Climb Rate iincrease for a Power increase

Discussion in 'Aviation' started by silence, Nov 30, 2013.

  1. silence

    silence Active Member

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    Is there a simple way to estimate a rate of climb increase for a power increase, given that the weight of the plane does not change?
     
  2. swampyankee

    swampyankee Active Member

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    Yes.

    Of course, to do it you need to know what the power requirement is for level flight at the speed for the best rate of climb and the efficiency of the propeller at the angle of attack for best rate of climb.

    To a zeroth approximation, assume that the propeller is 80% efficient, so 80% of the power increase goes into increasing rate of climb. So:

    Say the plane has a mass of 2,000 kg, and a best rate of climb of 10 m/s with an engine of 1000 kW. To raise any mass at 10 m/s against Earth's gravity takes 98 W/kg (power is force times velocity), so the useful power use for climb for this airplane is 196 kW. Divide by 0.8 for prop efficiency, so 245 kW is being used for climb and the rest (800 kW - 245 kW = 555 kW) is required for level flight. Raise the engine power to 1050 kW and hold everything else constant: 840 - 555 = 285 kW available for climb. Throw out 20% of this due to propeller inefficiencies, leaves 228 kW available for climb. 228 kW/19600 N (the aircraft's weight) and the new rate of climb is 11.6 m/s.
     
  3. silence

    silence Active Member

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    Math hurts.... But thank you; I think I got it.
     
  4. GregP

    GregP Well-Known Member

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    #4 GregP, Dec 1, 2013
    Last edited: Dec 1, 2013
    Hi Swampyankee,

    The rate of climb is 10 m/s, which is a velocity. The acceleration due to gravity is 9.8 m/s^2, so your number, 98, is velocity times acceleration, which does NOT equate to force. Force equals mass times acceleration, or F = ma and work is Force times distance times the cosine of the angle between the force and the displacement (W = Fd cos angle). If the plane is climbing at 10 m/s constant velocity, the acceleration is ZERO. In your analysis above I don't see any calculation for force. Also, W/kg is power divided by mass (assumes W = Watts), which isn't equal to anything I know of. Mass has very little to do with power as you can tell from the equation below in the last sentence. If W above equals work, then Power = W/t (n-m/s), not W/m or W/kg. Work would be foot pounds (or pound feet) or newton-meters (or meter-newtons).

    However, Physics is challenging and fun, so I can say I've made those mistakes, too, from time to time. So, no dirt from me. Good quick reply.

    Better try again. You can have your chance without further comment from me if it is correct. Oh, and power equals force times AVERAGE velocity, not force time instantaneous velocity. Basically, P = W/t = Fd/t = F * v-bar, where W = work.
     
  5. drgondog

    drgondog Well-Known Member

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    Simple answer - at a given altitude max ROC = maximum Excess Power/Weight where Excess Power = (Power available - Power Required)/Weight

    at a given forward speed; Excess Power = Thrust x Velocity - Drag x Velocity

    Expressed in Hp: Max ROC = 550*(Pa-Pr)/W in f/s.

    To calculate the Power Required for a given Velocity at a given altitude:
    First T=D; = 1/2*rho*V*V*S*CD and W=L; = 1/2*rho*V*V*S*CL-----> T/W=CD/CL

    Pr=Tr*V when T=D in level un-accelerated flight

    but Tr/W = CD/CL. Therefore Tr=W/(CL/CD) : For that Velocity CL=W/(.5*rho*V*V*S)
    and CD=W/(.5**rho*V*V*S)

    This drops thrust out of the equation so now Pr= (W/(CL/CD))*V where now V = SQRT[(2*W)/(rho*S*CL)]

    and CD= CDo+CL*CL/(Pi*AR*e)

    Then for level flight when L=W; V=SQRT[(2*W)/(rho*S*CL)]

    Remembering Pr=Tr*V and T=W/(CL/CD)

    Pr=W/(CL/CD)*SQRT[2W/(rho*S*CL)]


    So from here you have the base altitude and base velocity to insert for New ROC when extra Hp available.

    Swampyankee is correct that propeller efficiency is an important factor, but you can easily get lost in the efficiency carts when confronted with fixed pitch/variable speed props versus fixed speed/variable pitch
     
  6. silence

    silence Active Member

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    *tears apart the medicine cabinet looking for the ibuprofen*
     
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  7. silence

    silence Active Member

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    so:

    Pr = power required at altitude to maintain level flight(hp)
    Pa = power available at altitude (>= Pr) (HP)
    W = weight (pounds)
    rho = gravitational acceleration (32.2 f/s^2) - used as a divisor to convert weight to mass
    V = horizontal velocity (f/s)
    S = ???
    CD = ???
    CL = ???

