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- 1. SOLUTIONS TO THE PROBLEMS in TRANSPORT PHEOMENA Second Edition (2002) by R. Byron Bird Warren E. Stewart Edwin N. Lightfoot Department of Chemical Engineering University of Wisconsin Madison, Wisconsin 53706 USA This solutions manual has been prepared by the authors of the textbook for use by professors teaching courses in transport phenomena. It contains the solutions to 500 of the unsolved problems in the textbook. No part of this material may be reproduced in any form, electronic, mechanical, photocopying, recording, scanning, or otherwise. JOHN WILEY 8: SONS, Inc. New York, New York
- 2. 1A.1 Estimation of dense-gas víscosíty. a. Table E.1 gives Tc = 126.2 K, pc = 33.5 atm, and , ac = 180 X 10's g/ cm-s for Ng. The reduced conditions for the viscosity estimation are then: Pr = p/ pc = (1000 + 14.7)/33.5 x 14.7 = 2.06 T, = T/ Tc = (27315 + (68 - 32)/1.8)/126.2 = 2.32 At this reduced state, Fig. 1.3-1 gives u, = 1.15. Hence, the predicted viscosity is ; z = ;tr/ Mc = 1.15 X 180 >< 10"” = 2.07>< 10"* g/ cm~s. This result is then converted into the requested units by use of Table F.3-4: p = 2.07 x 10'** x 6.7197 x 10'? = 1.4 x 10'* lbm/ ft-s
- 3. 1A.2 Estímation of the viscosity of methyl fluoríde. a. . CH3F has M = 16.04-1_:008+19.00 = 34.03 g/ g-mole, Tc = 4.55+273.15 = 277.70 K, pc = 58.0 atm, and V. , = 34.03/0.300 = 113.4 cm3/g-mole. The critical viscosity is then estimated as [lc = 61.6(34.03 x 277.70)1/2(113.4)-2/3 = 255.6 micropoise from Eq. 1.3-1a, and u. ; = 7.70(34.03)1/”(58.0)? /3(277.7)'1/° = 263.5 micropoise from Eq. 1.3-1b. The reduced conditions for the viscosity estimate are T, = (370 + 273.15) / 277.70 = 2.32, p, = 120/58.0 = 2.07, and the predicted , ur from Fig. 1.3-1 is 1.1. The resulting predicted viscosity is y = ;um =1.1 x 255.6 X 10-6 = 2.8 X 10"* g/ cm-s via Eq.1.3-1a, or 1.1 x 263.5 x 10'** = 2.9 >< 10'4g/ cm~s via Eq.1.3-1b.
- 4. 1A.3 Computation of the viscosities of gases at low density. Equation 1.4-14, With molecular parameters from Table E.1 and collision integrals from Table E.2, gives the following results: For 02; M = 32.00, a = 3.433Â, e/ K = 113 K. Then at 20°C, kT/ e = 29315/113 = 2.594 and 9,, : 1.086. Equation 1.4-14 then gives , /3700 x”2:93.15 (3.433)2 x 1.086 z: 2.02 x 10-4 g/ cm-s = 2.02 x 10-5 Pa-s = 2.02 X 10-2 mPa-s. p, = 2.6693 x 10-5 The reported value in Table 1.1-S is 2.04 X 10'2 mPa›s. For Ng: M = 28.01, a = 3.667Â, e/ K = 99.8 K. Then at 20°C, kT/ e = 293.15/99.8 = 2.937 and Q” = 1.0447. Equation 1.4-14 then gives , /2801 x”293.15 (35672 x 1.0447 = 1.72 >< 10-** g/ cm-s = 1.72 x 10-5 Pa-s = 1.72 x 10-2 mPa-s. [1 = 2.6693 x 10-5 The reported value in Table 1.1-3 is 1.75 X 10'? mPa-s. For CH4, M = 16.04, a z 3.7s0Â, e/ K = 154 K. Then at 20°C, kT/ s = 29315/154 = 1.904 and 9,, = 1.197. Equation 1.4-14 then gives , /1”'6'. o4 x"2“9“3.15 (3.7so)2 x 1.197 = 1.07 x 10-4 g/ cm-s = 1.07 x 10-5 Pa-s = 1.07 X 10-2 mPa-s. 11 = 2.6693 >< 10-5 The reported value in Table 1.1-3 is 1.09 X 10"”, mPa-s. l-B
- 5. 1A.4 Gas-mixture viscosities at low density. The data for this problem are as follows: Component M p, poise X 106 1(H2) 2.016 88.4 2(CCl2F2) 120.92 124.0 Insertion of these data into Eq. 1.4-16 gives the foloowing coefñcients for mixtures of H2 and Freon-12 at this temperature: T11=T22=L0 1 2.016 -1/2 88.4 V” 120.92 V* 2 <I>12=- 1+1- 1+ - : - , /§ 120.92 124.0 2.016 =3.934 q) _i 1+120.92 “V” 1+ 124.0 V” 2.016 1/4 2 2* ' «ã 2.016 88.4 120.92 =0.092o Equation 1.4-15 then gives the predicted mixture viscosities: (E1: 1-:172 0.00 0.25 0.50 0.75 1.00 21: Z2= A: : B: : A+B= Z$BQIB zmúõh? 3131/11/21 37/12/22 Pmix X 106 Nobmpoise X 106 3.934 1.000 0.0 124.0 (124.0) 124.0 3.200 0.773 6.9 120.3 127.2 128.1 2.467 0.546 18.1 113.6 131.7 131.9 1.734 0.319 38.2 97.2 135.4 135.1 1.000 0.092 88.4 0.0 (88.4) 88.4
- 6. 1A.5 Viscositíes of chlorine-air mixtures at low density. Equation 1.4-14 and Tables E.1, E2 give the following _viscosities at 75°F(= 273.15 + (75 - 32)/1.8 = 297.03 K) and 1 atm: For component 1, (C12), M1 = 70.91, a¡ = 4.115Â, el/ K = 357K; hence, KT/ e¡ = 29703/357 = 0.832 and 9m¡ = 1.754, and s/70.91 X 297.03 = 2.6693 10-5-í “l X (4.115)2 x 1.754 = 1.304 x 10-4 g/ cm-s = 0.01304 cp. For component 2, (air), M2 = 28.97, a = 3.61711, (Sl/ K = 97.0K; hence, KT/ el = 29703/970 = 3.062 and 9,. , = 1.033, and s/28.97 >< 297.03 = 2.6693 10-5--_- “2 X (3.017)2 x 1.033 = 1.832 x 10-4 g/ cm-s = 0.01832 cp. Eq. 1.4-16 then gives the following coefficients for Eq. 1.4-15 at this temperature: eu = @g2 = 1.0 1 7o 91 *V* 0 01304 V” 28 97 “4 2 @i2 = : 1 + : '- 1 + i 4- , /§ 23.97 0.01832 70.91 = 0.5339 1 28 97 “V” 0 01832 V” 70.91 V** 2 «›»= =<1++> +<5 > e) x/ ã 70.91 0.01304 23.97 = 1.8360 Equation 1.445 then gives the predicted mixture viscosities: : v1: 21: 22: A: : B: : A+B= 1 - 932 É 151955119 É 90139213 5171/41/21 902/12/ E2 Nmíxycp- X 106 0.00 0.5339 1.000 0.0 0.01832 0.0183 0.25 0.6504 1.2090 0005012 0.011.365 0.0164 0.50 0.7670 1.4180 0008501 0006460 0.0150 0.75 0.8835 1.6270 0011070 0002815 0.0139 1.00 1.000 1.8360 0.01304 0.0 0.0130
- 7. 1A.6 Estimation of liquid viscosity. a. . The calculated Values for Eq. 1.5-9 at O°C 'and 100°C are as follows: T, K ~ p, g/ cma V = M/ p, cm3/g-mole AÚWTH cal/ g-mole: 897.5 x 18.016 x 252.16/453.59 41711111, / RT = 8989/1.98721 / T exp 0.408AÚ, ,,, ,T, /RT 17 h / 17, g/ cm-s Predicted liquid viscosity, g/ cm-s 273.15 0.9998 18.01 8989. 16.560 859.6 2.22 X 10"* 0.19 b. The predicted values for Eq. 1.5-11 at O°C and 100°C are: , e TLK N h / V, g / cm-s exp(3.8T¡, /T) Predicted liquid viscosity, g/ cm-s Summary of results: Temperature, °C Observed viscosity, centipoise[= ]g/ cm-s X 100 Prediction of Eq. 1.5-9 Prediction of Eq. 1.5-11 273.15 2.22 X 10'** 179.7 0.0398 0 1.787 19. 3.98 373.15 0.9584 18.80 8989. 12.120 140.5 2.12 X 10-4 0.0298 373.15 2.12 X 10"* 44.70 0.0095 100 0.2821 2.98 0.95 Both equations give poor predictions. This is not surprising, since the empirical formulas in Eqs. 1.5-8 et seq. are inaccurate for water andfor other associated liquids.
