Difference between revisions of "Specific impulse"

Revision as of 09:55, 21 October 2013

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

${\displaystyle I_{sp}={\frac {F_{T}}{\dot {m}}}}$

By multiplying this value with g₀ it is possible to change the unit to only seconds avoiding conversion issues between the SI and customary units. 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. 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.

${\displaystyle I_{sp,g_{0}}={\frac {F_{T}}{{\dot {m}}\cdot g_{0}}}={\frac {I_{sp}}{g_{0}}}}$

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 other one it has to be converted.

Multiple engines

${\displaystyle I_{sp}={\frac {\sum \limits _{i}F_{T_{i}}}{\sum \limits _{i}{\dot {m}}_{i}}}={\frac {\sum \limits _{i}F_{T_{i}}}{\sum \limits _{i}{\frac {F_{i}}{I_{sp_{i}}}}}}}$

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 it is more efficient and the specific impulse also rises. The specific impulse can only be calculated using this method for reaction engines and not jet engines as those work on another principle.

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.

Conversion factor

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 9.81 m/s² which is used in real world. To determine the factor the following formula can be used:

${\displaystyle g_{I_{sp}}={\frac {F_{T}}{({\dot {m}}_{f}\rho _{f}+{\dot {m}}_{o}\rho _{o})I_{sp,g_{0}}}}}$

The formula uses oxidizer and liquid fuel, but does work for all engines. When using one Rockomax "Mainsail" Liquid Engine on the launch pad at full throttle uses 48.96 units of liquid fuel and 59.84 units of oxidizer with an weight specific impulse of 280.8 seconds and a thrust of 1.5 MN. This gives a conversion factor between 9.81873052 m/s and 9.8205356 m/s, assuming that the density of liquid fuel and oxidizer is 5 kg per unit.^[1] The value isn't exact, because the exact specific impulse and mass flows aren't known. It can be assumed that the conversion factor is about 9.82 m/s² making the engines a bit more efficient than expected.^[2]

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:

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

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.

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

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

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

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

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

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

Notes

1. ↑ Gathered from File:Isp conversion factor.png.
2. ↑ In File:10X Xenon.png 10 PB-ION Electric Propulsion Systems are running at full power and consuming a lower value that 10× the theoretical value calculated with 9.81 m/s².