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.
Best Answer
As dmckee said, the problem is complex. Fortunately, there are some simplifications that can be made without much damage to the accuracy. I hope I can present them in an accesible manner.
My guess is that since You have constant inlet and ambient temperatures and a target outlet temperature, what You are looking for are the pipe's dimensions and properties (type of material).
At the beginning You can neglect the radiation from the surface - at 500K it barely exists.
As it was mentioned, the easiest way to solve this problem is to compute an algorithm using software like MatLab or Wolfram Mathematica. Also get familiar with dimensionless numbers used in heat transfer theory. Two of them are more important than others - Nusselt number and Reynolds number. Nusselt number is vital for determining forced convective heat transfer coefficients, there are many empirical correlations for Nusselt number available, but they usually depend on the value of Reynolds number. Reynolds number defines whether the flow is laminar, transitional or turbulent. For turbulent flows in horizontal pipe's Gnielinski correlation for convective heat transfer coefficient is quite good. Calculating the heat transfer coefficients You will often have to define the wall temperature. It is generally unknown, it differs not only along the pipe's length but also in the radial direction - it is hotter on the inner surface of the pipe. Since the pipe's walls are usually thin, in Your calculations You can assume that the temperature is uniform along the radius, but You cannot assume that it is uniform along it's length. That's why the easiest way is to solve it iteratively, calculating the fluid's properties after each, let's say 5 cm of the pipe. Each time You can set the wall temperature to $$\frac {T_{\text{fluid}} + T_{\text{amb}}}{2}$$, which isn't very bad approximation. You must remember though, that $T_{fluid}$ is 500K only at the first step, every next step it should be a result of the previous calculation. You carry on with the calculations unless the temperature of the fluid reaches 330K. Then You count the number of steps, multiply it by 5cm, and the result is the required pipe's length. Of course, at the beginning, pipe's inner and outer diameters shall be assumed as well as the pipe's material (it affects the value of thermal conductivity). Now that You have calculated the length You have to decide if it suits You, and alternatively change the pipe's diameters or material and repeat the calculations until the pipe's length satisfies You.