- 8. 1A.7 Molecular velocity and mean free path. From eq. 1.4-1, the mean molecular velocity in O2 at 273.2 K is _ [BRT /8 X 8.31451 X 107 X 273.2 4 U- m_ CHI/ S From eq. 1.4-3, the mean free path in O2 at 1 atm and 273.2 K is RT 82.0578 X 273.2 , x = _m = -__-__wí-__ Jimzzpzv @na x 10-8)2 x 1 x 6.02214 x 1023 = 9.3 >< 10's cm Hence, the ratio of the mean free path to the molecular diameter is (9.3 >< 10-4 / 3 X 10-8) = 3.1 X 104 under these conditions. At liquid states, on the other hand, the corresponding ratio would be on the order of unity or even less. l-7
- 9. 1B.1 Velocity profiles and stress components a. 1x, = ty, = -_ub, and all other 1,. , are zero. pvxv, = pbzyz, and all other pzgv, are zero. b. rx, = ty, = -2,ub, and all other 1,¡ are zero. pvxv, =pb2y2, pvxv, :pvy x = pb2xy, pvyvy zpbzxz, and all other 100,72, are zero. c. All 1,¡ are zero pvxvx : pbzyz/ Pvxvy : pol/ Ux = -Pbzxl/ r pvyvy : pbzxz and all other pviv, are zero. d. rx, = rw = ,ub, tz, = -2,ub, and all others are zero. the components of pvv may be given in the matrix: pvxvx = ipbzxz pvxvy = zpbzxy pvxvz = -âpbzxz pvv = pvyvx = âpbzxy pvyv, = âpbzyz pvyv, = -àpbzyz pvzvx z _àpbzxz Pvzvy = _â-pbzyz pvzvz : pbzzz
- 10. 1B.2 A fluid in a state of rigid rotation a. A particle within a rigid body rotating with an angular velocity vector w has a velocity given by v = [w X r]. If the angular velocity vector is in the +z-direction, then there are two nonzero velocity components given by 0x = -wzy and v, = +w, x. Hence the magnitude of the angular velocity Vector is b in Problem 1B.1(c). b. For the velocity components of Problem 1B.1(c), 87¡ 822 +y+âv5=0 and : y-avx 8x ôy 8x ôy c. In Eq. 1.2-4, we selected only the linear symmetric combinations of derivatives of the velocity, so that in pure rotation there Would be no viscous forces present. In (b) We see that the antisymmetric combination is nonzero in a purely rotational motion. =2b
- 11. 1B.3 Viscosity of suspensions Expanding the Mooney expression, we get (with 8 = qb/ cpo) Herr=1+ §$ +_1_ g4' 2+l g4' 3+_ , uo l-e 2! 1-8 3! l-ô =1+§<p(1+5+e2+---)+§q>2(1+2e+---)+%5-q>3(1+---)+--- 5 25 5 l 125 25 l 5 1 = l+íÓ+Ózíí+ííoj+Ó3(í+í%+§%j+"' The first two terms match exactly with the first two terms in Eq. 1B.3-l. We can make the third term match exactly, by setting 25 5 1 _+--=7.17 whence q) :0.618 s 2% ° and the coefficient of 4P becomes 125+25 1 5 1 _ - - =2 .2 43 40.618+20.382 O 6 If we try oo = 0.70, the coefficients of 452 and 453 become 6.70 and 17.6 respectively. This gives a somewhat better find of Vand's data. HU
- 12. 1C.1 Some consequences of the Maxwell-Boltzmann equation a. The mean speed is ü: lÃouse-mlâ/ ZKTLÍH: 21<T l: 53552616: 21<T à _ (8KT IÉuZe-muZ/ ZKTdu m Içêze-: Zdê m &V; nm b. First rewrite Eq. 1C.1-4 as m o. , _ 2 _Í uxe"""3/2KTdux _Í e muy/ zKTdu The integral over ux in the numerator of the first factor is zero because the integrand is the product of a factor "ux" (an odd function of the integration variable about ux = O) and an exponential function (an even function), and the range of integration extends equally far in the positive and negative directions. c. The mean kinetic energy per molecule is , myl/ zKTdu ZKT 'Ííêafgdê m ISãe-êzd: ' H5 mjãau4e 1 _2 1 _mu : - m_- 2 2 ao __ 2 2 '[0 u2e mu / ZKTdu and is thus àKT for each degree of freedom. I-H
- 13. 1C.2 The wall collision frequency When we change to dimensionless variables in the second line of Eq. 1C.2-l, we get z = trai” ? U5êesieâgüíesêtdê)Fàíííetdê) z Hl2ÃTl3/ZlRÍTTâlÊ-TWJlÉW/ E] z n â l-IZ
- 14. 1C.3 The pressure in an ideal gas a. The dimensions of the quantities in Eq. 1C.3-1 are s [= ] L2 ux [= ] L/ t At [= ] t m M f [= ] a/ L3›(L/ t)'3 duxduyduz' [= ] (L/ t)3 Using these units, one finds that the _expression on the right of Eq. 1C.3-1 has units of M/ Ltz (which are the same as the units of force per area). b. Combining Eqs. 1C.1-l and 1C.3-1 we get 222 3/2 p : ! Ninja uíe/ fflhxi-tlxuf) 2KTduxduyduz 3/2 m oo _ Í ac _ 2 eo _ 2 : znníñj uíe mux/2kTdux e muy/ ZKTduy _IO e muz/2KTduz 75K °° °° 3/2 = 211m( E” : Zerçzdê newP/ 'l . E; ÇZeTZpÍÇ = 2nm(&)(l)3/2 ) = n1<T 7?¡ Í-3
- 15. 1D.1 Uniform rotation of a fluid a. For the special case that w = ôzw, we get v = [w x r] = Z, .2¡Zks, .¡kõ¡w¡x, , = w(8¡s132y + õzemx) = w(-ô¡y + 82x) Then using Eqs. A.6-1, 2, 13 and 14, we can get the velocity components in cylindrical coordinates v, = (v-õ, ) = w((-õxy+õyx)-õ, )= w(-ycos6+ xsin 6) = w(-rsin6cos6 + xcos6sin 6) = O v9 = (v 69): w((-õxy+õyx)-ô6)= w((-y)(-sin 6)+xcos 6) = w(rsin6sin6 + rcos6cos 6) = wr Therefore, the angular velocity of every point in the fluid is v9 / r = w, which is a constant, and there is no radial velocity. This is the way a rigid body rotates at constant angular velocity. b. The vector operations are (using the abbreviated notation of §A.9 and the Einstein summation convention) (V-v)= â,. v,. =â. e. w x : s- w õ. :e =0 l tmn m n tmn m In nmn w m [Vv]¡¡= [V[w><r])ü=6.s. w x : a w õ. =s. .wm l ; mn m n ; mn m tn ; m1 = "ívvlp = "ívvlã and from this last result we see that VV + (Vv)l = O. c. The results above indicate that for a fluid is a state of pure rotation, the tensor 'c is identically zero. That is, there are no viscous stresses present in the fluid. This was the assertion made just before Eq. 1.2-4. l* M-
- 16. 1D.2 Force on a surface of arbitrary orientation. a. We can specify the surface area and the Orientation of the surface of AOBC as ndS. To project this surface onto the yz-plane, We take the dot product with ôx, so that the area of AOBC is ( n »õx )dS. b. The force per unit area on three triangles perpendicular to the three coordinate axes are Force on AOBC : õxnn +õy7rxy +ôzrtxz Force on AOCA = õxnyx +õy7rw +ôz7ryz V Force on AOAB = ôxnzx + ô yrczy + õzrtu c. Force balance on the volume OABC is then n, ,ds= (ô 7¡ +ôy7rxy+ô 7¡ )(n-õx)dS +íõ: xnyx +õynyy +Z5:zy, )(n -ôy)dS +(ôx7zzx+õy7rzy+õ 7¡ )(n-ô, )ds z zz OI' 1:" = [n-ôxôxnxxh[n-ôxôynWjHn-õxõznxz] +[n-ôyõx7zyxj+[n-õyõynw]+[n-õyõz7ryzj +[n-5,5x7zu]+[n-a 5 7: ]+[n. s,5,n, ,] zyzy : :ilzjljn-õiôgtükfn-n] [45
- 17. 2A.1 Thickness of a falling film. a. The volume How rate w / p per unit wall width W is obtained from Eq. 2.225: _w_ _ uRe _ (1.0037 x 10-2)(1o) pW _ 4 _ 4 Here the kinematic viscosity u for liquid water at 20°C Was obtained from Table 1.1- 2. Since 1 ft=12><2.54 cm, 1 hr :3600 s, and 1 gal=231.00 in3 X(2.54cm/ in)3=3785.4 cms (see Appendix F), the result in the requested units is _ = 2.509 >< 10-2 em? /s 1 3785.4 = 0.727 U. S. gal/ hr-ft _w- = 0.02509 em? /s >< PW gal/ cm2 >< 30.48 cm/ ft >< 3600 s/ hr b. The film thickness is calculated from Eqs. 22-25 and 22-22 as 31/ w 1/3 6 = -- (9 cos B PW) 31/ uRe 1/3 = (gcosBT) _ (3 x 1.0037 x 10-2 (980.665)(1.0) = 0.00361 in. 1/3 (2.509 >< 104)) = 0009167 cm
- 18. 2A.2 Determination of-capillary radius by How measurement. Assuming the flow to be laminar, we solve Eq. 23-21 for the capillary radius: R_ 4 &uLw _ 4 81/Lw _ 7rpA'P _ «A73 Insertion of the data in mks units gives R : 4 8(4.03 x 10-5)(0.5002)(2.997 x 10-3) (3.1416)(4.829 x 105) = í'/3.186 x 10-13 = 7.51 >< 10'** m = 7.51 x 10°* cm As a check on this result, we calculate the corresponding Reynolds number: Re: D(v, )p Z 4w : 2w p 7rD, u «Ri/ p 2 (2.997 x 10-3) = ;(7.51 x 10-4)(4.03 >< 10-5)(0.9552 >< 103) This value supports our assumption of laminar flow. Since the entrance length, Le = 0.035DRe = 0.35 cm is less than L, the enhance-effect correction to R is at most of the order of | [1 ~ (Le/ L)]1/4 - lj, or 0.2 percent of R in the present example. = 66.0 Difficulties with this method include: (1) Inability to account for departures from a straight, circular cylindrical wall geometry. (2) Inability to account for in*- advertent spatial and temporal variations of temperature, hence of the fluid density and viscosity. A simpler method is to measure the length L and mass m of a small Slug of liquid mercury (or another liquid of known density) injected into the tube, and calculate the mean radius R of the slug as (m / [p7rL])1/ 2, on the assumption that the slug is a right circular cylinder. This method allows comparisons of mean R values for various intervals of the tube length.
- 19. ~2A.3 Volume rate of flow thrugh an annulus. Assuming the flow to be laminar, we use Eq. 2.4-17 to calculate the volume flow rate w / p, with the specifications n = 0.495/11 = 0.45 [l = 136.8 (1b, ,,/ ft-hr)(1 hr/3600s) = 3.80 x 10-2 11o, . / ft-s ('Po ~ 'P¡_) = (5.39 psi)(4.6330 x 103 poundals/ ftz/ psi) = 2.497 x 104 11), , / ft- - -sz R = 1.1111.: 1.1/12 ft Here Appendix F has been used for the conversions of units. With these specifica- tions, Eq. 2.4-17 gives w _(7r)(2.497 104)(1.1/12)4 4 (1_(0.45)2)2 í “ (s)(3.s0Xx 10-2)(27) [(1'(°'45) )' ln(1/0.45) l (1_0.2025)2 1141/0495) l = (0.6748)[0.1625] = 0.110 03 / s = (0.49242) [(1 - 0.04101) - As a check on our assumption of laminar flow, we calculate the Reynolds num- ber: 2R(1 - n)(v, )p _ 2w ; L _ 7rRu(1 + n) _ 2(0.110)(80.3) " (3.1416)(1.1/12)(3.80 x 10-2)(1.45) This value is well within the laminar range, so our assumption of laminar flow is confirmed. Re: = 1110
- 20. 2A.4 Loss of Catalyst particles in stack gas. a. . Riearrangement of Eq. 2.6-17 gives the terminal velocity v: ,= 134m ~ p)g/15%# in which D is the sphere diameter. Particles settling at v¡ greater than the centerline gas velocity will not go up the_ stack. Hence, the -value of D that corresponds to v¡ = 1.0 ft/ s will be the maximum diameter of particles that can be lost in the stack gas of the present system. Conversions of data to cgs units give v¡ : (1 ft/ s)(12 X 2.54 cm/ ft) = 30.48 cm/ s p = (0.0451b, ,./ ft3)(453.59 g/ lbm)((l2 x 254)* ft3/cm3) = 7.2 X 10"* g/ cm3 D = ispv, z _ (18)(0.00026)(30.48) m” ' (p, - p)g (1.2 - 7.2 X 10'4)(980.7) = 61.21 x 10-4 = 1.1 x 10** cm = 110 microns b. Equation 2.6-17 was derived for Re« 1, but holds approximately up to Re=1. For the system at hand, Hence, = Dvtp = (1.1 x 10-2)(30.5)(7.2 x 10-4) Re p (000026) = 0.93 Hence, the result in a. is approximately correct. Methods are given in Chapter 6 for solving problems of this type without the creeping-flow assumption. .2-4
- 21. 2B.1 Different choice of coordinates for the falling film problem Set up a momentum balance as before, and obtain the differential equation dg; = pg cos B Since no momentum is transferred at x = ô, then at that plane rf, =0. This boundary condition enables us to find that C¡ = -pgõ cos B, and the momentum flux distribution is Tí, = -pgô cos B(1 - Note that the momentum flux is in the negative x -direction. Insertion of Newton's law of viscosity 7,7, = -/ ,L(dv, /dx ) into the foregoing equation gives the differential equation for the velocity distribution: dv, z pgõcosB [1_xj dx u 6 This first-order differential equation can be integrated to give vz-(M)(s1a) The constant C2 is zero, because v, = O at x = O. We note that x and x are related by x/ ô = 1 - (x/ ô). When this is substituted into the velocity distribution above, we get vzáâãâtêlílrãlsât-zêlãlzl] which can be rearranged to give Eq. 2.2-18.
