# A model for constant temperature of water in a container

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

$$V\rho C_p\frac{dT}{dt}=q\rho C_p(T_{NEW}-T)-kV\rho C_p(T-T_a)$$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.
• 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. – Patrick Jan 29 '16 at 23:01