examf - EGN3353C Fluid Mechanics Spring 2010 Final Exam...

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Unformatted text preview: EGN3353C Fluid Mechanics Spring 2010 ! Final Exam LAST NAME:___________________________ FIRST NAME:________________ SECTION:______ INSTRUCTIONS: ! Closed book, closed notes, calculator allowed. Formula Sheet Energy Equation • Control-volume form d ˙ ˙ Qnet in + Wshaft,net in = dt ￿ CV ρe d∀ + ￿ CS ￿ p V2 +u+ + gz ρ 2 ￿ ￿n ρVr · ￿ dA. For steady flow, evaluating control-surface integral for single inflow and outflow gives ￿ ￿ 2 2 ˙ net in + Wshaft,net in = m pout − pin + uout − uin + Vout − Vin + g (zout − zin ) . ˙ Q ˙ ρout ρin 2 (1) (2) • Energy-density form [J/kg]: qnet in + wshaft,net in = 2 V 2 − Vin pout pin − + uout − uin + out + g (zout − zin ). ρout ρin 2 (3) • Head form [m]: wshaft,net in V2 pin pout V2 + + in + zin = + out + zout + hL , g ρin g 2g ρout g 2g where hL is the head loss hL = uout − uin − qnet in . g (4) (5) Isentropic Flow Subscript 0 and superscript ∗ denote stagnation (total) and sonic conditions, respectively γ−1 2 T0 =1+ M T 2 γ ￿γ ￿￿ ￿ p0 T0 γ −1 γ − 1 2 γ −1 = = 1+ M p T 2 ￿ ￿1 ￿ ￿1 ρ0 T0 γ −1 γ − 1 2 γ −1 = = 1+ M ρ T 2 ￿￿ γ +1 ￿ ￿ A 1 2 γ − 1 2 2(γ −1) = 1+ M A∗ M γ+1 2 ￿ ￿ γ +1 √ 2(γ −1) p 2 0γ∗ ∗∗ ∗ √ m = ρuA = ρ a A = ˙ A γ+1 RT0 (6) (7) (8) (9) (10) Normal Shock Waves Subscripts 1 and 2 denote conditions ahead of and behind the shock wave, respectively 2 ρ2 u1 (γ + 1)M1 = = 2 ρ1 u2 2 + (γ − 1)M1 ￿ p2 2γ ￿ 2 =1+ M1 − 1 p1 γ+1 ￿ 2 2 + (γ − 1)M1 M2 = 2 2γ M1 − (γ − 1) 1 (11) (12) (13) EGN3353C Fluid Mechanics Spring 2010 ! Final Exam T A B TA . I 3 E a i al c O n e - d i m e n s i o n s e n t r o p ic o m p r e s s i bf l o wf u n c t i o n so r a n i d e a l le f g a sw i t hk : L 4 '+ ' + k_ I 2 k- | Ma *u') ^\-r(fr-r) *u') 2 k- | ^\-r ' * Z *u') Ma* 0 0.1 4.2 0.3 0.4 0.5 0.6 o.7 0.8 0.9 1.0 .\-kl(e-r) 0 0.1094 o.2182 0.3257 0.4313 0.5345 0.6348 0.7318 0.8251 0.9146 1.0000 1.1 83 5 1.2999 L4254 1.5360 1.6330 r . 7r 7 9 r.7922 1.8571 1.9i40 1.9640 2.236r 2.2495 r.2 I.4 1.6 1.8 2.O 2.2 2.4 2.6 2.8 3,0 5.0 CC oz PlPo AIA* co 5.8218 2.9635 2.0351 1.5901 1.3398 1.1882 r.0944 1.0382 1.0089 i.0000 1.0304 I.IT49 r.2502 1.4390 1.6875 2.0050 2.403I 2.8964 3.5001 4.2346 25.000 c( p lp o 1.0000 0.9930 0.9725 0.9395 0.8956 0.8430 0.7840 o.7209 0.6560 0.5913 0.5283 o.4124 o.3142 0.2353 o.1740 0.r278 0.0935 0.0684 0.0501 0.0368 2 0.027 0.0019 0 1.0000 0.9950 0.9803 0.9564 0.9243 0.8852 0.8405 0.7916 0.7400 0.6870 0.6339 0.531r 0.4374 0.3557 0.2868 0.2300 0.1841 0.1472 0.1179 0.0946 0.0760 0.01i3 00 TlTo 1.0000 0.9980 0.9921 0.9823 0.9690 0.9524 0.9328 0.9107 0.8865 0.8606 0.8333 O.7764 0.7I84 0.6614 0.6068 0.5556 0.5081 0.4647 A.4252 0.3894 0.3571 0.1667 T9-_ ( f t + 1 ) l TABTE-14 A O n e - d i m e n s i o n a lr m as h o c k u n c t i o n fso r a n i d e a g a sw i t h k : o l f l zkual- k+ | Mat zkMal - k + I k+ | ( f t + l)Mal vl v2 2+(k-l)Maf + kMal + kMal i.0 1.1 z/Pt 1.2 1.3 T2/Tl 1.4 +Ma1&-r) .9 2.0 1 . 5 C) +Mal(k-r) 1.6 ,'-) t.7 + a ' 1 1 M a i @ 1)/2 r ) / t 2 ( f t>l | T r.s 1 . 8 u r l l + M a i ( k- I ) / 2] ( k + 'a 1.9 I + kMal tl + tvta3& l)/21r'rw-1l 2 . O I | + kMal E t.o 2 . I U 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 3.0 4.O 5.0 oc a 6 Maz 1.0000 0.9118 0.8422 0.7860 0.7397 0.7011 0.6684 0.6405 0.6165 0.5956 O.577 4 0 . 5 6i 3 0.547I 0.5344 0.5231 0.5130 0.5039 0.4956 0.4882 0.4874 0.4752 0.4350 0.4752 0.3780 P2lPl 1.0000 r.2450 1.5133 1.8050 2.t200 2.4583 2.8200 3.2050 3.6133 4.0450 4.5000 4.9783 5.4800 6.0050 6.5533 7.1250 7.7200 8.3383 8.9800 9.6450 10.3333 18.5000 29.000 oo PzlPt 1.0000 1.1691 1.3416 L5157 r.6897 L8621 2.03t7 2.1977 2.3592 2.5157 2.6667 2.8179 2.9512 3.0845 3.2rr9 3.3333 3.4490 3.5590 3.6636 3.7629 3.857I 4.5714 5.0000 6.0000 2 T2lT1 1.0000 1.0649 1.1280 i. i909 r.2547 1. 