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22 Dec 16
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While I was toping off the tub the other day for my daughters bath (The tub was partially filled at some cooler temp) I felt like figuring out how long it should take in general before the entire bath became the temperature of the replenishing stream. I got a little stuck on this one, but I belive I now have the proper function for the conditions. I'm just looking for some verification ( or denial) from anyone more confident than myself. Here goes.
The energy stored within a volume of water is given by:
E_st = m_cv * c_p * T ...Eq(1)
m_cv = mass of water
c_p = specific heat of water
T = temperature of water
Performing a power balance on the control volume ( assuming no heat loss by typical means)
d ( E_st )/dt = d(E_in)/dt ...Eq(2)
The water stored in the control volume is both changing mass and temperature, hence the stored energy is changing with both parameters. Differentiation of Equation (1) ( assuming constant specific heat) yields:
d(E_st)/dt = d( m_cv)/dt * c_p * T + m_cv * c_p * d(T)/dt ...Eq(2) LHS
Now for Eq(2) RHS
The power brought into the tub is assumed constant, and is given by:
d(m)/dt * c_p * T_in ...Eq(2) RHS
d(m_cv)/dt * c_p * T + m_cv * c_p * d(T)/dt = d(m)/dt * c_p * T_in ...Eq(2)
For simplification: d(m)/dt = constant = m'
Now applying a mass rate balance to the control volume:
d(m_cv)/dt = m' ...Eq(3)
(which states the rate of mass gained in the tub is the rate at which mass enters the tub by way of the faucet in this case)
Substitue (3) -> (2) and dividing out c_p
m' * T + m_cv * d(T)/dt = m' * T_in --->
m_cv * d(T)/dt = m' * T_in - m' * T --->
m_cv * d(T)/dt = m' * ( T_in - T ) ...Eq( 2' )
m_cv is a function of time "t". From Eq(3) with initial conditions t= 0, m_cv = m_o ( mass of water in tub @ t=0 )
m_cv = m' * t + m_o ...Eq(4)
Sub Eq(4) --> Eq( 2' )
( m' * t + m_o ) * d(T)/dt = m' * ( T_in - T ) ...Eq(5)
Separate:
d(T) / ( T_in - T ) = m' / ( m' * t + m_o )*dt ...Eq( 5' )
Change variables:
Φ = T_in - T
dΦ = -dT
U = m' * t + m_o
dU = m' * dt
Substitute:
- dΦ / Φ = dU / U ...Eq(6)
Integrate:
Φ_o / Φ = U / U_o ...Eq(7)
where,
Φ_o = T_in - T_o
U_o = m_o
Substitute into Eq(7) and solve for "T" (the instantaneous temperature of the tub)
T(t) = T_in - ( m_o / ( m' * t + m_o ) )*( T_in - T_o ) ...Desired Relationship
Sanity Check: T(0) = T_o & T( t--> ∞ ) = T_in
It seem to make sense, hoping I didn't flub something up too bad.
22 Dec 16
Originally posted by joe shmoNever.
While I was toping off the tub the other day for my daughters bath (The tub was partially filled at some cooler temp) I felt like figuring out how long it should take in general before the entire bath became the temperature of the replenishing stream.
You must have your description wrong.
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22 Dec 16
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Originally posted by twhiteheadMy description is technically wrong. ( I wrote this as a problem to explore...I admittedly didn't know the exact result was "never" at the time I was pondering it). However, there will be a finite time when the bath temp is arbitrarily close to the replenishing stream temp. The equation can define that time. That and solving an unfamiliar problem was truly the point of the exercise. As I said, I posted it here for confirmation because it was not a familiar problem I have encountered.
Never.
You must have your description wrong.
Are you confirming that the function is correct?
22 Dec 16
Originally posted by joe shmoYou're using calculus when you don't need to this is linear. Suppose your reservoir has temperature T(t) at time t. At time t = 0 it contains mass M of water and has temperature T(0) = T_c. Water is added at an (absolute) temperature of T_h. Suppose the flow-rate F = dm/dt is a constant so mass m is added every second. The total thermal energy of the bath is mcT, where c is the heat capacity. So the mass of water added after some time T is:
My description is technically wrong. ( I wrote this as a problem to explore...I admittedly didn't know the exact result was "never" at the time I was pondering it). However, there will be a finite time when the bath temp is arbitrarily close to the replenishing stream temp. The equation can define that time. That and solving an unfamiliar problem was tru ...[text shortened]... not a familiar problem I have encountered.
Are you confirming that the function is correct?
m = F*t
The total mass is M + m = M + F*t.
