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Question ID: 330281

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  • Question ID: 330281

    Question:

    "(Refer to Flight Planning Manual MRJT 1 Figure 4.5.3.1 and Figure 4.3.1b)
    Given: twin jet aeroplane, Zero Fuel Mass 50000 kg, Landing mass at alternate 52000 kg, Final reserve fuel 2000 kg, Alternate fuel 1000 kg, Flight to destination: Distance 720 NM, True Course 030°, W/V 340°/30 kt, Long range cruise, FL 330, Outside air temperature -30°C
    Find: Estimated trip fuel and time with simplified flight planning"

    Explanation:

    "The question references ?SIMPLIFIED PLANNING? ? FIG 4.3.1B
    To use this, we need WIND COMPONENT ? so use your wiz wheel – for this you need WIND, TRACK and TAS

    ? TAS - FIG 4.5.3.2, Detailed Planning, 0.74 MACH, FL330 (Section 4 page 39)
    TAS 430 KT
    NOTE 2C Increase TAS 1 KT per degree above ISA
    ISA at FL330 = -50°C, OAT -30°C,
    TEMP CONDITION = ISA +20°C, TAS = 430 + 20 = 450 KT

    Now WIZ wheel, GS = 430 KT, so there is a 20 KT HEADWIND component

    ? Now reference FIG 4.3.1B – we have WIND, PRESSURE ALTITUDE,
    TEMPERATURE, just need to calculate LANDING MASS at destination
    LM (at DEST)= ZFM + ALTERNATE FUEL + FINAL RESERVE = 53 000 kg

    Now ? FIG 4.3.1B ? (Refer to IMAGE) 4850 kg, 1.9 hr ? 1 hr 54 min "


    Why does the explanation use the graph for 0.74 MACH, when the question says to use the LRC graph?

    - Lexie

  • #2
    Re: Question ID: 330281

    I am at a loss on this one - don't know why I used 0.74M ... answer has been corrected - has been passed on to support.

    Tom

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