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Exam+2+Solutions

Exam+2+Solutions - SSN Spiing 03 Exam#2 1 Write ALL...

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Unformatted text preview: SSN: ' Spiing 03 Exam #2 1. Write ALL equations in their most general form and then begin simplifying. State/list all your assumptions when you are simplifying the equations. , 2. Check your units! 3. 'Box your answers and provide units. Ph sical Pro net-ties of Water _ 2 , II— 9790 Nm 62AM l._ 998kg/m 1.94slugs/fi .- 0- - .- 0- m 0.00131 Ns/m 001 N-s/m 2 73x10-5 lbfes/ ‘ ' 0.073 N/m 073 N/m 0 05 lbf/fi 1230 N/m abs 2340 N/m abs 0.178 sia Temperature Conversion: T (R) = 460 + T (°F) _ . T (°K) = 273 + T (°C) Gravitational Constant: g = 9.81 m/sec,2 = 32.2 l’t/sec2 lbf= mass (slugs) x g (fi/secz) ‘ N = mass (kg) x g(m/sec2) In cylindrical coordinates for a circular cross-section: ‘dA = 21ndr In cartesian coordinates (x-y) dA=dxdy Short Answer (2 points each) * (a f t O s the velocity of flow increases 1. According to Bernoulli’s equation for a wa W the deer » , dim/r 3‘13 A E, wwm 2. I Give a verbal description of the following term 35; IpdV ’ km. MR 3 C/QM‘wxit sf Me,“ Affigg ”1» CV 3. Why (in mathematical terms) is flow out of a control volume always positive? «am—3...) . \i _ [’1 ) C; 51mg \l h 3“; 7"‘C-E-2206FiuidMeehaifies/Willson , - » i - ., —— , ~--»--- 71» 4. When applying Euler’s equation, 2 is always positive in what direction? 5. Are terms on the right hand side of the control volume equafiom'Eulerian or Langragiau? Eulen'an W o 6. (15 points) The pressure gage at pt. 1 measures 16 mmHzO. A pitot probe indicates 24 / ' mm of water. Calculate the velocity of the air. The density of the air is 1.203 kg/mS. Pf. \Lo \rwmkr Q1... " 24 6:”er €6‘ir “‘23 ll 17. v t P" —.1 + ’l’ ’L L {' VD de \bf _ zeihuu/r t. 1 t. ‘g'a'fii L? a [4.47 ' t .r \ 264,3 Q A 7 = , L \I“? Xumfiwoah ( E \) 7 (15 points) A velocity field' In a parucular flow 18 given by V: 20y? i- 2(1ny 111 m/sec. ()0 1‘5? _ ._%_--_. - a. Calculate the velocity at the point (1, -1 2). . ‘ \1 :70011 : 2001(7) J s20f4—"203 % 1/ ./ b.0CaIcu1ate the acceleration at the point (1 -1, 2) D3111$4L+w 191 (£111) {9:260 7X ax: 0)] 420% 9024037 '- ‘0 a =9 +1193+v91+ %. .3 91:,22 9x 93 . arm 214% (W ”3701 '201) "9116799 1 <7 4 9:; >3 ' :1 939’ 31’ ’QZW/fi ‘91- )lj é=L )2;n ._ 14;}, ‘ 8. (20 points) Water flows through ' ntal pipe bend and exits into the atmosphere (seefigm‘eheiow):'fhefiewrateis . ISee-———~~—— ,, ~ wvv——-—~——-—-— buzg’Jf» L/ \/l = 0%)“! 26/19/51 A mm‘ a. (7 points) Calculate the pressure [email protected] V 1 0'3 X L! / ' a 2 v‘ m m 2512”qu f ?\/ a a 0-) ) mflmf; “(“613 - L b. (13 points) Calculate the force in each of the rods olding the pipe bend in position. -F,(+ 54le7l_(0‘2/5)L= — vl-CQ ~6-nwame W; Y W \/ e \‘VV‘ ‘4’ d a é L! - :31 , I . z . / Uyfio‘ V x 2 5! [é [hi/H3 9. (15 points) A tank of liquid (S=O.9) that is 4 f: in diameter and 1 ft high is rigidly fixed to a rotating arm having a 3 ft radius (F3 11). The tank rotates about the axis such that the rwioeifiLafpt€is 75 fine: Ifthepressufeat—Aésrzopsfi—whafisflae-WGQ—m——— » r , Z Y; K} . 6» 55-h; —— z "3”! stu 3‘ 32-2 W” a): +S/5- 7. ugad/g/ / / [1: ’ /. ‘/ 2"- 20 , Iwrg = F6 + fé./5x0,g—,,,fl§(’sz_ 2 / ,1 19 (1'0 points) Ifthe piston and water are accelerated upward at a rate of 0.5g, what will be me 121635136 31 a dnpth of2 fim water column? (mmgsyaxozéxaz; 3/3 = LBL _ 9W2) 32' 3'2. ,ZSL'LZfi : "23‘, \‘ / ‘ az. '7 ‘21; 5 3533.134 , (5.141324) 94 AT ”‘3 70M— 2- JH) L..:::[email protected]/ (sea A 11. (15 points) Water flows 1n and out of a device as shown In the figure below. Calculate the rate of change of the mass of water (i. e ,'dm/dt) 1n the device. (Q; VA= 304;; (11304118? 56 : I. 117 FAQ/Se ( ,1 1 New is MCéegi’ 1,2, ‘9“ afi grin camuxzvxhaa chose. Hewfiueo, Ho‘s" wee/1‘5 Mass fs'acCuwoiquxcj :11 ”PL: 5354mm, BONUS (2 points) Why does the dmgys/dt term equal zero when deriving the conservation of mass (i. e., continuity) equation? . \/ haemosz q 8535wa how 9 $.59: Gimwvd' o‘F @MQSS afld M: COUA Mufir Chaim-$2 ...
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