The heat exchanger is a very important device used in many real world applications in which heat must be transferred from one medium to another. In many cases, the two mediums are separated by a solid wall, although in some cases the two mediums are in direct contact with each other, so that mixing occurs. For example, in some applications steam is injected into water in order to heat it up.

Some common applications where a heat exchanger is used are: refrigeration, air conditioning, space heating, and power plants, to name a few. There are of course many more.

There is a large variety of heat exchanger configurations, but most can be categorized into one of three types. The three types are: Parallel-flow or counterflow configuration, cross-flow configuration, and shell-and-tube configuration.

Parallel flow and counterflow configurations are shown in the two figures below. Both figures show a simple concentric tube arrangement in which one of the fluids flows on the inner tube and the other fluid flows in the annular gap (between the tubes). The figures show the hot fluid as being inside the inner tube, and the cold fluid as being inside the annular gap. In the parallel flow configuration both the hot and cold fluids flow in the same direction. In the counterflow configuration the fluids flow in opposite directions. These will be discussed in more detail later.

Heat transfer is usually better when a flow moves across tubes than along their length. Hence, cross-flow is often the preferred flow direction, and tends to be better than parallel flow or counterflow configurations. Cross flow configurations are shown in the three figures below. The first figure shows cross flow over an unfinned array of tubes. The second and third figures show cross flow over a finned array of tubes, with square fins and circular fins (as shown). The purpose of fins is to increase heat transfer between the hot and cold mediums. This will be discussed in more detail later.

Shell-and-tube configurations are shown in the three figures below. One of the fluids flows through the inside of the shell and the other fluid flows through tubes passing through the inside of the shell, thereby enabling heat transfer between the two fluids. Baffles are added to enhance the convection coefficient, which increases heat transfer between the two fluids. The baffles serve to induce turbulent mixing and a cross-flow component, both of which increase the convection coefficient. The first figure shows one shell pass and two tube passes. The second figure shows two shell passes and four tube passes. The third figure shows a more detailed drawing of a shell-and-tube heat exchanger with one shell pass and one tube pass.

Source: http://en.wikipedia.org/wiki/Heat_exchanger. Author: http://commons.wikimedia.org/wiki/User:H_Padleckas

In the next section we will explain how fins are used to increase the heat transfer in a heat exchanger, thus boosting its effectiveness.

A fin can be thought of as an extension of a surface. It adds additional surface area, which enables additional heat flow to or from the medium the fin is in contact with, by way of convection. To illustrate in quantitative terms the usefulness of a fin consider the following schematic showing a pin fin protruding out of a base surface at surface temperature

Where:

Assume steady state heat flow where the energy that enters the differential element equals the energy that exits the differential element. This is a valid assumption for steady state operating conditions.

We can write the energy balance as

The left side of the above equation is the heat energy entering the differential element. The right side of the above equation is the heat energy exiting the differential element.

By Newton’s law of cooling,

where

Substitute the above equation into equation (1). We get

Rearrange the above equation to give

Divide both sides of the above equation by

For

From Fourier’s law,

where

Substitute the above equation into equation (2), and simplify. This gives us the final general differential equation for one-dimensional steady state heat transfer from an extended surface (given below). Using this equation we can solve for the temperature distribution

Note that although

To get an idea of the degree to which a fin can increase heat transfer let's assume the pin fin discussed here is of constant cross-sectional area, where

Since the above is a second order differential equation, we need two boundary conditions in terms of

Thus, Fourier’s law at

so that

We can now solve equation (4) for the temperature distribution

By Fourier’s law,

Hence, solving for

where

and

Note that equation (5) represents the heat transfer from the base of the surface (with area

To see how much the fin increases heat transfer calculate the following ratio:

which becomes

To illustrate by example, assume the pin fin is circular so that

We get

Given this high ratio, it's clearly very useful to add fins to increase heat transfer from a surface. The alternative way to increase heat transfer is by increasing

Note that the heat transfer rate

Fins are especially important for situations where the convecting medium is air or some gas (with lower

Also note that

Fins can have a variety of shapes. For example, they can be pin fins protruding from a surface (as just described), or annular fins around a tube used to enhance heat flow into or out of the fluid flowing through the tube. The figure below illustrates such a fin.

If fins are attached to a wall with a metallurgical or adhesive joint, a significant thermal contact resistance may exist at the interface. This can be accounted for by a fin correction factor, discussed in a later section on the overall heat transfer coefficient.

