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Intercooler Maths

Working out what an intercooler will do

by Julian Edgar

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When a turbo or supercharger compresses air, the air is heated. While this hot air can be fed straight into the intake of the engine, there are two disadvantages in taking this approach.

Firstly, warm air has less density than cool air - this means that it weighs less. It's important to know that it's the mass of air breathed by the engine that determines power, not the volume. So if the engine is being fed warm, high pressure air, the maximum power possible is significantly lower than if it is inhaling cold, high pressure air.

The second problem with an engine breathing warm air is that the likelihood of detonation is increased. Detonation is a process of unstable combustion, where the flame front does not move progressively through the combustion chamber. Instead, the air/fuel mixture explodes into action. When this occurs, damage to the pistons, rings or head can very quickly happen.

If the temperature of the air can be reduced following the turbo or supercharger, the engine will have the potential to safely develop a higher power output. Intercoolers are used to produce this temperature drop.

Intake Air Temperatures

How much hotter the air gets as it is being compressed depends on the pressure ratio (how much it is being compressed) and the efficiency of the compressor. This means that the theoretical outlet temperature can be calculated if three factors are known: the inlet air temperature, the compressor efficiency, and the pressure ratio.

Before this can be done, the temperatures and pressures need to be expressed in the right units. Firstly, temperatures need to be converted to Kelvin (K), a measurement of absolute temperature.

K =°C + 273.15

A temperature of 35°C is therefore the same as 308.15K (or 308K for our purposes).

Boost pressures also need to be converted to pressure ratios. Note that 1 Bar = 14.5 psi.

Boost Pressure in Bar + 1
Pressure ratio = -------------------------------------

A boost of 1.5 Bar therefore becomes a pressure ratio of 2.5.

Let's have a look at an example.

If the inlet air temperature to a turbo is 20 °C (293K) and the boost pressure is 1.1 Bar (pressure ratio = 2.1) the theoretical outlet temperature will be:

Theoretical outlet temp = 293 x (2.1)0.286

= 293 x 1.236

= 362K (89 °C)

This means that there is a temperature rise of 69 °C (89 ° - 20 °= 69 °).

However, this doesn't take into account that the compressor efficiency will be less than 100 per cent. If we assume a compressor efficiency of 70 per cent (typical for a good turbo):

69 °
Actual temp increase = ------
= 98.6 °C

This is a temperature increase of 98.6 ° which when added to the ambient temp of 20° means that the actual outlet temp will be:

20 + 98.6 = 118.6 °(119 °C when rounded)

While the theory is fine, there are a number of factors that affect the accuracy of the calculated figure. Firstly, it is difficult to accurately estimate the efficiency of the compressor. And even if such a figure is available, it doesn't necessarily apply to all the different airflows that the compressor is capable of producing. In other words, there will be some combinations of airflow and boost pressure where the compressor is working at peak efficiency - and other areas where it isn't. While a well-matched compressor should be at peak efficiency most of the time, in some situations it will be working at less than optimum efficiency. This will change the outlet air temperature, usually for the worse.

Secondly, the turbo- or supercharged car engine is not working under steady-state conditions. A typical forced induction road car might be on boost for only 5 per cent of the time, and even when it is on boost, it is perhaps for only 20 seconds at a stretch. (But higher in both aspects for turbo diesels.) In most petrol engine turbo cars, longer periods of high boost occur only when hill-climbing, towing or driving at maximum speed. While all of the engine systems should be designed with the full load capability in mind, in reality very few cars will ever experience this. This factor means that the heat-sink capability of the intake system must be considered.

If the inlet air temperature of the engine in cruise condition is 20 degrees C above ambient, then on a 25 degreeday the inlet air temp will be 45 degrees C. After 30 minutes or so of running, all of the different components of the intake system will have stabilised at around this temperature. If the engine then comes on boost and there is a sudden rise in the temp of the air being introduced to this system, the temperature of the turbo compressor cover (or blower housing), inlet duct, throttle body, plenum chamber, and inlet runners will all increase. These components increase in temp because they are removing heat from the intake air and so limit the magnitude of the initial rise in the actual intake air temperature. As a result, the infrequent short bursts of boost used in a typical road-driven forced-induction car often produce a lower initial intake air temperature than expected.

This doesn't mean that intercooling is not worthwhile (it certainly is) but that the theory of the temperature increase doesn't always match reality. However, taking a theoretical, calculated approach at least gives a general idea of the temperature increase that is likely to be experienced.

