#### apc3161

Hello all, I have a question about thrust, velocity, and power. I'm reading "elements of propulsion AIAA series" now in my spare time, and have stumbled across something that I can't quite figure out.

η_P=(T*V)/(Q_in ) '

or

overall efficiency = (Thrust * velocity )/ Q'_in

where Q'_in is the rate of thermal energy being release by the burning fuel.

So here is where I'm stuck. Imagine there were two identical rockets in space. And they were both ignited at the same time, the only difference is that rocket A had an initial velocity while rocket B did not.

That equation above (or for that matter, the equation of Power = velocity x force), would state that even though the rockets were burning the same fuel at the same rate, and had the same thrust, rocket A would have a higher power rate, just because its velocity was originally higher. Something about that doesn't seem rate.

Also, with regard to the efficiency equation, couldn't you have an efficiency of over 100%? I mean assume the rocket was traveling through space at 8,000 m/s with the rockets off, and then they were ignited at their lowest setting (assume this setting is so small that it is almost negligible, this make Q_in' very small). So in this case, even though the rocket was barely doing anything, because the initial velocity was so high, the power output would be enormous based on power=force*velocity. And if the initial velocity was high enough (make it any number), the product of thrust*velocity would be much greater than Q'_in, putting the overall efficiency at greater than 100%.

Can anyone clarify this to me, explain to me what I'm not understanding.

Also, how does one import pretty equations into this forum so that you don't have to type out ugly equations?

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#### Mech_Engineer

Gold Member
I think the velocity in the equation is velocity of the propellant/ehxuast, not the velocity of the vehicle. When you multiply the thrust of the propellant by its velocity, you get the propellant power (energy/time). When you divide the propellant thrust power by the total thermal power being released you get the thermal efficiency of the process.

For pretty equations, you can use LATEX. There are a few tutorials on this forum about how to use it.

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#### apc3161

I think the velocity in the equation is velocity of the propellant/ehxuast, not the velocity of the vehicle. When you multiply the thrust of the propellant by its velocity, you get the propellant power (energy/time). When you divide the propellant thrust power by the total thermal power being released you get the thermal efficiency of the process.

For pretty equations, you can use LATEX. There are a few tutorials on this forum about how to use it.
The example problem definitely uses the speed of the aircraft (239 m/s) as the V in this equation, not the exhaust velocity (1200 m/s).

#### apc3161

As stupid as it sounds, I think my problem lies with the fact that I have a hard time accepting that F * S = W

Force, distance, work.

I almost want to see that F * S_res = W, where S_res is only the distance moved that the force was directly responsible for.

I just dont understand how F*V=P, because in such a case, the initial velocity can be huge (from some other force), yet the F at that moment can be small, and the power output can be large. The power output would be large, not as a result of the force being currently applied, but because of the initial speed. That doesn't seem to make any sense.

#### rbeale98

remember that thrust and net force on the vehicle are not the same. for steady state conditions, thrust=drag (horizontal motion) or thrust=drag+weight (vertical motion) and net force = 0 (constant velocity condition), thus no work is done on the vehicle.

in the case you describe where initial velocity is huge and thrust is small, the F in F*S=W is the net force, mainly due to drag and negative with respect to the change in S. this is not steady state, and net work is negative.

#### apc3161

remember that thrust and net force on the vehicle are not the same. for steady state conditions, thrust=drag (horizontal motion) or thrust=drag+weight (vertical motion) and net force = 0 (constant velocity condition), thus no work is done on the vehicle.

in the case you describe where initial velocity is huge and thrust is small, the F in F*S=W is the net force, mainly due to drag and negative with respect to the change in S. this is not steady state, and net work is negative.
Thanks. So then I still have this question which confuses me.

Assume there is a rocket in vacuum of space. No drag, no gravity, etc. Assume the rocket turns on its thrusters with a thrust T. So the power is V*T. Doesn't that mean that as time passes, the power output of the rocket engine is increasing as it speeds up? Does that make sense? And seeing as the amount of energy per second being burned by the rocket is

h*m' (kJ/s)

h= heating value of fuel, kJ/kg
m'=mass flow of fuel kg/s

So the energy per second being released from the burning fuel is constant. But eventually the velocity of the rocket will be high enough so that worked being done on it per second, is higher than the energy being released per second by the burning fuel. Essentially, the "power" of the rocket is higher than the "power" from the combustion process. That confuses me a lot.

#### rbeale98

you got me, i have no idea

post your explanation when you figure it out

#### Bob S

The power input Q is energy per second which equals mass flow/sec times heating value:

Q = 1000 *h * m' [ note: 1000 * kJ/kg * kg/sec = J/sec]

[I have deliberately converted to Joules per sec by multiplying by 1000 so Q is mks units]
=============
Force times distance = energy or work; E = T * x [note:Newton-meters, where T is in Newtons]

Force times distance/sec = output power; P = T * v [note: Newton-meters per sec]

Where v is the exhaust velocity of the rocket propellant relative to the rocket, which is constant, even in interstellar space..
==============
1 joule/sec = 1 Newton-meter/sec = 1 watt [note: mks system of units]
==================

Consider interstellar travel, hence no gravity or air drag.

Then efficiency (fraction) = P/Q =T * v /(1000 * h * m') [note:units are T (Newtons);v (meters/sec);h (kJ/kg); and m' (kg/sec)]

I hope this helps.

#### apc3161

Thanks for the reply Bob. Using the relative exit velocity would definitely seem to take care of the efficiency greater than 100% dilemma. (although it seems to create a few more questions, but I'll address those later.

I have attached an example problem from the book I am reading on the subject ("Elements of Propulsion, Gas Turbines and Rockets, AIAA Education Series".

As you can see, the example problem uses V_o (velocity of the vehicle, not the exhaust velocity).

This is confusing me. Do you think they made a typo? I would seem that the answer is definitely yes.

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#### aerospaceut10

Need to make a clarification:

the u in the Thrust Power term is in fact the velocity of the vehicle with respect to the atmosphere (or none, if in space)

Edit:

I just noticed a discrepancy. What I do know for a fact is that the velociy of question is the relative velocity of the vehicle, but that's for an air breathing engine. I could be wrong but it's possible that it's exhaust velocity.

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#### xxChrisxx

Its the velocitry of the aircraft that is V.

It would not have more efficiency if the inital V were higher becuase the Qin value would pre propoertionally higher also. It requires more fuel to sustain the higher inital velocity.

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#### rbeale98

don't be surprised to find a mistake in that book, i found a few in it before. it's a pretty big book

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