3 2 0 2 1.3880 1.4583 i.5316 L6079 r.6875 r.7705 1.8569 I.9468 2.0403 2.1375 2.2383 2.3429 2.4512 2.5632 2.6790 4.0469 5.8000 oo I.4 Po2l or P P o 2P l l 1.0000 0.9989 0.9928 r.8929 2.1328 2.4075 0.9794 0.9582 o.9298 0.8952 0.8557 o.8127 0.7674 0.7209 0.67 2 4 0.6281 0.5833 0.5401 0,4990 0.4601 0.4236 0.3895 0.3577 0.3283 0 .1 3 8 8 0.0617 2.7136 3.0492 3.4133 3.8050 4.2238 4.6695 5.1418 5.6404 6.1654 6.7165 7.2937 7.8969 8.5261 9.1813 9.8624 10.5694 rr.3022 12.0610 2 1. 0 6 8 1 32.6335 0cc rt Po 4 T Tx k+l I +ftM (}lf.a(l \t +tl P_ t+k P* 1+ftM L :p* v'*pl _(1 EGN3353C Fluid Mechanics Spring 2010 ! 1. (5 points) A concrete dam of height H and width W holds back a water reservoir of depth h, see the figure on the right. Water that seeps through the ground below the dam creates a linear pressure distribution as shown in the figure. Assume that ρw=1000 kg/m3 and ρc=2400 kg/m3. Given that a=ρwgh/10, b=ρwgh/20, W=2 m, h=10 m, and g=9.81 m/s2, determine the minimum dam height H that would prevent toppling of the dam. (You can assume that the dam has unit length normal to the page.) Final Exam W ρc h H ρw Dam b Ground Continue on the back of this page if necessary. a EGN3353C Fluid Mechanics Spring 2010 ! Final Exam 2. (5 points) Consider the low-speed wind tunnel shown on the right. It uses a fan to draw in air from the atmosphere (station ➀) and accelerate the flow into the test section. The 5 1 2 4 3 6 flow exhausts into the atmosphere (station ➅). Test section Flow direction Atmospheric conditions are ρ∞=1.2 kg/m3 and Fan p∞=105 Pa. Losses are negligible except across the fan, where a head loss of hL=10 m is incurred. Hydrostatic changes in pressure can be ignored. Given that u1≈0, u2=5 m/s, A2=10 m2, A3/A2=1/5, A4/A3=3, A5/A4=1, and assuming steady conditions, determine the fan shaft power in Watts. Work to four decimal places throughout. Continue on the back of this page if necessary. EGN3353C Fluid Mechanics Spring 2010 ! 3. (5 points) A so-called jet pump, see the figure on p1 the right, is a device in which a fluid stream of speed u1 is accelerated by a jet of fluid with velocity uj>u1. At the plane where the two streams meet, the static pressure is constant at p1 over Aj the entire cross section. Some distance downstream, mixing produces a uniform velocity u2 and static pressure p2. Assuming incompressible flow, derive an expression for the pressure difference p2-p1 in terms of u1, uj, and Aj/A2 (i.e., without u2). Final Exam p2 uj A2 u1 1 Continue on the back of this page if necessary. u2 ! Recall dynamic similarity between a model EGN3353C Fluid Mechanics Spring 2010 ! match. " How do we determine the ! ' s ? Final Exam 4. (5 points) The detonation of an explosive leads to a spherical shock wave in the atmosphere. Use the method of repeating variables to generate a dimensionless relationship for the radius R of the shock wave as a function of the energy E released during the detonation, the density ρ of the explosive material, and time t. Step 1 dimen that a i.e., it and a Step 3 dimen come Step 4 each Continue on the back of this page if necessary. EGN3353C Fluid Mechanics Spring 2010 ! Final Exam 5. (5 points) A reservoir with a stagnation pressure of p0=1.2 MPa and a stagnation temperature of T0=300 K provides air to a converging-diverging nozzle. The ratio of exit area to throat area is Ae/At=1.1. Assume that the gas constant is R=287 J/(kg K) and that the ratio of specific heats is γ=1.4. Calculate the back pressures that correspond to the following conditions: (a) when the flow is just choked but still subsonic throughout the diverging portion of the nozzle, (b) when a shock wave is located right at the exit of the nozzle, and (c) when the flow is perfectly expanded, i.e., the nozzle is at its supersonic design condition. (d) Compute the flow velocity at the exit of the nozzle for condition (c). Round to one decimal place when looking up values for A/A* in Table A-13. Continue on the back of this page if necessary. ...
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This note was uploaded on 09/05/2011 for the course EGN 3353C taught by Professor Lear during the Spring '07 term at University of Florida.

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