Total thermal energy = (M + m)*c*T(t)
Initial thermal energy = M * c * T(0) = M*c*T_c
energy of added water = F*t*c*T_h
So:
(M + Ft)*c*T(t) = M*c*T_c + F*t*c*T_h
=> T(t) = (M*T_c + T_h*F*t) / (M + F*t)
Let T_h = T_c + θ
Then:
T(t) = T_c + (θ*F*t / (M + F*t))
The time taken to get from T_c to (T_c + T_h)/2 is then found from:
1/2 = F*t / (M + F*t)
or t = M/F
In other words one has to add the same mass of water to get the average temperature. So the time taken to get to within (1/2)^n is n*M/F.
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22 Dec 16
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Originally posted by DeepThoughtAs far as I can tell mine and your derivations are equal statements ( you didn't explicitly state that...which had me concerned )? The important thing for me is that I applied the calculus correctly, not that I used a sledge hammer to swat a fly. Sure, had I been taking an exam that long winded solution would have cost me time, but personally, now that I'm out of college I prefer to use the heavier machinery to keep it moving. Worst case I'm inefficient on this one?
You're using calculus when you don't need to this is linear. Suppose your reservoir has temperature T(t) at time t. At time t = 0 it contains mass M of water and has temperature T(0) = T_c. Water is added at an (absolute) temperature of T_h. Suppose the flow-rate F = dm/dt is a constant so mass m is added every second. The total thermal energy of th ...[text shortened]... s of water to get the average temperature. So the time taken to get to within (1/2)^n is n*M/F.
23 Dec 16
Originally posted by apathistYou pop in just to take a jab at me because you are upset that I called you out for making false claims about me.
You pop in, tell someone they're wrong, without explaining yourself. See how people have to guess what you mean.
This was all foretold.
Bet you can't back that up with an actual reference of the foretelling.
As for 'explaining myself' I think it was obvious what I meant to the poster. He is asking when the bath temperature will reach the temperature of the incoming water. It is obvious that that can never happen. What he should be asking is when the bath temperature will reach a particular temperature (lower than the temperature of the incoming water) either in absolute terms or as a proportion of the temps involved.
23 Dec 16
Originally posted by joe shmoSure but the choice of how close you want it significantly affects the time taken to get there, so what you are getting is a formula for how close you have got after a given time rather than a formula for when you have got there.
However, there will be a finite time when the bath temp is arbitrarily close to the replenishing stream temp.
You might also want to ask what the temperature will be when the bath is full as it is not realistic to be filling the bath beyond that. And possibly more important, you might want to ask at what temperature of the original water is it better to let that out first before adding more. For example, if your daughter can bath in a quarter full tub but it will take more than a quarter to get to a reasonable temperature.
23 Dec 16
1 edit
Originally posted by joe shmoThe difficulty with using sledge hammers to crack nuts is that the chances of making a mistake diverge. What you are really doing here is taking a weighted average. So we have (M + m)T = MT_c + mT_h. I found your working difficult to follow because of the problems of notation in a text only interface and latched onto the minus sign in the eventual result (you measure down from the hot temperature, I up from the cold one).
As far as I can tell mine and your derivations are equal statements ( you didn't explicitly state that...which had me concerned )? The important thing for me is that I applied the calculus correctly, not that I used a sledge hammer to swat a fly. Sure, had I been taking an exam that long winded solution would have cost me time, but personally, now that I'm ...[text shortened]... prefer to use the heavier machinery to keep it moving. Worst case I'm inefficient on this one?
If you want a problem when you do need some calculus (I think) then work out what the flow rate required to get the (increasingly full) bath to some target temperature assuming the rate of loss of heat from the bath = const. * A * (T_bath - T_surroundings). A is the surface area of the bath water, if you like you can assume the bath loses energy by evaporation only so that for a bath with vertical sides the area from which heat can be lost doesn't change and the rate of heat loss is proportional to the temperature difference only. If you just want to keep the bath at a steady temperature then calculus is not required.
Incidentally because I was thinking and typing the last line of my earlier post is wrong, I should really do these things on paper before posting. To get the temperature difference to within (1/2)^n takes (2^n)*M/F seconds.
The only reason to solve simple problems with powerful techniques is to learn how to use the technique. Always keep everything as simple as possible - or suffer.
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23 Dec 16
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Originally posted by twhitehead"As for 'explaining myself' I think it was obvious what I meant to the poster."
You pop in just to take a jab at me because you are upset that I called you out for making false claims about me.
[b]This was all foretold.
Bet you can't back that up with an actual reference of the foretelling.
