Specific impulse/ru

Удельный импульс (англ. "specific impulse") определяет эффективность двигателя. Он связан с тягой и расходом топлива. Его единицы измерения это либо метры в секунду, либо просто секунды.

Формула

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

Где:

• $I_{sp}$ - это удельный импульс в метрах в секунду;
• $F_{T}$ - тяга в Ньютонах;
• ${\dot {m}}$ - расход топлива в кг/с

Пи перемножении этого значения с g₀ можно изменить единицы измерения исключительно на секунды, избегая проблем преобразования между единицами СИ и обычными единицами измерения. Значение g₀ ведет себя в качестве коэффициента преобразования и не изменяется, при изменении силы тяжести для космического аппарата. Обычно оба значения называют удельным импульсом и обозначают I_sp. Названия I_{sp, g0} используются здесь только для того, чтобы разъяснить, что оба значения - это не одно и то же. Это значение иногда вызывают весовым удельным импульсом. 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.

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

Где:

• $I_{sp,g_{0}}$ is the specific impulse in seconds
• $g_{0}$ is the surface gravity in the required unit (обычно 9.81 м/с^{2<sup />)</div>}

The formula using the it (например, для Δv) has to specify what unit it does expect and if the value is defined in the other one it has to be converted.

Несколько двигателей

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

Где:

• $I_{sp}$ 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 (например килоньютоны) and the specific impulse have all the same unit (например секунды). 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.

Влияние физических законов

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. In the real world this is usually reversed: The fuel consumption stays always the same but the thrust is increasing over time, because it is easier to have a steady fuel flow. Because a higher thrust with the same fuel consumption it is more efficient 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 is lower than the exhaust speed usually, because some of the fuel consumed isn't used for propelling directly, but runs the turbopumps to fuel the engine.

Коэффициент преобразования

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:

$g_{I_{sp}}={\frac {F_{T}}{\sum \limits _{i}({\dot {m}}_{i}\rho _{i})I_{sp,g_{0}}}}$

Где:

• $F_{T}$ is the thrust of the engine
• ${\dot {m}}_{i}$ is the mass flow of the fuel components
• $\rho _{i}$ is the density of the fuel components
• $I_{sp,g_{0}}$ is the weight specific impulse

For example liquid fuel engines have the fuel components are oxidizer and liquid fuel. 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. By reducing the engine's weight specific impulse it is possible to get higher mass flow rates improving the conversion factor to a value between 9.81994836 m/s and 9.82006181 m/s by using a thrust of only 85 kN and a weight specific impulse of 1 second for all pressures. The value isn't exact, because the exact 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.

Пример

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.

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

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.

Обратите внимание

• Термины и определения;
• Удельный импульс в Википедии;
• Specific impulse on Wikipedia.

Примечания