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  • Equation Dimensions in Gas Turbines

    Paper Notes - Chapter 14, AGK Lesson 29

    In both, you say that Thrust = Mass x (Velocity Difference) + Pressure Thrust.

    This is slightly misleading - it's actually mass flow to keep the dimensions correct.

    Mass Flow (kg.s-1) - i.e. kilograms / second
    Velocity (m.s-1)

    Multiplying the units together leads to Thrust = kg.m.s-2

  • #2
    Re: Equation Dimensions in Gas Turbines

    I've been lying awake trying to figure this one out. If we agree on F = ma and use Mass, Length and Time for our dimensional notation then thrust has dimensions ML/Tsq., or you could write it M/T, mass flow, times L/T, velocity.

    In the example,(Velocity Difference) is a rate term and has dimensions L/Tsq and the example, although confusing, is dimensionally correct

    Dick

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    • #3
      Re: Equation Dimensions in Gas Turbines

      In the example,(Velocity Difference) is a rate term and has dimensions L/Tsq and the example, although confusing, is dimensionally correct.
      I think that you are mistaken Dick.

      Velocity is certainly a rate. It is the rate of change of position.

      "Velocity difference" is a difference in that rate, but it is not an acceleration rate.

      To illustrate what I mean let's consider a change in velocity from zero ft/sec to 100 ft/sec.

      This is a velocity difference of +100 ft/sec.

      If that change took place over a period of 1 second it would represent an acceleration rate of 100 ft/sec/sec.

      But what if it took place over a period of 10 seconds. The velocity change would still be +100 ft/sec, but the acceleration rate would be only 10 ft/sec/sec.

      So if we are to use the equation F = MA we cannot use the ?velocity difference? to represent the A.

      The Rolls Royce book "The Jet Engine" gives the following equation for reaction reaction thrust:

      Reaction Thrust = W(V-Vj) /g

      Where

      W = The air weight flow rate.
      V = Inlet velocity.
      Vj = Outlet (Jet) velocity.
      g = gravitational acceleration.

      If we divide weight by g we get Mass, so if we divide weight flow rate by g we get mass flow rate.

      The equation then becomes:

      Reaction Thrust = Mass flow rate x velocity difference

      If we then add the pressure thrust element we get:

      Total Thrust = ( Mass flow rate x Velocity difference ) + PA
      Last edited by Loan Arranger; 29-12-2014, 13:29. Reason: To correct a factual inaccuracy.

      Comment


      • #4
        Re: Equation Dimensions in Gas Turbines

        Dick,

        I agree with your first sentence, but not the second. Velocity difference is a rate - but rate of change of position, not rate of change of velocity.

        As a basic example, the velocity difference of two cars (e.g. 120 mph and 100 mph is 20 mph) is still expressed in mph - not mph per hour.

        The Steely-Eyed missile men of NASA (General Thrust Equation about half way down the page - http://www.grc.nasa.gov/WWW/k-12/airplane/thrsteq.html) tend to go with the mass flow explanation.

        Comment


        • #5
          Re: Equation Dimensions in Gas Turbines

          Dear me!

          "we cannot use the ?velocity difference? to represent the A"

          "Reaction Thrust = W(V-Vj) /g"

          The elephant in the room is that we are discussing a jet engine with defined entry and exit criteria and V -V1 is the change of velocity across the engine, which is to say it is the acceleration of the airflow across the engine

          Dick

          Comment


          • #6
            Re: Equation Dimensions in Gas Turbines

            Dick,

            The elephant in the room is that we are discussing a jet engine with defined entry and exit criteria and V -V1 is the change of velocity across the engine, which is to say it is the acceleration of the airflow across the engine.
            You appear to be arguing that the time taken for the air to pass through the engine and to achieve that velocity increase has no bearing on the thrust produced. If this is true then a kg of air passing through an engine in 1 second and achieving a velocity increase of 100 m/sec, would produce the same thrust if it took a week to pass through the engine, provided it achieved the same velocity increase? Do you really believe that?