As for 'explaining myself' I think it was obvious what I meant to the poster. He is asking when the bath temperature will reach the te ...[text shortened]... rature of the incoming water) either in absolute terms or as a proportion of the temps involved.[/b]
If I knew the answer, I wouldn't have bothered to derive the function... My real concern was whether or not the relationship for Temperature vs Time was correct ( I too arrived at the result of "never" in response to the specific question when I was complete with the analysis )...
Let me put it to you this way... I was playing the part of the professor and pupil with equal enthusiasm. For instance, lets say a professor poses the exact question in a lecture. twhitehead confidently answers out loud: "Never"! I don't know what university you attended, but in my experience 100/100 times that response was followed up with something along the line of "Ok...now we are going to prove it".
"It is obvious that that can never happen"
If the answer to the question was so "obvious" then why did you not simply confirm or deny the relationships validity in any of your replies? I directly asked you that question, you completely ignored it, and now you are trying to tell me what I should be asking instead.
Now, I'm going to tell you to get bent. If you can't provide a mathematical proof ( as Deep Thought took the time to do...or even a general explanation) to what should have been a "clear exerscise" in mathematical physics then why are you even bothering to reply?
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23 Dec 16
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Originally posted by DeepThought"The difficulty with using sledge hammers to crack nuts is that the chances of making a mistake diverge. What you are really doing here is taking a weighted average. So we have (M + m)T = MT_c + mT_h. I found your working difficult to follow because of the problems of notation in a text only interface and latched onto the minus sign in the eventual result (you measure down from the hot temperature, I up from the cold one)."
The difficulty with using sledge hammers to crack nuts is that the chances of making a mistake diverge. What you are really doing here is taking a weighted average. So we have (M + m)T = MT_c + mT_h. I found your working difficult to follow because of the problems of notation in a text only interface and latched onto the minus sign in the eventual res ...[text shortened]... is to learn how to use the technique. Always keep everything as simple as possible - or suffer.
Yeah, notation is alway a challenge in here. Also, I undestand your position about using the correct tool for the task at hand. Its just that personally when I think rates of change ( i.e. rates of change in volume, rates of change in temperature I somewhat robotically go to calculus).
"The only reason to solve simple problems with powerful techniques is to learn how to use the technique. Always keep everything as simple as possible - or suffer"
Personally I feel that I am still learning calculus. In this case, the question was really just the means (albeit foolishly) to an exercise in the subject. I may try and solve the problem you framed ( either or both ways) when I get some time. The problem however is that I become somewhat obsessed with rendering a solution. I imagine you wouldn't have posed them to me, if you weren't willing to help out ( should I get stuck). so I'll try.
BTW. I took a jab at you in a thread the other day. I don't know if you saw it or not, but if you did...I'm sorry (I took it personal, I should have just shrugged it off, as you probably did my reply). You help me out a lot on here with problems...And I'm greatful for it. Whether that be your reasounding good will, or a neurosis which does not allow you to pass by the ignorant is irrelevant to me 🙂 ...Thanks
23 Dec 16
Originally posted by joe shmoBecause the exact relationship is not so obvious and I have not checked your work. Besides, DeepThought who is clearly better at such things than I has already done so.
If the answer to the question was so "obvious" then why did you not simply confirm or deny the relationships validity in any of your replies?
I directly asked you that question, you completely ignored it, and now you are trying to tell me what I should be asking instead.
DeepThought had already answered your question in detail. I thought it unnecessary.
Now, I'm going to tell you to get bent. If you can't provide a mathematical proof ( as Deep Thought took the time to do...or even a general explanation) to what should have been a "clear exerscise" in mathematical physics then why are you even bothering to reply?
When you mix two liquids with one hotter than the other, it is obvious that the resulting overall temperature will be somewhere in between. Seriously, if that is not obvious to you, then forget calculus. No mathematics required, practically no physics (just a basic understanding of heat) and common sense. You should always double check problems with a bit of common sense.
Let me give an analogy. If you have a bucket of Red paint, how much white paint must you add till the paint in the bucket is perfectly white? Do you need a mathematical proof to answer?
24 Dec 16
Originally posted by twhitehead...Science Forum, "Curiosity on Mars, secret announcement due?"
Bet you can't back that up with an actual reference of the foretelling....
Originally posted by apathist
...
Look, just go find another thread where you can pop in, tell someone they are wrong, and then attack their attempts to explain themselves, without ever actually putting your honest views up for counter-attack. You are good at that. Then hyper-focus in this direction or that while evading uncomfortable rebuttals until people realize how smart you are and finally you can get some sleep.