# Aerospace Why do Jet Engines Have a Compressor?

1. Oct 17, 2008

So like it says. It may be a silly question, but I am doing some reading on jet engines (on Wiki) and I am having trouble finding an explanation as to why we want to increase the pressure of the air before it goes into combustion. Does this just another way of increasing the thrust?

Thanks,
Casey

2. Oct 17, 2008

### jaap de vries

That is actually a very good question. In the following link an attempt is made in answering this very basic question.

http://www.aerospaceweb.org/question/propulsion/q0033.shtml

The main reason given in the link above is quote: "If uncompressed, the air-fuel mixture won't burn and the engine can't generate any thrust". As a combustion engineer I don't like this explanation (the German V1 rocket did not have a compressor I believe). I see no reason why a fuel air mixture uncompressed would not burn.
The whole idea behind a jet engine is that you accelerate the fluid going through it with as driving force a pressure gradient. The pressure in a jet engine is the highest right after the last stage of compression (for axial compressors). The combustion then takes place with minimum pressure loss and the the expansion begins through the turbines and finally through the nozzle . The high pressure created by the compressor forces the fluid to go out through the back. In other words, it creates the favorable pressure gradient used to accelerate the flow and produce thrust. By drawing out a Brayton cycle you can also show the effect this has on the efficiency.

3. Oct 18, 2008

### Danger

Good post. The V1, though, used a valved pulse-jet. Compression wasn't needed because the valves closed off the front of the combustion chamber upon firing. That made it essentially a rocket.

4. Oct 18, 2008

### Phrak

For the same reason you get better power out of an internal combustion engine with a higher compression ratio. This is really a question in thermodynamics.

Danger, the buzz bomb used a tuned exhaust. After the exhaust had pulsed out the back, the low pressure returned a portion of it to toward the combustion chamber. This compressed the air fuel mixture that had entered during the low pressure phase.

5. Oct 18, 2008

### Averagesupernova

I would say the main reason for the compressor is power density in any engine.

6. Oct 18, 2008

### Danger

From everything that I read when researching pulse-jets (which, admittedly, was more than 35 years ago), the original exhausts weren't tuned at all. Nobody expected the 'vacuum' intake effect. Something like 80% of the fresh intake charge got sucked in through the tailpipe, but that wasn't part of the design plan.
Subsequent designs capitalized upon that feature, and eventually the valve box was eliminated when people figured out how to properly tune the system so the pulses effectively closed the front of the chamber through pressurization.
The reason that the V1 was called the 'buzz bomb' was because of the noise made by the intake valves flapping.

7. Oct 18, 2008

### turbo

Casey, a compressor on a jet engine prevents back-flow and it acts like a supercharger on a car's IC engine, delivering an increased load of oxygen so that more fuel can be burned more efficiently.

8. Oct 18, 2008

### Staff: Mentor

Ramjets don't require a dynamic (turbo) compressor, and instead use the forward motion of the engine to compress the air in the inlet.

Thrust comes from the mass flowrate and velocity.

The compressor increases the pressure of the atmosphere and momentum of the air flow. The air (oxidizer) combines with the fuel, which burns and provides energy to the flow. The combusion does two things - the combutsion process breaks fuel molecules to make more molecules and provides heat which increases the temperature which decreases the density which causes an increase in velocity of the flow.

The compressor 'pushes' the flow into the combusion chamber, and the flow passes out the exhaust with a much high velocity than it entered - and that combined with the mass increase from the fuel - causes the thrust.

Think of the continuity (mass flow) equation. Mass flow in = mass flow out, otherwise the mass in the control volume (jet engine) would increase or decrease.

The momentum equation: momentum out > momentum in, and the change in momentum gives the thrust (force).

Energy (in flow) equation: Work-Energy out = work in + energy (thermal) produced

9. Oct 19, 2008

### Andre

In addition to all those excellent answers, there is also the air density problem. The higher the flight, the less dense the atmosphere, the less drag and the least fuel to be used. But at those densities there is not a lot of oxygen to burn and we really need high compression rates to sustain a nice little flame in that engine.

10. Oct 22, 2008

### jaap de vries

Fair enough, my point was that the argument that uncompressed fuel air does not burn makes no sense to me.

11. Oct 22, 2008

### FredGarvin

DING DING DING!! We have a winner. Well said Astro. Concise and right on.

To initiate flow, there has to be a favorable pressure graidient in the engine. The compressor's job isn't to make sure it keeps flow moving in a certian direction, although most of the time it does that as a by product of increasing the density (although in surge it definitely stops doing that). Also, as Jaap pointed out, the idea that an air-fuel mixture wouldn't burn uncompressed is just silly.

The whole point is that thrust is directly proportional to the mass flow through the engine. The more mass flow, the more thrust.

12. Oct 22, 2008

### Phrak

Interesting. I wonder if it would have worked at all without the accidental compression. I have part of a pulse jest--minus the tail pipe. It's amazingly simple: a set of reeds, a block of aluminum with holes for the intake, a fuel venturi, a starter spark plug and combustion chamber/exhaust tube.

13. Oct 22, 2008

### Danger

Neat! Original, or a reproduction?

14. Oct 22, 2008

### Phrak

Like I said, this is question in thermodynamics.

In a jet turbine engine, a 4 cycle gasoline engine, diesel, or coal dust turbine an element of combustable fluid goes through a process of compression, ignition and expansion to obtain work.

I could find a believable answer that gave power efficiency as a function of pressures. What's the power efficiency ratio of a compressed vs. uncompressed working fluid? Is it 4 to 1, 5 to 1, twenty to 1?

15. Oct 22, 2008

### FredGarvin

From a thermo point of view, the cycle efficiency is directly tied to the compressor's pressure ratio. The propulsive efficiency drops off as pressure ratio increases but the thermal efficiency increases with pressure ratio. This makes the overall efficiency increase with pressure ratio. If you look through some literature like Hill and Peterson, you can see plots of this very topic that illustrate this point.

16. Oct 22, 2008

### Staff: Mentor

As Andre wrote - when density of the mixture is too low, there is not enough energy produced by burning fuel to sustain the fire.

Simple thought experiment - we have a mixtures of oxygen and fuel of varying densities. On the one end we have mixture that we know will burn, on the other end we have vacuum (wih only traces of oxygen and fuel) - same mixture, but it will not burn. Somewhere in the middle must be the point where we move from "burning" to "not burning".

17. Oct 23, 2008

### Phrak

What is propulsion efficiency?

The efficiency of an Otto cycle, eta is given as

$$\eta = 1 - \left( \frac{V_{top}}{V_{bottom}} \right) ^{0.27}$$

as given in this link, http://courses.washington.edu/me341/oct22v2.htm that I can't generate a hyperlink for.

A volumetric compression ratio of 12 spits out 49% efficiency.

A ratio of one gives zero, which makes no sense, or course.

18. Oct 23, 2008

### FredGarvin

For a basic definition, the propulsive efficiency is

$$\eta_p=\frac{T*u}{m\left[(1+f)(u_e^2/2)-u^2 /2\right]}$$

Where:

T = thrust
u= vehicle velocity
ue = exit velocity
f = fuel air ratio

It is also considered to be

$$\eta_p = \frac{\eta_{overall}}{\eta_{thermal}}$$

Note: the term $$T*u$$ is often refered to as the thrust power.

Here's another reference on this with a reduced form:
http://en.wikipedia.org/wiki/Propulsive_efficiency
http://web.mit.edu/16.unified/www/FALL/thermodynamics/notes/node81.html

Last edited by a moderator: Apr 19, 2017
19. Oct 23, 2008