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Problem

I am trying to create an equation to calculate fuel temperature based on ambient temperature, heat exchange rate, and mass. The idea is very simple; when the mass is heated, depending on how big or how small it is – temperature will change and adapt to the ambient temperature. The items of interest is ambient temperature, regulated temperature (feedback to the integrator), mass, and heat exchange. So far, my understanding is that it is something similar to:

$$T = T_\mathrm{amb}\frac1m\cdot\int\dot{Q}\,\mathrm dt$$

I suck at physics so I am quite sure that this formula of mine is as good as rubbish. Could someone please advise if there is a good source somewhere?

NOTE

It will also be good if any equation needs volume because I can use volume of the container too.

UPDATE

after the first answer from @Wojciech, I have simplified the formula to be:

$$T = h\frac1C\frac1m\cdot\int(T_\mathrm{amb}-T)\,\mathrm dt$$

Is there a better solution? I am assuming that the units are:

1) $h$ = heat exchange ($\mathrm W$)

2) $C$ = specific heat capacity ($\mathrm{kJ/(kg\ ^\circ C)}$)

3) $m$ = mass ($\mathrm{kg}$)

Is this a plausible formula?

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  • $\begingroup$ I assume that Your fuel is in some sort of a tank. The temperature of fuel in a tank won't be uniform, so You will have to make many simplifications to find a general formula. Where did You get your equation from? $\endgroup$ – Wojciech Mar 7 '14 at 12:28
  • $\begingroup$ @Wojciech I made it up, I only need to model the tank temperature detector. My understanding was that heat exhange will be integrated to some form of K/degrees which I then need to add it with ambient temperature....or something similar.....the heat exchange could very well be a rate or purely a temperature value....I just need a very simply temperature calculcator from those values. $\endgroup$ – hagubear Mar 7 '14 at 12:36
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    $\begingroup$ I'm not particularly good at making up physical formulas, and I don't think it is a good way to model anything. $\endgroup$ – Wojciech Mar 7 '14 at 12:51
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    $\begingroup$ Your equation does not make sense. $\dot{Q}dt$ has units of Watt/m$^2$=kg m$^2$/m$^2$/s$^3$=kg/s$^3$, thus your right side has units of K/s$^3$ while the left side has units of K. $\endgroup$ – Kyle Kanos Mar 7 '14 at 14:24
  • $\begingroup$ @KyleKanos I guess that's why I suck! I just need to find a proper equation :( $\endgroup$ – hagubear Mar 7 '14 at 16:22
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Instead:

$$ \frac{dQ}{dt} = hA(T -T_{amb}) $$ and $$Q = c_pmT$$

where A is the surface area of the interface of the two temeperatures, h is the heat transfer coefficient and $c_p$ is the per-mass specific heat capacity.

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  • $\begingroup$ how about volume? If I had the volume known, can this equation be different? $\endgroup$ – hagubear Mar 10 '14 at 9:24
  • $\begingroup$ $T$ in the second equation is fuel temperature rise, right? @hagubear if You have the volume and density known, You can easily calculate the mass. $\endgroup$ – Wojciech Mar 10 '14 at 9:29
  • $\begingroup$ @Wojciech I know the mass, heat transfer coefficient (Heat Inflow?!!), and $Tamb$. I want to ignore area from the calculation. Also, I know that there is something called volumetric heat capacity J/K/m3?. I know that it is quite annoying, but my colleagues have not given me a spec of what unit this so called "Heat Inflow" is. If it is heat flux (W/m2/K), the I know the mass and it will be the 2nd Formula in DavePhD's answer. But the temperature initially wil be negative as the T will be smaller than Tamb....aarrggh...doesn't make sense...In any case, i would like not to use area (A). $\endgroup$ – hagubear Mar 10 '14 at 9:41
  • $\begingroup$ Heat transfer coefficient is h in W/m^2/K in the first formula. I think that by heat inflow You mean dQ or simply Q from the first formula also, and its units are J. I can't think of a way to exclude area from these equations. You can't create Your own formulas based on what data You have! :) $\endgroup$ – Wojciech Mar 10 '14 at 9:50
  • $\begingroup$ The sign of temperature difference will only indicate the direction of the heat flow, so You can ignore that. $\endgroup$ – Wojciech Mar 10 '14 at 9:52

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