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LS3: Combustion & Propellants
Original Author: Matt Morningstar '21, matt_m@mit.edu
Lecture Zoom Recording
Specific Impulse and the Rocket Equation, Revisited
Remember when we told you that Eq. 3 (see below) was the equation for the delta-v of a rocket?
We kinda lied. This equation makes a few assumptions that aren’t always realistic. First, it assumes that all the propellants go through the combustion chamber. This is often not the case in rocket engines, as some of the propellants are combusted separately to power the pumps, and then dumped overboard. Second, it assumes a matched nozzle (aka exit pressure = atmospheric pressure).
In reality, Eq. 4 is the more accurate rocket equation:
Where Isp is specific impulse (and g0 is 9.8 m/s^2). Specific impulse is a very important quantity that measures rocket engine efficiency. It is defined as the impulse produced per unit weight of propellant consumed. Put another way, it is proportional to the thrust of the engine divided by the mass flow rate of the propellants:
This quantity is analogous to the gas mileage of the rocket engine. It tells us how much bang (impulse) we get for our buck (a unit mass of propellant).
If we take our thrust equation and divide it by the mass flow rate, we can create an explicit equation for Isp:
It is important to note that the mass flow rate in the numerator is not necessarily the same as the denominator. The denominator refers to the total mass flow rate, which includes any propellants that might be used elsewhere in the engine and dumped overboard. The mass flow rate in the numerator refers only to propellants that go through the combustion chamber (subscript cc). If we assume that the two mass flow rates are the same (i.e. all propellants flow through the combustion chamber) and assume our nozzle is matched, we can see that Isp takes on a much simpler form:
This simplified form is what informed the original rocket equation we showed you. In this case, Isp is directly proportional to exhaust velocity. So, we didn’t really lie. Exhaust velocity still comes out as a very important quantity and the primary driver of Isp.
Exhaust Velocity
As we discovered in LSET 1, exhaust/exit velocity is an extremely important quantity for rocket engine performance. How can we predict what the exit velocity is, and how do we design our engine to achieve a high Ve?
Recall the isentropic relation introduced in Topic 2a:
As a reminder, this equation relates the stagnation & static pressures to the mach number. The greater the mach number of a flow, the lower the static temperature (as Tc is considered constant throughout the isentropic flow). We know how to take our Mach number and get velocity from the following equation:
If we take this expression for Mach number, substitute it into the previous equation, and do some algebra, we can find an explicit expression for velocity as a function of R, Tc (combustion/stagnation temperature), and Te (static temperature at the exit).
Based on this, we can say that the velocity at the exit is proportional to the square root of the following:
2R *
/ (
Tc - Te
This is a somewhat interesting result. The exhaust velocity is proportional to some constants (a function of propellant choice), and the change in temperature from combustion chamber to exit.
This is great and all, but we don’t really know Te, and we’d like to express exhaust velocity as a function of things we know. Based on our previously discussed isentropic flow relations, we can take the two of the equations we discussed, do some algebra, and show that a change in temperature within an isentropic flow can be related to a change in pressure via the following equation (I’m not going to go through that derivation here, but it will probably be in the saved presentation notes):
If express Te as Tc * (Te/Tc), and use the above equation to solve for (Te/Tc), we arrive at this final form:
We now have exhaust velocity expressed in terms of variables that we know (or can roughly assume). We now see that exhaust velocity is proportional to the square root of the following 3 things:
R *
This is simply a function of what your exhaust products are, which is a function of your propellant choice (and mixture ratio). We can express R as Runiversal / M, where M is the molecular weight of the exhaust products. Don’t confuse this with Mach number (I’ve italicized molecular weight to distinguish the two). (Runiversal is the constant value that you may have seen in high school w/ PV = nRT, Ru = 8314.4 J/(kmol.K)). This tells us that a propellant choice that yields exhaust products with a lower molecular weight is more efficient.
Tc
This is the temperature of combustion, which is a function of propellant choice (and mixture ratio). A propellant choice that yields a higher combustion temperature is more efficient.
[ 1 - (Pe/Pc)^((
This term is a little more complex. If our Pe/Pc is zero, then this term becomes 1 (maximum theoretical efficiency). However, in general, this term tells us that the larger our Pe/Pc ratio becomes, the less efficient we get. (As an example, if our = 1.2, Pe = 14.7 psi, and Pc = 1000 psi, this term becomes roughly ½. A 50% drop from maximum theoretical efficiency!) Rockets are most efficient when their nozzles are ‘matched’: the exit pressure equals the ambient pressure (you can find this result yourself by doing a lot of algebra with equations you already know). Thus, our Pe/Pc ratio is really a Pa/Pc ratio, where Pa is the ambient pressure. The only ‘knob’ we can really turn here is Pc. This equation tells us that a higher chamber pressure yields a more efficient engine, especially if our ambient pressure is higher. If our ambient pressure is very low (e.g. vacuum), then our Pe/Pc ratio will already be very low, and thus increasing Pc will not have a very sizable effect.
