Simple formula for liquid heat transfer I'm trying to do a simple simulation of a solar panel coupled through some piping to a boiler, with the aid of a pump. My input constants could be something like: volume of liquid inside the panel, volume of liquid inside the boiler, pump transfer rate.
For simplicity I can assume that connective piping does not absorb any heat, so panel and boiler are coupled next to each other through some hole and thermally insulated from another.Possible constants: ambient temperature, heat loss (depends on ambient temperature).
The variables: panel temperature, boiler temperature, whether the pump is running or not.
What I'm doing right now is like linear estimation, applying some energy to the panel (raises the heat linearly), running the pump transfers linearly energy from one vessel to another (temperature increase/decrease is inversely proportional to volume of liquid), heat transfer to boiler raises its energy linearly.
I would like to have some thing like: estimating the rise and fall of temperature in both vessels given the above constants and variables. It doesn't have to be scientifically exact, but I know that linear aproximation is wrong as the temperature increase is somehow inversely proportional to absolute temperature (i.e. it 'tapers').
The end result will be like: apply xxx units (Joules) of energy to solar panel for 1 minute > temperature rises by y degrees; start pump (open valve) for 1 minute -> temperature in boiler goes up by z degrees, temperature in panel goes down by w degrees.
Any graphs or formulas will be helpful, I can continue from there. This is not a homework or scientific study, just something for personal use.
Also some explanations about differences between water and oil energy absorbtion would be nice.
 A: Some notes to get you on your way:
Power in full sunlight at sea level $\approx 1kW/m^2$ - remember to adjust for relative angle of sun and solar panel.
Specific heat capacity of water: 4.186 J/g/K
Typical collector efficiency: somewhere in the range 50-90%.
Let's do a quick crude static calculation. It sounds like you're going to do a dynamic microsim - great stuff, go for it; if we can make a static estimate first, that'll give you an idea of how the numbers connect to each other, and give you a ballpark number to check the output of your simulation against.
Let's take the sun to be shining directly on the panel, along the normal to the plate, so the power hitting the plate is  $1000 W/m^2$. Let's assume a 60% efficiency, so we're putting $600 W/m^2$ into the heat transfer fluid. Let's ignore any anti-freeze for the moment, and just assume our heat-transfer fluid is pure water, thus having a specific heat capacity of 4.186 J/g/K. Now, if you had 1 litre of water collecting heat per square metre, then you'd raise its temperature by 
$$\frac{600 W}{1000 g \times 4.186 J /g/K} \approx 0.14 K/s$$
Now, here's a quick calculation for a whole day. Let's take a ball-park figure of 6 full sun-hours (that's 16 hours of summer daylight, derated by guesstimate to account for the variation of angle between panel and sun), and a 200-litre thermal storage tank; then our increase in temperature of the tank in a day, from a single square metre of collector would be, to the first order, be:
$$\Delta T_{tank} = \frac{0.14 K/s  \times 3600 s/h \times 6 h}{200} \approx 15 K$$
So a $4m^2$ system would give you $\Delta T_{tank} \approx 60 K$ . Then that would get derated, based on system losses, including thermal losses from the collector plate and the thermal store. 
Heat loss from the panel itself will (broadly) be proportional to the difference between the heat-transfer fluid, and the ambient air temperature - that gives the tapering you mentioned.
And you'll need a figure for the rate at which heat can be exchanged between the heat-transfer fluid that passes through the collector, and the thermal store.
(See also Appendix H of SAP 2009 which starts on pdf p73 - but bear in mind that that's a coarse static approximation of a solar thermal system in the UK, not a dynamic simulation - but it has some figures to get you started on collector efficiency and thermal losses)