- 22. 2B.2 Altemate procedure for solving flow problems Substituting Eq. 2.2-14 into Eq. 2.2-10 gives d dv, dzv, p cosB ; (1a dx ) : pgcosñ or dx, = - g u Integrate twice with respect to x (see Eq. C.1-10) and get v, =-pgC0SBx2+C¡x+C2 2M Then use the no-slip boundary condition that v, = O at x = ô , and the zero momentum flux boundary condition that dv, /dx = 0 at x = O. The second gives C¡ =0, and the first gives C2 = (pgco_sB/2/. z)õ2. Substitution of these constants into the general solution and rearranging then gives Eq. 2.2-18.
- 23. 2B.3 Laminar flow in a narrow slit a. The momentum balance leads to i, :(6045) dx x2 L (to 4%) and rn: x+C¡ dv, dx Substitution of Newton's law 1,, = -u into the above gives z _('ñ°0-'(PL)x2 +_C_, = -_--x+- or v = dx ML _u 2,uL , u x+CT Use of the no-slip boundary conditions at x = iB gives the expres- sions in Eq. 2B.3-1 and 2. One can also see that C¡ = O directly, since we know that the velocity distribution must be syrnmetrical about the plane x = 0. b. The maximum velocity is at the middle of the slit and is (v, -0°, )B2 z) (xy z-ííí- h Z :1- - , Imax 2M, and ence v B z, max 'U The ratio of the average to the maximum velocity is then m, IYIÍBP-(x/ Bfldxdv:130-4044_ vz, max j? j: dxdy dê c. The mass rate of flow is (v, - r, )B2 z 3 (ro - 1°, )pB3w w = p(2BW)<vz>= p(2BW)(%) zluL 3 HL d. In Eq. 2.5-22, set both viscosities equal to u. set b equal to B, and multiply by B Wp.
- 24. 2B.4 Laminar slit flow with a moving wall ("plane Couette flow") Start with the velocity distribution from part (a) of Problem 2B.3 (in terms of the integration constants). Determine C1 and C, . from the boundary conditions that v, = 0 at x = -B, and v, = V at x = B. This leads to v--@; -,í°; fí[-0T]+%0+; J This expression can be differentiated with respect to x and then Newton's law of viscosity 1,, =-u(dv, /dx) can be used to get the expression for the stress tensor. Notice that the velocity distribution is no longer symmetric about the midplane, so that C¡ : t O. 2-8
- 25. 2B.5 Interrelation of slit and annulus fonnulas From Eq. 2.4-17 we get 8,uL _ 4 (va-sff w wí-'mwli(““” Win1117? 2 = (1-1+4s-6s2+4e3-e4)+ (1-14-28-82 -s-àsz -âã3 -ãs4---- (482 -463 + a4) : :+452 +433 +§s4+--› = (4e-6e2 +483 -e4)-(4a-6e2 +§e3 -§e4+---) (48-682 +463 454)- This gives, finally, a result in agreement with Eq. 2B.5-l _ : :(05 -7P, _)R4p( 3 4 w 8_uL s -âe ul»
- 26. 2B.6 Flow of a film on the outside of a circular tube a. A momentum balance on the film gives düTn) _ d dv: _ dr +pgr-0 or ME(T dr)+pgr-0 The latter may be integrated to give p 2 01:- f; +C1lnr+C2 Next use the boundary conditions that at r = R, vz = 0 (no slip) and that at r = aR, dv, /dr = 0. When the integration constants have been found, we get for the velocity distribution 2 2 vz= pgR [1~(L] +2azlnl] R R 4# b. The mass rate of flow in the film is then w = fãwfpvzrdrdâ = ZnRZpJ: vzêdê in which a dimensionless radial coordinate §= r/R has been introduced. Then 2 R4 a = "Pá j¡ (1- g2 + 2.12 1n§)§d§ z 7tpZgR4 2# = np2gR4 3M w (g2 _g4 +2a2[-§§2 ggzmq) a 1 (-1 + 4412 - 3x14 + 4a** hm) c. If we set a = 1 + 8 (where 8 is small) and expand in powers of 6 using §C.2, we get 2-10
- 27. 2 4 2 4 3 w = _-_”p gR 83 + 0034)) z _k 3K 8 8# 3# This is in agreement With Eq. 2.2-21 if we make the identifications W = 27zR and ô = SR (and furthermore consider only the case that cos B = 1. 2-1¡
- 28. 2B.7 Annular flow with inner cylinder moving axially a. The momentum. balance is the same as that in Eq. 2.3-11 or Eq. 2.4-2, but with the pressure-difference term omitted. We can substitute Newton's law of viscosity into this equation to get dv” = &,whence v, =-&lnr+C2 or &= -D1lnI-+D2 dr r u v0 R 'M That is, we select new integration constants, so that they are dimensionless. These integration constants are determined from the no-slip conditions. at the cylindrical surfaces: v, (KR) = v0 and v¡ (R) = O. The constants of integration are D2 = O and D¡ = -l/ ln K. This leads then directly to the result given in the book. b. The mass rate of flow is 1: R2 w = ffkpvzrdrdâ = Znpígm f: (ln§)§d§ = mpfi (àêz Inf -§§2)| := mpfi (-§K2 lnK-à(1- K2» which is equivalent to the answer in the text. c. The force on a length L of the rod r= I:I: "(+~í: t] which gives the expression in the book. d. When we replace K by 1 - e and expand in a Taylor series, we get KRdOdz = znKRLuvo (if) r= KR 1 = ZNL/ .QÍJO F=2L- í-í-í--í ? t ( N)vo_(a+àez+âea+%s4___) 8 (]__%g_àg2.. .) To get this last result one has to do a long division involving the polynomial in the next-to-last step.
- 29. 2B.8 Analysis of a capíllary flowmeter Designate the water by fluid "I" and the carbon tetrachloride by "II". Label the distance from B to C as "I". The mass rate of flow in the tube section "AB" is given by z "(321 "PB)R4PI z “KVA "pB)+pIghlR4pI SML. SML Since the fluid in the manometer is not moving, the pressures at D and E must be equal; hence PA +PISh+PISÍ +P18H = PB *PISÍ *PIISH from which we get PA ' PB +P1gh : (911 “PÓSH Insertion of this into the first equation above gives the expression for the mass rate of flow in terms of the difference in the densities of the two fluids, the acceleration of gravity, and the height H . 243
- 30. 2B.9 Low-density phenomena in compressible tube flow When we replace no-slip boundary condition of Eq. 2.3-17 by Eq. 2B.9-1, we get C ÁJVÍJJÊÃ 2_ 4,uL 2_LtL so that the velocity distribution in the tube is v = (P0'PL)R2[1_(TJ2]+(VO'PL)RÇ z 4yL T 2_uL Next we write the expression for w, but consider only the flow through a length dz of the tube: w t I: n¡: p(z)v, (r, z)rdrd0 = Znlgzíãzjlíqjjãvzêdê 8 where we have introduced the ideal gas law, with Rg being the gas constant (we use a subscript g here to dístinguish the gas constant from the tube radius). We have also introduced a dimensionless radial coordinate. When “we introduce the velocity distribution above, we get 27zR4 M w= É 4p RgT 4 y RgT dz Rp This is now integrated over the length of the tube, keeping mind that the mass flow rate w is constant over the entire length L n11** M 4; _Íowdz = + 1 d 2 O J0[-pd-: ][<1-: Hãlêdé Z-l4-
- 31. w: ”(P0'Í7L)R4[&j[P0 +m RgT 2 z #090 _PL)R4 pavgM 8,uL which leads then to Eq. 2B.9-2. lr “Páj R 45o Rpavg l
- 32. 2B.10 Incompressible flow in a slightly tapered tube a. The radius at any downstream distance is R(z) = R0 + (R, - R0)(z/ L) b. Changing the independent variable proceeds as follows: w= "Ktcaiei "Rwtlílt *w 8,u dR dz 8p dR L c. First we rearrange the equation in (b) to get 124w L _1_ dR np RL-ROR** Then we integrate this equation to get v, 8uw L RL 1 “' = à** Im ( “P RL ' Raja R4 whence we can get the pressure difference in terms of the mass rate of flow . P _ao = SuwL RÉ-Rf' ° L 37zp Ro-RL Next we solve to get the mass flow rate w: 37r(ñ°0-? L)p Ro-RL = n(í°o-ñ°, _)Râp í_ Ro-RL 8,uL Rg3-Rg3 8,uL Rg Rg3-Rg3 This is the result, with the first factor being the solution for a straight tube, the second factor being a correction factor. It would be better to write the correction factor as "1- X ", so that the quantity X gives the deviation from straight-tube behavior. The quantity X is then
- 33. X_1_i_ Ro-RL _ _3l1'(RL/ Ro)l_1_3l1'(RL/ Ro)l(RL/ Ro)3 ' Ra* Râ-Rã' (Ro/ RL)3'1 " 1_(RL/ RO)3 _ _ 3(RL/ R0)3 = =1+(RL/ R0)+(RL/ R0)2-3(R¡_/ R0)3 1+(RL/ R0)+(RL/ R0)2 1+(RL/ R0)+(RL/ R0)2 which then leads to the desired result in Eq. 2B.10-B. 2-:7
- 34. 2B.11 The cone-and-plate viscometer a. In a parallel-plate system with rectilinear flow, the velocity distribution is just Ux / vo = y/ b, where b is the plate spacing and v0 is the velocity of the upper plate. We now make the following correspondences between the parallel-plate system and the cone- plate system (using 9 as the usual variable in spherical coordinates measured downward from the z-axis, and u¡ as the variable measured upward from the plate surface): vx<-›v$; voe-war; b<-›rsinu/0; y<-›rsinwzrw= r(à7t-9) When this correspondence is made, Eq. 2B.11-l results. b. From Eq. B.1-19, we get for the force per unit area in the qb-direction on a face perpendicular to the 9-dírection __, ,gei[g_ji_, iüi_yzi(âz<ifz: ü] r 89 sin9 ~ Here we have used the fact that the angle between the cone and plate is so tiny that 9 is very nearly àn so that sin 9 is very close to unity. c. The torque is obtained by integrating the force times lever arm over the entire plate area: _ 27: R __ #S2 R __ pá! R3 T= 'Io Jo (TW 'rMaus/ zrdrdw' #dr _mííjlí . ll/ o which leads to Eq. 2B.11-3. 2-18
- 35. 2B.12 Flow of a fluid in a network of tubes At A the pipe splits into three pipes, and at the next set of junctions the fluid flows equally in six pipes, and then at the next set of junctions the fluid flows back into three pipes, and finally at B the fluid is all retumed to a single pipe. Call the modified pressure at the junctions where three pipes split into six pipes ?3_,6, and that where six pipes join to form three pipes ?6_,3. Then in each of the first set of three pipes w N? EFA R4!) 8,uLw °r l°^4°3r°= ãñ In each of the batch of six pipes w nñ°_, -?P_, R4p SyLw : :Qó-mü; °r °°3+6“°°~3=z; ¡r4; and in each of the final batch of 3 pipes 10_ = "GPs-ea ' ¡°¡3)R4P ol_ ñ, _ . P z 8/. LLw 3 8uL "H3 B 37zR4p When all the pressure differences are added together, the unknown quantities ? M6 and $643 cancel out, and we get __ 4 . PA_¡, B=8;LLw(5) Or w=37t(lPA ? B)R p 7rR4p ã zopL ? KM
- 36. 2C.1 Performance of an electric dust Collector a. First we solve the problem of the vertical motion of the particle as it falls under the action of the electromagnetic field. The equation of motion for the particle (without gravitational acceleration or Stokes drag) is This equation may be integrated with the initial conditions that x= x0 and dx/ dt=0 at t=0,togive _ eôtz x: x° 2m From this we can get the time t¡ required for the particle to fall to x = -B: t _ 2m(B+x0) f' es Next we look at the horizontal motion. From Eq. 2B.3-2 and the expression for x(t), we find that (with *um = (po - pL)B2 /2,uL d _ 2 _d _ x -(e6t2/2m) 2 -f--vmíl-(Ê) J-â-vm[l-( o B j ] This may be integrated to give 1 x eôtz 262t4 L= f§dz= vmf; ¡[1-B-2(xâ- ° +6 2 jdt] m 4m _ (P0 ' PL) 2 2 xtleôtf 8262? “É” *ob* 3 '2 2 y m Om
- 37. Next, square this expression and then insert the expression for t¡ above to get 2( - )2mB+x L2=-11°§ÊL-, /-(~eí-°-)(3B-2x0)(1a+x0) Then, in order to remove the radical, we square this, thereby getting 8( - )2m L4 = (3B-2x0)2(B+x0)3 Next, we set dL4/dx0 equal to zero, and this yields 4 values for x0: àB, g3, B, and B. It is only the first of these that is physically acceptable. When that value is put back into the expression for L, we get finally L a = Bívp-_müâ V4 mm 35 uzeô
- 38. 2C.2 Residence-time distribution in tube flow ' a. A fluid element at a radius r within the tube will require a time t to reach the tube outlet m: L = L zw) vmax[l- (r/ Rf] All the fluid with a radius less than r will have left the tube at this time. Hence the fraction of the flow that will have left the tube is F(, )=L&rd: í9-2(gípy* _Íãnjf vzrdrd9 T R R When the first equation is solved for (r/ R)2 and substituted into the expression for F (t), we get Wi*till-í*vialrlvittlz*íaâwl b, The mean residence time is then obtained by solving the last equation in (a) for t. and Substituting into the Eq. 2C.2-1: Umax O “1 _ F vmax (vz) L dF 2L L tm= jãtdP= fl Q-'Ll
- 39. 2C.3 Velocity distribution in a tube The derivation in §2.3 is valid up through Eq. 2.3-15. If the viscosity is dependent on the radial coordinate, however, Eq. 2.3-16 is inappropriate. Instead we get __(¡Po'¡°L) ' 7 ~ 'Uz- IO T+C2 Application of the no-slip boundary condition at the tube wall gives __(l°0“3°L) R 7 - 0- J›0 T+C2 This may be solved for the integration constant, and the velocity distribution is then v : É--Çdoli-? JIK--df 7 or v = ÊPÕO°JÕOJRZ Jilídy Í 2 2L r #(77) Z 2L vp(g7) This is the same as Eq. 2.3-18 if the viscosity is a constant. Next we get an expression for the average velocity (vz) = = 'Êílf vzrdr = 2I§vzydy o o = QRÉÀÉISIÇH-Íyjdíydv = QÉÍÉISÊJÊydvd? :Wo "ÊLNZZJ-Ãíííjízzdy v¡ _ lylky/ Nldy Then we find the dimensionless ratio <vz> _ Igor/ upa' Qrl?
