I am building a preheater for a maple syrup evaporator and am going to use the steam generated by the heating process to pre-heat the incoming sap from, say 5 degrees C to (hopefully) something on the order of 80-90 C. (the goal is to heat the sap/fluid before it gets into the evaporating pan so that we are more efficient - the steam used is "free" - it is a waste product of the process and capturing the heat and subsequent condensation for hot water is desirable)

My flow rate through the copper tubes will be on the order of 35 to 56 liters per hour (10 to 15 gallons per hour) (this is for the end points of the assembly)

I only know the outside diameter of the copper pipes - they will be 1/2" OD.
There will be 5 parallel lines inside a hood and at each end a manifold - the incoming cold sap comes into the low side manifold and makes its way up to the high side manifold in one of the 5 pipes. The manifolds are 20 inches long and are perpendicular to the other pipes which are 40 inches long.

The heat exchanger will sit in a hood to provide a little back pressure for the steam and to collect the steam and force it to contact the pipes. The boiling surface area of the sap underneath the heat exchanger is larger than the heat exchanger and the hood but most of the steam rises through the hood and will transfer some heat to the copper which in turn heats the fluid.

So how can I calculate the some likely heat exchange/output temperature?

It seems to me that the issues are:

  • efficiency of heat transfer from steam to copper
  • efficiency of heat transfer from copper to fluid
  • how much time the fluid is in the lines

I guess I am looking for an upper bounds and lower bounds (I know I won't get the fluid over 100 C - that's about all I know for sure - well - I guess I can also be sure that the temperature will rise some - but not sure how much.)

Many people report output temps on the order of 95C but I am skeptical.

What can I do to improve the heat transfer to maximize the scavenged heat? I can try to find fins from a baseboard radiator, however this is for food product - so I am unsure if I will do that.

Some pictures of something I would like to do:




Wood is used as the heat source and one can assume a constant fire/heat source that will be providing 10 to 15 gallons of evaporation per hour from a 24" by 60" pan.

EDIT (answering a comment)

The wall thickness of the copper pipe is 0.569 inches.

There is only one fluid being heated - that is the sap from a maple tree. It is water with sugar in it - and by boiling it we can remove most of the water until it turns to maple syrup.

The level of the fluid (sap) in the evaporator pans is kept constant by a float valve. The sap in the pan boils - that is where the steam comes from. So the rate of evaporation will be the same (approximately) as the flow through the heating tubes.

The heat is generated by a wood fueled fire in a firebox just below the evaporator pans that hold the sap.

Maple sap is held in a storage tank above the whole contraption. Gravity feeds it into the evaporator (a 5'x2' pan) through the preheater/heat exchanger. The fluid flow/level is regulated by a float valve. Steam is a by-product of the process - and using the wasted heat from the steam (allowing it to condense on the copper pips and transferring heat then drawing that condensation away so as not to have to re-heat it) makes the process a little more efficient. One study that was done shows that there is approximately 15% improvement in boiling rate (best case) for preheating the sap before it goes directly into the pan that gets heated.

  • $\begingroup$ Can you measure the weight of a spare pipe segment of a known length so we can infer the thickness? And I want to know more about how the steam is produced, as in, what temperature does the syrup leave the thing before you (presumably) cool it down by boiling water off it. You will have a heat rate at which you cool it and a heat rate at which you preheat it, which will be less. At atmospheric pressure you will be limited by the steam production rate and the heat transfer rates. I'm trying to think how non-condenseable gases will affect it too. I'll get back to this later. $\endgroup$ – Alan Rominger Jun 29 '11 at 2:10
  • $\begingroup$ Edited question in response $\endgroup$ – Tim Jun 29 '11 at 4:09

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.

  1. The steam is being provided at a faster rate than it is being condensed
  2. 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.

  • $\begingroup$ Great - thank you. We can assume the fluid is water - it is sugary water, but still water. The boiling temp of the most concentrated liquid is about 2C higher than that of water. The steam (from all practical observation behaves as in case 1 - more steam than is needed - so there is saturation). Again, thanks for your time! $\endgroup$ – Tim Jun 29 '11 at 14:34
  • $\begingroup$ One kg of steam delivers about 540 kcal on condensation, heating one kg od feed from 20 to 100 C takes about 80 kcal. So, the only thing to avoid is having too much cold air drawn in with the steam. The 15 % saving is consistent with this figures. Saving more would afford two or more steps of evaporation with staggered working pressure. There was a funny post on maple syrup making here some years ago. $\endgroup$ – Georg Feb 3 '15 at 19:08

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