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.
I will assume your heat exchanger uses the common counter flow principle - that is, the direction of flow of the "cold" water opposes that of the "hot" water, so the hottest water in the heating loop (entering the heat exchanger) is in touch with the hottest water of the pool loop (just before exiting the heat exchanger).
The heat flow across the exchanger is proportional to the temperature difference. Since the "input temperature" is fixed at 180 F, the only variable is the temperature of the "sink" - the pool water. The colder the pool water, the greater the heat flow.
At the input of the heat exchanger, the temperature is the temperature of the pool; at the exit, it will be somewhat warmer. The slower the water flows, the more heat it will pick up, and the hotter the water that re-enters the pool. However, the hotter the pool water in the exchanger, the smaller the thermal gradient, and therefore the smaller the heat flux into the pool water.
The water will heat most rapidly if the pool water runs quickly - this keeps the temperature difference greatest.
There is just one caveat: the power of the pump moving the water. If the pump is working harder to move water through a constricted valve, it would generate a little bit more power; if the water flow is set up so heat from the pump is dumped to the water, you will get a small amount of additional heating; but I don't believe that would ever offset the benefit of the faster gradient.
One other consideration: what happens to the surface of your pool. This relates to the way the output of your heat exchanges returns to the pool. If you have a jet that dumps deep inside the pool, there will be little disturbance at the surface; if it's aimed at the surface, you will cause some "stirring". As you may know, the greatest heat loss from a pool happens through evaporation - so if there is anything in your setup that increases evaporation as a function of flow rate through the heat exchanger, that will affect the total heating time.
If it were my pool, I would probably rig up a thermocouple and a data logger, and look at the evolution of temperature. Turn the flow rate up and down every two hours or so, and see if you can observe a change in heating rate on the temperature trace.
I am sorry - according to the laws of physics, your pool guy is wrong. Open that valve, and let the water flow!
Best Answer
You want to make sure that the heat flow is always between fluids at the nearest possible temperature, to minimize the entropy production from the flow of heat. The best method is to flow the cold coolant in the opposite direction as the hot coolant, so that as the cold coolant gets hotter, it is transferring heat to hotter hot-coolant. If you adjust the pipes right, and make them long, you can do the entire circuit with as close to zero entropy generation as you like. This is the principle by which ducks can send blood to their feet and do so without losing any significant body heat in the feet, although they are immersed in cold water.
For the temperature profile, if you have the hot water be 100 degrees, and the cold water be 0 degrees, and you have a long linear pipe where they touch, then the profile can be exactly linear with a 1 degree difference in temperature at all points, assuming that that heat diffusion constant for the metal is constant over the range of temperature, and the specific heat of water is constant for the range of temperature, and both of these are close enough approximations for a practical heat exchanger.
If the pipe runs from 0 to L, the profile for hot water temperature is:
$T_H(x)= {100x\over L}$
The profile for the cold water
$T_C(x) = {100x\over L} - 1$
so that the difference between them is always 1 degree. You adjust the flow rate so that the heat transfer moves C units of heat energy in a length $L/100$, where C is the specific heat of water, and then the exchanger works with this profile. You can adjust the temperatures to be as close to each other as you like, and the entropy gain from the heat flow is:
$ \Delta S = {Q\over T^2} \delta T \approx \Delta S_0 {1\over 230}$
Where you use the absolute temperature T in the denominator, so that in this system you only make about 1 percent of the entropy you would if you let the hot and cold water transfer heat by direct contact without an exchanger.
These profiles are universal attractors, so if you just set up the appropriate flow rate, you will approach the linear profile with time.