- 40. 2C.4 Falling-cylinder viscometer a. Equation 2.4-2 is valid for this problem, but the pressure difference is not known. When Newton's law of viscosity is substituted into Eq. 2.4-2 we get vz : _ ¡Po _FL 4yL The two constants of integration and the (unknown) pressure difference can be obtained from two boundary conditions and a mass-conservation condition: At r = KR, vz = -v0; at r = R, vz = 0 ; and _ÍÊIIÉR vzrdrd9 = 7x( KR)2 vo. This states that the fluid displaced by the falling cylinder must be compensated for by a net motion upward through the annular slit. These three unknown constants may be obtained from these conditions (lengthy! ) and the result is & _ (1- §2)-(1+ K2)ln(1/§) v0 (1- K2)-(1+K2)ln(1/K) ]r2 -C¡lnr+C2 b. The force acting downward on the cylindrical slug of height H is (po - p) g - n( KR)2 H . The difference in the pressures acting on the top and bottom of the slug is an upward force (eo - íPH)-7r(KR)2 = -ff(d'tP/ dz)1z-7r(KR)2 _ 4voH . 71:(KR)2 R2[(1- K2)- (1+ K2)ln(l/ K)] In addition, there is an upward force associated with the frictional drag by the fluid 27¡(KR)H(_Trz›lr= KR = 2n<xR›Hu[dzf; ) MR _ 2K-(1+K2)(1/K) - 27IK77oHH[(1_ K2)_ (1 + Kama/ Ko 9: 'L4'
- 41. When these are equated and the result solved for the viscosity, the expression in Eq. 22C.4-2 is obtained. c. Next put K = 1- e, and expand in powers of e, keeping terms up to 65. Use §C.2, and obtain Eq. 2C.4-3.
- 42. 2C.5 Falling film on a conical surface a. A mass balance on a ring of liquid contained between s and s + As gives (27zs(sinfl)ô(s)(v))| s - (27rs(sin[3)ô(s)(v))| = O s+As Letting As -› O then gives d d v $(sô(v)) = 0 whence, from Eq. 2.2-20 63) = O Equation 2.2-20 is valid strictly for a flat plate with constant film thickness, but we apply it here approximately to a different geometry. b. When the equation in (a) is integrated, we get 363 = C, in which C is an integration constant. This constant is determined by requiring that the mass flow down the conical surface be the same as that flowing up the central tube (i. e., w). We hence write (width of film) x (thickness of film) x (mass flow rate), and then use Eq. 2.2-20: w = (27zssinB) - ô -p(v) = (Znssinñ) (ET/ g -p[ ] s Bu From this we get C; C = Buw = 3_uw (Znsinfh-(pzgcosñ) np2gsin2B The film thickness as a function of the distance down the cone from the apex is thus õ z 3,uw ¡zpzgs sin 2B 2-20
- 43. 2C.6 Rotating cone pump a. Inner cone not rotating. For sufficiently small values of B, the flow will resemble very much that for a thin slit (see Problem 2B.3), for which the mass rate of flow is given as the answer to part (b). This formula may be adapted to the flow in the annular space of height dz. as follows, if the inner cone is not rotating and if the gravitational force is not included: L2 _EE w~3( dz] where we have made the' identification (po -pL )/ L-› -dp/ dz and W -› 27tr = 27zzsin9. Across any plane z = constant, the mass flow rate will be a constant. Hence the above equation can be integrated to give B3(2rrzsin 9)p L : :aftl-ãlniz' Psm 1 _ 72d 3 'TT-F_ 47zB ps1n9 , uw IL, dz 3 L¡ í 0T P1'P2 b. Effect of the gravitational force and the centrifugal force The result in (a) may be modified to include the effect of gravitational acceleration g and the angular velocity Q of the inner cone. The gravity force in the z-direction (per unit volume) is given by Fzzzvlz = -pg cosB. The centrifugal force (per unit volume) acting in the middle of the slit will be, approximately, Fem = p(§Qr)2 / r = à pãlzz sin B, where r is the distance from the centerline of the cones to the middle of the slit. The component of this force in the z- direction is then Fcentlz = àpgzz sinz B. Then the first equation in (a) can be modified to give B3(27IzsinB)p weâ_(_ÍlÊ+§p§22zsm2B-pgcos[3] L dz This equation can be integrated to give
- 44. _4B3'B 12-2 2_2_ _ w- [(p1-p2)+(§ps2 sm B)(L2 L1) (PSCOSBXI-'z 1.1)] Many assumptions have been used to get this solution: (1) laminar flow (turbulent flow analog is not difficult to work out); (2) curvature effects have been neglected (correction for this is easy to do); (3) entrance effects have been ignored (this can probably be handled approximately by introducing an "equivalent length"); (4) instanta- neous accommodation of velocity profiles to the changing cross- section (it would be difficult to correct for this in a simple way). 7:28
- 45. 2C.7 A simple rate-of-climb indicator a. Consider two planes of area S parallel to the earth's surface at heightsz and z + Az. The pressure force in the z-direction acting on the plane at height z and that acting at the plane at height z + Az will be just the mass of air in the layer of height Az: Sp| z - Sp| Z+AZ = pgSAz Division by SAz and then letting Az-›0 gives the differential equation di-- di-- PM dz_ pg Or dz_ [RgTjg which describes the decrease in the atmospheric pressure with increased elevation; here Rg is the gas constant. b. Let p, - be the pressure inside the Bourdon element and po be the pressure outside (i. e., the ambient atmospheric pressure). We now write an equation of conservation of mass for the entire instrument: _ n09,- - PO)R4 d . . __ 75(Pi'Po)R4 dt ' °' dt. spL pm' Vdt pi z pavg Here mm¡ is the total mass of air within the system (Bourdon element plus capíllary tube), wo is the mass rate of flow of the air exiting to the outside, p¡ is the density of the air inside the Bourdon elements, and pzvg is the arithmetic average of the inlet and outlet densities within the capíllary tube (see Eq. 2.3-29). The third form of the mass balance written above has made use of the ideal gas law, p = pRgT / M . If we neglect changes in the arithmetic average pressure pzvg and use the abbreviation B = 7rR4pzvg /8,uL V, we can integrate the mass balance above and get p¡ = e'B' BpoeBtdt + C ) “2-29
- 46. To get po (t) we make use of the fact that there is a constant upward velocity, so that -At dp dp dz p M o 0 = °. _=_ _v_ . E-A h = dt dz dt” [RgTjg v¡ p” W ence p” me Then the mass-balance equation becomes -Bt o -Ai Bt BPS -At -Bt p¡= e (jBpoe e dt+C)= e +Ce Determine the constant of integration, C, from the initial condition that p? = p? at t = O. Then _ 0Be_At*Ae_Bt _P¡"P u_ A -(B-A) p¡-P0 and P-Ê-l-ñ(1-e t) In the limit that t-› oo, we get fir B»A _é z vzMg_ svpL P-› 4 B RzT ? TR pzvg Hence for pa z pzvz, the gauge pressure is _ = 8g) MsV p¡ PO UZÍER4 Hence the pressure difference approaches an asymptotic value that varies only slightly with altitude. c. To get the relaxation time, note that P = EÍLAQ - e'(B'A)' )L^› P_(1 - Em) whence 2550
- 47. P -P _ _ = e Bt E e t/ tre¡ eo It is necessary to have t assumption that B»A. so that rel S/ JLV nR4 1 trel = E z Pavg «100 to insure the plausibility of the
- 48. 2D.1 Rolling-ball viscometer The rolling-ball viscometer consists of an inclined _tube containing a sphere whose diameter is but slightly smaller than the internal diameter of the tube. The fluid viscosity is determined by observing the speed with which the ball rolls down the tube, when the latter is filled_ with liquid. We want to interrelate the Viscosity and the terminal velocity of the rolling ball. The flow between the sphere and the cylinder can be treated locally as slit flow (see Problem 2B.3) and hence the only hydro- dynamic result we need is d_r z l2n<vz> (t) dz 02 But we must allow the slit width o' to vary with 9 and z. From the figure we see that R2 = (R ~ r)2 + (r' + c)2 - 2(R - r)(r' + a)cos 9 where r' = x/ rz - 22 . Solving for o' we get o": -r' + (R - r)cos9 +R, /1+ [(1a - r)/ R]2(~sin2 e) The second term under the square-root sign will be very small for the tightly fitting sphere-cylinder system and will hence be neglected. Furthermore we replace lrz - 22 by VRZ - 22 and add compensating terms R-w/ Rz -22 +(«/ R2 -zz -Vrz-zz) R-r R-w/ Rz-zz +(R-r)--; -(R-r)(z/ R)2--] o'= (R-r) cos9+ R-r R-x/ RZ-zz] R-r = (R - r)[cos 9 + z (R - r)[(cos 9 + 1) +
- 49. _J 2_ 2 =2(R-r)(cos2g+--u R z j 2(R - r) The omission of the term containing (z/ R)2 and the higher-order terms is possible, since the greatest contribution to the viscous drag occurs at the plane z = 0, and hence less accuracy is required for regions of larger z. Note that the above result gives correctly a: O atz= O, 9=7t, and o'=2(R-r) atz= O, 9:0. Next we assert that dp / dz will be independent of 9, which is probably a good approximation. Then according to (*) (vz) must have the form <v, >=B(z)a2 (e) Next, the volume rate of flow across any plane z will be Q = IfZ<v, >a(e, z›Rde= Racer: : o<e, z›13de = 8RB(z)(R - r)3 jfyjcosz ge + arde = 8RB(z)(R - r)3 I(a) inwhich a= (R-/ R2 -z2)/2(R-r). The volume rate of flow Q at all cross-sections Will be the same, and its value will be, to a very good approximation Q = nR2v0, where v0 is the translational speed of the rolling ball. Equating the two expressions for Q gives 7rRv0. “Ztmm (atá-ü) Combining (*), (**), and (***) we get dp 37ruRvó dz = 2(R- r)3 [(01) The total pressure drop across the slit is then 24")
- 50. + Ap: J-_ udp dz Rdp __d zg ___ Rdz Z 0 dz dada into which we have to insert dz/ da. Virtually no error is introduced by making the upper limit infinite. From the definition of a zz = -4(R-r)20c2 +4R(R-r)a The first term on the right is smaller than the second, at least for small z. Then dz -- «/ R(R-r)da / x/ ã, and the pressure drop expression becomes (With 52 = a) Ap = 2,/ R(R-r)j; °%àda= 4q/ R(R-r)_Í; °§Éd§ =4 R<R-r›§¡',5§~'Í~: §:J, (;2)d: =§('§ã2§)§Í-Í-¡ where I: ZL: I(;2)d§= â[«/ Í-: /%(«/ TÕ+2)V2], O.531 The pressure drop multiplied by the tube cross-section must, according to an overall force balance, be equal to the net force acting on the sphere by gravity and buoyancy §7tR3(pS - p)gsin[3 = 7tR2Ap where ps and p are the densities of the sphere and fluid respectively. Combining the last three results gives the equation for the viscosity _aí-Í "o 4 R705 -p)8sínB(R-r)5” R 14W
- 51. 2D.2 Drainage of liquíds d. The unsteady mass balance is ; §<paw›= (p<v, >wõ) Z - (p(vz)Wô) z+Az Divideby pWAz and take the limit as Az -› O, to get Eq. 2D.2-1. b. Then use Eq. 2.2-22 to get Eq. 2D.2-1: aa = pg 863 __pgõ2Q§ T9? _í-à: u dz which is a first-order partial differential equation. c. First let A = Jpg/ ,u , so that the equation in (b) becomes: âA= A29A dt dz Inspection of the equation suggests that A = qlz/ t, which can be seen to satisfy the differential equation exactly. Therefore Eq. 2D.2-3 follows at once. This equation has a reasonable form, since for long times the boundary layer is thin, whereas for short times the boundary layer is thick.
