Turning a real world problem into a calculus problem

[cookiemnstr510510]

Homework Statement

I have posted a picture of the HW problem below.

Relevant Equations

potentially could use:
Accumulation=Input-Output+Generation-Consumption

So I am a bit confused on how to get started. So far my thought process is we have water flowing in and water evaporating from the pool. The part that I think we are interested in is the leakage. The leakage has the rate it is flowing out per unit time. I will call it change in volume, or dV, per unit time, or dt.
My initial thought is:
dV/dt=Q_L from above equation and integrate both sides, but I don't have the variable "t" on the right hand side of the equation. If someone could walk me through a logical way of turning this word problem into a derivative that would be great.
Something else I would like to clarify: Are we trying to turn this into a differential equation? The course I am in uses DE's and I honestly don't know.

Thanks so much!

 

[BvU]

$${dV\over dt}=Q_L$$

Is a genuine differential equation. The fact that the righthand side does not depend on $t$ makes it a fairly easy one, but a DE nonetheless. However, you forget two other $Q$.

And in

Accumulation=Input-Output-Generation+Consumption

I suspect a sign error: if someone swallows water, that does not increase he holdup of the pool ... and the sign for 'generation' likewise seems erroneous.

 

[haruspex]

you forget two other Q.

Maybe not forgotten, just canceled out? Don't we know that , after filling, Q_h=Q_e?

 

[BvU]

A whole winter has gone by ... If $Q_e = Q_{h, \text{last year}}$ even getting it filled is a problem.

 

[haruspex]

A whole winter has gone by ... If $Q_e = Q_{h, \text{last year}}$ even getting it filled is a problem.

If the same procedure is followed as in the first year, it is first filled at whatever rate will do then throttled back to match Q_e.

 

[BvU]

Maybe not forgotten, just canceled out?

Let's ask around in sesame street ... @cookiemnstr510510 ?

 

[cookiemnstr510510]

Is a genuine differential equation. The fact that the righthand side does not depend on $t$ makes it a fairly easy one, but a DE nonetheless. However, you forget two other $Q$.

And in I suspect a sign error: if someone swallows water, that does not increase he holdup of the pool ... and the sign for 'generation' likewise seems erroneous.

Hi BvU!
So I have spent some time looking at the problem and I now have a few more questions.
When setting up my D.E. I have learned that what I am doing with this problem is first: setting up boundaries. My boundaries are the swimming pool. I then want to know how the mass of the water is changing over a certain time period. I then choose my time period to be very close to zero so I can use differential calculus.
Let me show you with some work:

I will keep everything in units of mass. We are given flow rates of things (volumes), I will denote the density of liquid water ρ_l and vapor water ρ_v

ΔVρ_l=(Q_hρ_l-Q_eρ_v-Q_lρ_l)Δt
I now let Δt-->0
dVρ_l/dt=Q_hρ_l-Q_eρ_v-Q_lρ_l
now I think we have a separable D.E. so I can separate and integrate both sides.

I asked around and got some info from classmates that I am now trying to make sense of. They are saying that I should turn Vρ_l to the equivalent equation of πr²hρ_l. And my bounds of integration would be h=0 to h=? since we want an equation in terms of height.
Is the reason we need to turn our volume into πr²h so that we have the variable h, which is what the question is asking?
I was just going to leave it in terms of V...but haven't tried that yet.

 

[cookiemnstr510510]

Let's ask around in sesame street ... @cookiemnstr510510 ?

No, I did forget/didn't understand where to put the other Q's at that point

 

[BvU]

Kudos for mentioning $\rho_v$ -- I had completely missed it and assumed $Q_e$ in m³/h water

So far you have $$\rho_L\, dV/dt= \rho_L\, Q_h - \rho_v \, Q_e - \rho_L \,Q_L$$ with $Q_h$ and $Q_e$ independent of $V$, but $Q_L$ not. Either you do $V = \pi R^2 h$ and $Q_L = k_L\,2\pi Rh$ to write an equation in terms of $h$, or you leave $V$ and take $Q_L = k_L\,2\pi R \displaystyle {V\over \pi R^2}\$.
Since the exercise asks for $h(t)$ the former is a good choice.
So is starting with $h(0) = 0$.
And the integration goes from 0 to $t$ to get $h$ as a function of $t$.

