final exam eqns

final exam eqns - 0.0683 slug (= 2.2 lb) = 1 kg 3.28 ft = 1...

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Unformatted text preview: 0.0683 slug (= 2.2 lb) = 1 kg 3.28 ft = 1 m 0.224 lb (= 0.102 kg) = 1 N T S = 59 o F = 15 o C = 519 o R = 288 o K 1.8 o R = 1.0 o K ( _ ) o R = ( _ ) o F + 460 ( _ ) o K = ( _ ) o C + 273 p S = 2116 lb/ft 2 = 1.013 x 10 5 N/m 2 = 1 atm s = 0.002378 lb sec 2 /ft 4 = 1.225 kg/m 3 a s = 1116 ft/s = 340 m/s T(h) = 15 - .0065h (T in o C, h in meters) R air =1716 ft lb/(slug o R)=287 N m/(kg o K) air = 1.4 s = 3.7373 x 10-7 lb sec/ft 2 = 1.7894 x 10-5 kg/(m sec) c p, air = 6006 ft lb/(slug o R) = 1004 N m/(kg o K) p c R 1 = - F ma = & & F = GMm/R 2 G = 6.67 x 10-11 m 3 /(kg s 2 ) g o = 32.2 ft/s 2 = 9.81 m/s 2 p = RT dp = - g dh G dp = - g o dh p 2- p 1 = - g (h 2- h 1 ) AV = m = constant dp = - VdV M = V/a a 2 = RT q = V 2 / 2 2 2 1 2 1 2 V V p p p q p = constant 2 2 + = + = + = 1 g (h h ) RT 1 p e p-- = g aR 1 1 p T ( ) p T- = E G E G R h ( )h R h = + 1 2 2 2 1 1 1 p T p T - & & = = 2 2 1 2 p 1 p 2 p...
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This note was uploaded on 03/19/2012 for the course AERO 201 taught by Professor Strganac during the Fall '05 term at Texas A&M.

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