- 52. 3A.1 Torque required to turn a fríction hearing. Equation 35-31 describes the torque required to turn an Outer cylinder at an angular velocity QO. The corresponding expression for the torque required to turn an inner rotating cylinder at an angular velocity Q¡ is given by a formally similar expression, 2 T, = 47rpQ, -R2L( "° 5 1 - ! i2 derivable in like manner from the corresponding velocity profile in Eq. 35-32. The specifications for this problem (converted into SI units via Appendix F) are: 1.000 2 n _ ? O55 _ 0998004, n _ 0996012 n? 0996012 (1 - n? ) ' 0003988 ' 2495 , l = (200 cp)(10*3 kg/ m-s/ cp = 0.200 kg/ m-s Q, - = (200 rpm)(1 min/ 60 s)(27r radians/ revolution) = 201/3 radians/ s R2 = (1 in2)(1 m/39.37 m? = 0,000645 m2 L = 2 in = 2/39.37 m = 0.0508 m p = (50 16,, /ft3)(0.45359 kg/ lbm(39.370/12 ft / m)3 = 800.9 kg/ m3 Hence, the required torque is T, = (41r)(0.200 kg/ m-s)(207r/3radians/ s)(0.000645m2)(0.0508 m)(249.8) = 0.431 kg-mz/ sz = 0.32 ft-lb¡ and the power required is P = TÁ), - = (0.32 lb¡-ft)(2O7r/3 s'1)(3600 s/ hr)(5.0505 x 10"' hp-hr/ lbf-ft) = 0.012 hp In these calculations we have tacitly assumed the How to be stable and laminar. To test this assumption, we formulate a transition criterion based on the critical angular velocity expression given under Fig. 3.6-2: . 2 _ 3/2 Re : =_ < about 41.3 for n z l. p . Insertion of numerical values for the present system gives e z (207r/3 radians/ s)(800.9 kg/ m3)(0.000645 m2)(1 - 0.998O04)3/2 (0.200 kg/ m-s This Re value is well below the transition value of 41.3 for this geometry; therefore, the foregoing predictions of T, and P are realistic. R = 0.0048 3*¡
- 53. 3A.2 Friction loss in bearings. The power expended to overcome the bearing friction is 2 P = 11,9,- = 4WQ3R2L( '° ) 1-/522 in which L is the total hearing length of 2 X 20 X 1 2: 40 ft for the two shafts. The specifications for this problem (converted to SI units via Appendix F) are: 16 2 = _m = . - = 0998751 'i' 16 + 2 x 0.005 O 999375* K n? 0998751 (1 - n? ) T 0001249 ' 7996 , u = (5000 cp)(l0"3 kg/ m-s/ cp) = 5 kg/ m-s O¡ = (50/60 rev/ s)(27r radians/ rev) = 57r/3 radians/ s R2 = (8/3937 m)2 = 004129 m2 L = 40 ft = (40 >< 12/39.37 m): 12.2 m With these values, the calculated power requirement is P = (47r)(5 kg/ m-s)(57r/3 rad/ s)2(0.04129 >< 12.2 m3)(799.6) = 6.938 x 105 kg-m2/s3 This result is then expressed in horsepower by use of Table F.3-3: P = (6.938 x 105 kg-m2/s3)(3.7251 x 10-7 hp-hr-[kg-m2/s2]_1)(3600 s/ hr) = 930 hp Thus, the fraction of the available power that is lost in bearing friction is 930/ (4000+ 4000) = 0.116.
- 54. 3A.3 Elfect of altitude on air pressure. For a. stationary atmosphere (i. e., no wind currents), the vertical component of the equation of motion gives dp _ The air is treated as an ideal gas, = M. p RT with M z 29, and with temperature in °R given by T(z) = 530 - 0.003z at elevation z ft 'above Lake Superior. The pressure pg at zg = 2023 - 602 = 142,1 ft above lake level is to be calculated, given that p¡ = 750 mm Hg at z = O. The foregoing equations give dlnp _ _ Mg dz _ R(530 - 0.003z) Integration gives M 1421 111072011) = “Rijo _É 1 ” R 0.003 o ' _ Mg 1 525.737 " 0.00312 n 530 dz (530 - 0.003z z=1421 ln[530 - 0.003z] z=0 Insertion of numerical values in lbm-ft-s units gives (29 15,. /lb-mol)(32.17 ft/ sz) (0.003 R/ ft)(4.9686 x 104 lbm-ftz/ sz-lb-mol-R) = ;00505 ln(pg/ p¡) = ln [525.737/over530] Hence, . pg = p¡ exp(-0.0505) = 750 >< 0.9507 = 713 mm Hg Since the fractional change in P is small, one gets a good approximation (and a quicker solution) by neglecting it. That method gives pg = 712 mm Hg.
- 55. 3A.4 Viscosity determnation with a rotating-cylinder viscometer. Here it is desirable to use a sufficiently high torque that the precision of viscosity determinations is limited mainly by that of the measurement of angular velocity. A torque of 104 dyn›cm, corresponding to a torque uncertainty of 1%, appears reasonable if the resulting Reynolds number is in the stable laminar range. The geometric specifications of the viscometer are: R = 2.25 cm; NR = 2.00 cm n = 2.00/2.25 = 0888889; K2 = 0790123 1 - n? = 0209877; (KEY = 4.00 cm2 L = 4 cm; R2 = 5.0625 cm2 The angular velocity corresponding to this torque value is: _ T, (1 - x2) _ (104 g-cm2/s2)(0.209877) 9° _l 47rp(nR)7L _ 41r(0.57 g/ cm-s)(4 cm2)(4 cm) = 18.3 radians/ s The Reynolds number at this condition is: _ 0,12% _ (18.3)(5.0625)(1.29) " , u ' 0.57 Re = 210 Accordng to Fig. 3.6-2, this Re value is well within the stable laminar range; there- fore, a torque of 104 dyn~cm is acceptable.
- 56. 3A;5 Fabrícation of a parabolic mirror. Equation 3.6-44 gives the shape of the free surface as z-z- 92- 2 0- 2g7' The required derivatives of this function at the axis of rotation are dz d2z 92 Setting the desired focal length equal to half the radius of curvature of the mirror surface at 'r = 0, and using Eq. 3A.5-1, We obtain f = ây/ W Thus, the required angular velocity to produce a mirror with focal length f = 100 cm at standard terrestrial gravity is _É_ 0_ 2¡ _ 980.665 cm/ s2 v (2)(100 cm) = 2.214 radians/ s which corresponds to 6011/2# = 21.1 revolutions per minute.
- 57. 3A.6 Scale-up of an agitated tank. The specifications for the operation in the large tank (Tank I) are N¡ = 120 rpm; p¡ = 13.5 cp; p¡ = 0.9 g/ cm3 and the tank is to be operated with an uncovered liquid surface. To allow direct prediction of the operation of Tank I from experiments in the smaller system (Tank II), the systems must be geometrically similar and must run at the same values of Re and Fr. To meet the latter requirement, Eqs. 3.740,41 must be satisfied. Equation 3.7-41 requires D¡1N¡2¡ = DIN? when, as usual, the gravitational fields for the two systems are equal. Then the model must operate at N" = Nn/ DI/ DI¡ = l20/1_0 z 380 rpm and Eq. 3.7-40 requires Dn 2 N11 2 u = _- -- 11 VI D] N] = (l3.5/0.9)(0.l)2(/10) = 0.474 cp From Table 1.1-l, we see that this value of un corresponds closely to the value for liquid water at 60°C. Thus, the model should opeerate at ;380 rpm, with liquid water at very nearly 60°C.
