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.
Efficiency of one heat pump in cooling can be defined by expression
$$\eta = \frac{Q_C}{W},$$
that is heat that is taken from cooler reservoir divided by the work put into the pump.
If you have two pumps in parallel efficiency shall be the same, as you will have heat twice as large and work twice as large
$$\eta = \frac{2Q_C}{2W}.$$
If you however have two pumps in sequence equation reads as
$$\eta = \frac{Q_C}{2W}.$$
So in order to obtain at least same efficiency you should extract heat twice as large by each pump. So the efficiency of each pump should be twice as large for about half of the temperature change. So far so easy.
In the next step we must take into consideration exact cyclical thermodynamical processes. There are plenty of them that you can use and no can be exactly theoretically calculated. In such cases it is useful to observe the most efficient thermodynamical process, that is Carnot cycle and extract conclusions from it.
Efficiency of the heat engine based on Carnot cycle can be shown to be
$$\eta = \frac{T_C}{T_H-T_C}.$$
In case of two pumps in the first iteration intermediate temperature $T' = \frac{T_C+T_H}{2}$ in exactly in the middle. Efficiency of two pumps shall be
$$\eta_1 = \frac{T_C}{T'-T_C}, \eta_2 = \frac{T'}{T_H-T'}$$
Obviously $\eta_1 = 2 \eta$ and $\eta_2 > 2 \eta$, therefore two sequential pumps in cooling should be more efficient.
It is interesting that two sequential pumps in warming should be less efficient using the same arguments! I cannot find the error in the reasoning, so before accepting the answer, please wait some time that others check and give their comments.
Best Answer
It sounds as if you are describing a countercurrent heat exchanger.
The theoretical efficiency of these can reach 1, though note that for heat exchangers efficiency doesn't mean the same as for heat engines i.e. heat converted to work. For heat exchangers an efficiency of 1 just means the incoming air is heated to the same temperature as the air in the house and likewise for the outgoing air.