Air Density Changes

In the example above there was a temperature increase from 20 ° to 119°. But how does this temperature change affect the all-important density of the air? The density of air depends on two factors - its temperature and pressure. The drop in density due to increased temperature is directly proportional to the ratio of the temperatures, when they are expressed in Kelvin. So the drop in density of the air at 119 °C (392K) versus 20 °C (293K) is found by:

Density difference = -----
= 0.75

In other words, the temperature increase would have caused a drop in density by 25 per cent, had the air still been at the same pressure. But of course its pressure is now higher because the turbo or supercharger has compressed it!

If the air pressure is doubled without a change in temperature, the density is doubled. To work out the effect of both the loss of density because of the temp rise and the increase in density because of boost pressure, the two factors are multiplied.

Inlet Air Temp (K)
Increase in Air density = -------------------------- x pressure ratio
Outlet Air Temp (K)
= ------ x 2 = 1.49

In other words, the increase in density has been 49 per cent at 1 Bar boost with a compressor outlet air temperature of 119 °C. This means that, all things being equal, the engine can develop 49 per cent more power.

Intercooler Efficiency

An intercooler will do two things - it will lower the temperature of the intake air and at the same time cause a slight drop in boost pressure. The latter comes from the restriction to flow caused by the intercooler. Some restriction is unavoidable because the flow through an efficient intercooler core needs to be turbulent if a lot of the air is to come in contact with the heat exchanger surfaces. However, if the pressure drop is too high, power will suffer. A pressure drop of 1-2 psi can be considered acceptable if it is accompanied by good intercooler efficiency.

Intercooler efficiency is a measurement of how effective the intercooler is at reducing the inlet air temperature. If the intercooler reduces the temperature of the air exiting the compressor to ambient, the intercooler will be 100 per cent efficient. It will also be a bloody marvel, because no conventional intercooler can actually achieve this!

Intercooler efficiency is given by:

Actual Temperature Drop
Intercooler Efficiency = ---------------------------------------------------
Maximum Possible Temperature Drop

For example, if on a 20 °C day the outlet air temp of the turbo is 110 °C and the temperature of the air after it has passed through the intercooler is 45 °C then:

110 ° - 45 °
Intercooler Efficiency =---------------
110 °- 20 °
= -----
= 0.72 (or 72 per cent)

Final Calculations

Let's take an example that puts together all of the figures. You fit an intercooler that is 72 per cent efficient and has a pressure drop of 0.14 Bar (2 psi) to an engine running 1.5 Bar (22 psi) boost developed by a turbo compressor that is 70 per cent efficient. It's a 25 ° Celsius day.

Firstly, the theoretical compressor outlet air temp:

298K x (2.5 pressure ratio)0.286= 387K = 114 °C

114 °C - 25 ° aC = a temperature increase of 89 °C

89 °C
------------------------------- = 128°C temperature increase
compressor efficiency

128 °C temp increase + 25 °C ambient = 153 °C air temperature coming out of the turbo!

The maximum possible temperature decrease through an intercooler: 153 °C - 25 °C = 128 °C

An intercooler that is 72 per cent efficient will give an actual temperature drop of:

128 °C x 0.72 = 92 °C

A turbo outlet temp of 153 °C - 92 °C intercooler drop = actual engine inlet air temp of 61 °C.

Boost pressure was 1.5 Bar but with the pressure drop through the intercooler it becomes:

- 0.14 = 1.36 Bar.

1.36 + 1
----------- = a pressure ratio of 2.36

So with an intercooler we have a temp of 61 °C at a boost pressure ratio of 2.36, compared with a non-intercooled 153 °C at a pressure ratio of 2.5.

Intercooled:-------x 2.36 pressure ratio = 2.1 times non-turbo air density

Non-Intercooled: -------- x 2.5 pressure ratio = 1.75 times non-turbo air density

If the engine developed 100kW (134hp) in naturally aspirated form, with 1.5 Bar (22 psi) boost it could develop a theoretical 175kW (235hp) without an intercooler and 210kW (281hp) with an intercooler. On this example, fitting the very big intercooler is good for a 20 per cent power gain. However, again reality intrudes - the picture is often even better than this. Any forced induction car seems to go disproportionately harder when the ambient temp is low - much more so than the physics suggest. The same applies to a car equipped with a very good intercooler - throttle response is superior and power markedly so.


The maths is straightforward and clearly shows that maximum intercooler efficiency matched with the lowest pressure drop will result in best power.

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