Continuously draining tank (constant input\dependant output)

  • Thread starter Thread starter Mike T
  • Start date Start date
  • Tags Tags
    Output Tank
Mike T
Messages
2
Reaction score
0

Homework Statement


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 \frac{dI}{dt} and drains with a flow rate \frac{dO}{dt} 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.

Homework Equations



netflow = \frac{dI}{dt} - \frac{dO}{dt}

3. The Attempt at a Solution

So I've assumed that I should come up with a an equation in the form h(t) 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. h = I(x) - O(x), 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.
 
Physics news on Phys.org
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?
 
You are given that the bath tub drain at a rate "directly proportional to \sqrt{h}". How do you write that as an equation?
 
Mike T said:

Homework Statement


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 \frac{dI}{dt} and drains with a flow rate \frac{dO}{dt} 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.

Homework Equations



netflow = \frac{dI}{dt} - \frac{dO}{dt}

3. The Attempt at a Solution

So I've assumed that I should come up with a an equation in the form h(t) 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. h = I(x) - O(x), 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.

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:
Ray Vickson said:
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+Δt)(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, Δt>0\Delta t > 0 is a very small time increment.)

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 \triangle t the net flow in is r \triangle t.

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, r \triangle t as the basis for calculating the outflow volume in this time? This makes sense (if I've got it correct!) for a small period, \triangle t, 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:
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top