SSME energy and power relationships

The Space Shuttle main engines are rockets that use liquid hydrogen and oxygen propellants. The ejection of propellants represents a “flow” of momentum; as a consequence of the law of conservation of momentum an equal and opposite force and momentum change must occur for the engine which is ejecting the propellants. This is the fundamental principle of rocket propulsion.

Although the law of conservation of momentum explains the change in momentum of the vehicle, engine, and propellants, it does not explain where the energy is coming from. If one was to calculate the maximum chemical energy available from the reaction of the propellants; it would be seen that the rate of work done on the vehicle can, and often does exceed the rate of chemical energy release of the propellant reaction(s). This leads one to the question; what is the source of this “extra” energy.

Energy is required to accelerate the propellants (relative to the rocket frame of reference). Two important sources of energy are available from the propellants of a moving rocket:

Chemical energy

The first is the internal chemical energy which is released during the chemical reactions that takes place when the propellants are mixed together, are ignited and react. Typically the propellant reactions are combustion processes. Combustion of the SSME propellants H₂ (which is gaseous at the time of entry into the combustion chamber) and LO2 (which liquid at the time of entry into the combustion chamber) releases, on average, 241,800 joules per mole of reaction products H₂O). The energy released during the formation of a product is known as the energy of formation; for H₂O the energy of formation is −241.8 kilojoules per mole (negative because energy is released during the formation process).

The H₂/O₂ reaction produces water (H₂O). For every 2 moles of H₂ (4 g, 2 g/mol) and 1 mole of O₂ (32 g, 32 g/mol), 2 moles of H₂O (36 g, 18 g/mol) are produced with the accompanied release of 483 kilojoules of energy (mostly in the form of heat). Thus, for every mole of O₂ reacted with hydrogen (2 moles of hydrogen are required), 483.6 kilojoules of energy will be released. For every mole of H₂ reacted ½ mole of H₂O will be produced along with the release of 241.8 kilojoules of energy.

The SSME runs “fuel rich” thus there is not enough O₂ to consume all of the H₂ during the combustion process. Many rocket engines run fuel rich when the molecular mass of the “fuel” propellant is significantly smaller that that of the oxidizer. The reason for this is because the nozzle efficiency is maximized when the molecular weight of the exhaust gases is minimized (because of turbulence).

Flow rates

At FPL (104.5% RPL) the oxygen flow rate into the SSME is about 935 lb/s (424 kg/s), while the hydrogen flow rate is about 155 lb/s (70.3 kg/s).

Since the SSME is operated fuel rich not all of the hydrogen will be consumed, but all of the oxygen will be consumed in the combustion process. The energy release per mole (32 g) of O₂ is 483.6 kilojoules, thus at a flow rate of 424.33 kg/s (424,330 g/s or 13,260.3 mol/s) we have an energy release rate of approximately 6.4127 GJ/s (GW).

Energy in the exhaust gas stream

The exhaust gases carry almost all of the energy produced during the combustion process. This energy is in many forms; kinetic, thermal, acoustic, optical, vibration, potential, etc. The majority of the energy in the exhaust stream is in the form of kinetic energy with the remaining portion mostly in the form of heat carried in the exhaust gases.

The greater the fraction of the thermal energy released during the combustion process that is converted into kinetic energy, of the ejected propellant stream, the more efficient the engine will be at producing thrust. The SSME exhaust gas velocity is about 4444 m/s, with a total exhaust stream flow rate of about 492 kg/s; thus the kinetic energy ($K\_E=frac\{1\}\{2\}m\{v\_e\}^2$) of the exhaust stream is approximately 4.85 GJ/s which is about 76% of the energy released during the reaction of the propellants. This energy is the total enthalpy flow rate of the fluid output of the combustion chamber.

Kinetic energy of the un-reacted propellants

The second source of energy is the kinetic energy of the propellants prior to combustion. Since the propellants are accelerated as the space craft is accelerated they possess a fraction (proportional to their mass relative to the total space craft mass) of the kinetic energy transferred to the spacecraft from the propulsion system. This kinetic energy is very significant; for example when the shuttle is moving at 4500 m/s the total kinetic energy of the fluid flow into each SSME (relative to the Earth Reference frame) is in approximately 5 GJ/s.

