Review_problems_Exam1_2010

# Review_problems_Exam1_2010 - 2—73 Consider laminar ﬂow...

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Unformatted text preview: 2—73 Consider laminar ﬂow of a Newtonian ﬂuid of viscos- ity to between two parallel plates. The flow is one-dimen- sional, and the velocity proﬁle is given as My) = 4umax [y/h — (gt/hf], where y is the vertical coordinate from the bottom surface, It is the distance between the two plates, and is the maximum flow velocity that occurs at midplane. MITIEIX Develop a relation for the drag force exerted on both plates by the fluid in the ﬂow direction per unit area of the plates. My) : 41; 32/11 — (y/hﬁ] max[ 2—73 The velocity proﬁle for laminar one-dimensional flow between two parallel plates is given. A relation for friction drag force exerted on the plates per unit area of the plates is to be obtained. Assumptions 1 The ﬂow between the plates is one-dimensional. 2 The fluid is Newtonian. Analyst's The velocity proﬁle is given by My) 2 411mm, [1' f h — [if It )2 ] HC‘l') = 4am“ [1'].th —(:.1'fh)3] Where It is the distance between the two plates. _1' is the vertical distance from the bottom plate. and rim is the maximum flow velocity that occurs at midplane. The shear stress at the bottom surface can be expressed as '3 \ 3. If it It? r Ida w I 05‘ d, "1 2}“ _ 4jttinm _ it — 4min.“ dr . — 4 yum“ J J.:0 ‘- 3 17? he i y=o Because of synunetiy. the wall shear stress is identical at both bottom and top plates. Then the friction drag force exerted by the ﬂuid on the inner surface of the plates becomes Sham FD : ZTwApfate : h ‘ plate Therefore. the friction drag per unit plate area is Strum plate : h Discussion Note that the friction drag force acting on the plates is inversely proportional to the distance between plates. FD A 3—141E A semicircular 30-ft-diameter tunnel is to be built under a lSO-ft-deep, 800-ft-10ng lake, as shown in Fig. P3—l4lE. Determine the total hydrostatic force acting on the roof of the tunnel. Water T 150 ft ||<l t—soa—l Chapter 3 Pressure and F Im'd Statics 3—141E A semicircular tunnel is to be built under a lake. The total hydrostatic force acting on the roof of the tunnel is to be determined. Assumptions The atmospheric pressure acts on both sides of the tunnel. and thus it can be ignored in calculations for convenience. Properties ‘We take the density of water to be 62.4 lbm ft3 throughout. Amnfvst's \Ve consider the free body diagram of the liquid block enclosed by the circular surface of the tunnel and its vertical (on both sides) and horizontal projections. The hydrostatic forces acting on the vertical and horizontal plane stn'faces as well as the weight of the liquid block are determined as follows: Horizontal force on vertical surface (each side}: FH —Fx —PmA—,Lgl1CA—pg(s : R 2)A = (62.4lbm ﬁ3)(32.2ﬁ. 52 }(135+15 2ft}[15 fthOOft}[ llbf ] 32.2 lbrn - ft. s2 = 1.06? x10 3 lbf (on each side of the tunnel) Vertical force on horizontal surface (downward): F}. = PMA 2 pghcA = ,LgthPA llbf = 62.4lb ﬁ3 32.21%. 3 135a 30ft SOOft —,] ( m ll: 5 K ll: K )[322lbm-fts‘. = 2.022x1081bf \Veight of fluid block on each side within the control volume dmvmvm'd) .' W = mg = gl/ = igtal —m‘€2 4}(2000 ft} = (62.4 lbm ﬁ3)(31.2 ft s2)(15 ft}2(l-:I4}(SDO ﬁ}[ llbf ] 32.2 lbm - ft. s2 = 2.410 x 106 lbf [on each side) Therefore. the net downward vertical force is Ff. =F}. +2W= 2.023103 +2><0.02410><10g =2.{J7><10El lbf This is also the net force acting on the tunnel since the horizontal forces acting on the right and left side of the tunnel cancel each other since they are equal ad opposite. 