Click here to Purchase full Solution Manual at http://solutionmanuals.info 9–1. Crude oil flows through the 2-mm gap between the two fixed parallel plates due to a drop in pressure from A to A B B of 4 kPa. If the plates are 800 mm wide, determine the flow. 2 mm 0.5 m SOLUTION = 0 and U = 0. From Appendix A, m = 30.2 ( 10-3 ) N # s>m2 and dh Here, dx r = 880 kg>m3. a3b dp Q = - 12m dx 3 2 ( 10-3 ) m 4 3(0.8 m) - 4 ( 103 ) N>m2 = - £ § 12 3 30.2 ( 10-3 ) N # s>m2 4 0.5 m = 0.1413 ( 10-3 ) m3 >s = 0.141 ( 10-3 ) m3 >s Ans. To evaluate the maximum Reynolds number, we must first find the maximum velocity of the oil flow. a2 dp 3 2 ( 10-3 ) m 4 2 - 4 ( 103 ) N>m2 u max = - = - £ § = 0.132 m>s 8m dx 8 3 30.2 ( 10-3 ) N # s>m2 4 0.5 m Thus, the maximum Reynolds number is ru max a ( 880 kg>m3 )( 0.132 m>s ) 3 2 ( 10-3 ) m 4 Re = = = 7.719 6 1400(OK) m 30.2 ( 10-3 ) N # s>m2 Ans: 0.141 ( 10-3 ) m3 >s 930 Click here to Purchase full Solution Manual at http://solutionmanuals.info 9–2. Crude oil flows through the gap between the two fixed parallel plates due to a drop in pressure from A to B A B of 4 kPa. Determine the maximum velocity of the oil and 2 mm the shear stress on each plate. 0.5 m SOLUTION = 0 and U = 0. From Appendix A, m = 30.2 ( 10-3 ) N # s>m2 and dh Here, dx r = 880 kg>m3. a2 dp 3 2 ( 10-3 ) m 4 2 4 ( 103 ) N>m2 u max = - = - £ - § 8m dx 8 3 30.2 ( 10-3 ) m 4 0.5 m = 0.1324 m>s = 0.132 m>s Ans. Thus, the maximum Reynolds number is rUmax a ( 880 kg>m3 )( 0.1324 m>s ) 3 0.2 ( 10-3 ) m 4 Re = = = 7.719 6 1400 (OK) m 30.2 ( 10-3 ) N # s>m2 At the top and bottom plate, y = a = 2 ( 10-3 ) m, and y = 0, respectively. dp a 4 ( 103 ) N>m2 2 ( 10-3 ) m tt = ay - b = £ - § £ 2 ( 10-3 ) m - § dx 2 0.5 m 2 = - 8 Pa Ans. dp a 4 ( 103 ) N>m2 2 ( 10-3 ) m tb = ay - b = £ - § £0 - § dx 2 0.5 m 2 = 8 Pa Ans. Ans: u max = 0.132 m>s tt = - 8 Pa tb = 8 Pa 931 Click here to Purchase full Solution Manual at http://solutionmanuals.info 9–3. Air at T = 40° F flows with an average velocity of 1 — in. 15 ft>s past the charged plates of the air cleaner. The plates 8 are each 10 in. wide, and the gap between them is 1>8 in. If fully developed laminar flow develops, determine the A B pressure difference pB - pA between the inlet A and the outlet B. 24 in. SOLUTION From Appendix A, at T = 40° F, ra = 0.00247 slug>ft 3 and ma = 0.363 ( 10-6 ) lb # s>ft 2. The Reynolds number of the flow is 0.125 ( 0.00247 slug>ft 3 )( 15 ft>s ) a ft b raVa 12 Re = = ma 0.363 ( 10-6 ) lb # s>ft 2 = 1063 6 1400 (laminar flow) Here, the flow can be considered as horizontal flow caused by pressure difference between the inlet and outlet through two fixed parallel plates. Also the air is incom- pressible. Therefore, 2 0.125 2 a ft b a dp 12 pB - pA Vavg = - ; 15 ft>s = - c d 12ma dx -6 # 12 3 0.363 ( 10 ) lb s>ft 4 ( 24>12 ft ) 2 pB - pA = - 1.204 lb>ft 2 = - 1.20 lb>ft 2 