Optimizing return temperature in a brazed plate heat exchanger

I have a brazed plate heat exchanger similar to these. I can regulate flow and temperature of water on the primary side of the exchanger to achieve the desired temperature of water on secondary side. The desired temperature and flow on the secondary side are not under my control.

How do I optimize the flow and temperature on the primary side so that I get the lowest possible temperature for water exiting the heat exchanger, while still achieving the desired temperature on the secondary side?

Note: The transfer medium is chemically treated water, the maximum temperature is around 100 °C at 4 bar.

If I increase the flow to some maximum value I will also increase the heat loss through the pipe walls. However, increasing the flow rate would allow me to lower the supply temperature, which decreases the heat loss. How would I calculate the best flow rate and supply temperature combination to minimize heat loss through the pipe walls?

• Is the water on the primary side being heated or cooled (conversely, is the water on the secondary side being cooled or heated?) Feb 16 '15 at 19:45
• This is a heating system so the water is being heated. On the primary side I have a mixing valve and a frequency regulated pump, I can regulate the temperature and flow very accurately. Supply temperature on primary side is around 75°C - 95°C and normal demanded temperature on secondary side is 60°C-85°C. Feb 17 '15 at 6:26

As stated the problem is probably not well enough constrained.

Given the specification, the way you get minimum temperature out is to pump as much volume as you can through the primary and then regulate the input temperature to achieve desired output.

I.e. the more primary flow there is the lower delta temperature drop you need in the primary side. Pumping "flat out" may not have been what you had in mind.

Defining $T_{p,in}$ = Temperature primary in
$T_{s,out}$ = Temperature secondary out etc

$$kV_{in}(T_{p,in}-T_{p,out}) = V_{out}(T_{s,out}-T_{s,in})$$

k is an efficiency factor for your heat exchanger.

i.e. Energy removed from the primary stream is $V_{in}\Delta T_{in}$ and energy received by the secondary stream is $V_{out}\Delta T_{out}$, with a factor to account for losses and inefficiencies in the system.

Even with a perfectly insulated lossless system there will be a temperature delta above ideal due to the thermal transfer resistance of the exchanger.

I assume that heating due to flow / pumping losses in the heat exchanger are ignored (or otherwise they add to the energy available to transfer).

$T_{p,out}$ must always be greater than or equal to $T_{s,out}$ and may be greater than $T_{s,in}$ depending on design (but you hope not). For an ideal non counterflow exchanger $T_{p,out}$ is usually much greater than $T_{s,out}$.

If minimum $T_{p,out}$ is a genuine need then you want to use a counterflow heat exchanger. A counterflow exchanger allows $T_{p,out}$ to approach $T_{s,in}$ and $T_{s,out}$ to approach $T_{p,in}$.

• Note that the volumetric flow rates are only valid in these equations if the fluids are constant densities. If they're both liquid water at all points of the heat exchanger, that's not a terrible assumption, but it doesn't scale well to heat exchangers which use a gas or contain a phase change. Feb 17 '15 at 14:38
• OK looks like I was over thinking this problem. I was under the impression that the efficiency of the heat exchanger starts to drop if you increase the flow over a certain point but it looks like the higher the flow the better efficiency (up to a point). So my strategy will be to work with max flow and adjust the temperature until I reach the desired temperature on secondary side. Feb 17 '15 at 19:01
• Could you also check, additional question in my original post. Feb 17 '15 at 19:35

Russell covered this fairly well, but I want to expand on a couple topics he only briefly mentioned, and look at the analysis slightly differently.

First, you should have control over the direction of your flows as well. That is, you should be able to switch the inlet and outlet of the primary flow so that they're on the same side of the heat exchanger as the inlet and outlet of the secondary flow or vice versa. What will give you the lowest possible $T_{p,out}$ is a cross-flow arrangement (i.e., the primary inlet and secondary outlet on one side, and the primary outlet and secondary inlet on the other side.) There must be a temperature difference in the two streams at every contact point in the heat exchanger to have continuous heat transfer, but a cross flow arrangement allows the cold side output to be higher than the hot side output, which is not the case in a parallel flow arrangement, as shown in the image below. After this, finding the actual lowest possible $T_{p,out}$ is not as simple as doing a few calculations. You have to set up a system of equations that govern the behavior of the different sides of the heat exchanger as well as the boundary between the two flows and then use those to find the lowest primary outlet temperature using an optimization method. Optimization of thermal systems is a semester-long 400-level (undergrad senior, early grad student) college course. This is on the simpler side of those problems, but it's still not something trivial, because you either need a computer program to calculate a bunch of scenarios or use a more nuanced approach that takes less computing power but more thought and consideration, and is more in-depth than we can do here.

However, we can make a few general statements. The heat that leaves the primary side is the same heat that is transferred to the secondary side. Since we can't change anything about the secondary side, we can only change how we transfer the heat from the primary side. In general, that equation is

$$Q_p=\dot{m_p}(h_{p,out}-h_{p,in})$$

where $Q$ is the heat transfer, $\dot{m_p}$ is the mass flow rate, and $h$ is the enthalpy. Enthalpy can be tricky to define if you don't have much experience in thermodynamics, but it's a general way to define the energy in a flow based on temperature, phase, pressure, and volume. For example, it can quantify the energy change when water changes to steam at 100°C. The point is, if we can't change $Q_p$, what we want to do is generally lower $h_{p,out}$. To do this, we can either lower the input temperature (thus lowering the enthalpy at the input, but with the same mass flow rate, the enthalpy difference must be the same) or we can decrease the mass flow rate. This will increase the enthalpy difference, and if the input enthalpy remains the same, the output will get lower.

Note here that there's a limit to how low the output of the primary flow can be. It can't go lower than the input temperature of the secondary flow, otherwise the heat transfer would be reversed; this violates the 2nd Law of Thermodynamics, and in this house, we obey the laws of thermodynamics!.

There's also some diminishing returns, because as you decrease the mass flow, you're going to see less effective heat transfer, so you just might not get your desired output on the secondary side. My advice would be to lower the input temperature to just about the output temperature you need on the secondary side and keep the flow rate at a good pace.

• Thank you for your answare. I was under the impression that efficiency of the heat exchanger starts to drop if you increase the flow to much. Feb 17 '15 at 19:06
• The efficiency of the heat exchanger won't, but the efficiency of your overall system might. Increasing the flow will increase the pressure drop over the heat exchanger, which means you're going to be working harder on your pump. That's where the limitation on flow rate comes into play. Feb 17 '15 at 19:16
• Ok one more thought, If I increase flow to some maximum value. I will also increase the heat loss trough pipe walls, but on the other side I will lower it because I will operate with lower T supply. How would I calculate the best flow/T supply combination to minimaze heat loss trough pipe walls? Feb 17 '15 at 19:32
• I would treat that more as an effect than a design consideration. First, do everything you can through insulation. Some heat loss in the pipes is unavoidable, but the more heat you add, the faster you lose it, so there's only so much you can do efficiently and effectively. Modeling the pipes can be time consuming and complex, so it's probably just best to do it experimentally, increasing the supply temperature gradually to compensate for the losses. Feb 17 '15 at 19:49
• :-) "In this house ..." -> "1. You can't win. 2. You can't break even. 3. You can't get out of the game. " Feb 18 '15 at 14:01