- 58. 3A.7 Air entrainment in a draining tank. As this system is too complex for analytic treatment, we use dimensional analy- sis. We must establish operating conditions such that both systems satisfy the same dimensionless differential equations and boundary conditions. This means that the large and small systems must be geometrically similar, and that the Froude and Reynolds numbers must be respectively the same for each. Choose D (tank diameter) as characteristic length, 'and 7rD2) as charac- teristic velocity, where Q is the volumetric draw-off rate. Then 4 Re-íp- 16Q2 _ WD” Fr and _ T2 Dsg Subscripts L and S will be used to identify quantities associated with the large and small tanks, respectively. We take the gravitational field g to be the same for both. Then the requirement of equal Reynolds numbers gives <a6-: >-<, e-: ><: -:>-<: :s> and the requirement of equal Froude numbers gives Qs Ds 5/2 (a) ~ (m) Combining these requirements, we obtain (à) = (0.o2277)2/3 = 0.080 DL , Hence, D5 = (0.080)(60 a) = 4.8 ft Qs = (0.080)5/2(8O0 gel/ min) = 1.46 gal/ min Therefore: a. . The model tank should be 4.8 ft in diameter. b. Its draw-off tube should be 0.080 ft in diameter and 0.080 ft high. c. Its draw-off tube should have its axis 0.32 ft from the wall of the tank. Furthermore, if water at 68°F (20°C) is withdrawn from the model tank at 1.46 gal/ min, air entrainment will begin when the liquid level is (4.8/ 60) of the level at which entrainment would begin in the large tank at its withdrawal rate of 800 gal/ min. 3“7
- 59. 3B.1 Flow between concentric cylinders and spheres d. The derivation proceeds as in Example 3.6-3 up to Eq. 3.6- 26, which we choose to rewrite as v R 2 The boundary conditions are that U9(KR) = S2,KR and v9(R) = QGR. Putting these boundary conditions into the above equation for the angular velocity gives S2,= D1+D2i2 and K n, =D, +D, These equations can be solved for the integration constants n, 1/;8 n, 1 D = .QO 1 = QO K2 : QÍKZ-Qo 1 |1 1/;8 , l 1 K2-1 1 1 1 K2 1 o, x2 n, D _ li n, _x2 n, _x2(n, -n, ) "r 1/K2- 1 1 ' K2-1 1 1 1 K2 Hence the solution to the differential equation is v, _ (eo en? ) , (9 r 1-K2 1-K2 T' The z-components of the torques on the outer and inner cylinders are d T, = jájãü- 1,919”, Rd9dz = Z7ILR2[+/ ,trd-[U_B)J T T' r= R
- 60. (fÊÊdodQT-z-Igí) = -47tuL(Q, - o, ) : Ri: = 27rLR2uR 1- K n d T, = lj: (+T,6KR)r= KR KRd9dz= 2nL(›cR)2(-ur; (27°)] _ R (KRY = +47t, uL($2, - Q, )1 _ K2 b. In Example 3.6-5 it is shown how to get Eq. 3.6-53 for the velocity distribution. The boundary conditions are : v, (KR) = KRQ, sin9 and v, (R) = R9, sin 9. Equation 3.6-53 can be written in the form 3 222 : D1 14345) rsm9 r The constants can be obtained according to the method of (a) and the final expression is v, _(9a~9.-›<3), (sz, .-sz, ›(n: f rsm9_ 1-K3 1-K3 r The torques at the outer and inner cylinders are then r, = fã2l; '(-r, ,)r= R(Rsir19)R2 sin 9d9dq> W 7¡ d . . : j: _[0 [Hxrãvlfj] (Rsme)R2s1n9d9dçí› r= R R 3 = -8n¡.1(n, .-n, )í': 23 2a It . 2 . (KRW T, :jo _[0(+1,, ),= KR(KRs1n9)(KR) sm9d9drp= +87zp($2¡ -S2,)1_K3
- 61. 3B.2 Laminar flow in a triangular duct a. It is clear that the boundary conditions that v, = 0 at y = H and at y = iw/ Êx. Therefore the no-slip boundary conditions are satisfied. Next it has to be shown that the equation of motion 3° -ÍP d2v, d2v, O: OL L-l-M[ax2+ây2) is satisfied. Substituting the solution into the second-derivative terms, we get 2220-22 L2 _Oi 2 _ 3_ 2 2 p( 4MLH Max, +ay2 (3x y y 3Hx +Hy ) 1° -r = ( : LHL)(6y-6H-6y+2H) and this just exactly cancels the pressure-difference term. b. To get the mass rate of flow we integrate over half the cross-section and multiply by 2: 0° -02 w = 2p( ; MLHL _lã/ my - H)(3x2 - y2 )dxdy =2pl224)#Ll: LjJ22(y2H)(x3 -19126 7P -TP H -2y3 =2P( , fyLHLjlo (V'H)[ 33/2 Jdy «prvii-Hei-teew) 33/2,uLH 2-0 180;1L The average velocity is then the volume rate of flow (w/ P), divided by the cross-sectional area H 2 / «É so that (v >: (220 _pLM-Iz 2 60_uL 3-10
- 62. The maximum velocity will be at the tube center, or at x = 0 and y = y = 2H/3, so that z (2,0 'lpLlHz : E 9 <vz> vz, max
- 63. 3B.3 Laminar flow in a square duct a. The boundary conditions at x = iB and y = iB are seen to be satisfied by' direct Substitution into Eq. 3B.3-1. Next we have to see whether the differential equation p _w a2 z a2 z o= (__§›L_L), ,,(axg , 8:2) is satisfied. Substituting the derivatives from Eq. 3B.3-1 into this differential equation gives o= üs2etzuí%fàízeíãlziâzll Hence the differential equation is not satisfied. b. The expression for the mass flow rate from Eq. 3B.3-1 is given by 4 times the flow rate for one quadrant: w = 4. ¡; *¡; *[1-g-Tlli-(âzldxdy = MÊEÍBJJIJ<I-r›<vn0d: dn = &íihlãiq m1- @agr (e, - 1°, )B2p m2 H¡- 2 z 0.444(a°, - r, )B4p _uL
- 64. 3B.4 Creeping flow between two concentric spheres a. From Eq. B.4-3, there is only surviving term on the left side 1 d . rsmeãhv, sin9) = 0 whence v, sin 9 = u(r) b . From Eq. B.6-8 (omitting the left side for Creeping flow) the only surviving terms are 1 d? O: _íü+'u li(r2 01- 0:. _______. +u 1 i(r2dl) r d9 r2 dr dr r d9 r2 sin9 dr dr c. When the equation in (b) is multiplied by r sin 9, the left side is a function of 9 alone, and the right side contains only r. This means that both sides must be equal to some constant, which we call B. This gives Eqs. 3B.4-2 and 3. Integration of the pressure equation proceeds as follows: anàür-s) l , P2_, P1=B1nt cotzs tanàe : Bln , rlz : In-E or 5 sin9 tanàe From this we get the constant B _ 222-221 _ ? $421 B “ : É ' Ni lncot 5a Zlncotís Next we integrate the velocity equation ipzégjzü o, ,, =§15(I__%+c) dr dr 11 2,11 R r 2 where we have selected the constants of integration in such as way that they will be dimensionless: C1 = -K and C2 = -K- 1 (from the no-slip condition at the walls). When this solution is combined with the expression for B we get: 1611-9-111911-1-661111-11+691
- 65. which leads to Eq. 3B.4-S. d. The mass rate of flow must be the same through any cross- section. It is easiest to get w at 9:57: where v, =u(r)/ sin9= u(r)/ sinà7t= u(r): w = "mp6, | e=ânrdrdp = 27zR2pKu<§d§ _ _(3° -'ú° )R 1 _ _ul_ -zum, [wilncoíâgjx (1 §)+ 1 é &dá; : _. _k1- ?2)R . 4” 1mm, E 6 (1- K)3 (see Eq. 3B.4-6) - 2 344-
- 66. _ 3B.5 Parallel-disk viscometer d. The equation of continuity (Eq. B.6-5) just gives 0 = 0. Equation B.6-5 gives for the 9-component of the equation of motion d 1 d d2v = - --(rv ))+ 9 dr rdr 2 dz2 b. When the postulated form for the tangential component of the velocity is inserted into this, we get simply d 2 f / dz2 = O, which has to be solved with the boundary conditions that _f (0) = 0 and f (B ) = Q, which are just statements of the no-slip conditions on the wetted surfaces of the disks. It is easily shown that f = Qz/ B. and therefore that v, = Qrz/ B. c. The z-component of the torque exerted on the fluid by the upper rotating disk is n dv r, = j: jf(-rz, r)| z=Brdrd9= 271 ¡jpd-azijrl _ Q R 3 FBrdr-Znp --Ê-_lo r dr We can now do the r-integration and solve for the viscosity to get the desired formula, ,u = 2BT, /1111124 . 3-15
- 67. 3B.6 Circulating flow in an annulus a. Neglect of curvature gives for the z-component of the equation of motion di? d2vz Or d2v, _ É dz dr2 'u dr2 dz The left side is a function of r alone and the right side is a function of z alone, so that both sides must equal a constant. Therefore d2v, dr2 = C' with boundary conditions v, (KR) = v, and v, (R) = 0 Use the dimensionless variables q) = v, /vo and ê = r/ R; instead of the latter, it is more convenient to use Ç = (éf - K)/ (1- K) in this part of the problem. Then 2 $322)- = 2C¡ with boundary conditions <p(0) = 1 and 6(1) = 0 Integration of this equation gives fp = C152 'l' C251' C3 The boundary conditions give C3 = 1 and C2 - (1 + C1) so that $= C1Ç2'(1+C1)Ç+1 The mass-balance condition já qbdÇ = 0 gives C¡ = 3, and the velocity distribution is 6=3Ç2-4Ç+1 b. We use exactly the same procedure when the thin-slit ap- proximation is not made, but the algebraic manipulations are messier. The equation to be solved is
- 68. 1d( dvz) d? , _ *TE dr 'E with the same no-slip boundary conditions as before. Use the same dimensionless variables as in (a) and find that the solution has the form : pf-CIE +C2ln§+C3 The three constants of integration are determined from the no-slip boundary conditions q>(K)=1 and (p(1)= O, along with the mass- conservation condition n cpêdâdô = 0. The results are 1-[213/(14 K2)]ln(1/K)_ _ 1_ K2 C1: "C3 z ': ”“_”“'(1- K2)_(1+ amu/ xy 2 '"_”__'“””_(1- K2)-(1+ K2)ln(1/K) With these expressions for the integration constants, Eq. 3B.6-2 follows. 3-17
- 69. 3B.7 Momentum fluxes for creeping flow into a slot a. Inside the slot, the nonzero component of pvv is mxvx= pízââizltiãfi2 Outside the slot, the nonzero component of pvv are pv_ pwqgivfj_ ” ” ítWp (x2 +y2)“ y y "WP (x2 +y2)4 __ _ 2w 2 xsy pvxvy _pvyvx _ p (x2 + yg )4 b. Atx= -a, y=0 pvv= p2l2i "xl 7IWp a2 This quantity is positive as we would expect, since positive x-momentum is being transported in the x-direction. c. At x= -a, y-+a. _ 2w 2 1 pv*vy"pnTm 24a* This is negative, since at the point in question the y-momentum is negative and being transported in the x-direction. d. The total flow of kinetic energy in the slot is (if we use 17 = y / B): 3w 4BWp Jfjffepvãmdydn zwjfepviyzwpaw( 16 3w 3 : 35 4BWp The total flow of kinetic energy across the plane at x = -a is jjja- n3)dn 313
