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Specific Impulse and Exhaust Velocity

  1. Feb 22, 2009 #1
    I would like an explanation with this formulas for a rocket engine and any other engine to calculate its exhaust velocity and its specific impulse. I realized that this formulas neglect combustion temperature, engine pressure, speed of the fuel is injected, etc.
    -----------------------------
    This is the formula that I found in http://www.answers.com/topic/specific-impulse for the specific impulse:
    Fthrust = Isp * (Δm/Δt) * g0
    >>>
    -Fthrust = Force thrust in Newtons
    -Isp = Specific Impulse in seconds
    -(Δm/Δt) = is the mass flow rate in kg/s (lb/s), which is minus the time-rate of change of the vehicle's mass since propellant is being expelled.
    -g0 = is the acceleration at the Earth's surface, in m/s² (or ft/s²).

    This another variation to find the thrust of an object:
    Fthrust = ve * (Δm/Δt)

    This is the formula the exhaust velocity:
    ve = g0 * Isp
    -----------------------------
    I want to know know how do you find the change of mass and change of time in the the formulas, (Δm/Δt). Does this part of the formula translate as: (massf - massi)/(timef - timei)? And my last question is for g0 when it means acceleration at Earth's surface, is it 9m/s2 or is it 9m/s2 * something? I found more equations for an ideal rocket in NASA's page, http://www.grc.nasa.gov/WWW/K-12/rocket/rktpow.html. I would like to see some example using this formulas because you cannot really apply them if you don't know how they are applied. Thank you for your time. This is not a homework question.
     
  2. jcsd
  3. Feb 23, 2009 #2

    rcgldr

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    Maybe an explantion of the force part would help.

    Force = mass x acceleration
    Force = mass x (change in velocity) / (unit time)
    Force = mass / (unit time) x (change in velocity)
    Force = mass flow rate x change in velocity
     
  4. Feb 23, 2009 #3
    When you say the change in velocity, do you mean this: velocityfinal - velocityinitial? That's something that has been bugging me because I think that what it means, but I'm not sure because there isn't an example to see if its velocityfinal - velocityinitial.
     
  5. Feb 23, 2009 #4

    rcgldr

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    Yes, the total change in velocity.
     
  6. Feb 23, 2009 #5
    And Δm or change in mass is the same as the chane in velocity which is massf - massi? Also g0 the acceleration at Earth's surface is equal to 9.8m/s2? I want to make sure that I have these formulas first in check before I start to use them so I use them how they are suppose to be used.
     
  7. Feb 23, 2009 #6

    Astronuc

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  8. Feb 23, 2009 #7
    Can you give an example of g0 as well as the mass flow rate because its still isn't quite clear to me. Would it be easier to understand if I apply it into a real time object, like a space shuttle main engine, SRB, F-1, J-2, etc.
     
    Last edited: Feb 24, 2009
  9. Feb 24, 2009 #8
    I've read the information in the link that you send me, but I still need some clarification on mass flow rate along with the (Pe - P0) * Ae. To get the pressure for Pe and P0, you do Pressure = Force / Surface Area?
     
  10. Feb 24, 2009 #9

    Astronuc

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    Take a look at this nozzle simulator - http://www.lerc.nasa.gov/WWW/K-12/airplane/ienzl.html [Broken]

    Pe is the pressure in the flow at the nozzle exit, and Po is the free stream pressure, which is essentially 0, vacuum, in space.

    The use of go is arbitrary. IMO, it's an uncessary proportionality constant which converts mass to weight.

    With respect to mass flowrate, see -

    http://www.lerc.nasa.gov/WWW/K-12/airplane/thrsteq.html [Broken]
    http://www.lerc.nasa.gov/WWW/K-12/airplane/mflchk.html [Broken]

    It's just the amount of mass flowing across some reference cross-sectional area per unit time, e.g. 10 kg/s, and is also the measure of propellant consumption. One could measure it volumetrically, e.g. gallons per minute, pgm. Multiply the volumetric flowrate by density, one would get mass flow rate.
     
    Last edited by a moderator: May 4, 2017
  11. Feb 24, 2009 #10

    D H

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    My advice: Forget about specific impulse as being in units of seconds. IMO, its much better to think of it as having units of velocity. For one thing, that is much more in line with the name. "Specific" means divided by mass (of the exhaust, in this case), "impulse" means change in momentum. Specific impulse should logically have units of momentum/mass, or velocity. In fact, this is exactly how many who use the metric system express specific impulse. That specific impulse is typically expressed in seconds in the US is essentially a reflection on the goofy system of measurements used in the US -- as Astronuc already said, thrust per weight flowrate.

    A thruster with a specific impulse of 408 seconds used in orbit around the Pluto will have an effective exhaust velocity of about 4000 m/s. That effective exhaust velocity is very real. The 408 seconds is not.
     
  12. Feb 25, 2009 #11
    Now I get what you mean about specific impulse which is to tell how efficiant an engine is. A rocket engine isn't efficiant because it burns more fuel than an airplane will to produce thrust to lift off the ground. So when you say that a thruster has a specific impulse of 408 seconds that means that its consuming fuel just like a liquid fuel rocket would be consuming. A solid rocket booster will have an even lower specific impulse because they burn their fuel at a faster rate than a liquid fuel rocket then. So, a Me 163B Komet 1 will have an approximate specific impulse of 400 seconds because its a bi-propellant rocket, similar to the OMS of the Space Shuttle where the 2 propellants mix to produce thrust? Thanks for the clarification of specific impulse, DH.

    The URLs that you posted, Astronuc, I finally figured out how to calculate or the calculations of the mass flow rate that I found on the links and on my school notes as well. This is what I have for the mass flow rate:
    [tex]
    F = \frac{\Delta\ p}{\Delta\ t}
    [/tex]

    which is:

    [tex]
    F = \frac{mV2-mV1}{t2-t1}
    [/tex]
     
  13. Feb 26, 2009 #12
    So, is the mass flow rate given by that formula that I posted?
     
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