Exhaust Velocity Calculator

Exhaust Velocity Calculator:

Enter the values of thrust, F_(N), exit pressure, p_e(Pa), ambient pressure, p_a(Pa), exit area, A_(m2) and mass flow rate, mdot_(kg/s) to determine the value of exhaust velocity, V_(m/s).

Enter Thrust :		N
Enter Exit Pressure :		Pa
Enter Ambient Pressure :		Pa
Enter Exit Area :		m2
Enter Mass Flow Rate :		kg/s
Result – Exhaust Velocity :		m/s

Exhaust Velocity Formula:

Exhaust velocity is a crucial concept in rocket propulsion, referring to the speed at which exhaust leaves a rocket engine. This velocity is a key determinant of a rocket’s efficiency and capability, directly impacting its thrust and overall performance in space or atmospheric flight.

The higher the exhaust velocity, the more effective the rocket is at converting thermal energy into kinetic energy, thereby improving its thrust-to-weight ratio.

The exhaust velocity, V_(m/s) in metres per seconds is equal to the thrust, F_(N) in Newtons minus the product of the difference in exit pressure, p_e(Pa) in Pascals and ambient pressure, p_a(Pa) in Pascals times the exit area, A_(m2) in metres square all divided by the mass flow rate, mdot_(kg/s) in kilograms per seconds.

Exhaust velocity, V_(m/s) = F_(N) – (p_e(Pa) – p_a(Pa)) * A_(m2) / mdot_(kg/s)

V_(m/s) = exhaust velocity in metres per seconds, m/s.

F_(N) = thrust in Newtons, N.

p_e(Pa) = exit pressure in Pascals, Pa.

p_a(Pa) = ambient pressure in Pascals, Pa.

A_(m2) = exit area in square metres, m².

mdot_(kg/s) = mass flow rate in kilograms per seconds, kg/s.

Exhaust Velocity Calculation:

1.A rocket engine produces a thrust of 5000 N, the exit pressure is 101325 Pa (1 atm), the ambient pressure is 101325 Pa, the exit area is 0.1 m² and the mass flow rate is 2 kg/s.

Given: F_(N) = 5000N, p_e(Pa) = 101325Pa, p_a(Pa) = 101325, A_(m2) = 0.1m²,

mdot_(kg/s) = 2kg/s.

Exhaust velocity, V_(m/s) = F_(N) – (p_e(Pa) – p_a(Pa)) * A_(m2) / mdot_(kg/s)

V_(m/s) = 5000 – (101325-101325) * 0.1 / 2

V_(m/s) = 5000 / 2

V_(m/s) = 2500m/s.

2.Thrust is 10000 N, exit pressure exceeds ambient pressure by 5000 Pa, the exit area is 0.05 m², and the exhaust velocity is 1950m/s.

Given: F_(N) = 10000N, p_e(Pa) = 106325Pa, p_a(Pa) = 101325, A_(m2) = 0.05m²,

V_(m/s) = 1950m/s.

Exhaust velocity, V_(m/s) = F_(N) – (p_e(Pa) – p_a(Pa)) * A_(m2) / mdot_(kg/s)

mdot_(kg/s) = F_(N) – (p_e(Pa) – p_a(Pa)) * A_(m2) / V_(m/s)

mdot_(kg/s) = 10000 – (106325 – 101325) * 0.05 / 1950

mdot_(kg/s) = 10000 – (5000) * 0.05 / 1950

mdot_(kg/s) = 10000 – 250 / 1950

mdot_(kg/s) = 5kg/s.

 

Applications and Considerations:

• Calculating the exhaust velocity allows engineers to optimize rocket engine designs for specific missions, adjusting factors like nozzle shape and size to maximize performance.
• The relationship between thrust and exhaust velocity is fundamental in achieving desired acceleration and payload capacities.
• High exhaust velocities are generally preferred for space travel where efficiency must be maximized in a vacuum, whereas lower velocities might be suitable for certain atmospheric operations.