5—49E Water flows through a horizontal pipe at a rate of l gal/s. The pipe consists of two sections of diameters 4 in and 2 in with a smooth reducing section. The pressure difference between the two pipe sections is measured by a mercury manometer. Neglecting frictional effects, determine the dif— ferential height of mercury between the two pipe sections. Answer: 0.52 in 5—4912 Water ﬂows through a horizontal pipe that consists of two sections at a speciﬁed rate. The differential height of a mercury manometer placed between the two pipe sections is to be determined. y' Assumptions lThe ﬂow through the pipe is steady. incompressible. and irrotational with negligible frictional effects (so that the Belnoulli equation is applicable). 2 The losses in the reducing section are negligible. Properties The densities of mercury and water are pHg : 847 lbm-"ft3 and pw : 62.4 lbm-"ﬁ". Analysis We take points 1 and 2 along the centerline of the pipe over the two tubes of the manometer. Noting that :1 = :2. the Bernoulli equation between points 1 and 2 gives P V2 P V? ,VE—Vg 1:1::_2:‘::2 > Fla—p“: 1) (1) pg 23 pg 23 2 We let the differential height of the mercury manometer be I? and the distance between the centerline and the mercury level in the tube where mercury is raised be 5-. Then the pressure difference P2 7 P1 can also be expressed as a + ago-+10 = P; + pwss' + pHgga a P1 7P2 = (mg 7mg}: (2) Combining Eqs. (1) and (2) and solving for h. Pw (V22 — I712) 2 ,0»- (V22 *1/12) L722 — V12 33“ng _ 10w) — 2g(pHg pl“ _ D = (pHg —pw)gh > it Calculating the velocities and substituting. ' ' a II; ’ 1 3 \ V1: 1/ I 1: : igars2 [0.13368 ft J21-53WS Al :rDl star-=12 ft) lgal ' ' 1 .3 3‘ V2: V I (f : ga s2 [011368ﬁ 1—6'13ﬁfs A: at); :r{2.-"13 ft) lgal (613M)2 (1 <‘ft"s)2 . — “.1 h _ _ 0.0435 n = 0.52 in 2(32.2 ft.-‘"s2 )(847 62.4 71) Therefore. the differential height of the mercury column will be 0.52 in. P429 Consider a steady, two—dimensional, incompressible ﬂow of a newtonian ﬂuid with the velocity ﬁeld 11 : —2xy, V : yz — x2, and W: 0. (a) Does this ﬂow satisfy conservation of mass? (b) Find the pressure ﬁeld y) if the pressure at point (X : 0, y : 0) is equal to pa. Solution: Evaluate and check the incompressible continuity equation: A ‘1 A ()1: 0V (2w + — + = q a ’1 0X Q17 02 0 : —2_V+ 2y+ 0 E 0 Yes! Am. (a) (b) Find the pressure gradients from the Navier—Stokes X— and yrelations: (3:1 (3:1 811 5p ( (32.7.: (9211 (3‘2le puﬁ IVA 1W“ : a+r1L 2: 2: 0r: (7X (2y (2'2 0X (3.; y (3 (3 (3' p[ 2m Zyj : (f 33x 2x)]:— j” + ;.1(0+0+0), 0r: AP Earn/2+?) 0X ()X and, similarly for the _V—momentum relation, ﬂ ‘5 ﬂ ‘1 ( "‘2 y “‘2 _ ‘32 A (7' V (7 V (7' V (713 (7 l} (7 Ir“ 0‘ I” [)[uﬁ +Vﬂ I‘Wﬁ]— a IﬂL 7+ 9+ ,J, or: 0X QV CZ 0y 7X“ (3V (32‘ 5 (3' p[—2er(—2 x)+ (y: — X2)(2_V)] : — + y(—2 + 2 + 0), or: j” = —2p(X2_V+ f) 0y 0y The two gradients ﬁpx‘fﬁx and (3pfr3‘y may be integrated to ﬁnd p(x, y): (3p ( I; X“ . , . : n am : —2 ) ‘ +— + 1‘ y , then dlﬁffﬁﬂffafﬂ p r q 2 4) t.) a (if ‘1 df J . (Hp : —2p(.x*_yj +— : —2,0(X'_V+ 5x3), Imc‘nce — : —2p_;r3, or: 10") : —ﬂ—_V4 + C 0? (iv JV 2 HMS: p: —§(21{2y2 + X4 + _}"4)+ C: pa at (.1311!) : (0,0), or: C : pa Finally, the pressure ﬁeld for this ﬂow is given by l p : pa — 310(2x2y2 + x4 + y4) A115.(b) ...
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## This note was uploaded on 12/08/2011 for the course MECHANICS 57:020 taught by Professor Fredrickstern during the Fall '10 term at University of Iowa.

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Review_problems_Exam1_2010 - 2—73 Consider laminar ﬂow...

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