What you need to apply is the engineering equations for a heat exchanger. In the below equation, $\dot{Q}$ is the heat transfer rate, $UA$ is the heat transfer coefficient times the area and $\Delta T_m$ is the log mean temperature difference (LMTD).
$$\dot{Q}=U A \Delta T_m$$
For this problem there are 3 barriers to the heat transfer in series. You have the convective heat transfer coefficient of the syrup $h_c$, the convective heat transfer of the steam $h_s$, and that of the pipe metal $h_R$. Take $n$ to be the number of pipes, $r_i$ to be the inner radius of the pipe, and $r_0$ to be the outer radius of the pipe, and $R$ the heat transfer coefficient of the Copper.
$$UA = \frac{2\pi n}{\frac{1}{h_s r_i}+\frac{1}{R L} ln \frac{r_o}{r_i} +\frac{1}{h_c r_0} }$$
The LMTD is the average temperature difference, but for the specific case where one part of the heat exchanger is saturated, and thus constant temperature, just know that it is the following, where $T_{s}$ is the saturation temperature of the steam, or 100 degrees C. Then $T_h$ and $T_c$ are the temperatures after and before preheating respectively.
$$\Delta T_m = \frac{T_h-T_c}{ln\frac{T_h-T_s}{T_h-T_s} }$$
The hard part of the above equations is the $h_c$ and the $h_s$. You will probably use things like the Dittus-Boelter correlation. But it's more important that for the moment we address $\dot{Q}$ itself. For the described hood, I see two possibilities.
- The steam is being provided at a faster rate than it is being condensed
- The steam is being provided at a slower rate that it is being condensed
In the first case, you will see steam leaking out of the hood. In this case, the equilibrium operation see the air mostly evacuated because it is pushed out. In the other case, steam is present with air and the air reduces the heat transfer. In case #2 the given is the rate of steam condensation, which directly determines $\dot{Q}$ and $h_c$ adjusts to compensate. In case #1 $h_c$ is a given based on the assumption of your geometry and the atmospheric steam heat transfer properties.
For $h_s$ use Wikipedia:
$$h_s={{k_w}\over{D_H}}Nu$$
where
$k_w$ - thermal conductivity of the liquid
$D_H$ - $D_i$ - Hydraulic diameter (inner), Nu - Nusselt number
$Nu = {0.023} \cdot Re^{0.8} \cdot Pr^{n}$ (Dittus-Boelter correlation)
Pr - Prandtl number, Re - Reynolds number, n = 0.4 for heating (wall hotter than the bulk fluid) and 0.33 for cooling (wall cooler than the bulk fluid).
What I don't have right now are the numbers for applying the above for Maple Syrup and the approach for the steam side of the tubes. This is all I have time for right now, but I think this amount will still be helpful. I can try to look up these things later, maybe you can specify what you understand the least first.
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)
Best Answer
Suppose you have a mass $M$ of oil at a temperature $T$ and specific heat capacity $C_{oil}$. You pump water at a temperature of $T_0$ and flow rate of $f$ kg/sec, and we'll assume as you say that the heating of the water by the oil is effectively instant. The heat absorbed by the water per second is:
$$ \frac{dU}{dt} = f \space \left(T - T_0 \right) \space C_w $$
where $C_w$ is the specific heat of water. The corresponding temperature drop per second in the oil is:
$$ \frac{dT}{dt} = \frac{1}{M \space C_{oil}} \frac {dU}{dt} = - \frac{f \space C_w}{M \space C_{oil}} \left(T - T_0 \right) $$
and one quick integration later we get:
$$ T - T_0 = T_{init} \space \text{exp} \left( - \frac{f \space C_w}{M \space C_{oil}} t \right) $$
where $T_{init}$ is the initial temperature in your tank i.e. 55C.
You give values for $f$ = 0.0833 kg/sec and $T_{init}$ = 55C. The specific heat of water, $C_w$ is 4.2 kJ/kg and oil is around 2 kJ/kg depending on the exact type of oil. However you don't give a value for M, so it's not possible to give an actual figure for the cooling rate. However if you plug in the mass of oil in your tank you can calculate the temperature as a function of time.