Specific impulse

The specific impulse defines the efficiency of an engine. It is linked to the thrust and fuel consumption. The unit is either meters per second or only seconds.

Formula

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Where:

• $I_{sp}$ is the specific impulse in meters per second
• $F_{T}$ the thrust in newtons
• ${\dot {m}}$ the fuel consumption in kg/s

By multiplying this value with g₀ it is possible to change the unit to only seconds. The value g₀ behaves like a conversion factor and doesn't change when the gravity for the craft is changing. Usually both values are called specific impulse and are abbreviated by I_sp and the name I_sp,g0 is used here only to clarify that both values aren't the same. This value is sometimes called weight specific impulse.

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Where:

• $I_{sp,g_{0}}$ is the specific impulse in seconds
• $g_{0}$ is the surface gravity in the required unit (usually 9.81 meters per second squared)

The formula using the it (e.g. for Δv) has to specify what unit it does expect and if the value is defined in the one it has to be converted.

Multiple engines

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Where:

• $I_{sp}$ is the specific impulse in meters per second
• $I_{sp_{i}}$ the specific impulse of each engine in meters per second
• $F_{T_{i}}$ the thrust of each engine in newton
• ${\dot {m}}$ the fuel consumption in kg/s

When the fuel consumption is not used in this formula, it is only important that all thrust values have the same unit (e.g. kilonewtons) and the specific impulse have all the same unit (e.g. seconds). The result is then in the same unit as the specific impulses of the engines. If all engines have the same specific impulse the resulting specific impulse will be the same.

Physical background

In KSP the fuel consumption on most engines depend on the atmospheric pressure with the lowest consumption in vacuum. So the specific impulse is at the highest point in the vacuum. This is loosely the same behaviour as in the real world, but the reason is different. The fuel consumption is always the same but the thrust is increasing over time. Because the a higher thrust with the same fuel consumption is more efficient the specific impulse also rises.

To convert the specific impulse between the handy weight specific impulse and the physical usable specific impulse it had to be converted with g₀. It appears that this value isn't the real world 9.81 m/s²; instead it is a little higher value of 9.82 m/s² making the engines a bit more efficient than expected.

Although the unit of the specific impulse is a velocity, it usually isn't the exhaust speed. Because some of the fuel consumed isn't used for propelling directly, but runs the turbopumps to fuel the engine.

Example

The Kerbal X has six LV-T45 Liquid Fuel Engines with a specific impulse of 320 s in atmosphere and one Rockomax "Mainsail" Liquid Engine with a specific impulse of 280 s. The average specific impulse of all engines is then:

$I_{sp}={\frac {6\cdot 200{\text{kN}}+1500{\text{kN}}}{6\cdot {\frac {200{\text{kN}}}{320{\text{s}}}}+{\frac {1500{\text{kN}}}{280{\text{s}}}}}}=296.47{\text{s}}$

To convert this value into a physical usable value. Note that the conversion factor used here is the KSP relevant value of 9.82 m/s² and not the 9.81 m/s² used in real world science.

$I_{sp}=g_{I_{sp}}\cdot 296.47{\text{s}}=2911.34{\frac {\text{m}}{\text{s}}}$

This value can then be used to calculate the fuel consumption:

${\begin{aligned}I_{sp}&={\frac {6\cdot 200{\text{kN}}+1500{\text{kN}}}{\dot {m}}}=2911.34{\frac {\text{m}}{\text{s}}}\\{\dot {m}}&={\frac {6\cdot 200000{\text{N}}+1500000{\text{N}}}{2911.34{\frac {\text{m}}{\text{s}}}}}\\&=927.4{\frac {\text{kg}}{\text{s}}}\end{aligned}}$

Because the engines use liquid fuel and oxidizer with a density of 5000 kg/m³ it is possible to calculate the volume consumed.

$V={\frac {\dot {m}}{\rho }}={\frac {927.4{\frac {\text{kg}}{\text{s}}}}{5000{\frac {\text{kg}}{{\text{m}}^{3}}}}}=0.185{\frac {{\text{m}}^{3}}{\text{s}}}=185{\frac {\text{l}}{\text{s}}}$

Of course this values of the craft are valid for Kerbin's atmosphere. Because the air gets thinner with altitude the efficiency is rising and the fuel consumption is falling. Also because of the staging pattern engines will be dropped until reaching vacuum and thus changing the efficiency again. This time the efficiency lowers, because the higher efficient engines drop first leaving only the Mainsail engine with the lowest efficiency of all engines on the craft.

See also

• Terminology
• Specific impulse on Wikipedia