Ans. The negative sign indicates that the pressure drops from the inlet to outlet. Ans: pB - pA = - 1.20 lb>ft 2 932 Click here to Purchase full Solution Manual at http://solutionmanuals.info *9–4. Glue is applied to the surface of the plastic strip, which has a width of 200 mm, by pulling the strip 10 mm/s 40 mm through the container. Determine the force F that must F be applied to the tape if the tape moves at 10 mm>s. Take 40 mm rg = 730 kg>m3 and mg = 0.860 N # s>m2. 300 mm SOLUTION The Reynolds number of the flow is τA ( 730 kg>m3 ) c 10 ( 10-3 ) m>s d (0.04 m) F rgVa τA Re = = mg 0.860 N # s>m2 (a) = 0.3395 Since Re62300, steady laminar flow occurs. Also, the glue is incompressible. Here, the flow is horizontal which is caused by the top moving plastic strip. Therefore, the shear stress acting along the surface of the plastic strip can be determined using Eq. 9–14. Umg c 10 ( 10-3 ) m>s d ( 0.860 N # s>m2 ) t = = = 0.215 N>m2 a 0.04 m Consider the horizontal equilibrium of the FBD of the plastic strip in Fig. a, + ΣFx = 0; F - 2tA = 0 S F = 2tA = 2 ( 0.215 N>m2 ) (0.3 m)(0.2 m) = 0.0258 N Ans. 933 Click here to Purchase full Solution Manual at http://solutionmanuals.info 9–5. The 20-kg uniform plate is released and slides down the inclined plane. If an oil film under its surface is 0.2 mm 0.75 m thick, determine the terminal velocity of the plate along the plane. The plate has a width of 0.5 m. Take ro = 880 kg>m3 and mo = 0.0670 N # s>m2. 15 SOLUTION Here, the terminal velocity is constant. Thus, the plate is in equilibrium. Referring to the FBD of the plate shown in Fig. a, 20(9.81)N + ΣFx = 0; Fv - [20(9.81) N] sin 15° = 0 Fv = 50.78 N S 15° Thus, the shear stress acting on the bottom plate’s surface is Fr Fv 50.78 N tp = = = 135.41 N>m2 Ap (0.75 m)(0.5 m) We will assume that the steady laminar flow occurs and the oil is incompressible. N dp dh (a) Here = 0, = - sin u and U = u t. Then dx dx ut u0 a t = - g ay - b sin u a 2 For this case u = 15° and a = 0.2 ( 10-3 ) m. At y = a = 0.2 ( 10-3 ) m, t = tp = 135.41 N>m2. Substitute this data into the above equation u t ( 0.0670 N # s>m2 ) 0.2 ( 10-3 ) m 135.41 N>m2 = - ( 880 kg>m3 )( 9.81 m>s2 ) c 0.2 ( 10-3 ) m - d (sin 15°) 0.2 ( 10 -3 )m 2 u t = 0.4049 m>s = 0.405 m>s Ans. The Reynolds number is ru t a ( 880 kg>m3 )( 0.4049 m>s ) 3 0.2 ( 10-3 ) m 4 Re = = m0 0.0670 N # s>m2 = 1.06 6 1400 (laminar flow) Ans: 0.405 m>s 934 Click here to Purchase full Solution Manual at http://solutionmanuals.info 9–6. Using pins, the plug is attached to the cylinder such that there is a gap between the plug and the walls of 0.2 mm. If the pressure within the oil contained in the cylinder is 4 kPa, determine the initial volumetric flow of oil up the sides of the plug. Take ro = 880 kg>m3 and mo = 30.5 1 10-3 2 N # s>m2. 