            Regardless of whether we are talking about a jet engine or any other device or scenario, a velocity increase is not the same thing as an acceleration rate.

            A velocity increase can be expressed as a number of meters/second. But an acceleration rate would be expressed as meters/ per second/ per second.

            In your previous post you wrote the following:

            If we agree on F = ma and use Mass, Length and Time for our dimensional notation then thrust has dimensions ML/Tsq., or you could write it M/T, mass flow, times L/T, velocity.
            You were absolutely correct!

            Now look at what you said. Mass flow x velocity.

            But the OP indicated that your course notes say Mass x Velocity change.

            Mass x velocity change would be expressed in kg x M/sec = kgM/sec.

            A kgM/sec is not a unit of force.

            But Mass Flow x Velocity Change would be expressed as kg/sec x M/sec = kgM/sec/sec.

            A kgM/sec/sec is a NEWTON which is unit of force.

            Unfortunately in your original post, having correctly stated that "thrust = mass flow x Velocity" you made the error of arguing that the statement in your notes (Thrust = Mass x velocity Change) is correct. It is not.


            If you are still unconvinced then try the following:

            To illustrate what I mean let's consider a change in velocity from zero ft/sec to 100 ft/sec.

            If a 1 kg mass of air is accelerated from zero to 100 ft/sec in 1 second what is the force applied?

            Force = Mass x Acceleration

            We have a velocity increase of 100 M/sec achieved in 1 second, so the acceleration is 100 M/sec/sec.

            Force = Mass x Acceleration = 1 kg x 100 m/sec/sec = 100 kgM/sec/sec

            That is a force of 100 Newtons.

            Now let's try the same velocity change in a longer period.

            If a 1 kg mass of air is accelerated from zero to 100 ft/sec in 100 seconds what is the force applied?

            Force = Mass x Acceleration

            We have a velocity increase of 100 m/sec achieved in 100 seconds, so the acceleration is 1 m/sec/sec.

            Force = Mass x Acceleration = 1 kg x 1 M/sec/sec = 1 kgM/sec/sec

            That is a force of 1 Newtons.

            Forget the elephant and look at the numbers!
            Last edited by Loan Arranger; 29-12-2014, 14:49.

            Comment


            • #7
              Re: Equation Dimensions in Gas Turbines

              Originally posted by Dick W View Post
              The elephant in the room is that we are discussing a jet engine with defined entry and exit criteria and V -V1 is the change of velocity across the engine, which is to say it is the acceleration of the airflow across the engine
              Dick,

              I agree that (V - V1) is a change in velocity (and as per my second post this still gives you a velocity resultant) - but to turn that into an acceleration, you must have a time period for this acceleration to occur over. This is where the Mass Flow term comes in (i.e. Mass / Time). You apply the change in velocity to a amount of mass per time.

              These two measured in (L / T) and a amount of mass per second (M / T). Multiplying these two together, you now get a the Force (M L / T^2).

              Comment


              • #8
                Re: Equation Dimensions in Gas Turbines

                Jon, You are right!
                I think the formula shown in our notes originally comes from the RAF AP3456 which acknowledges that this is a simplified version of the full thrust equation as the majority of the thrust produced is a result of the momentum change in the gas stream.
                The LO for the subject requires the student to understand the simplified thrust formula so perhaps the original author decided to leave the formula as shown in our manual.
                No doubt the subject expert will revisit the topic when he returns next week.
                So we'll leave it there for now.
                Thank you for your input and sharp minded observation!

                Colin

                Comment


                • #9
                  Re: Equation Dimensions in Gas Turbines

                  Colin,

                  Thanks - makes sense now. All hail AP3456 - the main reason quite a few are undertaking the EASA exams due to its 'compact' nature!

                  Jon
                  Last edited by jon@jon-owen.co.uk; 29-12-2014, 18:49. Reason: Punctuation.

                  Comment

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