We all know about engine power. We’ll mention it when talking about our favourite car. And it seems obvious – the more power something has, the faster it can go.

But what is power exactly?

To answer that, we need to dive into some physics.

## The equation for power: power = work / time

Fundamentally, power is the work done, divided by the time taken. And work is equal to the force applied, times by the distance travelled. So the equation is:

Power = work / time = force x distance / time

For example, if an object needs a large force to make it move, and you push it a long way, then you have done a large amount of work. The time you take to push it that distance then determines your power output.

Because velocity (which is basically speed) is equal to distance divided by time, we can boil this equation down to power equals force times by velocity:

Power = force x velocity

This provides another way of looking at power – the faster something is moving, and the greater the driving force, the higher the power output.

## Engine power: horsepower vs. kilowatts

Quantifying engine power is done with a couple of different unit systems, which can make things confusing. The USA uses horsepower (hp) which is a system 18th Century engineer James Watt devised to benchmark newly invented steam engines against draft horses. One hp is equal to 550 pounds of force acting through one foot in one second.

Most of the world now use the metric system, with the watt (w) being the unit for power. The watt is named after James Watt for his contributions to the steam engine and the industrial revolution.

Being part of the metric system, the watt makes for easier calculations as it links up more coherently with other units of measurement. One watt is equal to one joule per second. And one joule (work done) is equal to one Newton (force) applied through one metre.

As one watt is a small amount of power, we typically quote engine power in terms of kilowatts (kW). One kilowatt is equal to one thousand watts.

## How much power does a car need?

We can apply the power equals velocity times force equation to cars. It tells us how much power an engine needs to make for a car to run at a certain speed. The faster the speed, and the greater the driving force pushing the car along at the wheels, the higher the power demand.

At constant speed, the driving force needs to equal the sum of the resistive forces. Aerodynamic drag is the main resistive force, and it grows exponentially with the car speed. This means the power demand of a car also increases exponentially with speed. For this reason, significant effort goes into the aerodynamic design of race cars to reduce drag and increase top speed.

Two other resistive forces are rolling resistance and gravity. Rolling resistance is a force made by tyre deformation. Gravity comes into play if a car is driving uphill, where some of the downward force will act in the car’s backward direction. Both of these require extra driving force, and therefore power, to overcome.

If an engine cannot provide enough driving force, then the car will slow down. Conversely, if a car needs to speed up, then the driving force needs to be greater than the resistive forces which increases the power demand.

## The PV chart - key to understanding engine power

So now we see how the concept of work and power applies to a car. But how do we apply it to an engine which is what powers the car?

Plotting cylinder volume by pressure of a working engine gives us a PV chart. The lines of this chart represent the cylinder condition at different points of the cycle as the piston moves up and down. And the enclosed area is equal to the work done per cycle. Before we explain how this is, let’s look at the engine processes on the PV chart.

For a 4-stroke Otto cycle, which is the cycle of most petrol/gasoline engines, the PV chart shows:

Intake: The engine draws fresh air and fuel into the cylinder while the cylinder volume increases and the pressure stays constant (or at least close to). The PV chart shows this as a horizontal line moving from low volume to high.

Compression: The piston moves up and squeezes the intake air and fuel as the cylinder volume decreases. The PV chart shows this process as a curved line moving from high volume and low pressure, to low volume and high pressure.

Combustion: The fuel and air react, resulting in a rapid expansion of gases which push the piston down. The chart shows this process as a vertical line, further increasing pressure from the compression stage. From there, a second line for this stage goes from low volume and high pressure, to high volume and low pressure. The third segment is a vertical line that shows a reduction in pressure as heat is taken away from the engine by the cooling system.

Exhaust: The piston pushes the exhaust gases out of the engine as it moves in an upward direction. The PV chart shows this process as the same line as the intake stroke, however, it is going in the opposite direction.

This chart is an idealised version of an engine cycle. Actual engine PV charts have a less regular shape for several reasons. But a chart like this is still generally good enough for looking at the performance of an engine.

## How to use the PV chart to calculate engine work

As I mentioned earlier, the enclosed area in the PV chart is equal to the work done by the engine. How is this so?

To demonstrate, let’s look at a simplified process where a cylinder goes from high volume to low volume while at constant pressure. If we plot this on a PV chart, it would be a line travelling horizontally from right to left.

Area = Pa x Δm^{3 }= N/m^{2 }x Δm^{3} = N x Δm

One pascal of pressure (Pa) is equal to one newton (N) of force applied to one square metre (m^{2}). When we multiply this to a change in volume (Δm^{3}), the metre exponents cancel out, and we are left with newtons x metres. This is once again our equation for work – force times by the change in distance!

## Using integral calculus for engine PV charts

Engine PV charts are more complex than the example process discussed above. The curved lines mean you cannot simply multiply the pressure by the change in volume to get the enclosed chart area and work done.

What we need to figure out the chart area is integral calculus. I will not cover this in detail, as calculus is a subject in itself and can be tricky. But, the process involves integrating between A and B in the chart below to calculate the work made during combustion. Then, by integrating between C and D, we get the work required by the engine to compress the fresh air and fuel. Subtracting the latter from the former, gives the net engine work of the cycle.

## Calculating engine power from work and RPM

Once we know the work done by the engine, we can factor in time to find the power. To do so, we first find the revs per second (RPS) by dividing the RPM by 60. Then we find the time taken per revolution by getting the inverse of the RPS. As 4-stroke engines complete a full cycle every two revolutions, multiplying the revolution time by two gives the cycle time.

Working in metric units, we can then divide the work done by the engine, by the cycle time to give the engine power. As an example:

**Engine work (chart enclosed area): **3000 J

**RPM:** 5000

**RPS:** 5000 / 60 = 83.3

**Time per revolution:** 1 / 83.3 = 0.012 sec

**Time per cycle:** 2 x 0.012 = 0.024

**Engine power:** 3000 / 0.024 = 125,000 W = 125 kW

**If you want to convert to horsepower:**

Horsepower = 125 x 1.341 = 168hp

## Hey, what about power = torque x RPM?

You might be thinking, “I thought power was torque x RPM”. Well, that is true too!

In physics, there are often many equations that explain the same thing. It all depends on the angle you are looking from. As discussed in our previous post, the power equations regarding engine torque (τ) are:

Metric units:

Power = τ x π x RPM / 30

Imperial units:

Horsepower = τ x RPM / 5252

With the engine PV chart, we are looking at the work and power exerted on the piston. Whereas, applying the torque x RPM equation to an engine gives the engine power at the crankshaft, which is driven by the piston. Putting any efficiency losses aside, the power of these will be the same. It’s just that they refer to different stages of the power transferring process.