Propellants
The first two terms that we discussed in the last section tell us that propellant choice has a huge effect on exhaust velocity, and thus engine efficiency. In fact, if we neglect the last term and talk only about theoretical maximum efficiency, the propellant choice is the only thing that affects exhaust velocity.
To summarize from the previous section, we want our propellants to have the following properties to maximize exhaust velocity:
High combustion temperature
Low molecular weight of our exhaust products
These two characteristics are the primary drivers for modern propellant choices. As you may have heard before, hydrogen and oxygen is the propellant choice with the highest theoretical efficiency. We can see why! Hydrogen and oxygen produce a very high combustion temperature, and hydrogen has a very low molecular weight, leading to a low molecular weight of the exhaust products.
Other factors that impact propellant choice:
Storable vs. Cryogenic
Many propellants (like oxygen, hydrogen, and methane) are not liquids at room temperature and are thus stored at extremely low temperatures. Dealing w/ these cryogenic propellants requires additional infrastructure and consideration, as they continually evaporate (“boil-off”) when tanks warm. On the other hand, other propellants like RP-1 (kerosene) and ethanol are liquid at room temperature (this is called a storable propellant).
Density
Some propellants are much less dense than others, requiring a larger tank volume to hold the same mass. These larger tanks are heavier, which reduces the overall efficiency of the vehicle. (This is a main reason why hydrogen is not often used as the fuel in modern vehicles, despite the fact that hydrogen/oxygen has the highest theoretical efficiency. Hydrogen has a very low density and requires huge tanks to store.
Common propellants in modern launch vehicles:
Oxidizers:
Liquid Oxygen (LOX/LO2) - Cryogenic
Nearly all modern launch vehicles use liquid oxygen as the oxidizer. It is the most efficient oxidizer.
Fuels:
Liquid Hydrogen (LH2) - Cryogenic
The most efficient fuel. Very low density. Used frequently in second stages of launch vehicles to achieve high exhaust velocities.
Examples: Saturn V’s J-2, Blue Origin’s BE-3, Aerojet Rocketdyne’s RS-68
Hydrocarbon Fuels
Methane (usually called LNG - ‘Liquid Natural Gas’) - Cryogenic
Historically not a common fuel, but has garnered tons of recent attention and development. No orbital rocket using methane has been flown to date. Can be created in situ on Mars.
Examples: SpaceX’s Raptor, Blue Origin’s BE-4, Relativity Space’s Aeon
RP-1 (Kerosene) - Storable
Very common propellant historically.
Examples: Saturn V’s F1, SpaceX’s Merlin, Rocket Lab’s Rutherford
Ethanol - Storable
Common propellant in the early days of rocketry.
Examples: Redstone engine, RS-88, ~Helios~
Mixture Ratio
The mixture ratio is the mass flow ratio of the oxidizer to the fuel:
The ratio at which the two propellants are injected into the combustion chamber has an important impact on the combustion characteristics.
Above is the chemical reaction for the combustion of hydrogen and oxygen. If we balance this equation, we can determine the required ratio of mols of oxygen to mols of hydrogen. By using the molar mass of each of the molecules, we can then determine the mass ratio of oxygen to hydrogen.
This is the stoichiometric mixture ratio, where all the reactants are combusted. The stoichiometric mixture ratio yields the highest combustion temperature because all of the reactants go through the combustion process and release heat (there are no molecules that are ‘along for the ride’).
Combustion temperature vs. equivalence ratio (aka normalized MR, 1.0 = stoichiometric)
If we use a higher mixture ratio than stoichiometric, it is considered oxygen-rich. If we use a lower mixture ratio than stoichiometric, it is considered fuel-rich.
As we discussed in the first section, a high combustion temperature is most efficient. So you might expect that rocket engines use a stoichiometric mixture ratio. Unfortunately, the combustion temperatures achieved at a stoichiometric mixture ratio are so hot that they exceed the limits of our material/cooling capabilities. As a result, nearly all rocket engines use a fuel-rich mixture ratio (ox-rich combustion presents a number of issues, which is why an ox-rich mixture ratio is not commonly used in the main combustion chamber). It is also true that a stoichiometric mixture ratio (even if we could withstand its temperature) does not always yield maximum Isp. In the case of hydrogen/oxygen, it is more optimal to operate fuel-rich, as it leaves more light H2 molecules in the exhaust.
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