Is there a 1:1 relationship between boost pressure and power increase?

Question

This question got me thinking: if you were to install a turbocharger on an engine, is there a straightforward relationship between the boost pressure and the amount of power you can expect?

For example: if the engine makes 100 kW naturally aspirated, and you install a turbo and set it to provide a max. boost of 0.5 bar, can you expect 150 kW max. power (i.e. new power output = original power * (boost pressure +1)) ? Or is the relationship more complicated?

Let's assume the engine is set up correctly to take advantage of the turbocharger, i.e. the injectors have enough capacity, and fuel/air mix remains the same.

Answer 1

Preamble

What does forced induction buy?

In a word, density.

Remember:

• For compressible fluids, pressure alone doesn't tell the full story

  But pressure and temperature together do.

  The old physics adage "hot air rises, cold air sinks" is a great example of this. Air at the same pressure but different densities at different temperatures.

• The internal combustion engine is a volumetric device

  What this implies is that every time the engine turns over and completes a cycle, the volume of air that is admitted into the combustion chamber(s) is fixed.

• Power depends on mass, not volume

  The power developed by the engine is proportional to the mass of air admitted into the combustion chamber and not its volume.

  So more dense = more air molecules per cylinder = moar power

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So, is the ratio 1:1?

No. Because physics said so.

Time to break out ye olde Evo example with the 85% efficiency turbocharger:

• At atmospheric conditions (14.7 psi, 25 °C)

  Air density = 1.184 kg/m^3

• With 22 psi of boost, air density doubles:

  Turbo discharge conditions: 36.7 psi, 92 °C

  Air density = 2.413 kg/m^3

These two data points alone show that a 2.5x increase in pressure yielded a 2x increase in density.

So the pressure-power relationship isn't 1:1.

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Hmm, but could the ratio be constant?

Again, the answer is no. Because physics said so.

Let's up the Evo's boost to 29.4 psi to check this. We'll maintain the same turbocharger efficiency (85%):

• @ 29.4 psi boost (so outlet pressure = 3x inlet pressure):

  Turbo discharge conditions = 44.1 psi, 155 °C

  Air density = 2.473 kg/m^3

So a 3x change in air pressure resulted in 2.08x density change. Clearly not linear, especially considering the result obtained with 22 psi boost.

Answer 2

tl;dr: no, a 1:1 ratio is only possible in imaginary perfect laboratory conditions.

Or is the relationship more complicated?

It's a bit more complicated but for perfectly understandable reasons.

NOTE: I'm intentionally leaving intercoolers and bags of ice out of the discussion below. They are germane to boost discussions but should be covered under a different question.

Let's assume the engine is set up correctly to take advantage of the turbocharger, i.e. the injectors have enough capacity, and fuel/air mix remains the same.

The most important missing assumption is a critical one: constant temperature.

Let's back up all the way to the core of the engine: the combustion. The air and fuel are mixing at an approximately 14:1 ratio, igniting, expanding and pressing outwards to make chemical potential energy into kinetic.

But what is that ratio really? It compares the molecules of air to the molecules of fuel. Get those out of balance and the combustion reaction is no longer at peak efficiency (note: we are going to see this word again).

Given that background, what does boost do? In theory, it's a molecule inserter: your boost mechanism is trying to go get more air molecules to which the engine will add an increased number of fuel molecules. Combust that augmented mixture with its increased amount of chemical energy and you'll get more kinetic energy, right?

Yes, but not as much as you might think. You've already run into Boyle's Law. Even. If you have a perfect air molecule scooper, just forcing those molecules into the engine is going to increase their temperature. The engine computer is going to have to correct for that temperature by adding more fuel (as a sort of coolant), retarding timing, etc. Failing to deal with this temperature will lead to put the engine on the knocking curve which eventually terminates in a disastrous transformation into an external combustion engine (i.e., important bits will come out).

It gets worse. Remember that perfect molecule scooping boost mechanism? Not possible. It also has an efficiency factor that is less than 100%. It's going to grab air and compress it but, unfortunately, it increases the temperature even faster than Boyle's Law (efficiency is less than 100%). This engages the other terms of the Law: density of the intake air will drop with temperature: it's both hotter and there are fewer molecules.

The result of all this back of the envelope hand waving is that, if you're really focused on wanting 50% more power, you're going to need more than 50% as much air and more than 50% more fuel.

In short, 100% efficiency is the theoretical maximum but is only achievable in Perfect World. That said, small boost systems can come much closer to 1:1 more easily than high boost.

Answer 3

Answer to question is basically YES.

I disagree in the manner the above have characterised this, Ur not wrong exactly just overly complicated and this is poor teaching practice, for a given volume / mass of gas at constant temp then doubling the pressure halves the volume ie inversely proportional ie pv=constant, so basically under these conditions U would be able to stuff twice as much air in, the fuel ratio being fixed then double the power, anyway that's the place to start, and of course your ratios are not constant when you using less than 100% efficiency and the temps not constant, anyway start with the simple perfect world then apply the application specifics eg flow turbulence esp due to metal / rubber hose stepping, heat due to compression of gas, intercoolers, hit n cold side pressure control BOV / gates, and so on and so forth, bang it on a dyno is time and money better spent than theorising endlessly, efficiency / optimisation is the game for most machines, getting more from a finite resource, more 'useful' work, thanks.