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  1. Subscriber joe shmo
    Strange Egg
    22 Dec '16 03:06 / 1 edit
    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.
  2. Standard member apathist
    looking for loot
    22 Dec '16 10:43
    The heat transfer is quick. Test with finger. Or the inside of wrist. Swirl water and retest. Start bathing process when happy.
  3. 22 Dec '16 11:07
    Originally posted by joe shmo
    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.
    Never.

    You must have your description wrong.
  4. Standard member sonhouse
    Fast and Curious
    22 Dec '16 13:19
    Originally posted by twhitehead
    Never.

    You must have your description wrong.
    You are saying the temperature variable is an asymtote?
  5. Subscriber joe shmo
    Strange Egg
    22 Dec '16 13:20 / 1 edit
    Originally posted by twhitehead
    Never.

    You must have your description wrong.
    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 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.

    Are you confirming that the function is correct?
  6. Standard member DeepThought
    Losing the Thread
    22 Dec '16 17:41
    Originally posted by joe shmo
    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?
    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 the bath is mcT, where c is the heat capacity. So the mass of water added after some time T is:

    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.
  7. Subscriber joe shmo
    Strange Egg
    22 Dec '16 19:31 / 2 edits
    Originally posted by DeepThought
    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.
    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 out of college I prefer to use the heavier machinery to keep it moving. Worst case I'm inefficient on this one?
  8. Standard member apathist
    looking for loot
    23 Dec '16 00:41
    Originally posted by twhitehead
    Never.

    You must have your description wrong.
    You pop in, tell someone they're wrong, without explaining yourself. See how people have to guess what you mean. This was all foretold.
  9. 23 Dec '16 11:39
    Originally posted by apathist
    You pop in, tell someone they're wrong, without explaining yourself. See how people have to guess what you mean.
    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.

    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.
  10. 23 Dec '16 11:44
    Originally posted by joe shmo
    However, there will be a finite time when the bath temp is arbitrarily close to the replenishing stream temp.
    Sure 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.

    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.
  11. Standard member DeepThought
    Losing the Thread
    23 Dec '16 14:05 / 1 edit
    Originally posted by joe shmo
    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?
    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).

    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.
  12. Subscriber joe shmo
    Strange Egg
    23 Dec '16 15:59 / 1 edit
    Originally posted by twhitehead
    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]
    "As for 'explaining myself' I think it was obvious what I meant to the poster."

    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?
  13. Subscriber joe shmo
    Strange Egg
    23 Dec '16 16:32 / 2 edits
    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 res ...[text shortened]... is to learn how to use the technique. Always keep everything as simple as possible - or suffer.
    "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)."

    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
  14. 23 Dec '16 17:11
    Originally posted by joe shmo
    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?
    Because 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.

    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?
  15. Standard member apathist
    looking for loot
    24 Dec '16 08:34
    Originally posted by twhitehead...
    Bet you can't back that up with an actual reference of the foretelling....
    Science Forum, "Curiosity on Mars, secret announcement due?"

    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.