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test2soln

# test2soln - SOLUTlOH,——-—'— MAE 261 Test#2 Print...

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Unformatted text preview: SOLUTlOH __________.,,——--—'-'—' MAE 261 Test #2 Print Your Name March 14. 2008 YOU DO NOT HAVE TO PROVIDE “GIVEN" and “FIND" .Open Notes and Textbook Staple this sheet on the top of your solution 1. A wing with AR = 6 and S=1.5 ft2 is tested in a wind tunnel with Vm= 100 file and standard sea level conditions in the test section. The test results were: Angle of Attack (deg) Lift (lb) Drag (lb) 15.50 1.000 Calculate: a) wing lift slope ' b) proﬁle drag coefﬁcient c) span efﬁciency factor For problems 2 and 3 use the following airplane characteristics: 5: 40 ft“; h= 20 ft; w= 1,000 lb; e= 0.70; cm: 0.020; cum = 1.7 engine of 90 hp @ sea level; I1 = 0.8; Calculate: 2. (a) Minimum thrust required for level, un«accelerated flight (b) Equivalent stall velocity (0) Maximum range for gliding ﬂight from 1,000 ft to sea level 3. (3) Maximum rate of climb at sea level (b) altitude for the absolute ceiling (c) approximate attitude for service ceiling PLEDGE: l have neither given nor received help on this test. w s g?— ‘ _, L» 15" T0136 [g5 Qxcxg m C .q~ﬂ~ : ' : 0 869' ) L {as 0,35%30 Hr) c m 00 9‘ (L :5. 3—5 .., dam“ “L d:—1‘“_ (0.86‘31’0 2 0.. 7243 "1 01. 10° __ Evie) - r10 WEAR [9) whoa-'10" C =2 3»: 2sz o 0960'? .. 3 D as mswﬁym 2. I i C 6* 13*) C3): Cd+_c_'-—-—-, 30,05'66‘7 WEAR CL20!369\ C: r; prciz 0,05‘601—0,90 5999 =: 0.05907 “GAR .L v. C 7— WCL :1 23’8”!) z o, 8 003 WAR (01051301) Tr[9>(¢,05"ao'n 2-7. (0,01,) C40 . Q 2 UL) Fm M TR K/C‘AWJ’QW C =m :Ugaoquxm «2 6632 0:9: 2c}, 2 0,04 ,0 TR.:: W ~319ML: :awu M KCL/chw (0 GER/mtg Q6) VesrdL: Z W 1(100oﬂm 6M JW 1:: \\\.’Z. {Av/M LC) R 2‘0 CL. W "'-—" ”Wm WgBOO-«Hj(0 665A :1 I6 380 IR @404») M a: N56908:!- ‘WLL, >721 31’40MLQO4 I... __. P :oR Rm) CL 2W: 3(o‘n\11(o.)ﬁﬂo )2! [4‘3 8 CD: 4.09/0 .2 0.0 W L“ V F% 20mm S?“ 0L. 409 (“0013???) )(f ”5% 2(35' 3 Aid/MC “V Gig/”“3601? mt “(0,3) 90kb>€§OL 1,, 15>,Lo,og>(l3s,3 WM) 5 5° 3 (9'3 “UR/Q LL: “ Egowfﬁ“( (G>:’t;m Ame“ H49 12ch :: 30'18VW 3.0131! Em) '5 Quit; W 1/ 3 ’4'908 49(135'34‘9’m) {3: ME: _ WWW” 7” W (09046- M __,_ 0.3839 {’st. (34¢ :, O 3555 0:091377ﬁgﬂ4‘)= a ”'5- [3-4 44123“ 1* «9‘3 \x 2. 29m +500 9'1st 9 n; 5 19 340 “8° ) # .th 91397 907% _. ) “g” (Q ~ w W \$19)“?th W ALL M-c. {35L m (>9: (25¢ : 0 . 44 2 5" 9.93H* 5:15.44- Tb W m (m wﬁ 340% 237106;} 7.5%: META—HOOK. ’3 loS‘éZ 2.7360 ‘6‘? L1“ :1 Z’Tlomﬁj + 500 14(53311 ~9,805") : 1.7) 360 f1" ...
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test2soln - SOLUTlOH,——-—'— MAE 261 Test#2 Print...

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