eqns test 1 10C

Eqns test 1 10C - 0.0683 slug(= 2.2 lbm = 1 kg 1.8 R = 1.0 K o o 3.281 ft = 1 m R = F 460 o o 0.224 lb(= 0.102 kgf = 1 N K = C 273 o o pS = 2116

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Unformatted text preview: 0.0683 slug (= 2.2 lbm) = 1 kg 1.8 R = 1.0 K o o 3.281 ft = 1 m ( _ ) R = ( _ ) F + 460 o o 0.224 lb (= 0.102 kgf) = 1 N ( _ ) K = ( _ ) C + 273 o o pS = 2116 lb/ft2 = 1.013 x 105 N/m2 = 1 atm s = 0.002378 lb sec /ft = 1.225 kg/m as = 1116 ft/s = 340 m/s 2 4 3 TS = 59oF = 15oC = 519oR = 288oK s = 3.7373 x 10-7 lb sec/ft = 1.7894 x 10-5 kg/(m sec) air = 1.4 cp = R -1 2 2 Rair = 1716 ft lb/(slug oR) = 287 N m/(kg oK) go = 32.2 ft/s2 = 9.81 m/s2 -11 3 2 cp, air = 6006 ft lb/(slug oR) = 1004 N m/(kg oK) F = ma F = GMm/R T(h) = 15 - .0065h (T in oC, h in meters) 0 (h - h1 ) p = e RT p1 G = 6.67 x 10 p T 0 = ( ) aR p1 T1 -g m /(kg s ) M=V/a h=( a2 = RT RE )h G R E + hG q= V 2 2 -g dp = -g dhG p = RT dp = -go dh p2 - p1 = -g (h2 - h1) dp = -V dV AV = m = constant p1 + V12 V 2 = p 2 + 2 = p + q = po 2 2 p2 = 2 p1 1 V2 V2 cp T1 + 1 = c p T2 + 2 = c p T0 2 2 p 02 2 ( + 1) 2 M1 = 2 p1 4 M1 - 2( - 1) T0 -1 2 = 1+ M1 T1 2 T = 2 T1 -1 -1 2 1 - + 2 M1 +1 dA dV = (M 2 - 1) A V A At 2 1 2 -1 2 = 2 1+ M M +1 2 +1 -1 Re x = V x Re L = V L w = dV dy y =0 ...
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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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