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I put some water in a container with initial temperature $T_0$ in a room, and the room's initial temperature is $T_a$. Now the container is filled to the maximum, so any more water coming in will result in an overflow.

I want to keep the water at initial temperature for a long time, preferably for inifinite amount of time. Also water entering the container flows on a constant rate and constant temperature.

Sorry for being not specific. So far I came up with the differential equation on the temperature of water at any point :

$$T(t)=T_a+(T_0-T_a)e^{-kt}$$ That is just Newton's law of cooling. Then I have found the following equation: The volumetric flow rate in a heating system can be expressed as

$$q=\frac{h}{c_\rho*\rho*(T_{new}-T_t)}$$

where

$q$ = volumetric flow rate

$h$ = heat flow rate

$c_\rho$ = specific heat capacity

$\rho$ = density

$T_{new}-T_t$ = temperature difference

so I thought I would solve it for $h$ and add to the equatio for $T(t)$ Hence $$T(t)=T_a+(T_0-T_a)e^{-kt}+q*c_\rho*\rho*(T_{new}-T_t)$$ From then I made $T_t$ the subject and then set LHS to $T_0$ because that is the temperature I want to keep. I have no idea if that is correct, probably I missed out a lot of crucial variables so I seek help. Is this correct approach?

If not, can you tell me what is? Regards, Patrick

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Your setup of the problem isn't correct. Let V be the volume of water in the container. Then the rate of accumulation of heat in the container is equal to the heat in minus the heat out. The correct equation for this is:

$$V\rho C_p\frac{dT}{dt}=q\rho C_p(T_{NEW}-T)-kV\rho C_p(T-T_a)$$This assumes that the tank is well-mixed so that the exit temperature is the same as the bulk temperature in the container. If we divide this equation by $V\rho C_p$, we obtain:$$\frac{dT}{dt}=\frac{(T_{NEW}-T)}{\tau}-k(T-T_a)$$where $\tau$ is the mean residence time in the tank, given by $\tau = V/q$. You can solve this differential equation for T as a function of time.

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  • $\begingroup$ Thank you very much for the help. I don't want to be pain, but do you have a source that I could relate to? It is a part of a project you see, so I can't just come up with an equation without any justification behind it. In other words how did you derive this differential equation $\endgroup$ – Patrick Jan 29 '16 at 22:20
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    $\begingroup$ It is a standard development for handling any stirred tank problem in heat transfer and chemical reactor design, for continuous flow systems. In reactor design, concentration replaces temperature. Look up Continuous Stirred Tank Reactors (CSTRs), or get a book on Reaction Engineering. Transport Phenomena by Bird, Stewart, and Lightfoot may also have this. I'm having trouble thinking of other references because I used this stuff so much in my professional career that it is just second nature to me. $\endgroup$ – Chet Miller Jan 29 '16 at 22:31
  • $\begingroup$ No problem, thanks for the references, There is only one thing I am missing from your equations. Obviously we can change the volume of new water coming in. I will try to be as clear as possible, I am a maths stduent you see, I dont do physics for years now. Essentially what Im trying to say that if the volume of water per second added is small, then this water must be significantly hot so the temperature of water in the container remains the same comparing to the situation when the volume per second is quite big. Please tell me you udnerstood what I meant. Shouldnt I consider that? $\endgroup$ – Patrick Jan 29 '16 at 22:39
  • $\begingroup$ That's all captured by the equation. Start out by getting the steady state solution to the equation, when dT/dt = 0. Then you can find the relationship between Tnew and tau such that the temperature in the tank remains constant at its initial value. $\endgroup$ – Chet Miller Jan 29 '16 at 22:49
  • $\begingroup$ Ok, it is getting a lot clearer now, my equation for $q$ is correct I imagine? You used the variables that I defined when posting the question? I can't really get my head around the flow rate yet. Different website represent it differently. $\endgroup$ – Patrick Jan 29 '16 at 23:01

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