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Continuously draining tank (constant input\dependant output)

  1. May 26, 2015 #1
    1. The problem statement, all variables and given/known data
    I've been asked to solve a problem where a bathtub is being filled at a constant rate, meanwhile water is draining out continuously. The bathtub fills with a constant flow rate of [itex] \frac{dI}{dt} [/itex] and drains with a flow rate [itex] \frac{dO}{dt} [/itex] which is directly proportional to √h where h is the height of the water.

    The question is to come up with an equation describing the height of the water over time, given this constant input and varied output. Use this to find out what time the bath water will overflow.

    2. Relevant equations

    [itex] netflow = \frac{dI}{dt} - \frac{dO}{dt} [/itex]

    3. The attempt at a solution

    So I've assumed that I should come up with a an equation in the form [itex]h(t)[/itex] where h is the height, and t is the time variable.

    I know the volume of the bath, so what height needs to be reached. I've tried integrating both of the flow rates, and then combining them e.g. [itex]h = I(x) - O(x)[/itex], and then solving for the initial condition h = bath height. This gives me the time to fill the bath (although I don't now how accurate it is!) - how I can I make this into a more general solution to find the height at any time t?

    I'm having great difficulty relating the input and output in this equation together, and would welcome any points on the general approach, as well as my current approach above.
     
  2. jcsd
  3. May 26, 2015 #2

    Orodruin

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    You need to write down a differential equation. Before you have done that you cannot integrate it properly. How do you relate the net flow to the rate of change in height?
     
  4. May 26, 2015 #3

    HallsofIvy

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    You are given that the bath tub drain at a rate "directly proportional to [itex]\sqrt{h}[/itex]". How do you write that as an equation?
     
  5. May 26, 2015 #4

    Ray Vickson

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    What is the relationship between the volume of water (V) and the height of water (h)? If it is a cylindrical bathtub (with vertical sides) the relationship is just V = Ah, where A = area of base. However, for a bathtub with sloping sides and rounded bottom, etc., the relationship could be a lot more complicated. So, you had better start by stating an assumption about the bathtub's shape.

    Then, if r = volumetric inflow rate (which you called dI/dt), what is the volume of inflow in the time interval ##(t,t + \Delta t)##, and what is the volume of outflow during this same interval? What is the NET inflow volume in that interval? (Here, ##\Delta t > 0## is a very small time increment.)
     
    Last edited: May 26, 2015
  6. May 27, 2015 #5
    Hey, thanks Ray - in this case, I'm assuming a bath tub has vertical edges, so the volume is directly proportional to the height helpfully. So in this time period [itex]\triangle t [/itex] the net flow in is [itex]r \triangle t [/itex].

    Calculating the net outflow in this time period is what I'm finding hard as it is dependant on the volume (or height) of the water. Ah, is this where I use the net inflow volume, [itex]r \triangle t[/itex] as the basis for calculating the outflow volume in this time? This makes sense (if I've got it correct!) for a small period, [itex]\triangle t[/itex], but it wouldn't for a large period. So I need to find the definite integral between v amd max volume?

    edit: I mean, integrate from t=0, to t=infinity of course, and then find at what value of t, the volume matches my expected result?
     
    Last edited: May 27, 2015
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