100 mm Assume the flow is similar to that between parallel plates since the gap size is very much smaller than the radius of the plug. 50 mm SOLUTION The flow is assumed to be steady laminar, and the oil is incompressible. Positive x axis is directed in the direction of flow which is vertically upwards. 0 r2 - r1 h2 - h1 (p + gh) = + ga b 0x L L 0 - 4 ( 103 ) N>m2 0.05 m - 0 = + ( 880 kg>m3 )( 9.81 m>s2 ) a b 0.05 m 0.05 m N>m2 = - 71.367 ( 103 ) m Substitute this value and U = 0 into 1 N>m2 u = 0 - J - 71.367 ( 103 ) R 3 0.2 ( 10-3 ) y - y2 4 2 3 30.5 ( 10-3 ) N # s>m2 4 m u = 1.1700 ( 106 ) 3 0.2 ( 10-3 ) y - y2 4 Here b = 2p(0.05 m) = 0.1p m. Then the flow rate is LA Q = udA 0.2(10-3) m L0 = 1.1700 ( 106 ) 3 0.2 ( 10-3 ) y - y2 4 (0.1p)dy y3 0.2(10 ) m -3 = 0.36755 ( 106 ) J0.1 ( 10-3 ) y2 - R` 3 0 = 0.4901 ( 10-6 ) m3 >s = 0.490 ( 10-6 ) m3 >s Ans. The average velocity is Q 0.4901 ( 10-6 ) m3 >s V = = = 7.800 ( 10-3 ) m>s A (0.1p m) 3 0.2 ( 10-3 ) m 4 The Reynolds number is rVa ( 880 kg>m2 ) 3 7.800 ( 10-3 ) m>s 4 3 0.2 ( 10-3 ) m 4 Re = = m0 30.5 ( 10-3 ) N # s>m2 = 0.0450 6 1400 (laminar flow) Ans: 0.490 ( 10-6 ) m3 >s 935 9–7. The boy has a mass of 50 kg and attempts to slide down the inclined plane. If a 0.3-mm-thick oil surface develops between his shoes and the surface, determine his terminal velocity down the incline. Both of his shoes have a total contact area of 0.0165 m2. Take ro = 900 kg>m3 and mo = 0.0638 N # s>m2. 2 SOLUTION Since the boy is required to move with terminal velocity (constant), a = 0. Referring to the free-body diagram of the boy in Fig. a, 50(9.81) N a= 0 + ΣFx = max; 350(9.81)N4 sin 2° - F = 0 d 2° x F = 17.118 N The shear stress on the oil layer in contact with the boy’s shoes is F 17.118 N t = = = 1037.47 Pa A 0.0165 m2 F dp dh N Here, = 0 and = - sin 2°. dx dx (a) Um d a t = + c (p + gh) d ay - b a dx 2 N U ( 0.0638 N # s>m2 ) 0.3 ( 10-3 ) m 1037.47 = + 30 + ( 900 kg>m3 )( 9.81 m>s2 )( -sin 2° ) 4 c 0.3 ( 10-3 ) m - d m2 0.3 ( 10-3 ) m 2 U = 4.879 m>s = 4.88 m>s Ans. Thus, the maximum Reynolds number is rUa ( 900 kg>m3 ) (4.879 m>s) 3 0.3 ( 10-3 ) m 4 Re = = = 20.65 6 1400 (OK) m 0.0638 N # s>m2 Click here to Purchase full Solution Manual at http://solutionmanuals.info Ans: 4.88 m>s 936 *9–8. The 2.5-lb plate, which is 8 in. wide, is placed on the 12 in. 3° inclined plane and released. If its terminal velocity is 0.2 ft>s, determine the approximate thickness of oil 3 underneath the plate. Take ro = 1.71 slug>ft 3 and mo = 0.632 1 10-3 2 lb # s>ft 2. SOLUTION + ΣF = max; (2.5 lb) sin 30° - F = 0 S 2.5 lb F = 0.1308 lb a=0 Thus, the shear stress on the oil layer in contact with the bottom surface of the plate 3° x (y = a) is P 0.1308 lb lb F t = = = 0.1963 2 A 12 8 ft a ft ba ft b 12 12 N dp dh (a) Here, = 0 and = - sin 3°. dx dx Um d a t = + c (p + gh) d ay - b a dx 2 lb ( 0.2 ft>s )( 0.632 ( 10-3 ) lb # s>ft 2 ) a 0.1963 = + 30 + ( 1.71 slug>ft 3 )( 32.2 ft>s2 ) ( - sin 3°) 4 aa - b ft 2 a 2 1.4409a2 + 0.1963a - 0.1264 ( 10-3 ) = 0 Solving for the positive root, a = 0.6410 ( 10-3 ) ft = 7.69 ( 10-3 ) in. Ans. Thus, the maximum Reynolds number is rUa ( 1.71 slug>ft 3 )( 0.2 ft>s ) 3 0.6410 ( 10-3 ) ft 4 Re = = = 0.3469 6 1400 (OK) m 0.632 ( 10-3 ) lb # s>ft Click here to Purchase full Solution Manual at http://solutionmanuals.info 937 9–9. The water tank has a rectangular crack on its side having a width of 100 mm and an average opening of 100 mm 0.1 mm. If laminar flow occurs through the crack, determine the volumetric flow of water through the crack. The water is at a temperature of T = 20° C. 2m 100 mm SOLUTION Assuming that steady laminar flow occurs, and the water is incompressible. The flow can be considered as horizontal flow through two stationary parallel plates driven by a pressure gradient. The pressure of water at the inlet of the crack is pin = rwgh = ( 998.3 kg>m3 )( 9.81 m>s2 ) (2 m) = 19.587 ( 103 ) N>m2 At the outlet, pout = patm = 0. Thus, the pressure gradient is dp pout - pin 0 - 19.587 ( 103 ) N>m2 N>m2 = = = - 195.87 ( 103 ) dx L 0.1 m m a3b dp 3 0.1 ( 10-3 ) m 4 3(0.1 m) N>m2 Q = - = - J -195.87 ( 10 -3 ) R 12m dx 12 3 1.00 ( 10-3 ) N # s>m2 4 m = 1.632 ( 10-6 ) m3 >s = 1.63 ( 10-6 ) m3 >s Ans. The average velocity is Q 1.632 ( 10-6 ) m3 >s V = = = 0.1632 m>s A 3 0.1 ( 10-3 ) m 4 (0.1 m) The Reynolds number is rVa ( 998.3 kg>m3 )( 0.1632 m>s )( 0.1 ( 10-3 ) m ) Re = = m 1.00 ( 10-3 ) N # s>m2 = 16.29 6 1400 (laminar flow) Click here to Purchase full Solution Manual at http://solutionmanuals.info Ans: 1.63 ( 10-6 ) m3 >s 938 9–10. A solar water heater consists of two flat plates that rest on the roof. Water enters at A and exits at B. If the A pressure drop from A to B is 60 Pa, determine the largest 3m gap a between the plates so that the flow remains laminar. For the calculation, assume the water has an average temperature of 40° C. a 7.5 B SOLUTION From Appendix A at T = 40° C, rw = 992.3 kg>m3 and mw = 0.659 ( 10-3 ) N # s>m2. It is required that steady laminar flow occurs. Also, the water is incompressible. Here 0 60 N>m2 0 - (3 m) sin 7.5° (p + gh) = - + ( 992.3 kg>m3 )( 9.81 m>s2 ) c d 0x 3m 3m N>m2 = - 1290.60 m 1 0 u = - c (p + gh) d ( ay - y2 ) 2mw 0x 1 N>m2 u = a - 1290.60 b ( ay - y2 ) 2 3 0.659 ( 10-3 ) N # s>m2 4 m u = 0.9792 ( 106 )( ay - y2 ) The flow rate can be determined from a L L0 Q = udA = 0.9792 ( 106 ) ( ay - y2 ) b dy = 0.1632 ( 106 ) a3b A The average velocity is Q 0.1632 ( 106 ) a3b V = = = 0.1632 ( 106 ) a2 A ab To satisfy the condition of laminar flow, we require Re = 1400 rwVa = 1400 mw ( 992.3 kg>m3 ) 3 0.1632 ( 106 ) a2 4 a = 1400 0.659 ( 10-3 ) N # s>m2 a = 1.786 ( 10-3 ) m = 1.79 mm Ans. Click here to Purchase full Solution Manual at http://solutionmanuals.info Ans: 1.79 mm 939
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