Exit Velocity Question

[Stephen Smyth]

Hello!

I'm currently working on a school project, and I'm researching nozzle design. As of now I am having issues with understanding if exit velocity varies with altitude, and if so..how exactly?

I know that exit velocity depends on nozzle geometry, and assuming we are using 1D flow, one could use isotropic relations to calculate it.That being said the only thing I can think of that could cause exit velocity to change is chamber temperature, and exit temperature.

Any explanation as to how exactly this might work would be great! Thanks for your time!

 

[Hooked On Rockets]

Hello!

I'm currently working on a school project, and I'm researching nozzle design. As of now I am having issues with understanding if exit velocity varies with altitude, and if so..how exactly?

As little as I know about rockets so far, but answering this first question.... the higher up your altitude, the less back pressure, due to less atmospheric pressure on the nozzle/propellant charge, there will be an increase in output velocity...regardless of temperature...provided the output pressure remains the same....

 

[OverTheTop]

To accelerate sub-sonic flows you constrict the flow. To accelerate supersonic flows you expand the flow. The throat of the motor has the flow at sonic speed. There is an equation that describes this effect (can't remember it offhand). That is the basics of the de Laval convergent-divergent nozzle.
https://en.wikipedia.org/wiki/De_Laval_nozzle

The expansion of the exiting flow depends on the atmospheric pressure around the nozzle and will be optimised at only one altitude. Exhaust expansion lower than that altitude is under-expanded (higher atmospheric pressure squeezes exhaust), and above it is over-expanded.

Exit velocity depends on the amount of expansion. That expansion is controlled by the nozzle and atmospheric pressure. Exit velocity will change with change of expansion.

 

[jsdemar]

There is an optimal expansion ratio given the ambient atmospheric pressure. The length of the nozzle and the opening diameter will create the same exhaust pressure as the outside air pressure. At sea level the optimal nozzle is shorter than one used in a second stage, for example. In Space, the optimal nozzle would be infinitely long.

The factor used to represent the efficiency of the nozzle is called the thrust coefficient, Cf. It is directly multiplied times the thust. A 10% improvement in Cf by simply improving the nozzle length and shape is a 10% improvement in total impulse of the motor without changing anything internal to the motor.

The simplest way to look at it is: you want all the mass to accelerate straight down with the highest velocity. If the nozzle is underexpanded or overexpanded, there is sideways vector component to the exhaust gases. The work done sideways against the atmosphere does not accelerate the rocket.

The best expansion ratio will depend on the propellant gas products, the chamber pressure, and the ambient atmospheric pressure. See bottom graph:
https://thrustgear.com/docs/nozzle_optimization.html

Since the rocket will not be at any one pressure (altitude) for the whole burn, the expansion ratio is a compromise.

 

[Chuck Rogers]

Stephen:


The complete equations are here in this free pdf download on the RASAero web site:

https://www.rasaero.com/dl_solid_motor.htm

Click on the three parts of "Performance Analysis of the Ideal Rocket Motor".


Right below it on the web page is Departures from Ideal Performance. But Performance Analysis of the Ideal Rocket Motor has all of the derivations which will answer your questions.


Charles E. (Chuck) Rogers
Rogers Aeroscience

 

[rocket_troy]

Hello!

I'm currently working on a school project, and I'm researching nozzle design. As of now I am having issues with understanding if exit velocity varies with altitude, and if so..how exactly?

I know that exit velocity depends on nozzle geometry, and assuming we are using 1D flow, one could use isotropic relations to calculate it.That being said the only thing I can think of that could cause exit velocity to change is chamber temperature, and exit temperature.

Any explanation as to how exactly this might work would be great! Thanks for your time!

Well, it’s probably possible to gather the exhaust velocity without using pressure terms given that

Ve = SQRT( ((2 * k) / (k - 1)) * (R * Tc / M) * (1 - (Pe / Pc)^((k-1)/k)))

Where:
k = ratio of specific heats of the gas
Tc = Chamber Temp
R = specific gas constant (IIRC depending on what units you’re using for M)
M = Molecular Weight of gas
Pe = Exit Pressure
Pc = Chamber Pressure


Using isentropic relations, we can substitute the pressure ratio of the equation with a temperature ratio given that

(Pe / Pc)^((k-1)/k) = Te/Tc


To provide us with

Ve = SQRT( ((2 * k) / (k - 1)) * (R * Tc / M) * (1 - Te/Tc))


How practical this is? You’ll still need to process pressures to gather Te anyway and the reason why the pressure ratio is in this equation is because the pressure ratio is a fundamental part of the exhaust delta V.

More fundamentally: Ve = SQRT(2*(Ho-He)) where Ho and He are the chamber and exhaust enthalpies respectively. Remember that pressure is a fundamental factor of enthalpy and whichever flavour of fundamental exhaust velocity you prefer (there are many); they’re all fundamentally based around enthalpy.

 

[aerostadt]

If you have a fixed geometry for the classical Delaval convergent-divergent nozzle, i.e., the area ratio defined by the throat area divided by the nozzle exit area, being fixed and the chamber pressure is above the critical pressure ratio the flow in the divergent section will be supersonic and thus the nozzle exit pressure is then also fixed. The nozzle exit pressure may not match the ambient (surrounding) pressure. The nozzle exit pressure may be higher than the ambient pressure, which means that the flow is under-expanded. The nozzle exit pressure may be lower than the ambient pressure, which means that the flow is over-expanded. If the nozzle exit pressure matches the ambient pressure, that is the optimum condition. In all these cases the nozzle exit velocity is fixed. Even if you decrease the ambient pressure by going up in altitude, the nozzle exit velocity is still the same, because it is supersonic flow and the exit Mach number is only determined by the area ratio. Now can the thrust increase if the altitude is increased? The answer is yes, because the thrust is the sum of the force produced by the momentum thrust, which is determined by the exhaust velocity, and the pressure thrust, which is the difference between the nozzle exit pressure and the ambient pressure. So, you might ask why worry about the nozzle exit pressure being mismatched with the ambient pressure. It turns out that the thrust is optimum, when the nozzle exit pressure is equal to the ambient pressure. In such a case the pressure thrust that we just talked about is zero, but this is made up by the momentum thrust and nozzle exit velocity being at their highest.