Hw 1 solutions - Llo According to information found in an...

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Unformatted text preview: Llo According to information found in an old hydraulics book. the energy loss per unit weight of fluid flowing through a nozzle connected to a hose can be estimated by the formula II = (0.04 to O.09)(D/d)“V3/2g where h is the energy loss per unit weight. D the hose diameter. d the nozzle tip diameter. V the fluid velocity in the hose, and g the acceleration of gravity. Do you think this equation is valid in any system of units? Explain. %= {0.054 169 0.0?) (SW; [4% m) Hagan? [L J =' [0.04 4-0 0.07] [L] 5/5“ and. 4cm 1;: 1h: 2344,1110), musl hat/e the Mme whens/05:5, fire 6&fi5/vnz’ firm {MI/750.0?) Mus/- be c/immsm'a/es. Thus] '77:: 6344.290); /3 a. jrnrrl/ bomogefle’avs 61441519), ‘fiuzi‘ :3 Valid /h an}; syn/rm 4 "HI-75¢ I. 6.3 A 25-mm-diameter shaft is pulled through a cylindrical hearing as shown in Fig. Pl.63 The lubricant that fills the 0.3-mm gap between the shaft and bearing is an oil having a kinematic viscosity of 8.0 x 10" m’ls and a specific gravity of 0.91. Determine the force P required to pull the shaft at a velocity of 3 m/s. Assume the ve- locity distribution in the gap is linear. FIGURE “.63 TA Z F; :0 ~— w- <- w- 77M: ’ [0:234 mm ,4 a no it (shut my). ,;. bum, ) = r01 4"" rue/“H, or 55.3) _ v T‘/‘ (34} via!th -/1_ So fhli P: 6“ who!) since fl 37/ 3 7/ {Say/3,095.“: ), P: (a 0xl0-”—"2)/0.4/ 2‘ Io: )/3-§-"}/77)/0, 02:3”)(05m) J (0. 00mm) 1' 28AM 1,71 A l2-in.-diametcr circular plate is placed over a fixed Rom.“ mm. bottom plate with a 0.1-in. gap between the two plates filled I" with glycerin as shown in Fig. PL?!- Determ‘me the torque tequimd to rotate the circular plate slowly at 2 rpm. Assume that the velocity distribution in the gap is linear and that the shear suess on the edge of the rotating plate is negligible. I FIGURE P1.71 75ryu¢,d°~7” due 1‘0 allurm'y firm: on P/at‘c I! ejun/ +0 ‘C' dv 47': Ir fact & when 44- zrrdr. Thug, R d 07': P T erdr R . an” 62.3” *1 Z. 0/}. JfIGIa ‘6}1'!’ on belle»: a‘rPl‘k O . —*V= t—w 5m 7y». g, m A .1 13%, been velouih, dtsfn'bufm'n (Ste-fijure) L I T: 4-9 = y.- : La—D £5— du s 5 77m: K v Veloé+ d' +‘L l' l l 9 I: h u [OM 7r 4: a s ZZL~{£ ”J’: 16‘— » Jr J t ) 4!”! 101.77! fire drug:- 70;?" 1f Md / If: i 7- 27’ (0.03/3 a": ‘77 731/){2445 '2 It) +200) = 0.0772 JIM; L78 L75 Determine the speed of sound at 20 °C in (a) air. (b) helium. and (e) natural gas. Express your answer in m/s. C = Vaé/ET (52. ’40) 114/11 T: 07m +373 = 293k : (4) 5r 1”, c =l’fl#o}(.2?£,1;;%<)éfik) = 34,3 5;: L ' {V J I )5» Ae/mm, C (A66)(Jo77‘;:K){243/¢) = [0/0 5'; (c) Far natuml as = A. J , C Wal)(5/s.?£;Z/.<)/2¢ak) : 444,2"— S 1.103 (See Fluids in the News article titled “Walking on water." Section 1.9.) (a) The water strider bug shown in Fig. H.103 is supported on the surface of a pond by surface tension acting along the interface between the water and the bug's legs. Determine the minimum lenth of this interface needed to support the bug. As- sume the bug weighs 10“ N and the surface tension force acts vertically upwards. (b) Repeat pan (a) if surface tension were to support a person weighing 750 N. IFIGURE P1.103 0-1 For eiurltlflttun aw: ace .. 71.) to.) “lg _ to "N , 1- 0. ‘ 7.?l’x'o.£_’y- all)“ weighi: ! [m 0”” Star-lace JWS/on = '3‘)“; rm flrv ienjflt of min-19m ...
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Hw 1 solutions - Llo According to information found in an...

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