Next, we will analyze the parallel flow heat exchanger for steady state flow. Steady state flow is a valid assumption for steady operating conditions and temperatures.

The figure below shows a schematic of a parallel-flow heat exchanger along with temperature distribution for the hot and cold fluids. We treat the heat exchanger as insulated around the outside so that the only heat transfer is between the two fluids.

As one would expect, the cold fluid (bottom curve) heats up and the hot fluid (top curve) cools down.

Where:

Assume steady state heat flow where the energy that enters the top differential element equals the energy that exits this element. We can write the energy balance as

The left side of the above equation is the energy entering the differential element. The right side of the above equation is the energy exiting the differential element.

Rewrite the above equation so that it becomes

By the First law of thermodynamics the left side of the above equation can be expressed as

Where:

If we assume constant specific heat

Note that we are treating

Also note that

Similarly, we can also apply an energy balance to the bottom differential element. Following the same procedure as before, we get

Where:

From Newton’s law of cooling

where

Combining equations (6)-(8) and using calculus to solve for

Where:

By direct observation, the above equation tells us that

It is worth noting that the total heat transfer rate

The two equations above can be used in addition to equation (9) to solve a problem involving parallel flow heat exchangers. The above two equations also apply for a counterflow heat exchanger. Note that numerical iteration may be necessary in equation (9) for cases where unknown inlet and/or outlet temperature(s) need to be solved for. However, with computers this is easy to do.

By direct observation, the above two equations tell us that

It follows that, for a given

The figure below shows a schematic of a counterflow heat exchanger along with temperature distribution for the hot and cold fluids. Once again, we treat the heat exchanger as insulated around the outside so that the only heat transfer is between the two fluids.

As one would expect, the cold fluid (bottom curve) heats up and the hot fluid (top curve) cools down.

The same steps are followed to derive

Note that a counterflow heat exchanger is more efficient than a parallel flow heat exchanger. It takes a smaller heat transfer surface area

Also note that in a counterflow heat exchanger

In principle, the maximum possible heat exchange would be achieved with a counterflow heat exhanger of infinite length. In such a heat exchanger the maximum possible temperature difference achieved (by one of the fluids) would be equal to

Knowing

For a parallel flow or counterflow heat exchanger, if one of the (hot or cold) fluid streams is condensing or evaporating, its temperature will remain approximately constant between inlet and outlet. This constant temperature can then be applied to the equations as

The overall heat transfer coefficient

Where:

The

where

Also,

where

Equation (10) comes from the fact that energy flow is constant (steady state) through the different mediums located between the tube fluid and the ambient. To show this, consider the schematic shown below. For visualization purposes let's imagine we have an imaginary differential strip (of thickness

We can write the following heat rate equations:

Where:

For steady state heat flow,

where

and

Note that

Since

Note that equation (11) has the same form as equation (8) (with

Where:

A positive value of

For the case of a tube with fins on the inside and/or outside we can still use equation (10) but with a correction factor

Where:

The area on the inside of the tube (fin area plus exposed base) is equal to

Once again,

The correction factors

Finally, we can add fouling factors to the above equation which accounts for deposits accumulating on the inner and outer surfaces, over time. These deposits can be a result of fluid impurities, rust, or chemical reactions between the fluid and wall material. The modified equation accounting for fouling factors then becomes

where

Since the heat transfer

Looking at the above equation you can only maximize

For example, lets say the fouling factors and

Now, substitute

Now let’s say that we add fins to the inside tube surface as well so that

Now, substitute

Estimating the convection coefficient for a variety of geometries, fluid types, and flow conditions, is a difficult problem, and is by no means simple to explain in brief terms. For this reason it is best to refer to dedicated sections in books that explain in detail how to estimate convection coefficients for different configurations, along with associated pressure drop for the fluid streams. For standard finned surfaces one may refer to the book

Some typical values of the convection coefficient

• Free convection for gases: 2-25 W/m

• Free convection for liquids: 50-1000 W/m

• Forced convection for gases: 25-250 W/m

• Forced convection for liquids: 50-20,000 W/m

• Convection with phase change (boiling or condensation): 2500-100,000 W/m

The reference for the above values is from:

We are given a counterflow heat exchanger with the following known values:

Assume that the heat transfer surface area on the cold and hot side are approximately equal.

