2021 109th day.
Today, let's consider how to calculate the delta-v of a spacecraft where
an electrical drive is used. In the case where the electricity is generated
by a mass-invariant system, for example by solar panels, radio-thermoelectric
generators, or by a nuclear reactor with closed-cycle cooling, the Tsiolkovsky
equation holds well.
However, in the case where the electrical generator is some sort of consumable,
this must be taken into account to calculate the effective specific impulse of
the drive. For this purpose, we need quantities I'll call Esp or specific
energy, which is a ratio of electrical energy divided by mass, that is, the
amount of energy produced per mass consumed by the electrical generator.
The specific energy can be calculated using quantities of volume flow rate Q,
density ρ of the consumed resource, and power P produced, by the following
equation:
Esp = P / (Q * ρ)
Energy in Kerbal Space Program is not measured in any unit other than the
lightning bolt symbol ϟ. So Esp is measured in ϟ/t or "bolts" per tonne. In
KSP, a single fuel cell provides 1.5 ϟ/s while consuming 0.00375 L/s of LFO.
The density of LF and Oxidizer are both 0.005 t/L, so the specific energy of
KSP's fuel cells is:
Esp = P / (Q * ρ)
= 1.5 / (0.00375 * 0.005)
= 80000 ϟ/t
So, now consider the electrical consumption of the electric drive. For KSP's
xenon-powered ion drive, we have a power P of 8.74 ϟ/s at full thrust F of
2 kN, and consumes a volume flow rate Q of 0.4856 litres of Xenon per second.
The density ρ of Xenon in KSP is 0.0001 t/L. The rated specific impulse of
this drive is an amazing 4200 seconds. However, when powered by fuel cells,
the electrical consumption of this drive is converted to a mass consumption:
ṁ1 = P / Esp
= 8.74 / 80000
= 0.00010925 t/s
The mass consumption of the Xenon is:
ṁ2 = Q * ρ
= 0.4856 * 0.0001
= 0.0004856 t/s
So the overall mass flow rate is
ṁ = 0.0004856 + 0.00010925
= 0.00059485 t/s
Since the thrust is 2 kN, the effective specific impulse can be calculated
by dividing thrust by mass flow rate:
Isp = F / ṁ
= 2 / 0.00059485
= 3362.2 s
For future reference, the consumption rationes of the three fuels by volume can
also be calculated. The volume flow rate of the LFO is:
Q = 8.74 / (1.5 / 0.00375)
= 0.02185 L/s
The ratio of Xenon to LFO is therefore:
0.4856 / 0.02185 = 22.2242563
Since the ratio of LF to Oxidizer is 9 to 11, we set the denominator to 20 and
calculate:
22.2242563 = 444.5 / 20
Volumetric ratio is thus: 444.5 Xe to 9 LF to 11 O.
-- Oren Watson.