How Is Thrust Calculated for a Jet Engine in a Lab Test?

[controlfreaks]

Homework Statement

A jet engine is being tested in the laboratory and it is found to consume 20 kg of air per second while using 0.2 kg of fuel per second. Given that the exit velocity of the gases is 500 m s-1, calculate the thrust generated by the engine.

Homework Equations

F = M(V-U) where F is thrust, M is mass and V and U are final and initial velocities.

The Attempt at a Solution

not sure what to be doing here as i don't know what's going in where. what goes in for U the initial velocity?

 

[Shooting Star]

Hi, welcome to PF.

Homework Statement

A jet engine is being tested in the laboratory and it is found to consume 20 kg of air per second while using 0.2 kg of fuel per second. Given that the exit velocity of the gases is 500 m s-1, calculate the thrust generated by the engine.

Homework Equations

F = M(V-U) where F is thrust, M is mass and V and U are final and initial velocities.

Can this equation be correct? The left hand side has the dimension of force while the right hand side has the dimension of momentum.

Let's do it in a very simple way.

In one second, how much mass is exiting the rocket? That is given.

So, what is the impulse on this mass?

We know that Impulse = Force X time.

What low should be used now?

 

[controlfreaks]

Hi Shooting Star!
Thanks for the reply!

how do we know how much mass is exiting the rocket?
we know how much fuel and air it consumes and the exit veolcity of the gases but not the mass exiting do we?

 

[Shooting Star]

Hi Shooting Star!
Thanks for the reply!

how do we know how much mass is exiting the rocket?
we know how much fuel and air it consumes and the exit veolcity of the gases but not the mass exiting do we?

In this case , "consumes" means that the fuel and the air mixes and combustion takes place and the end product exits as a hot gas , producing thrust on the docket . That is given in the very first sentence of the problem . Do you understand now ?

 

[controlfreaks]

ok i get what you mean there! total mass is 20.2Kg

so why do we need the impulse? and as far as the thrust goes how do we now find what this would be. I don't need you to solve for me, just let me kno what's happening with it.
cheers!

 

[controlfreaks]

bump!

can anyone help on this?
much appreciated!

 

[Shooting Star]

ok i get what you mean there! total mass is 20.2Kg

so why do we need the impulse? and as far as the thrust goes how do we now find what this would be. I don't need you to solve for me, just let me kno what's happening with it.
cheers!

The fuel and air mixture exit the rocket after burning and therefore produces a reaction force on the rocket. The impulse on the existing gas has been given.The impulse is the change in momentum and the time is one second. Use the equation I have given you. The force on the exiting gas must be equal to the thrust on the rocket, according to Newton's third law of motion.

 

[minger]

I think we're making this problem more difficult than it needs to be; and it seems that way because of a simple mistake. You are trying to use Conservation of Momentum for a fluid, but you forgot time rate.
<br /> \begin{equation}<br /> \begin{split}<br /> F \neq m(V_2 - V_1) \\<br /> F = \dot{m}(V_2-V_1)<br /> \end{split}<br /> \end{equation}<br />
As shooting star mentioned, ALWAYS check your units. If you're not getting Newtons, then you're not calculating force. Also, there may be a pressure component that should be included, if not, then you haev mass flow rate and initial and final velocities. Plug and chug.