The total extractable/available energy from the propellant flow as it flows through the SSME is the sum of the kinetic energy of the propellants, relative to the Earth reference frame, and the enthalpy of the flow due to energy added through combustion of the propellants.

The total extractable enthalpy rate can be determined from the kinetic energy flow rate i.e. the power of the exhaust gas stream when the engine is stationary. This power is well known through the relation:

:$P\_\{Enthalpy\}\; =\; P\_\{K\_Exhaust\_Stream\}\; =\; frac\{1\}\{2\}frac$dmdtv_e ^2

Where $v\_e$ = exhaust gas velocity, and dm/dt = the propellant flow rate out of the nozzle.

The kinetic energy flow rate of the fluid flowing into the combustion chamber is given by:

:$P\_\{K\_Pr\; opellants\_Into\_Combustion\_Chamber\}\; =\; frac\{1\}\{2\}frac$dmdtv_R ^2

Where $v\_R$ = velocity of the shuttle relative to the Earth reference frame.

The total available power is then the sum of these two:

:$P\_\{Total\_Extractable\}\; =\; frac\{1\}\{2\}frac$dmdt(v_R ^2 + v_e ^2 )

The total energy absorbed by the nozzle is the difference between the total extractable energy in the propellant flow and the total kinetic energy, relative to the Earth reference frame, of the exhaust fluid stream as it leaves the nozzle. Then we have the following relationship:

:$P\_\{K\_Pr\; opellants\_Absorbed\}\; =\; frac\{1\}\{2\}frac$dmdtleft( {v_R ^2 + v_e ^2 } ight) - frac{1}{2}fracdmdtleft( {v_R - v_e } ight)^2 = frac{1}{2}fracdmdtleft( {left( {v_R ^2 + v_e ^2 } ight) - left( {v_R - v_e } ight)^2 } ight)

:$P\_\{K\_Pr\; opellants\_Absorbed\}\; =\; frac\{1\}\{2\}frac$dmdtv_R v_e

Since$frac$dmleft( t ight)dtv_e left( t ight) = Thrustleft( t ight), we have:

:$P\_\{K\_Pr\; opellants\_Absorbed\}\; =\; Power\_\{Rocket\}\; left(\; t\; ight)\; =\; Tleft(\; t\; ight)v\_R\; left(\; t\; ight)$

This is the expected result.

When the shuttle is moving at 4500 m/s and assuming a thrust of 2.089 MN (470,000 lbf), each SSME is producing over 9.4 GW (12.6 million horsepower) which is more than 2 times the available enthalpy of the fluid flow output of the combustion chamber. The remaining energy being absorbed is due to “recovered” kinetic energy of the un-reacted propellants. Of course the kinetic energy of the propellants is due to work done on them earlier by the three SSMEs (and the boosters).

The main concept here is that kinetic energy has been stored in the propellants (i.e. part of the propulsion system) and is being released later. This concept is similar to energy storage in a flywheel or a spring.

Note that the maximum energy transfer to the propellants from the combustion process is equal to the sum of the heat of formations of the propellant products ($E\_\{F\_Total\}$) of mass M (conveniently the molar mass). If all of this energy could be transferred to the propellant flow stream (i.e. kinetic energy of the propellants flow) then the maximum achievable $v\_e$ is that associated with the kinetic energy of that flow:

:$E\_\{F\_Total\}\; =\; frac\{1\}\{2\}Mv\_e\; ^2$

:$v\_e\; =\; sqrt\; \{frac$2E_{F_Total} {M

For example the combustion process involving H₂ and O₂ produces water; the energy of formation of a mole of water (molar mass = 18 grams) is 241.8 kilojoules. If all of this heat energy could be converted into kinetic energy of that mole of water then the kinetic energy of the water would be equal to 241.8 kilojoules. Thus using the equation above we have:

:$v\_e\; =\; sqrt\; \{frac$2E_{F_Total} {M = sqrt {frac483,6000.018 = } 5183.3 mathrm{m}/mathrm{s}