sides like you tried in post #3, but you need to keep the h with the dh. So the integral becomes [itex]\int\limits_0^1 \frac{dh}{0.08-A\sqrt{2gh}}=\int\limits_0^T dt[/itex]
    (T is the time you wish to solve for)

    As for calculating the integral... I don't immediately see how to do it. I may not be able to help you with that part.
     
  8. Jan 30, 2015 #7
    This is also give me inspiration how to do this, of course dont need to be sorry for telling my mistake, thank you for that, I really mean it.
    Now I just need to solve the integral, I am close to see the answer. I will post again here once I get it, thank you so much !
    :):):)
     
  9. Feb 26, 2015 #8
    Here is the full answer, as I said before..
    An open cylinder of height 5ft and cross sectional area of 1 ft2 is initially empty. There is a small hole at the bottom of the cylinder with an area of 0.005 ft2. Water is drawn into the tank at a rate of 4.8ft3/min. At the same time water is discharged out of the cylinder through a small hole at the bottom of the tank. The outlet velocity may be estimated from Torricelli’s theorem. Estimate the time required for the water level in the tank to reach one foot depth.

    Water level in the tank is x at limit time t.
    Amount of water in = 4.8 ft3/min
    amount of water out = (vout) (A hole) = (vout) (0.005) = 0.04 x0.5
    v out= sqrt (2gh) = sqrt(2gx) = sqrt [(2)(32.2 ft/sec2)(x)]

    Then we setting up the model as :
    amount of water in - amount of water out = rate of accumulation
    (4.8/60) ft3/sec - v Ahole = d/dt (Atank x)
    0.08 - 0.04x0.5 = d/dt [(1)(x)]
    dx/dt = 0.08-0.04x0.5

    This equation can be separated as :
    dx/(0.08-0.04x0.5) = dt
    dx/ (2-x0.5) = 1/25 dt ...(1)

    To solve this, lets assume :
    z=2-x0.5
    -x0.5=z-2
    dz/dx=-0.5 (1/x0.5)
    dx = -2x0.5dz

    subtitute to equation (1), then we have :
    -2x0.5dz /z=0.04 dt
    2(z-2) dz/z=0.04 dt
    Integrating both sides :
    2z-4lnz = 0.04t+c
    2(2-x0.5)- 4ln(2-x0.5)=0.04t+c General Solution

    Subtitute initial condition x(0) = 0, then we have c=4(1-ln2)
    2(2-x0.5)- 4ln(2-x0.5)=0.04t+ 4 (1-ln2) Particular solution

    x = 1 ft
    Subtitute x=1 ft to particular solution, then we have t=19.3 sec
     
  10. May 10, 2017 #9
    Did you account for volume pressure/suction at the hole or is that not relevant?
     
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