Question: To my knowledge, bypass air produces 80% of total thrust. But I don't understand to how it does that. By accelerating the air, by increasing the speed or increasing the pressure of the air? Is it doing this by Bernoulli's principle or something like that?

Answer: 
The bypass air is accelerated by the fan at the front of the turbofan engine. This changes its velocity and therefore its momentum, which is the definition of a force (in this case: thrust):

$$ F = \frac{\text{d}}{\text{d}t} p = m \frac{\text{d}}{\text{d}t} v = m \cdot a $$

This thrust contributes to the total thrust of the engine. How much will depend on the bypass ratio of the engine and other parameters, but 80% is plausible for a high bypass turbofan. Bernoulli's principle has nothing to do with this.

You can see the accelerated airflow in the following animation:

[![Turbofan airflow](https://i.stack.imgur.com/0fvyl.gif)](https://i.stack.imgur.com/0fvyl.gif)  

(source: [Wikimedia](https://en.wikipedia.org/wiki/File:Turbofan3_Labelled.gif))

The air coming from the engine core will move even faster, but there is less of it in a high bypass turbofan resulting in less total thrust coming from the core:

> 
> The fan airflow, referred to as the cold air stream, is accelerated by the fan and passes through the engine remaining outside of the engine core. The cold air stream moves much slower than the hot stream gas flow passing through the engine core.
> 
> 
> 

([skybrary.aero](https://www.skybrary.aero/index.php/Turbofan_Engine))
A bypass fan provides thrust in the same way a propeller provides thrust: by increasing the energy content of the gas mass passing through the disk.

The added energy is most effectively converted into thrust by allowing it to expand until internal pressure is equal to ambient pressure, so all added energy is converted into kinetic energy. This expansion takes place in the cowling behind the fan.

But the initial energy addition is a mixture of both increased pressure and increased velocity: the gas will require time to accelerate to the outlet velocity.