Propellant Mass Flow Rate from Thrust and Isp

This is the sizing calculation that turns a thrust requirement into a tank. Combine $F = \dot m v_e$ with $v_e = I_{sp} g_0$ and you get $\dot m = F / (I_{sp} g_0)$, which takes the two numbers an engine is catalogued by — thrust and specific impulse — and returns how fast it drinks. Everything about the vehicle's plumbing follows from that: tank volume, pump capacity, feed line diameter, and the burn time the stage can sustain.

The magnitudes are worth feeling. A 1000 kN engine at 300 s specific impulse consumes $1{,}000{,}000 / (300 \times 9.80665) = 340$ kg/s — over 20 tonnes a minute, from one engine. A first stage with nine such engines is moving three tonnes of propellant every second, which is why turbopumps rather than pressure-fed tanks dominate launch vehicles: no tank light enough to fly could hold enough pressure to push propellant that fast.

The answer is the total flow, and splitting it between the two tanks needs the mixture ratio. At an oxidiser-to-fuel ratio of 2.5, the oxidiser takes 2.5 parts in every 3.5, so 340 kg/s becomes 243 kg/s of oxidiser and 97 kg/s of fuel. That ratio also sets the relative tank volumes once density is applied, and because liquid oxygen is denser than kerosene and far denser than liquid hydrogen, the tank that holds the most mass is rarely the tank that looks biggest.

Match the Isp to the condition. A sea-level thrust figure goes with a sea-level Isp and a vacuum thrust with a vacuum Isp; crossing them gives a flow rate wrong by the same 10 to 30% that separates the two figures, and it will be wrong in the flattering direction. And remember that $g_0$ here is the defined 9.80665 m/s², not the gravity at the launch site, the test stand or the destination — it is doing unit conversion, not physics.