Problem 5.52 - V 400 nmi hr 6080 ft nmi hr 3600 s V 675.56 ft s v 2500 ft min min 60 s v 41.67 ft s Therefore u 675.56 ft s 2 41.67 ft s 2 u 674.27

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Problem 5.52 [Difficulty: 2] Given: Instruments on board an aircraft flying through a cold front show ambient temperature dropping at 0.7 o F/min, air speed of 400 knots and 2500 ft/min rate of climb. Find: Solution: We will apply the concept of substantial derivative t T z T w y T v x T u Dt DT Governing Equation: (Substantial Derivative) Assumptions: (1) Two-dimensional motion (velocity not a function of z) (2) Steady flow (velocity is not a function of t) (3) Temperature is constant in y direction x T u Dt DT Based on the above assumptions, the substantial derivative reduces to: Finding the velocity components:
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Unformatted text preview: V 400 nmi hr 6080 ft nmi hr 3600 s V 675.56 ft s v 2500 ft min min 60 s v 41.67 ft s Therefore: u 675.56 ft s 2 41.67 ft s 2 u 674.27 ft s So the rate of change of temperature through the cold front is: δ T x 0.7 Δ °F min s 674.27 ft min 60 s 5280 ft mi δ T x 0.0914 Δ °F mi Rate of temperature change with respect to horizontal distance through cold front....
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This note was uploaded on 10/19/2011 for the course EGN 3353C taught by Professor Lear during the Fall '07 term at University of Florida.

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