Textbook Notes (368,245)
ENGG 2230 (2)
Chapter

# FluidMechWhite5eCh04.pdf

50 Pages
205 Views

School
Department
Engineering
Course
ENGG 2230
Professor
Bahram Gharabaghi
Semester
Winter

Description
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
More Less

Related notes for ENGG 2230
Me

OR

Join OneClass

Access over 10 million pages of study
documents for 1.3 million courses.

Join to view

OR

By registering, I agree to the Terms and Privacy Policies
Just a few more details

So we can recommend you notes for your school.