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*P1.55 The device in Fig. P1.54 is called a rotating disk viscometer  . Suppose that  R  5cm and h  1mm. If the torquerequired to rotate the disk at 900 r/min is 0.537 N  m,what is the viscosity of the fluid? If the uncertainty in eachparameter (  M  ,  R , h ,  ) is  1 percent,what is the overalluncertainty in the viscosity? *P1.56 The device in Fig. P1.56 is called a cone-plate viscometer  . The angle of the cone is very small,so that sin      ,and the gap is filled with the test liquid. The torque  M  to rotate the cone at a rate  is measured. Assuming a lin-ear velocity profile in the fluid film,derive an expressionfor fluid viscosity  as a function of (  M  ,  R ,  ,    ).      8 r  L 04 Q   p  Pipe end effects are neglected . Suppose our capillaryhas r  0  2mm and  L  25cm. The following flow rateand pressure drop data are obtained for a certain fluid: Q ,m 3  /h0.360.721.081.441.80   p ,kPa15931847712741851 What is the viscosity of the fluid?  Note: Only the first threepoints give the proper viscosity. What is peculiar about thelast two points,which were measured accurately? P1.59 A solid cylinder of diameter  D ,length  L ,and density   s falls due to gravity inside a tube of diameter  D 0 . The clear-ance,  D 0   D   D ,is filled with fluid of density   andviscosity  . Neglect the air above and below the cylinder.Derive a formula for the terminal fall velocity of the cylin-der. Apply your formula to the case of a steel cylinder,  D  2 cm,  D 0  2.04 cm,L  15 cm,with a film of SAE30 oil at 20 ° C. P1.60 For Prob. 1.52 suppose that P  0.1 hp when V   6 ft/s,  L  4.5 ft, b  22 in,and h  7/8 in. Estimate the vis-cosity of the oil,in kg/(m  s). If the uncertainty in eachparameter ( P ,  L , b , h , V  ) is  1 percent,what is the over-all uncertainty in the viscosity? *P1.61 An air-hockey puck has a mass of 50g and is 9cm in di-ameter. When placed on the air table,a 20 ° C air film,of 0.12-mm thickness,forms under the puck. The puck isstruck with an initial velocity of 10m/s. Assuming a lin-ear velocity distribution in the air film,how long will ittake the puck to ( a )slow down to 1m/s and ( b )stop com-pletely? Also,( c )how far along this extremely long tablewill the puck have traveled for condition ( a )? P1.62 The hydrogen bubbles which produced the velocity pro-files in Fig. 1.13 are quite small,  D  0.01 mm. If the hy-drogen-water interface is comparable to air-water and thewater temperature is 30 ° C estimate the excess pressurewithin the bubble. P1.63 Derive Eq.(1.37) by making a force balance on the fluidinterface in Fig. 1.9 c . P1.64 At 60 ° C the surface tension of mercury and water is 0.47and 0.0662 N/m,respectively. What capillary heightchanges will occur in these two fluids when they are incontact with air in a clean glass tube of diameter 0.4mm? P1.65 The system in Fig. P1.65 is used to calculate the pressure  p 1 in the tank by measuring the 15-cm height of liquid inthe 1-mm-diameter tube. The fluid is at 60 ° C (see Prob.1.64). Calculate the true fluid height in the tube and thepercent error due to capillarity if the fluid is ( a )water and( b )mercury.Problems  51  RR Clearance  h Oil Ω P1.54 Ω Fluid  R θ  θ *P1.57 For the cone-plate viscometer of Fig. P1.56,suppose that  R  6cm and    3 ° . If the torque required to rotate thecone at 600 r/min is 0.157 N  m,what is the viscosity of the fluid? If the uncertainty in each parameter (  M  ,  R ,   ,   ) is  1 percent,what is the overall uncertainty in the vis-cosity? *P1.58 The laminar-pipe-flow example of Prob. 1.12 can be usedto design a capillary viscometer . If Q is the volumeflow rate,  L is the pipe length,and   p is the pressure dropfrom entrance to exit,the theory of Chap. 6 yields a for-mula for viscosity: P1.56  30   Solutions Manual   ã   Fluid Mechanics, Fifth Edition   πµω  µω π θ θ  ==   o r43o0 ror:Torque M2rdrhsin2hsin   We may compute the cone’s slowing down from the angular momentum relation: 2ooo d3MI,where I(cone)mr,mcone massdt10 ω  =−==   Separating the variables, we may integrate: o wt4oo0 rddt,or: .2hIsin  Ans ω  πµ ω ω θ    =−         µωω  =− 2oo 5rtexp3mhsin   1.54* A disk of radius R rotates at angular velocity Ω  inside an oil container of viscosity  µ  , as in Fig. P1.54. Assuming a linear velocity profile and neglecting shear on the outer disk edges, derive an expres-sion for the viscous torque on the disk. Fig. P1.54 Solution: At any r ≤  R, the viscous shear τ    ≈    µ  Ω r/h on both sides of the disk. Thus, wR30 rd(torque)dM2rdA2r2rdr,hor:M4rdrh  Ans  µ τ  µ π  Ω===πΩ==.   4 Rh   µ   1.55  Apply the rotating-disk viscometer   of Prob. 1.54, to the particular case R =  5 cm, h =  1 mm, rotation rate 900 rev/min, measured torque M =  0.537 N·m. What is the fluid viscosity? If each parameter (M,R,h, Ω ) has uncertainty of ± 1%, what is the overall uncertainty of the measured viscosity? Solution: The analytical formula M =   πµ  Ω R 4  /h was derived in Prob. 1.54. Convert the rotation rate to rad/s: Ω =  (900 rev/min)(2 π   rad/rev ÷ 60 s/min) =  94.25 rad/s. Then, 44 hM(0.001 m)(0.537 Nm)kg or msR(94.25 rad/s)(0.05 m)  Ans  µ π π  ⋅    ===.    ⋅Ω 2 Ns0.29m      Chapter 1   ã   Introduction 31   For uncertainty, looking at the formula for  µ  , we have first powers in h, M, and Ω  and a fourth power in R. The overall uncertainty estimate [see Eq. (1.44) and Ref. 31] would be 1/22222hMR22221/2 SSSS(4S)[(0.01)(0.01)(0.01){4(0.01)}]0.044or:  Ans  µ  Ω   ≈+++   ≈+++≈. 4.4%   The uncertainty is dominated by the 4% error due to radius measurement. We might report the measured viscosity as  µ    ≈  0.29 ±  4.4% kg/m·s or 0.29 ±  0.013 kg/m·s. 1.56*  For the cone-plate viscometer in Fig. P1.56, the angle is very small, and the gap is filled with test liquid  µ  . Assuming a linear velocity profile, derive a formula for the viscosity  µ   in terms of the torque M and cone parameters. Fig. P1.56 Solution: For any radius r ≤  R, the liquid gap is h =  r tan θ  . Then w rdrd(Torque)dMdAr2rr,orrtancos τ µ π θ θ    Ω    ===        R320 22RMrdr,or:.sin3sin ns π µ π µ  µ θ θ  ΩΩ===   3 3Msin2R   1.57  Apply the cone-plate viscometer of Prob. 1.56 above to the special case R =  6 cm, θ    =  3 ° , M =  0.157 N   ⋅   m, and a rotation rate of 600 rev/min. What is the fluid viscosity? If each parameter (M,R, Ω , θ  ) has an uncertainty of ± 1%, what is the uncertainty of  µ  ? Solution: We derived a suitable linear-velocity-profile formula in Prob. 1.56. Convert the rotation rate to rad/s: Ω = (600 rev/min)(2 π   rad/rev ÷ 60 s/min) =  62.83 rad/s. Then, 33 3Msin3(0.157 Nm)sin(3)kg or .ms2R2(62.83 rad/s)(0.06 m)  Ans θ  µ π π  ⋅°    ===    ⋅Ω 2 Ns0.29m   For uncertainty, looking at the formula for  µ  , we have first powers in θ  , M, and Ω  and a third power in R. The overall uncertainty estimate [see Eq. (1.44) and Ref. 31] would be 1/22222MR22221/2 SSSS(3S)[(0.01)(0.01)(0.01){3(0.01)}]0.035,or:  Ans  µ θ  Ω   =+++   ≈+++=.. 35 %   The uncertainty is dominated by the 3% error due to radius measurement. We might report the measured viscosity as  µ    ≈  0.29 ± 3.5% kg/m·s or 0.29 ± 0.01 kg/m·s.  with manometer fluid   m . One side of the manometer is opento the air,while the other is connected to new tubing whichextends to pressure measurement location 1,some height  H  higher in elevation than the surface of the manometer liquid.For consistency,let   a be the density of the air in the room,   t  be the density of the gas inside the tube,    m be the den-sity of the manometer liquid,and h be the height differencebetween the two sides of the manometer. See Fig. P2.38.( a ) Find an expression for the gage pressure at the mea-surement point.  Note: When calculating gage pressure,usethe local atmospheric pressure at the elevation of the mea-surement point. You may assume that h    H  ; i.e.,assumethe gas in the entire left side of the manometer is of den-sity   t  . ( b ) Write an expression for the error caused by as-suming that the gas inside the tubing has the same densityas that of the surrounding air. ( c ) How much error (in Pa)is caused by ignoring this density difference for the fol-lowing conditions:    m   860 kg/m 3 ,    a   1.20 kg/m 3 ,   t    1.50 kg/m 3 ,  H    1.32 m,and h   0.58 cm? ( d  ) Canyou think of a simple way to avoid this error?is very large. If the inclined arm is fitted with graduations1 in apart,what should the angle   be if each graduationcorresponds to 1 lbf/ft 2 gage pressure for  p  A ?Problems  107P2.38 An interesting article appeared in the  AIAA Journal (vol. 30,no. 1,January 1992,pp. 279 – 280). The authors explain thatthe air inside fresh plastic tubing can be up to 25 percentmore dense than that of the surroundings,due to outgassingor other contaminants introduced at the time of manufacture.Most researchers,however,assume that the tubing is filledwith room air at standard air density,which can lead to sig-nificant errors when using this kind of tubing to measurepressures. To illustrate this,consider a U-tube manometer h (1)(2)30  2 m P2.35P2.39 An 8-cm-diameter piston compresses manometer oil intoan inclined 7-mm-diameter tube,as shown in Fig. P2.39.When a weight W  is added to the top of the piston,the oilrises an additional distance of 10 cm up the tube,as shown.How large is the weight,in N? P2.40 A pump slowly introduces mercury into the bottom of theclosed tank in Fig. P2.40. At the instant shown,the airpressure  p  B   80 kPa. The pump stops when the air pres-sure rises to 110 kPa. All fluids remain at 20 ° C. What willbe the manometer reading h at that time,in cm,if it is con-nected to standard sea-level ambient air  p atm ? 50 cm50 cmOilSG = 0.8WaterSG = 1.0  L  P2.36 1 inReservoir θ   D = 516 in  p  A P2.37 h H  1U-tube manometer   m  t (tubing gas)  a (air)  p a at location 1  p 1 P2.38

Jul 22, 2017

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