    T= thrust (ft-pounds) ???
    D - drag (ft-pounds) ???

    Man, its been a LONG time since I messed with dynamics.
     
  8. GregP

    GregP Well-Known Member

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    #8 GregP, Dec 3, 2013
    Last edited: Dec 3, 2013
    S = wing area (sq ft or sq m)
    CD is coefficient of Drag
    CL is coefficient of lift (not CD0, which is CD at zero lift)
    rho is air density (.0023769 slugs/ft^3 or .001225 g/cm^3 or whatever conversion you need from these ... at sea level and 15°C or 59°F)
    Thrust is force in pounds or newtons
    drag is also force in pounds or newtons

    I use English units and simply convert to metric units when necessary, but either calculation is the same. Get the units right and do it in Excel or whatever math app you use.
     
  9. silence

    silence Active Member

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    Ah, I originally thought rho was air density - shoulda trusted my instincts!

    I actually prefer metric, which I guess nullifies my US Citizenship....

    heh: slugs - I remember slugs ... damn buggers!
     
  10. drgondog

    drgondog Well-Known Member

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    Got most of it right
     
  11. drgondog

    drgondog Well-Known Member

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    I'm too old to change, it hurts my head to reshuffle into metrics because I don't easily 'weigh' the results for accuracy upon converting.
     
  12. RCAFson

    RCAFson Well-Known Member

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  13. professorcurly

    professorcurly New Member

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    If I may ask a related question, is there a point where adding additional power stops giving you extra performance? For example, comparing the F8F-1 vs the Rare Bear. The F8F-1 has a ~2400hp engine, the Rare Bear has ~4000-4500hp, but the best time-to-10000ft record for them is only a few seconds apart. I've been trying to wrap my head around that one for a while.
     
  14. GregP

    GregP Well-Known Member

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    You know, it seems like it SHOULD be simple. But when you get into the actual aerodynamics, the estimate of change in climb rate due to a power increase is NOT so simple.If you are mathematically inclined it isn't bad. If you aren't, is is difficult. In the real world, you almost never get a simple change in power. Sometimes you get a few percent increase, but many times it comes with both or either of a weight or drag penalty. VERY rarely is the power increase without one or the other or both, but is DOES happen sometimes.

    Sometimes you even get a power increase, AND a drag decrease.

    Think of the Bell X-1 and the Bell X-1E. The X-1 weighed 7,000 pounds empty; 12,225 pounds loaded, and had 6,000 pounds of thrust, and went 957 mph. The Bell X-1E weighed weighrd 6,850 pounds empty (and had aboutn 10% less wing area), 14.750 pounds loaded, and went 1.450 mph.

    The basic difference was aerodymics ... despite looking substantially the same.

    Increase in speed for a power increase is pretty straightforward, but climb rate makes you think a bit, doesn't it? You also might get a nice change in climb rate by a simple change in propellers! Think of the P-47 with the narrow props versus the paddle-bladed prop. The difference was due to the aerodynamics of airfoils ... the prop, not power.

    Thanks for the posts above, Bill.
     
  15. Aozora

    Aozora Well-Known Member

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    #15 Aozora, Dec 5, 2013
    Last edited: Dec 5, 2013
    From the Aerodynamics 101 handbook; Climb rate vs Power:

    [​IMG]

    It's so simple...*sigh!*

    Also from Aerodynamics 101; General Design Principles and Good Aerodynamics:



    [​IMG]

    *No longer needs Ibuprofen*
     
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  16. drgondog

    drgondog Well-Known Member

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    Swampyankee touched on looking at props and prop efficiencies as another factor to improve ROC.

    That will force you to look at T=eta*HP/V and look in-depth to RPM, number of blades, blade diameter and calculate the activity factors to ultimately yield the optimum 'eta' for low speeds (~160-180mph for WWII fighters) while considering configs for climb..
     
  17. JtD

    JtD Member

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    Good summary, but you applied prop efficiency twice. A 50 kW engine power increase at 80% prop efficiency on a 2000kg plane gives
    40kW / 2000kg / 9.81m/s² = 2.04 m/s
    increase in climb rate, the 20% difference to your 1.6 coming from the mentioned double efficiency application.
     
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