- 70. W 0° o: : _[0 _ÍÍN (àpvz )vxdydz = ¡pWJÍJvÍ + v; )vxdy _¡ 2w 3 + a9 a7y2 _105 2w 3 7¡ variam) a The integrais appearing here can be found in integral tables. We conclude that the total flow of kinetic energy across the plane at x = -a _is not the same as that in the slit. As a-à eo, theflow of kinetic energy tends toward zero, since the fluid velocity tends to zero as' x -›'-oo. This emphasizes that kinetic energy is not conserved. e. Eq. 3B.7-1 clearly satisfies the equation of continuity, since é for incompressible flow (Blix/ dx) + (ãvy / + (âvz/ âz) = O. When the derivatives are calculated from Eqs. 3B.7-Z, 3, and 4, it is found that these expressions also satisfy the incompressible equation of continuity as well. f. From Eqs. 3B.7-2 and B.1-1 we get T _ _zy âvx _ 2,1( 2w ) 3x2 _ 4x4 = _ 4w, u xx 3x ; go "WP (x2 + yz )2 (x2 + y2)3 Fo "WPXZ dv 2w 3x2 4x4 'rxx= -2u x :2/4?) -------í-- =0 3x x= O “WP (x2 + y2)2 (x2 + yz )3 no The second of these is an illustration of Example 3.1-1. g. From Eqs. 3B.7-2, 3B.7-3, and B.1-4, we get, after evaluating the derivatives «r n âvx 9V Tyx Tyd' 0 x=0 3-rq
- 71. 3B.8 Velocity distribution for creeping flow toward a slot a. For the given postulates, the equation of continuity gives lim), )““= '0 from which it follows that v, = 1 f (9) r âr r Since the flow is symmetric about 9 = 0, df / d9 = 0 at 9 = O; and since the fluid velocity is zero at 9 = ià 71:, it follows that f = O at 9 = :à n. b. The components of the equation of motion given in Eqs. B.6-4 and 5, appropriately Simplified are 2 0=-§; °-+'uii and 0=-âí+2_uí dr r_3d92 99 rz d9 c. When the first equation is differentiated with respect to 9 and the second with respect to r and the two results subtracted we get Eq. 3B.8-l. d. The equation in Eq. 3B.8-1 can be integrated once to get d2 í6-j: +4f= C1 A particular integral is fp¡ = àCl, and the complementary function is (according to Eq. C.1-3) fa; = C2 cos 29 + C3 sin29 . The complete solution is then the sum of these two functions. e. The integration constants are determined from the boundary conditions. It is found that C3 = O, and that &Cl = C2. Then from 2 fdâ = ~2WpC2 *mcosz @de = -WpC27I wz-Wplm/ zv rd9=-Wpf: :;/ ;M -zr/2 ' 2 we get C2 = -w/ Wpn and the velocity distribution is given by Eq. 3B.8-2. f. From the velocity distribution and the equations obtained in (b) we can get 34°
- 72. 93° y 4w . _a7 = nwpycosz 9 - smz 9) and hence vhíz( 2"' )(cos29-sin29)+13(9) v) r nWp Furthermore av 2p df (zu) 2w 2 _. __= _.. _ 39:_ __ a6 r¡ de or r¡ "Wp cos 9+G(r) Here F and G are arbitrary functions of their arguments. The second expression for the modified pressure can be rewritten as 3° = -f-z( nwp)(cos2 e + (1 - sinz 9)) + G(r) = mgí( : Vzgpkcosz 9 - sinz 9) + (*) By comparing the two (*) expressions for the modified pressure, we see that they are the same except for the functions P and H . Since the first is a function of 9 alone and the second a function of r alone, they must both be equal to a constant, which we call 'PW This is the value of the modified pressure at r = eo. g. The total normal stress exerted on the wall at 9: 7z/2 is' (when one uses the result of Example 3.1-l) "salvar/ z = (p + 799 )|9_-. ;;/2= (ao _pgh)|9=7g/2 2,uw Zuw rzwprz 7rWpr2 = i°w+ -pgh= p__+ h. From Eq. B.1-11
- 73. + ; a6 : learn 2w -pt[0 -( "Wprz )(-2cos 9sin 9)] The first term is zero since v9 / r= f since v9 =0 was one of the postulates. The second term is zero, as can be seen by using Eq. 3B.8- 2, and the fact that cosã = 0. This is agreement with the result in Problem 3B.7(g). i. Since the z-component of the velocity is zero, we, can expand the velocity vector in either the cylindrical coordinate system or the Cartesian system thus = O e= n/2 ' v = õ,v, +õ9v9 : ôxvx +õyvy Since v9 = O was one of the postulates, when we take the dot product of this equation with 5,, we get the x-component of the velocity 2w 3 nWpr 2wx3 2wx cos3 9 = - - - WWPT4 - nWp(x2 + y2)2 v, =(õ, -ô, )=v, cos9=- Similarly for the y-component of the velocity . 2w . Zwxzy v, = (ô, -ô, ) = v, sin9 = - nwprcosz 9sm9 = - nwpr, _ _ 2wx2y nWp(x2 + y2)2 These results are in agreement with Eqs. 3B.7-2 and 3. 3 'TL
- 74. 3B.9 Slow transverse flow around a cylinder a. At the cylinder surface we get by using Eq. B.1-11 _ Cyva, cos 9 plnR = p_ R - pgRsin 9 9 v9 19v, Cuvwsin9 fralda** fa: 7 *na : T r=9 Also from Example 3.1-l we know that 7,, [r= R = O. b. If n is the outwardly directed unit normal vector for the cylinder, then the force per unit area acting on the surface is - [n 41:] evaluated at the surface. But n = ô,, so that the x-component of the force per unit area at every point is _(81 'lôr The pressure and stress terms are evaluated thus (using Eq. A.6-13): _(5x . [5, r= R = _(5): .[5, . (.. .+5r59Tr9+. ..)]) : _(õx 'õ9)Tr9l¡= R = Tr9lr= R r= R E-(õx 'lõr (Pô “m r= R r= R z _(õx r= R = -pLzR cos 9 r= R These expressions lead to Eq. 3B.9-5. c. The total force on a length L of the cylinder is then P, = jájã”(_p| r=R cos 9 + 19|”, sin 9)Rd9dz d9 C _, _ 29 . Cv, ,,sin29 = RLI02"(-p, cos9+ m) ; os +pgRsm9cos9+lt-: í) = RL( C*1*: '°°)j02”d9 = 27rC, uv, ,L The first and third integrals in the next-to-last line are zero since the integrands are odd. 3 »23
- 75. r r dr _ 9:99:13¡ 2_ 3B.10 Radial flow between parallel disks a. The continuity equation is, for v, = v, (r, z) , from Eq. B.4-2 1.a_ ; ar from which (rv, ) = 0 rv, = 4›(z) The equation of motion is obtained from Eq. B.6-4 pv avi' = -É+Iu 0+ azvr ' 9r dr 9r r9r ' 922 b. When the results for the equation of continuity in (a) are used in the equation of motion, we get Eq. 3B.10-1. c. With the Creeping flow assumption, Eq. 3B.10-1 gives d? _ dão . Tí-MÉ fromwluch r-d7~B . d2$ and since the left side is a function of r alone and the right side a function of z alone, and therefore both sides must be equal to a constant, B. When the pressure equation is integrated from r¡ to rz, we get z 92-9, ln(r2/r¡) d. When this result is substituted into the qb equation we get whence B 'i r . 929 91-? , . r -v 22 í= ___. _____ frgmwfuch $= _:19í2___ dzz uln(r2/r¡) uln(r2/r¡) 2 The integration constants are obtained from applying two boundary conditions. We could require that q) = O at z = ib, and thereby determine the integration constants. Another method is to recognize that the flow is symmetric about z = O, and use as one of the boundary conditions dçb/ dz = 0 at z = 0. Either method will give C1 = O, and then C2 is easily obtained. The final result is + Clz + C2 344
- 76. (F1 - ? P2 )b2 z 2 cb = --- 1- - 2,u ln(r2/r¡) b Division by r then gives Eq. '3B.10-3. e. The mass rate of flow at any cylindrical surface in the system must be the same. Select the surface at r = rz and obtain _ , , + _ (r -9 )b2 w - I_: pv, |r= r2 dzr2d9- 27tpb-2- LK1 - §2)d§ The integral gives 2 / 3, so that Eq. 3B.10-4 is obtained. 3-15
- 77. 3B.11 Radial flow between two coaxial cylinders a. From Eq. B.4-2 we get for this flow, with v, (r), C whence v, = - where C is a constant r dr , r At r = R, v, (R) = C/ R so that C = Rv, (R). b. The relations in Eq. 3B. ll-1 follow immediately from Eqs. B.6-4, 5, and 6 for the velocity profile v, (r) in (a). c. Integration from r to R gives 1 1 *ipczlríl 1 R 'IF(R) - õ°(r) = pCí-ZÍ) T This gives, making use of the meaning of C obtained in (a) r<r›-r<R›= àp[Rv, (R›]2(¡§-§)= àp[v, <12›]2)1-(5)2) r d. The only nonzero components are (from Eqs. B.1-8 to 13) d , 1 , 1 1,, = -Zuír- = +2,LLCr-2; 199 = -2907 = -2uCr-2
- 78. 3B.12 Pressure distribution in incompressible fluids The equation of 'motion method to get the pressure distribution is correct. On the other hand, the second method gives nonsense, as one can see from Fig. 3.5-1. For an incompressible fluid (the Vertical straight line), specifying the density does not give any information about the pressure. 3-27
- 79. 3B.13 Flow of a fluid through a sudden contraction a. For an incompressible fluid, Eq. 3.5-12 becomes 99079 'U12)+(P2 '91)+9g(h2 'h1)=0 m' 99079 '771Z)+(; P2 “ñ°1)=0 If "1" is the large tube and "2" the small one, then the fluid velocity in "2" must be greater than in "1." Then the modified pressure in "2" must be less than that in "1." Thus the modified pressure decreases as the fluid moves from the large cross-section region to the small cross-section region, in agreement with experimental observations. b. For an ideal gas, Eq. 3.5-12 becomes RT §p(v§ ~v'¡")+-1íln: -Í+pg(h2 -h¡) = 0 The pressure and elevation terms may no longer be combined. If the elevation does not change, the pressure decreases as the fluid moves into the contracted part of the tube. 3.2x
- 80. 3B.14 Torricelli's equation for efflux from a tank From Eq. 3.5-12 we get 1 àlvezfflux _ + 507mm _ patm ) + . - = 0 Here it has been assumed that the velocity at the surface is Virtually zero, that the pressure is atmospheric at both "1" and "2", and that the datum plane for the height is at the exit tube. When the above equation is solved for the efflux velocity we get Torricelli's equation.
