How are heat exchangers normally calculated?

Question

Most of the calculations that I find in books are about calculating the heat transfer rate, which I'm not sure why I would care about. What I imagine is the case in a real world situation is that I have a fluid that I want to cool down from temperature A to temperature B, and I want to find an exchanger that can do the job. What I know are the inlet temperatures and flowrates of both fluids.

Most calculations require both inlet and outlet temperatures to be entered, but if I'm looking to buy an exchanger, I can't know what the outlet temperatures are going to be.... I only know my inlet temperatures... so I'm a bit confused.

Answer 1

Your specification is that you know the two inlet temperatures, $T_{h,i}$ (hot in) and $T_{c,i}$ (cold in). Interestingly, you also say that you want to cool the hot stream to a defined temperature $T_{h,o}$. So, you have at least three defined parameters for your "real world" case.

Let's take also as a given that you are specifying the fluids for the hot and cold fluids. This means, you are specifying whether the fluids are water, air, oil, steam, or whatever. When you do this, because you have the inlet temperatures of the fluids, you also specifying the starting points for the specific heat capacities of your fluids, $\tilde{C}_{p,h/c}$ (J/kg K). Take the specific heat capacities as constant over the heat exchanger to start.

Two approaches can be taken next. In one case, you also specify the mass flow rate of at least one stream. Let's suppose that, because you have specified that you want to cool the hot stream, you also specify how much flow rate $\dot{m}_c$ (kg/s) you need to cool.

Immediately, you see that you can calculate the required heat demand (W)

$$ \dot{q} = \dot{m}_h\tilde{C}_{p,h}\left(T_{h,i} - T_{h,o}\right) $$

Now, you are in position to determine the mass flow rate and outlet temperature of the cold inlet stream.

$$ \dot{q} = \dot{m}_c\tilde{C}_{p,c}\left(T_{c,o} - T_{c,i}\right) $$

Essentially, you are in a position to play around with the three parameters $T_{c,o}$, $\dot{m}_h$, and $\dot{m}_c$. When you specify any one of the three, the other two are defined by the two heat balance equations.

But what about the TYPE and SIZE of the heat exchanger? The type is a choice such as shell-and-tube or complex with flows as parallel or cross-flow. The size is expressed as the area $A$ (m$^2$).

Assume that you have solved the heat transfer equations. The simplest method to set the type and size is the log mean temperature difference (LMTD)

$$\dot{q} = f\ U\ A\ \Delta_{LMTD} T $$

In this, $f$ is a factor based on the type of exchanger, $U$ is the overall heat transfer coefficient (W/m$^2$ K), and $\Delta_{LMTD} T$ is the log mean temperature difference (K). The value of $U$ will depend on the construction of the exchanger, the type of exchanger, and the mass flow rates.

In some cases, the problem is open-ended. We are only given the inlet temperatures and the fluids. We are after both outlet temperatures as a function of flow rates, type of exchanger, and size of exchanger. The more respected approach in this case is the number of transfer units (NTU) method.

$$ \dot{q}_{act} = \epsilon\ \tilde{\dot{C}}_{p,min}\left(T_{h,i} - T_{c,i}\right) $$

Here, $\epsilon$ is an effectiveness that depends on the type of exchanger, $U$, and $A$, while $\tilde{\dot{C}}_{p,min}$ (W/K) is the smallest value combining $\dot{m}$ (kg/s) and $\tilde{C}_p$ (W/kg K) for the fluids.

Summary details are provided effectively at this Wikipedia link. Full details and graphs are provide at this link. Finally, a comparison of LMTD and NTU methods is provide at this link.

Answer 2

Most of the calculations that I find in books are about calculating the heat transfer rate, which I'm not sure why I would care about.

Let's see if that's true.

What I imagine is the case in a real world situation is that I have a fluid that I want to cool down from temperature A to temperature B, and I want to find an exchanger that can do the job. What I know are the inlet temperatures and flowrates of both fluids.

So you have the flow (mass per unit time) and the ΔT (temperature difference A - B). Using the specific heat capacity (SHC) you can calculate the energy to be removed per unit time - that's joules/second (J/s) or watts (W). Now you know the required heat transfer / exchange rate and you have a specification for your exchanger.

Most calculations require both inlet and outlet temperatures to be entered, but if I'm looking to buy an exchanger, I can't know what the outlet temperatures are going to be.... I only know my inlet temperatures... so I'm a bit confused.

You use the inlet and outlet temperatures of the fluid to be cooled. That's A and B in your example. You specify the outlet temperature.

Answer 3

My answer will focus on your statement:

Most of the calculations that I find in books are about calculating the heat transfer rate, which I'm not sure why I would care about.

Assume you have a hot(ter) fluid with temperature $t_{h1}$, that exits at temperature $t_{h2}$, and a cool(er) fluid that comes in temperature $t_{c1}$, that exits at temperature $t_{c2}$. Then you know the following:

• $t_{h1}>t_{h2}>$ the exit temperature of the hot is always less than the entry temperature because the hot looses some heat
• $t_{c1}<t_{c2}$ the exit temperature of the cool(er) is always more than the entry temperature because the cool(er) fluid gains some heat
• the obvious $t_{h1}>t_{c1}$ (otherwise there would be no heat exchange or it would have the opposite direction).

(Note: depending on the type of heat exchanger you might also know that $t_{h_2}>t_{c_2}$, but that is not true for all exchanger).

The key point from above, is that heat energy is lost from the hot(ter) fluid and the same heat energy is gained by the cool(er) fluid.

The amount of heat energy (actually its power because its energy in the unit of time) that is transferred is:

$$\color{red}{\text {heat energy lost from hot}} = \color{green}{\text {energy transferred}}= \color{blue}{\text{ heat energy gained from cool}}$$

Because when a material which flows with a rate $\dot m $ changes temperature from $t_1$ to $t_2$ then the the heat energy change for that material is $\dot{m}\cdot C_{p} \cdot (t_{1}- t_{2})$, the above equality becomes.

$$\color{red}{\left|\dot{m}_h\cdot C_{p,h} \cdot (t_{h1}- t_{h2})\right| } = \color{green}{\text {energy transferred}}= \color{blue}{\left|\dot{m}_c\cdot C_{p,c} \cdot (t_{c2}- t_{c1})\right|}$$

That is the reason that there is so much interest in the heat energy transferred. Because it defines the limits that the heat exchanger can operate and it will define also

• the minimum temperature of the hot, and
• the maximum temperature of the cooler fluid.