You consider $Q_h$ as input (but to get the thing filled before the summer is over, it should be bigger than the $Q_h$ after filling last year ) .

A bit strange that the exercise gives numerical values for a few of the variables, but not enough for a complete calculation (e.g. $Q_e$).

 

[haruspex]

but to get the thing filled before the summer is over, it should be bigger than

It is very confusing the way the description reuses Q_h. It cannot be the same Q_h as in the first year since that equalled Q_e, and we are told the Q_h in the second year is greater than that.
In the first year, it is first filled at some unknown but sufficient rate. After filling it is throttled back to match the evaporation rate. Were it at the lower rate through out it would not filled then either.

In the second year, it is unclear whether the same procedure is followed. I think I was wrong to assume so. Looks like we should take it as filling from empty with input constant and somewhat greater than the constant evaporation. For purposes of analysis, it is only the difference of the two that matters.
And it may well be that it never fills this time.

 

[Chestermiller]

I get $$A_{top}\frac{dh}{dt}=Q_h-Q_e-k_L(2\pi R)h$$This is a first order linear ODE in h.

 

[BvU]

I get $$A_{top}\frac{dh}{dt}=Q_h-Q_e-k_L(2\pi R)h$$This is a first order linear ODE in h.

There is the small matter of $\ {\rho_v\over\rho_L} \$ in front of $Q_e$, but apart from that I agree (as in post #9)

 

[Chestermiller]

There is the small matter of $\ {\rho_v\over\rho_L} \$ in front of $Q_e$, but apart from that I agree (as in post #9)

To me, it is obvious that $Q_e$ is supposed to represent the rate at which liquid water is evaporating. In my judgment, for a problem presented to a student at this level, this is the only thing it could mean.

 

[BvU]

Same here, but the wording

appears to state otherwise

 

[Chestermiller]

Same here, but the wording

View attachment 256241

appears to state otherwise

Ooops. I missed that. But it makes no sense, unless it is at the interface temperature and equilibrium vapor pressure (which are not specified) and which the student at this level would not know how to determine anyway.

 

[PhDeezNutz]

Forgive me if I’m wrong. But due to conservation of volume wouldn’t the evaporation rate of liquid water equal the volumetric gas flow rate? I.e. the “surroundings gain volume at the same rate the system loses volume”.

 

[BvU]

No. 1 kg of liquid water at 1 atm and 20 $^\circ$C is 0.001 m³.
It evaporates to 57.8 m³ of water vapour.

And I never heard of a 'conservation of volume' law

 

[PhDeezNutz]

No. 1 kg of liquid water at 1 atm and 20 $^\circ$C is 0.001 m³.
It evaporates to 57.8 m³ of water vapour.

And I never heard of a 'conservation of volume' law

I messed up pretty badly.

 

[PhDeezNutz]

No. 1 kg of liquid water at 1 atm and 20 $^\circ$C is 0.001 m³.
It evaporates to 57.8 m³ of water vapour.

And I never heard of a 'conservation of volume' law

So would $\frac{\rho_g}{\rho_l} = \frac{m}{V_g} \frac{V_l}{m} = \frac{0.0001}{57.8} = 1.73 * 10^{-6}$??

For a final differential equation of (the h by the Q is a subscript)

$\frac{dh}{dt} = \frac{Q_h}{\pi R^2} - \left( 1.73 * 10^{-6} \right) k_e - k_l \left(\frac{2}{R} \right)h$which can then be solved using the method found here
http://tutorial.math.lamar.edu/Classes/DE/Linear.aspx

$\frac{dy}{dt}+ p\left( t \right)y = g\left( t \right)$

$y\left(t \right) = \frac{\int \mu \left( t \right) g\left( t \right) dt + c}{\mu \left(t \right)}$

where

$\mu \left( t \right) = e^{\int p\left(t\right) dt}$

If I'm spouting nonsense and detracting from the thread/hindering progress let me know and I will stop.