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*P4.95 Two immiscible liquids of [ L 4 V : : h equal thickness / are being sheared . 1ty v e between a fixed and a moving plate, y h Pis as in Fig. P4.95. Gravity is neglected, v X - > | Fixed and there is no variation with x. - Fig. P4 .95 Find an expression for (a) the velocity at the interface; and (b) the shear stress in each fluid. Assume steady laminar flow. Solution: Treat this as a Ch. 4 problem (not Ch. 1), use continuity and Navier-Stokes: _— du dv dv . Continuity : = - o 0+ —= 0 : thus v =const = 0 forno -=slip at the walls X )y This tells us that there 1s no velocity v, hence we need only consider u(y) in Navier-Stokes: du du ap a’u @ d*u U— + V—) = =— + + or: 0+0=0+ 0+ p|.2( ox ay) ox I"I.2( axz ('9)’2 ) ‘u].z( ).-‘2 ) Thus w =a + by 4-62 Solutions Manual * Fluid Mechanics, Eighth Edition The velocity profiles are linear in y but have a different slope in each layer. Let u; be the velocity at the interface. (a) The shear stress 1s the same in each layer: Uy V=u, Solve for u; = By Ans.(a) h h My + o (b) In terms of the upper plate velocity, IV, the shear stress is r o= (¥ Ans.(b) uy+uy h

In this problem, water enters the turbine nozzles at 800 kPa absolute with £ s dow velocity. And exits at a pressure of 100 kPa, the maximum velocity %6 which water can be accelerated by the nozzles before striking the turbine blades can be determined as following: Assume that point 1 is at the inlet of the nozzle and point 2 is at the exit of the nozzle(as shown in the fig.) By applying Bernoullis equation between the points (1) and (2) as following: Where Vi = 0, and 2; = 29, by substituting in the Bernoullis equation as following: By substituting with given values as following: | [800%10% 100+ 10 Vo =/ 2 L \ 1000 1000 =37.4166 m/s The maximum velocity to which water can be accelerated by the nozzles before striking the turbine blades, V5 = 37.4166 m/s

Possible drag reduction in pipes = Shear Thinning = Newtonian «== Shear Thickening yH VIV Figure 3. Velocity profiles for different fluids 2 10. Bernoulli’s Equation is as follows: f + VT + gZ = Constant a) (4 pts) Bemoulli’s 1738 treatise Hydrodynamica contains many excellent sketches of flow patterns related to his frictionless relation. One, however, redrawn below, seems physically misleading. Identify what is wrong with the figure and explain why. Assume flow to be | P_[—L\,e_ l/\/U%A/\A' e +o (Y /.[\ < Wvo be esquod to Har (10« Wy ™, ele vok on (1) | [ com we Bermouds Hdrom (x) - (2) lo show Hrek Dne b\,e.\a\/u\' must be fhe saxme fvom (A) () v

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Solution Homework O1 P1.45 A block of weight W slides down an inclined plane while lubricated by a thin film of oil, as in Fig P1.45. The film contact area is A and its thickness is h. Assuming a linear velocity distribution in the film, derive an expression for the “terminal” (zero-acceleration) velocity V of the block. Find the terminal velocity of the block if the block mass is 6 kg, A=35 cmz, 9=15°, and the film is 1-mm- thick SAE 30 oil at 20°C. Liquid film of thickness /i J v Block contact } P arca A 1.45 A block of weight W slides down an inclined plane on a thin film of oi1l, as in Fig. P1.45 at right. The film contact area is A and its thickness h. Assuming a linear Oil film, velocity distribution in the film, derive an thickness h analytic expression for the terminal velocity V of the block. Fig. P1.45 Solution: Let “x” be down the incline, in the direction of V. By “terminal™ velocity we mean that there 1s no acceleration. Assume a linear viscous velocity distribution in the film below the block. Then a force balance in the X direction gives: : : Vv 2FE =Wsinf-rA=W smf)—( ,u;] A=ma, =0, hW sin @ HA Ans

P3.29 In elementary compressible-flow theory (Chap. 9), compressed air will exhaust from a small hole in a tank at the mass flow rate m = Cp, where p is the air density in the tank and C is a constant. If p, is the initial density in a tank of volume v, derive a formula for the density change p(r) after the hole is opened. Apply your formula to the following case: a spherical tank of diameter 50 cm, with initial pressure 300 kPa and temperature 100°C, and a hole whose initial exhaust rate is 0.01 kg/s. Find the time required for the tank density to drop by 50 percent. AlIR Volume v 2 Se Density p (1) Solution: For a control volume enclosing the tank and the exit jet, we obtain 0=%(fpdv)+n’1w,, or: \P(:l—f=-mw,=—Cp, Chapter 3 * Integral Relations for a Control Volume 3-21 P t or: f d—p=—£fdt, or: ﬂ--exp[—gt] Ans. po p v (’ pﬂ v Now apply this formula to the given data. If po = 300 kPa and To = 100°C = 373°K, then Po = p/RT = (300.000)/[287(373)] = 2.80 kg/m". This establishes the constant “C": 3 . =Cp, =001 *&_ C(2.80 k-%) or C=0.00357 == for this hole. S m’ S The tank volume is v = (7/6)D’ = (7/6)(0.5 m)’ = 0.0654 m’. Then we require 0.00357 0.0654 p/pu-O.S-cxp[- t] if t=13s Ans.

P4.29 Consider a steady, two-dimensional, incompressible flow of a REWIONIAN fluid with the velocity field u = —2xy, v = y> — xZ, and w = 0. (a) Does this flow satisfy conservation of mass? (b) Find the pressure field p(x, y) if the pressure at point (x =0,y =0) is equal to pa. Solution: Evaluate and check the incompressible continuity equation: ﬂ+fi+ﬂ =0=-2y+2y+0=0 Yes! Ans. (a) ax dy 0z (b) Find the pressure gradients from the Navier-Stokes x- and y-relations: ( du du (?u) dp [ Fu Fu P u) plu—+v—tw—| =+ | —5+—+—5|, OF dx dy az dx t ax* dy- a7 J Pl=2x(=2y)+ (" - ¥ )(-2x)] = -%’i +u(0+0+0), or %’1 ==2p(xy + X') x X Chapter 4 +« Differential Relations for a Fluid Particle 4-19 and, similarly for the y-momentum relation, (a‘ v m] ap (v @y /Fv\ plu—+v—s+w + T+ Jx ay az pl=2x(=2x)+ (¥ = x)(2y)] = -?+ w(=2+4+2+0), or %= =2p(x°y+y) dy y The two gradients ¢ p/d xand d p/d y may be integrated to find p(x, y): 2.2 4 pP= f op d\’i\ Const ™ 21)( : +l;—} + f(v), thendif f erentiat. o = -2p(x2 _\-‘)+ﬂ = -2p(x2 V4 )'3 ), whence d—f = -2p_v3 ,oor: fly)= -B)A +C ay dy dy 2 Thus: ps-%(ZJt’zy2 +x4+y4)+C=pa at (x,y)=(0,0), or: C=p, Finally, the pressure field for this flow is given by p=p, - %p(szyz +x4+ y‘) Ans.(b)"

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