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Problem 18 Assumptions: Problem 18. (38 Points) Water flows steadily through « splitter as show In he igure with the following data: 4 Steuey incompressible slows p = 1000 ky/a! ‘+ Uniform velocity at each section (1-D at ports) @ Volume flow rate at inlet, ¥i 08, n*fa ‘+ Neglet weigh and wall shear + Volume flow rate st outlet 2, # =0.05 m/s + Gange pressures are used. ‘© Diameters: D; = Ds em, Da 120kPa, P on Data / gcometry: 10kPa, Ps = 80kP. © Gage presrure readlage: Fa Osis, Ve=005m'/s = Va 3m" /s ‘Determine the external force in the 2— and y— directions needed to hold the splitter fixed, discegarding Dy Dy —Ot2m ‘the weight effects : 0.0510 Hy = 120K a, 1 = 8D, 1% — 80a Let the inlet (1) be along the pipe centerline at @ = 40° above +2 toward the janetion, outlet (2) be 4, and outlet (3) be 2 Areas and average velocities _ xD} w(0.12"? 7 Ay Ay oonsim?, — A,— = 0.001 9 W008 Ya _ 0 ua rorm/s, Vi a, dons = TNS TOIT Mass flow rate: Ja _ 003 As O.OT1ST 1S mf Jin, = pi, = AUK, thy = ADI tig = ADK /s Vicon: + sin 6 j Yi=Vai Linear momentum (steady) SFL Taek Excermal foros on the li CV: SR-A+K whore = (Uti + Ryd) 1s the net force exerted by the walls/supports on the tid, and £5 is the net pressure force on the Momentum-flux term lon = sna + tat — cam9y = G0V)7 (30)(2.65)2 + (50) (25.5)7 Dav = 80(7.07) {co 79.63 + (12732)3 N nV) = SOVi(cos# i + sin6 j) 40°3) B+ (808.7)3 N Tims -353.9)3 + (909.5}9 N Pressure forces Use A = —SipAA with outward unit normals: fay = —(cus9i sin 07), a By = praia = 4 p.Ar(v0s 64 4 sin 03) fia ——medah. Fa ——ra Aad Combine: . - F, = (pr Ay cos @ — pyAg)é + (pr Ay sits — ppAg)y Numerically: ge = 120000(0,011:51) cos 4 — 89000(0,011:51) Fyy = 120000(0.011:31) sin 40® — 90000(0.001963) = 695.7 N ON = B= (134.9) + (695.7)j N Solve for support force components 9=131.9——A888N. Ry — 900.5 — 695. 4.89 x 10 N fo), Ry 214 x 10 N (ap) Resultant magnitud = (488.8)? + (215.8)? = 5.34 x 10°N Ve Note: fl is the force exerted by the supports/walls on the flukl; 1 8 also the external force required to hold the device five in the shanm axes (oq a the splitter). wl opposite to the fhuid force on Problem 19. (38 Points) Consider stendy, incompressible, laminar, planat/coiumuar, fully developed: Couette flow of 2 Newtonian fuid in the narrow gap between twa infinite perlll plates, as shown below. ‘The top plate 's moving at speed V, and the bottom plate is moving with -2¥ end there is a constant ‘opplled pressure gradient in tho w direction given by ap a “The distenca between these tiv plotes ish, and gravity acte in the negative y-diraction, hk Fluid: p, p. tt, oa oe (a) (25 Points) Find the velocity and pressure fleld. List your assumptions aud indicate them in the equations you use (©) (8 Points) Find velocity and shoar stress in the mile plane et y= $ (b) Velocity and shear stress at y — h/2 Midplane velocity: av h av (3) -wv Problem 19 Assumptions (as stated): steady, incompressible, laminar, planar/columnar, fully developed Couette flow of a Newtonian fluid. Let x be along the plates and y be normal to the plates with =O at the bottom plate and y= h at the top plate. u=uly), v= 0, p4e=const, gacts in = Boundary conditions: (0) =-2V, fh) v (a) Velocity and pressure fields g-momentum: with w= u(y) aud v = 0, au ay? op , du an Fae? Integrate twice: 1 1 (4 - “= (2) vsCw4G u(0) = Cz = —2V Apply BCs ‘Thns, With 2 = ‘y) we 3 u(y) = 2 4 oat oi ulh—y) yemomentum: with v =) and no vertical acceleration, pe or _ ay ay 8 Combine with $4 = —3 (constant): Po — 30 ~ egy where pp is a reference constant. With 22 = —3: Shear stress: Differentiate t(y): du BV _ A (am) yg dy hk & (i) th 29) 3V«di‘ fap = ron 5 (P) oan At y=h/2 
