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ENGG 2230 (2)


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ENGG 2230
Bahram Gharabaghi

Chapter 4Differential Relations for a Fluid Particle 41 An idealized velocity field is given by the formula 2424txtyxzVijk Is this flow field steady or unsteady Is it two or threedimensional At the point x y z1 1 0 compute a the acceleration vector and b any unit vector normal to the acceleration Solution a The flow is unsteady because time t appears explicitly in the components b The flow is threedimensional because all three velocity components are nonzero c Evaluate by laborious differentiation the acceleration vector at x y z1 1 0 d uuuuu22uvw4x4tx4t2ty04xz04x16txdttxyzd vvvvv224 uvw4ty4tx02ty2t4xz04ty4tydttxyzdwwwww22uvw04tx4z2ty04xz4x16txz16xzdttxyzdV242or4x16tx4ty4ty16txz16xzijkdtdV23414t4t1t0 cij kat x y z1 1 0 we obtain Ans dtd At 1 1 0 there are many unit vectors normal to dVdt One obvious one is k Ans42 Flow through the converging nozzle in Fig P42 can be approximated by the onedimensional velocity distribution 2x 10 0uVwoLa Find a general expression for the fluid acceleration in the nozzle b For thespecific case V10 fts and L6 in oFig P42 compute the acceleration in gs at theentrance and at the exit 259 Chapter 4Differential Relations for a Fluid ParticleSolution Here we have only the single onedimensional convective acceleration 222Vduux2Vx2oo 1 a uVA n s1odtxLLLL22102ftdux 610 140014ForLandVxwith x in feet o612612sd t22At x0 dudt400 fts 12 gs at xL05 ft dudt1200 fts 37 gs Ans b43 A twodimensional velocity field is given by 22Vxyxi2xyyj in arbitrary units At x y1 2 compute a the accelerations a and a b the xyvelocity component in the direction 40 c the direction of maximum velocity and d the direction of maximum acceleration Solution a Do each component of acceleration duuu22uvxyx2x12xyy2yaxdtxy dvvv22 uvxyx2y2xyy2x1aydtxyAt x y1 2 we obtain a18i and a26j Ans a xyb At x y1 2 V2i6j A unit vector along a 40 line would be ncos40ij Then the velocity component along a 40 line issin40V 2 6 c o s 4 0 s i n 4 0b Ans Vnijij539 units 40402212a1826316 units at 553 Ans c d c The maximum acceleration is max 2344 Suppose that the temperature field T4x3y in arbitrary units is associated dTdt at x y2 1 with the velocity field of Prob 43 Compute the rate of change Solution For steady twodimensional flow the rate of change of temperature is dTTT222 829uvxyx xx y yydtxyAt x y2 1 dTdt51659125 units Ans260 Solutions ManualFluid Mechanics Fifth Edition 45 The velocity field near a stagnation point see Example 110 may be written in the form UxUyoouv andare constantsUL oLLL15 m a Show that the acceleration vector is purely radial b For the particular case 2x y1 m 1 m is 25 ms what is the value of U if the acceleration at oSolution a For twodimensional steady flow the acceleration components are 2UUduuuxyoo uvUU 0 xoo2dtxyLLLL 2UUdvvvxyoo uvU 0 Uyoo2 dtxyLLLL2222Therefore the resultant ULxyULpurely radial aAns ai j r oo2b For the given resultant acceleration of 25 ms at x y1 m 1 m we obtain 22UUmmoora25 2 msolve forUbAns 63o222ssL15 m 46 Assume that flow in the converging nozzle of Fig P42 has the form V12xLi Compute a the fluid acceleration at xL and b the time required for a Vofluid particle to travel from x0 to xL Solution From Prob 42 the general acceleration was computed to be 222V6V2duuxoo1 a ua t x L A n s dtxLLLb The trajectory of a fluid particle is computed from the fact that udxdt Lt2dxxdx1uVo rV d too12dtLxL00 Lln3 bortAns0L2Vo
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