# How much efficiency is lost from a fixed De Laval nozzle when modified for atmospheric use?

Fixed nozzles designed for vacuum are larger/broader than those designed for atmospheric or combined air/space use, which have less flanged nozzles due to the drag a larger flange would incur in the atmosphere.

How much efficiency is conceded by reducing the flange of the nozzle for use at ground launch, compared with a nozzle designed for use purely in a (near) vacuum?

### Edit:

Rephrasing the question:

How much less efficient is a fixed nozzle designed for optimal sea-level launch efficiency when used in a vacuum when compared with a fixed nozzle designed for optimal in-vacuum efficiency?

• I think this is a duplicate of this question: space.stackexchange.com/questions/14820/… Jul 11, 2016 at 3:16
• @OrganicMarble That appears to be the other way around, overexpanded rather than the underexpansion an atmosphere-optimized nozzle will have in space. Jul 11, 2016 at 3:43
• Same equation, no? Jul 11, 2016 at 3:44
• The nozzle size optimization isn't drag-related. Jul 11, 2016 at 4:09
• The matter mentioned in the question linked as a duplicate is 'untapped potential in the gas before exiting the nozzle' when they are underexpanded. Though the equation is the same, that seems like a physical phenomenon that is different than overexpanded nozzles. So, i'd like more input from people with more knowledge, but i'm leaning towards leaving this open. Jul 14, 2016 at 14:04

We can use the equation

$$F = q V_e + (P_e-P_a) A_e$$

Where $P_e$ is the exit plane pressure, $P_a$ is ambient pressure, and $A_e$ is the exit plane area. $qV_e$ without the correction term gives the thrust when the exit plane pressure matches ambient. Here $q$ is mass flow and $V_e$ is exit velocity.

For a nozzle optimized for sea level, $P_e$ = $P_a$ so there is no exit plane pressure effect thrust loss at liftoff.

Once the vehicle reaches vacuum, the term in parentheses would be equal to $P_e$. Per the equation you can see that this gives additional thrust equal to the product of $P_e$ $A_e$. However, this exhaust would be severely underexpanded and would experience losses due to a pattern of expansion waves at the nozzle exit.

A picture from here Underexpansion is inevitable since an infinitely long De Laval nozzle would be required to get the exit plane pressure to zero. (This means that your "fixed nozzle designed for optimal in-vacuum efficiency" is always a compromise - it gets cut off due to weight, packaging considerations, etc.) Underexpansion is also more desirable than overexpansion which can result in flow separation and severe shock losses, hence the use of dual-bell nozzles and other schemes to try to match the exit plane and ambient pressures better. If these fancy schemes are not practical, some compromise design exit plane pressure is used.

The linked presentation does a great job of explaining all this with real world examples and mitigation strategies.

• Shouldn't it be $P_e$ in the third paragraph, not $P_a$? Jul 16, 2016 at 13:58
• Pe = Pa (if Pa = ground level P) but your suggestion is clearer, editing. Jul 16, 2016 at 14:33

This might be a duplicate as Organic Marble suggests. But I think there is a misconception about the Laval Nozzle. The Purpose of the Nozzle is to achieve pressure equilibrium between Exhaust gases and the surrounding pressure.

If they are not in equelibrium, thrust is lost.

Think about the surrounding pressure as a ring directly connected to the nozzle. If the ring is smaller then your nozzle, energy is lost to move around the ring, if it is bigger, energy is lost filling in the gap between nozzle and ring.

Since bigger nozzles are needed for lower pressure surroundings, the higher drag of a nozzle greater then the rocket body would never occur if engineered for efficiency of exhaust gasses.