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At what rate does a Soyuz propulsion system (KDTU-80) consume it's propellants? I am interested in the mass consumption rate of both the main engine (e.g. an s5.80) and a single DPO thruster.

Please specify if its the total propellant mass being consumed, or just that of the fuel or oxidizer (the ratios are readily available, but not the mass flow rate).

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Estimating exact propellant consumption rate (mass flow rate) from easily available sources such as the Wikipedia page of KTDU-80 might be a bit imprecise, since we'll be using the thrust and Isp values to find mass flow rate, and in practice, it is done the other way around, as this answer by Russell Borogove suggests.

Now if rough figures are acceptable to you, we can find the mass flow rate from the definition of Specific Impulse:

$$ F_{thrust} = g_0 . I_{sp} . \dot{m} $$ $$ \therefore \dot{m} = \frac{F_{thrust}}{g_0 . I_{sp}} $$

Now using the values from KTDU 80's Wikipedia page,

For S5.80: $$ \dot{m} = \frac{2950}{9.81×302} ≈ 0.9957~~kg/sec $$

For DPO-B: $$ \dot{m} = \frac{129.16}{9.81×291} ≈ 0.0453~~kg/sec $$

And for DPO-M: $$ \dot{m} = \frac{25}{9.81×285} ≈ 0.0089~~kg/sec $$

Hope this helps! The answers to this question by uhoh discuss the applicability of the expression used above. Those might be helpful further.

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    $\begingroup$ Awesome, thanks! That's the mass flow rate of both fuel and oxidizer, right? $\endgroup$
    – Innovine
    Commented Jun 9, 2022 at 17:29
  • $\begingroup$ Yes, I think so...basically this assumes that all the products of combustion flow through the nozzle, producing thrust. The flow rate of that mass is what we see in the expression $\endgroup$ Commented Jun 9, 2022 at 21:45

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