# Applying Power concepts to a Rocket?

1. Nov 7, 2013

### Jorghi123

I've recently learned about the "Tsiolkovsky rocket equation" which can be used to relate initial mass, final mass, rate of fuel change, velocity change, etc.. Except I'm having difficulty applying it to Power.

How do you find the instantaneous Power of the engine, fuel, and the total power of the engine?

2. Nov 8, 2013

### rcgldr

The fuel has a chemical potential energy, which after losses (such as heat), is converted into the total energy of the rocket and the plume of spent fuel ejected by the rocket engine.

One way of calculating the power is to assume an initial state where the rocket is not moving and all of the energy is going into the ejected mass.

update as mentioned in the next post, power equals increase in energy per unit time

The power will equal the 1/2 mass flow per unit of time (m dot), times the exit velocity^2 of the mass.

Last edited: Nov 8, 2013
3. Nov 8, 2013

### K^2

Thrust is equal to mass flow rate times flow velocity, not power. Power is more complicated.

For starters, power is frame-dependent. In the frame of the rocket, we can only talk about power output in terms of work done on the exhaust. Naturally, that's half of the mass flow rate times the flow velocity squared. $P = \frac{1}{2}\dot{m}v_p^2$ Just like kinetic energy, but we are using mass flow rate instead of mass to get power output. This quantity, however, is pretty useless.

It is much more useful to talk about work being done on the rocket itself from a perspective of an external inertial frame. In this case, power of the rocket is thrust times rocket's velocity.

$$P = F_p v = \dot{m}v_p^2 ln\left(\frac{m_0}{m}\right)$$

Here, $m$ is the current mass of the rocket, $m_0$ is initial mass, $v_p = g I_{SP}$ is exhaust velocity, and $\dot{m}$ is the mass flow rate.

The interesting bit here is that power obviously increases as the rocket speeds up. This is known as the Oberth Effect. Furthermore, once the rocket is really going, this power output can be significantly higher than available chemical energy of the fuel! How is this possible? Well, once the rocket is traveling at high speeds, fuel also has kinetic energy. So you get higher power output of the rocket engine by tapping into kinetic energy of the fuel you are burning, which, in turn, was built up by burning fuel when the rocket was going slow.

4. Nov 8, 2013

### rcgldr

My understanding is that if the energy of the ejected plume of fuel and the energy of the rocket are both included as the total mechanical energy of the system, then the power output from the engine is independent of the frame of reference, as long as the frame of reference is inertial.

5. Nov 8, 2013

### D H

Staff Emeritus
That is exactly the quantity that K^2 described as pretty useless.