Find the outlet temperature of the hot and cold fluid and the rate of heat transfer (

There are three unknowns to solve for. They are

The first equation to use represents the heat transfer rate (

The second equation to use represents the heat transfer rate (

The third equation to use is given by equation (9), with the following variables defined as follows for a counterflow heat exchanger:

To calculate

Solving numerically, we get

To choose a suitable heat exchanger for a certain application requires a level of knowledge and experience. The information presented here along with the books referenced here is certainly a good start. When designing or choosing a heat exchanger there is no single "correct" solution. Different types of heat exchangers can work equally well.

There are different ways to optimize heat exchanger design. Part of the optimization usually requires that the tube wall thickness be as small as possible (while still being strong enough), and the thermal conductivity of the tube material be as high as possible. This ensures that the thermal resistance of the tube wall is as low as possible which aids in heat transfer. Also, we wish to keep the pressure drop of the hot and cold fluid streams between inlet and outlet as small as possible. However, to increase the rate of heat exchange between the two streams we must increase the convection coefficient (

Given the inherent mathematical complexity of heat exchanger design (such as estimating convection coefficients for complex flow patterns), it is often necessary to use heat exchanger software to aid in the design process, and to do so in a time efficient way. This is evident when one considers all the different design parameters to be evaluated when coming up with a "best" design. Some examples of common design parameters to be taken into account are:

• Flow rate of both fluid streams

• Inlet and outlet temperatures of both streams

• Operating pressure of both streams

• Allowable pressure drop of both streams

• Fouling resistance for both streams

• Physical properties of both streams

• Type and configuration of heat exchanger

• Tube sizes, number of tubes, number of baffles (if applicable), baffle size, baffle spacing

• Number of fins (if applicable), fin size, fin spacing, and the fan or blower power necessary to force air (or another medium) through the fins

• Material types used in the heat exchanger

• Size limitations

• Others, depending on design requirements

Clearly, heat exchanger design is a multivariable problem which does not usually lend itself to a simple solution. Several design iterations may be necessary before settling on a good design.

I went through my own heat exchanger design efforts when building a homemade air conditioner some time ago. It was a fun project which gave me a good practical understanding of how heat exchangers work. My first build attempt is shown in the two pictures below.

The picture above shows the inside of the unit. Air blows through the annular space and around a cold ice-water filled metal pail, causing the air to cool down before exiting through a hole at the top. Temperature measurements revealed that the air came out about 2.5 degrees Celsius cooler than it was going in. The heat exchange taking place is between the ice-water and the air. Naturally I want the heat exchange to be as high as possible so that the air comes out as cool as possible. So the task of figuring out how to do this is a heat exchanger problem.

For my second attempt I added vertical wooden rods in the flow stream which helped to induce turbulent mixing which boosted heat transfer between the ice-water and air stream. This is shown in the picture below. The result was that the air came out about 3 degrees Celsius cooler, which is a slight improvement.

Not being satisfied with the result I decided to change the design completely. In this design (shown below) I made a U-shaped channel, which worked much better. Temperature measurements revealed that the exiting air was about 10 degrees Celsius cooler than it was going in, which is a great improvement over the earlier design.

The picture above shows the U-shaped channel I made out of thin aluminum sheet, wood, and lots of silicone sealant around the edges so that water doesn't leak in when immersed in the ice-water. The channel has about a 1.8 cm gap through which the air flows. The width of the channel is 25 cm, and the length of the channel (i.e. the flow length along the U-shape) is about 60 cm. The U-shaped channel sits in a cooler of ice-water and the fan sits on top of the duct which has a hole in it so that the fan sits in snugly.

However, the household fan has difficulty blowing air through the narrow and long gap of the U-channel. The greatly improved heat transfer (and increased cooling of the air) resulting from the narrower and longer gap means that significant back pressure is created, which results in the fan having a much harder time pushing the air through the channel. The fan speed has to be set at the highest setting to get a decent rate of air flow through the channel. This is the tradeoff mentioned earlier between greater pressure drop and the increased heat transfer that results. Physically speaking, the fan must overcome the resistance to air flow associated with the pressure drop caused by the narrow and long gap the air must travel through. In this case it would be better to use a blower of some sort which is designed to push air in the presence of back pressure.

Return to

Return to

If you need one-on-one help with physics, get help from a Chegg Tutor! 30 minutes free for new users.