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Unformatted text preview: University of California Davis Department of Mechanical and Aerospace Engineering ENG 103 Fluid Mechanics Fall 2011 Professor Kennedy Chapter 1 25 Introduction Homework 1 P1.41 An aluminum cylinder weighing 30 N, 6 cm in diameter and 40 cm long, is falling concentrically through a long vertical sleeve of diameter 6.04 cm. The clearance is filled with SAE 50 oil at 20 C. Estimate the terminal (zero acceleration) fall velocity. Neglect air drag and assume a linear velocity distribution in the oil. [HINT: You are given diameters, not radii.] Solution: From Table A.3 for SAE 50 oil, = 0.86 kg/m-s. The clearance is the difference in radii: 3.02 – 3.0 cm = 0.02 cm = 0.0002 m. At terminal velocity, the cylinder weight must balance the viscous drag on the cylinder surface: V )( DL) , where C clearance rsleeve rcylinder C kg V or : 30 N [0.86 )( )] (0.06 m)(0.40 m) m s 0.0002 m Solve for V 0.0925 m / s Ans. ________________________________________________________________________ W 1.42 wall Awall ( Some experimental values of T, °K: , kg/m s: 200 1.50E–5 of helium at 1 atm are as follows: 400 2.43E–5 600 3.20E–5 800 3.88E–5 1000 4.50E–5 1200 5.08E–5 Fit these values to either (a) a power-law, or (b) a Sutherland law, Eq. (1.30a,b). Solution: (a) The power-law is straightforward: put the values of and T into, say, an Excel graph, take logarithms, plot them, and make a linear curve-fit. The result is: Power-law curve-fit: He 1.505E 5 TK 200 K 0.68 Ans. (a) The accuracy is less than 1%. (b) For the Sutherland fit, we can emulate Prob. 1.41 and perform the least-squares summation, E [ i – o (T i /200)1.5(200 S)/(T i S)]2 and 34 Solutions Manual P1.55 Fluid Mechanics, Sixth Edition A block of weight W is being pulled W over a table by another weight W o , as shown in h Fig. P1.55. Find an algebraic formula for the steady velocity U of the block if it slides on an oil film of thickness h and viscosity . The block Wo bottom area A is in contact with the oil. Neglect the cord weight and the pulley friction. Fig. P1.55 Solution: This problem is a lot easier to solve than to set up and sketch. For steady motion, Fx,block 0 A U CW hapter 2 ( Pressure Distribution in a Fluid ) A Wo o h 81 2 W1 Hence P SolveAsmall U poil for (40744) o h 222 lbf . Ans 4A 12 there is no acceleration, and the falling weightP/16 222/16 14 lbfesistance of the oil film: Therefore the handle force required is F balances the viscous r Ans. The block weight W has no effect on steady horizontal motion except to smush the oil film. 2.21 In Fig. P2.21 all fluids are at 20 C. 1Gage A or the350 kPa absoviscometer in .56* F reads cone-plate lute. Determine Fig. P1.56, the angle is very small, and the (a) the height h in cm; and (b) the reading gap is filled with test liquid . Assuming a of gage B in kPa absolute. linear velocity profile, derive a formula for the viscosity in terms of the torque M Solution: Apply Fig. P1.56 and cone parameters.the hydrostatic formula from the air to gage A: Solution: For any radius r R, the liquid gap is h r tan . Thenig. P2.21 F p A pair h r dr 180000 (9790)h r d(Torque) dM dA w133100(0.8) 350000 Pa,r, or 2r r tan cos Ans. (a) R Solve for h 3 6.49 m 2 2 R 3M sin M r 2 dr , or: Ans. 3sin Then, with h known,sin can evaluate the pressure at gage B: R 3 we 0 2 pB 180000 + 9790(6.49 0.80) = 251000 Pa 251 kPa Ans. (b) 2.22 The fuel gage for an auto gas tank reads proportional to the bottom gage pressure as in Fig. P2.22. If the tank accidentally contains 2 cm of water plus gasoline, how many centimeters “h” of air remain when the gage reads “full” in error? Fig. P2.22 2.45 Determine the gage pressure at point A in Fig. P2.45, in pascals. Is it higher or lower than Patmosphere ? Solution: Take 9790 N m3 for water and 133100 N m3 for mercury. Write the hydrostatic formula between the atmosphere and point A: patm (0.85)(9790)(0.4 m) (133100)(0.15 m) (12)(0.30 m) (9790)(0.45 m) p A , Chapter 2 (10000)(15) (288.2h 2 )[(h/2) csc 60 MB 109 Pressure Distribution in a Fluid (h/6) csc 60 ] 4898(7.5cos60 ) 0, 3 F or: 110.9h 150000 18369,ig.hP2.45 (131631/110.9)1/3 10.6 ft or: p A patm 12200 Pa 12200 Pa (vacuum) Ans. Ans. 2.63 The tank in Fig. P2.63 has a 4-cmdiameter plug which will pop out if the hydrostatic force on it reaches 25 N. For 20 C fluids, what will be the reading h on the manometer when this happens? Solution: The water depth when the plug pops out is F 25 N h CG A or h CG (9790)h CG (0.04)2 4 Fig. P2.63 2.032 m It makes little numerical difference, but the mercury-water interface is a little deeper than this, by the amount (0.02 sin 50 ) of plug-depth, plus 2 cm of tube length. Thus patm (9790)(2.032 0.02 sin 50 or: h 0.152 m 0.02) (133100)h Ans. patm , P2.97 The contractor ran out of gunite mixture and finished the deep corner, of a 5-m-wide swimming pool, with a quarter-circle 2m water piece of PVC pipe, labeled AB in A Fig. P2.97. Compute the (a) horizontal and 1m Fig. P2.97 B (b) vertical water forces on the curved panel AB. Solution: For water take = 9790 N/m3. (a) The horizontal force relates to the vertical projection of the curved panel AB: FH , AB hCG Aprojected N (9790 m3 )(2.5 m)[(1m)(5m)] 122, 000 N Ans.(a) (b) The vertical force is the weight of water above panel AB: FV (9790 168 N m 3 )[(2m)(1m) Solutions Manual 4 (1 m) 2 ](5 m) 136, 000 N Ans.(b) Fluid Mechanics, Fifth Edition 2.147 The tank of water in Fig. P2.147 accelerates uniformly by rolling without friction down the 30 inclined plane. What is the angle of the free surface? Can you explain this interesting result? Solution: If frictionless, F W sin ma along the incline and thus a g sin 30 0.5g. Thus tan ax g az Fig. P2.147 0.5g cos30 ; solve for g 0.5g sin 30 30 ! Ans. The free surface aligns itself exactly parallel with the 30 incline. P2.148 A child is holding a string onto which is attached a helium-filled balloon. (a) The child is standing still and suddenly accelerates forward. In a frame of reference moving with the child, which way will the balloon tilt, forward or backward? Explain. (b) The child is now sitting in a car that is stopped at a red light. The helium-filled balloon is not in contact with any part of the car (seats, ceiling, etc.) but is held in place by the string, which is held by the child. All the windows in the car are closed. When the traffic light turns green, the car accelerates forward. In a frame of reference moving with the car and child, which way will the balloon tilt, forward or backward? Explain. (c) Purchase or borrow a helium-filled balloon. Conduct a scientific experiment to see if Q. 8 ...
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This note was uploaded on 12/18/2011 for the course ENG 103 taught by Professor Chattot during the Fall '08 term at UC Davis.

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