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Forgive me I'm new and learning all this.

I want to know is the horsepower determined by how much pressure is in the conbustion chamber? I asked cause i seen away to increase horsepower is to add a cold air intake. Which would make the oxygen cooler argo more heat will transfer to it which will increase pressure. Or am i completely wrong?

  • What year make and model are you maintaining or repairing? – cory May 24 '17 at 12:58
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Cold air intakes increase power by reducing the air intake temperature which gives a higher (greater) density of the air charge : more available air means a better combustion and more power.

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Yes, of course (in theory). An internal combustion engine is a heat engine. That means it just works by combustion of fuel/air mixture which creates heat energy --> the pressure created by the heat and expanding gases create mechanical energy + waste energy --> these are "pulled" out of the engine through the drive train and the radiator/heater core/drive belts/air cooling/etc.

But that's very simplistic. Yes, while holding all things equal, in theory combustion of more air/fuel will of course create more heat --> more pressure --> more mechanical power--Nitrous does this in the combustion chamber.

There are, however, a number of factors which create horsepower: displacement being the biggest one. Mechanical efficiency is another (i.e. in the new Honda Civic they put lighter piston heads to increase power and efficiency). RPM is yet another.

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Power output is determined by the amount of fuel that is burned in a combustion cycle. This in turn depends on the amount of oxygen available to burn the fuel with. The amount of oxygen is influenced by the presence of turbo/supercharging, the compression ratio the amount of air resistance in the intake, and the air temperature.

A cold air intake reduces the air temperature, which makes the air density higher, allowing more oxygen to fit in the combustion chamber.

The heat transfer to the colder intake air does not make a difference.

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Without getting into the specific math, horsepower is basically determined by both torque and how fast that torque is allowed to be applied.  Torque (in this case) is the force that chemical combustion creates on the pistons, which is determined by things like compression ratios (how wide the combustion chambers are, how tall or short they are, types of pistons, etc.) and combustion efficiency.  Combustion efficiency is determined by how thoroughly the relevant air and fuel molecules bond before being ignited, which is aided through proper compression.  This is where stoichiometric balance comes in.

Stoichiometric balance in a combustion engine is the proper mixture of air and fuel volume, specifically oxygen and hydrocarbon content, necessary for optimum combustion.  The amount of fuel that is fed to the chambers will depend on how oxygenated the air that is drawn in is.  But keep in mind that there is only so fast that air volume can be drawn in so the more oxygenated (cold) the air is, the faster fuel can be delivered, which will allow faster combustion.  The faster combustion allows the torque to be applied at a faster rate. 

The cold air intake will increase the rate of air-fuel mixture drawn in because: 1. It's more oxygenated than when it's otherwise stationed close to the hot engine and 2. It's usually designed to allow more air in anyway, regardless of greater oxygen content.

Keep in mind that combustion also produces chemical waste and, if the rate at which air-fuel mixture is drawn in increases, the emission of its waste production will necessarily need to increase as well. 

So, to sum up, horsepower will be determined by how much torque is applied to create the rotational force and how fast it's allowed to be applied.  Because of this, many, many things determine horsepower.  The pressure in a combustion chamber will vary by actual air-fuel mixture but it's at its best with stoichiometric balance. A cold air-intake improves rate of torque by allowing the appropriate mixture to be drawn in, combusted and emitted at a faster rate.

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