Why does force decrease with increase in velocity in case power was constant?

Question

Suppose an internal combustion engine burns the same amount of fuel every cycle (regardless of engine or car speed ) that means it creates the same pressure every cycle and the force on the piston due to gas pressure is $\Delta P\cdot A$ (let's assume this process is isobaric and the whole fuel mixture ignites simultaneously for sake of simplicity). how is force the same according to the previous equation while the P=FV equation says otherwise?

this implies that the car accelerates at a constant rate and power keeps increasing but that doesn't happen in real life that's why we use forced induction or just increase the amount of burnt fuel to increase power, also cars on dynos reach max power at a certain rpm so it doesn't look like a thing due to air resistance.

Am I missing something?

Note: I'm not talking about internal combustion engines specifically I just found it to be the best example to demonstrate my reasoning and know that I said force and velocitybut changing to their rotatinal counterparts won't change anything about the physical rules or affect the reasoning.

Answer 1

Engines don't create force (or torque) all by themselves. There must be a load against it. Just like you can push several pounds of force against a wall, but you cannot develop pounds of force in the air.

As the velocity increases, the ability of the road to be an effective load decreases (in the absence of air resistance, which this question seems to be assuming).

Imagine standing next to a playground merry-go-round. It's pretty hefty, so it takes a lot of force to accelerate much. Initially you have no difficulty suppling a large force. But as you push, it gets up to speed and spins quite quickly. You'll find you can no longer supply much force at all to the equipment since it is moving so fast.

The same thing happens between the car and the road. If you are in a low gear, the engine is turning very rapidly and the gas expansion in the cylinder is not producing the same pressure (because the volume of the cylinder is larger at the same point in the cycle). Or you can shift into a high gear, but now the leverage of the transmission means that your forces (torques) created on the ground are lower.

Answer 2

how is force the same according to the previous equation while the P=FV equation says otherwise?

You need to be careful about just throwing equations around. You are confusing two completely separate velocities. One is the velocity of the piston inside the engine, and the other is the velocity of the vehicle on the road.

Focusing momentarily on the velocity of the piston. The faster that the piston expands the less pressure is exerted on the piston. So it is not correct that $\Delta P \cdot A$ is constant when varying piston speed. Also $\Delta P \cdot A$ is not power, it is energy. This is energy per cycle, so as the cycles go faster the power will increase even with constant energy. This is what makes the engine have a peak power RPM. At lower RPM you are producing fewer cycles at a given energy, and at higher RPM you are producing less energy per cycle.

Now, focusing on the velocity of the vehicle. Suppose that we have an ideal continuously variable transmission so that the engine can run at one fixed optimal power. The transmission will be continually varying the gearing ratio between the engine and the road. As it changes the gearing ratio that will reduce the force at the road. So you will get reduced $F$ at the wheel as $V$ increases in order to maintain a constant $P$.