NLF215F considerations, Cl for different conditions

My earlier post about the NLF215F simulations with XFLR5, the related parameters for aircraft would be in the use case (one iteration of thinking):

– low altitude cruise:
* altitude = 12000 ft
* W/S = 22 lbs/sqft
* Clcruise = 0.41
* NLF215F flap in the -10 degrees position, gap seals closed

– high altitude cruise:
* altitude = 36000 ft
* W/S = 22 lbs/sqft
* Clcruise = 0.96
* NLF215F flap in the 0 degree position, gap seals closed

– extreme high altitude cruise
* some fuel burned already -> W/S reduced to 21 lbs/sqft
* altitude = 46000 ft
* W/S = 21 lbs/sqft
* Clcruise = 1.48
* NLF215F flap in the 0 degrees position, gap seals closed

– approach
* 1 slot open
* W/S = 15 lbs/sqft
* altitude = 1000 ft
* Cl = 1.1, V = 75 kts (at gross weight, W/S 22 lbs/sqft)
* Cl = 1.1, V = 65 kts (when fuel tanks nearly empty, W/S 15 lbs/sqft)
* NLF215F flap in the +10 degrees position, 1 slot open

– landing
* 2 slots open

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    • dodlithr
    • February 7th, 2009

    Very interesting. Now what is your wing area (span) at which weight to use that concept?

    BTW: Here is a small Wing calculation skript. You can not enter any airfoil to it, but it gives some preliminary estimates for some ?average? shape.

    http://www.aa.washington.edu/courses/aa101/WebTools/Wing-Area.shtml

    • dodlithr
    • February 7th, 2009

    Very interesting. Now what is your wing area (span) at which weight to use that concept?

    BTW: Here is a small Wing calculation skript. You can not enter any airfoil to it, but it gives some preliminary estimates for some ?average? shape.

    http://www.aa.washington.edu/courses/aa101/WebTools/Wing-Area.shtml

    • Karoliina Salminen
    • February 7th, 2009

    The specifications vary between iterations. During this round of iteration, I used Wg = 880 kg, Cr = 1 m, Ct = 0.5 m, AR = 14, S = 8 m2, B = 10 m, Vs = 54 kts, W/S = 22 lbs /sqft.

    Everything is calculated with my own program, it calculates a whole lot of more than the simple web page you gave the link.

    • Karoliina Salminen
    • February 7th, 2009

    The specifications vary between iterations. During this round of iteration, I used Wg = 880 kg, Cr = 1 m, Ct = 0.5 m, AR = 14, S = 8 m2, B = 10 m, Vs = 54 kts, W/S = 22 lbs /sqft.

    Everything is calculated with my own program, it calculates a whole lot of more than the simple web page you gave the link.

    • Exo Cruiser
    • February 8th, 2009

    Ok let’s do some checking with the script.

    http://www.aa.washington.edu/courses/aa101/WebTools/Wing-Area.shtml

    W = 880 kg = 1940 lbs
    Airport Altitude = 0 (SL)
    Landing Speed = 54 kts = 100 km/h = 62 mph

    Then I choose maximum high lift devices (best possible wing in that concept) and I should maybe get near 8 m^2 wing area.

    And it prints out that the wing area should be minimium:

    S = 142 ft^2 = 13.2 m^2

    That is 65% more than 8 m^2.

    How do you estimate the wing-fuselage maximum lift CLmax?

    • Exo Cruiser
    • February 8th, 2009

    Ok let’s do some checking with the script.

    http://www.aa.washington.edu/courses/aa101/WebTools/Wing-Area.shtml

    W = 880 kg = 1940 lbs
    Airport Altitude = 0 (SL)
    Landing Speed = 54 kts = 100 km/h = 62 mph

    Then I choose maximum high lift devices (best possible wing in that concept) and I should maybe get near 8 m^2 wing area.

    And it prints out that the wing area should be minimium:

    S = 142 ft^2 = 13.2 m^2

    That is 65% more than 8 m^2.

    How do you estimate the wing-fuselage maximum lift CLmax?

    • Exo Cruiser
    • February 8th, 2009

    Additional to that:

    AR = b^2/S

    b=10
    S=8

    AR = b^2/S = 12.5

    You are using 14..

    W=1940
    Su=86.1
    loading=W/Su=22.5

    You are using 22, which is about the same.

    • Exo Cruiser
    • February 8th, 2009

    Additional to that:

    AR = b^2/S

    b=10
    S=8

    AR = b^2/S = 12.5

    You are using 14..

    W=1940
    Su=86.1
    loading=W/Su=22.5

    You are using 22, which is about the same.

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