- 81. 3B.15 Shape of free surface in tangential annular flow a. The velocity distribution is given by Eq. 3.6-32, and the equations corresponding to Eqs. 3.6438 and 39 are: (ã*: 'ííf]2%[(ê)“2+(; :-T] Integration of these equations gives (see Eqs. 3.640, 41, and 42) n, ZR 2 R 2 2 ¡màíl-KKZ) [(7) -4lnr+(â) ~pgz+C Now let p = pau, at r = R and z = zR, where zR is the height of the liquid at the outer-cylinder surface. Then we can Write at r = R and Z = ZR SL-KZR 1 - K2 2 Patria( ) [-1-41rIR+1]-pgzR+C which is the equation that determines C . When we subtract the last equation from the equation for p, we get 2 2 p-patm (-à_4m§+§2) _pglz-ZR) The equation of the liquid surface is then the locus of all points for which p = pm, or 1 sz, 212 2 1 ZR“Z= ã[ llçlçz] b. When the outer cylinder is rotating, we can use Eq. 3.6-29 for the velocity distribution to get
- 82. _QOKRZIrZ 1KR2 Win92) 'ZIMíÍTl “P32” Then, we select r = R and z = zR as the point to determine C. rapaz 2 1 1 2 1 9atm=9(1_K2) -21I1R'§(K)2:| °98ZR+C Subtracting, we get P'Patm = §(? f:§)2[(â)2 (32 -4lnê (É) +(K)2]+pg(Zn -Zl From this equation we can get Eq. 3B.15-2 by setting the left side of the equation equal to zero and solving for zR - z. 3-3'
- 83. 3B.16 Flow in a slit with uníform cross flow From Eq. B.6-1, for this problem We have dv“= -:-(? °_? L)+yd2í“ or ü- Ê+I= O dy L dy P00 dnZ in which n = y/ B, «p = v, ,/[(a°o - 'PL)B2/uL], and A = Boop/ y. This equation has to be solved with the no-slip condition at n = 0,1. We write the solution as the sum of a complementary function and a particular integral. The equation 'for the complementary function is . . C with solution c¡ = 11H" + C2 dzÓcr do __-A CF -_-o dnz dn By inspection, the particular integral is 45H = n/ A. Application of the boundary conditions then gives the constants of integration. The final solution is then (with A = Boop/ u) i= .L(n-e””*] or = .(°°°”“°L)B2l[1_e”/ B*1j A e** - 1 b. The mass rate of flow in the x-direction is then _ 3 (B°O-'PL)B3Wpl(l_l+ 1 ) ; LL A 2 A e^-1 c By making a Taylor-series expansion about A = O, from (a) we get q) = &(1; - n2)+ O(A). When A -› O, this result can be shown to be equivalent to Eq. 2B.3-Z. Similarly, A Taylor-series expansion about A: O yields from the result in (b) Agp-; 1 ) [('ú°O-'P, _)WB3p/ yL] A 2 A e^-1
- 84. 1 1 1 1 z_ --_-+ A 2 A (1+A+§A2+§A3+-~)-1 1 1 1 1 _. .__. _+ 2 A A(1+§A+§A2+-~) 1 1 1 1-§A+àA2+-~] A - ---+ A 2 A A = É+O(A) But the "B" in this problem is twice the "B" of problem 2B.3. If we switch to the "B" of Problem 2B.3, it is found that the answers agree exactly. d. For the coordinate system here, we select as the dimensionless quantities Hz, V: vx , :M: b (ñ°0~ñ°L)b2/uL a _u Then the differential equation and boundary conditions are V(i1) = O The solution is then the sum of a complementary function and a particular integral (as before in (c)) V= íe“Y+C2+-X a ü Application of the boundary conditions then leads to -e“Y + cosha + Ysinha asinha V: Then the average value of this over the cross-section of flow is
- 85. _ÍÍVdY 1 _ij-NY + cosha + Ysinha)dY : _(151) sinha + cosha asinha asinha <v>= +1 "' _lay 2 Then we can form the ratio given in Eq. 3B.16-3: l: e” -Ysinha-cosha (V) (1/a)sinha - cosha A As a check on this we can go directly from Eqs. 3B.16-1 and 2 to Eq. 3B.16-3. From the first two equations we get Ay/ B_ x_ Ux _ã _2A[(e^-1)y-B(e^”“-1)] (Ux) _ w/ WB _ à-. .? 81414 ' B[(A- 2)(e^ -1)+2A] Next, we make the Connections between the notations in the two different approaches: y= z+b; B=2b; A=2a (the "y" of part (c) is called "z" here, and Ç = z/ b). Then . vx a[(e2°' - 1)(z + b) - 2b(e(“/ bx”") - 1)] (Ux) b[(a - 1)(e2“ - 1) + 2a] z (em - 1)(ç + 1) - 2(eé? e" - 1) = (ea - e** )(~ç + 1) - 2(e°'? - e-a) (1/oc)[(a - 1)(e2“ - 1) + 2a] (1/a)[(a - 1)(e°' ~ fa) + 2ae'°' z de” 'e-a)+(ea +e"“)-2e“ç = Çsinhoncosha-eaç (1/a)[a(e“ + e'“)- (ea -e'“ COSha - (1/a)sinha
- 86. 3C.1 Parallel-disk compression viscometer a. The equation of continuity of Eq. B.4-2 for incompressible fluids, taking into account the Symmetry about the z-axis is just Eq. 3C.1-6. The equation of motion in Eq. 3C.1-7 comes from Eq. B.6-4 ignoring the hydrostatic pressure, the inertial force terms, and ornitting the terms that are small. b. Equation 3C.1-7 can be integrated with respect to z to give 1 d v, :E2522 +Clz+Cz The constant C2 is found to be zero from the boundary condition in Eq. 3C.1-8, and C¡ is found from Eq. 3C.1~9. c. Integrating of Eq. 3C.1-6 with respect to z from O to H Hi3_ . Lie _ -_-vo l 0 ralr 2” drz(z H)]dz- _Ío dv, Performing the integrations then gives -filipdij z U 12g r dr dr ° d. Integration of the equation in (c) then yields 12m) r¡ " p: The integration constant C¡ must be zero, since the pressure is finite at the center of the disks, and C2 is determined from Eq. 3C.1-10. Equation 3C.1-13 is thus obtained. e. The force on the upper plate is then 2 2 2 m) = j: j: [1 - ]rdrd6 = 27:18 ; (1 ~ : Q5111: The integral is 1 / 4, and this leads to the result in Eq. 3C.1-M.
- 87. f. In this situation, the radius of the glob of liquid R(t) and the instantaneous disk separation H (t) are related to the sample volume Vby V = 7T[R(t)]2 H (t). Then the force acting on the upper disk is m) = i: nlRm [1 - (J-jz ]rdrd9 ° [H<t›]3 R(t) z 37ruv0[R(t)]4 z, 3,uv0V2 2[H<t>]'°” 2›r[H<t›]5 g. If, in Eq. 3C.1-14 we replace v0 by - dH/ dt, we then have an ordinary, separable differential equation for H (t). Integration gives HdH 2P t ° ídt= wñí 37c_uR4 ° whence 1 1 4m = _+ [H<t›]2 Hã snuR**
- 88. 3C.2 Normal stresses at solid surfaces for compressible fluids First write the equation of continuity for a compressible fluid as a 1 ãlnp r "W ' V) ' ; W 'Vpl The normal stress on a surface perpendicular to the z-axis is Tzzlz= o = (-2u a” «su - ›<)<v~v›] âz z=0 = 2p avx+ü+íhip+l(v-Vp) +(lp-K)-í1np+l(v-Vp) 8x ây ât p o 3 ât p z=0 â = (gp + xíãlnp) z=0 The terms containing v drop out by the no-slip condition at the surface, and their derivatives with respect to x and y drop out on the surface as explained in Example 3.1-l. This result shows that the normal stresses at surfaces are zero for compressible flow if the flow is at steady state.
- 89. 3C.3 Defonnation of a fluid line The curve at any time t is 9(r, t) = (vg/ Qt, Which in tangential annular flow is (from Eq. 3.6-32) 2(R2/r3)§2¡t dr 6(r, t)= [ ]Q¡t and 116:- 2 (l/ K) -1 (l/ K) -1 The differential element of length along the curve is given by 2 4(R/ r)4(s2,«t) ((1/›<)2 - 1)2 The total length of the curve is then 4(R/ r)4(27tN)2 2 2 drzRf: 1+ d§ ((1/I<) -1) V ((1/›<) -1) é** To get a rough, order-of-magnitude estimate assume that N is large and then the "1" can be neglected and the integral performed analytically (dZ)2 = (dr)2 + (rd6)2 = (dr)2 1+ + Z= JlidÍ= IÍR 1 ~47INK l Í 1+1( (limit of large N)
- 90. 3C.4 Altemative methods of solving the Couette viscometer problem by use of angular momentum concepts a. By making an angular momentum balance (actually the z- component of the angular momentum balance) over an annular region of thickness Ar and height L we obtain O (2mL) - (me )| r - (27r(r + Ar)L) - (11,9 )| r+Ar Dividing through by 27tLAr and rearranging we get (rzrm) 2 - r r) r+Ar < 79 . '=0 Ar whence É( 21,, )=o the second form resulting from taking the limit as Ar -› O. Then using Eq. B.1-11 for the stress-tensor component, we have i si fa) - drlr dr(r j-O whence U9 = C27' From this Eq 3.6-2O follows. b. Here we start with Eq. 3.4-1, which simplifies to the following for the symmetric stress tensor [v. (rxz)*]= o The z-component of this equation is 3-37
- 91. 1d frx1]; )=0 or -í-àé(rírxrlzr)= o where, in cylindrical coordinates, r = õ, r + õzz. We now work out the cross product, which is [r x 1]” = em¡ (ô, r + 822)¡ r]. , : agarra +sz9r(O)r, , I 1 Hence the equation of change for angular momentum simplifies to 3%( bel” and the development proceeds further as in (a).
- 92. 3C.5 Two-phase interfacial boundary conditions a. This result follows at once from Eq. 3C.5-l, when the viscous-stress-tensor terms are omitted. b. To get the right side of Eq. 3C.5-B, it is evident that Eq. 3C.5-l had to be multiplied by 1/ plvã . The interfacial-tension term in Eq. 3C.5-s is then nit+tli ° 11~1e+e111 °1 R¡ R2 lovãpl R¡ R2 lovâp¡ The tenns involving the viscous stress tensor are lLllz_“_ll_nl_j_l_l-_fEjl_lí7iz_[nrxqu u' 12 12 12 P770 P170 P770 7o [ni_Till= _un[nlÍYH]: _FII[HI_7IIJBQ= _[HIin] un BI: p'vã p'vâ plvã lo lovop" p' And finally, the pressure terms are converted to modified pressure terms plus terms involving the gravitational acceleration HIV' -ro +p'g(h-ho) _ P" "Po +1o“s(h-ho)] _nx 3(h"'o)(PI “Pnl I2 12 I2 , DUO DUO DUO ntslit]-nitlt“;1”"Ji We see that the Reynolds numbers for the two phases, the Weber number (Eq. 3.7-12), and the Froude number (Eq. 3.7-11) appear as well as the density ratios for the two phases. 5-14
- 93. 3D.1 Derivation of the equation of motion from Newton's second law of motion a. Equation 3D.1-1 is the statement that the time rate-of- change of momentum is equal to the sum of the surface forces and the gravity forces acting on a small blob of fluid. When the Leibniz formula (Eq. A.5-5) is applied to the left side of Eq. 3D.1-1, we get d â a - fpvdV= _i -pvdV+ jl(pv)(v-n)dS= _l -pvdV+ I[n-pvv]dS dt va) vo) a¡ su) V(t) at - su) a . = j -pvdV+ j[v-pw]dv (usmgEq. A.5-3) V(t) V(t) The term containing the stress tensor in Eq. 3D.1-1 can also be rewritten as a volume integral using Eq. A.5-3 to give a _ j ãpvdV= -Jl[V-pvv]dV-I[V-1I: ]dV+ _ÍpgdV V(t> V(t) VU) V(t) Since the choice of the blob volume was arbitrary, all the volume integral operations may be removed, and we obtain the equation of motion of Eq. 3.2-9. b. If the blob is fixed, then we can write a momentum balance over the blob as follows: ; ÊIpvdV= -J. [n-pxrv]dS-_Í[n-1t]dS+JpgdV This states that the rate of increase of momentum within the fixed volume equals the rate of increase because of convective transport, the rate of increase because of molecular momentum transport, and the force acting on the system by gravity. The time derivative can be taken inside, since the volume is fixed, and the surface integrals can be converted to volume integrals. The result is an equation . containing only volume integrals over the fixed volume: jzãpvdvhnvpwyzv-nv-zqdwjpgdv V V V V : sw
- 94. Since the volume was chosen arbitrarily, the volume integrals can be removed, and, once again, the result in Eq. 3.2-9 is obtained.
- 95. 3D.2 The equation of change for vorticity Method I: Start with the Navier-Stokes equation in the 'D/ Dt-form, but rearranged thus: 8V 1 -at-= -[v-Vv]-; Vp+ vV2v+g = -Vàvz +[vx[Vxv]]-%Vp+ vV2v+g Next we take the curl and introduce the vorticity w = [V x v] íw-= [Vx[vxw]]+ vVzw a1 01' aw í= [V-wv]-[V-vw]+vV2w Then using Eq. A.4-24 and the fact that (V -v) = 0 for incompressible fluids and (V -w) = 0 always (since the divergence of a curl is always zero, we get Eq. 3D.2-l. Method II: Start now with the Navier-Stokes equation in â/ ãt -form = -[V -vv]-lVp'+ vV2v+g à ât p Take the curl of this equation and introduce the vorticity to get à? = -[V x [V + vVzw 01' à? = [ez(vv-vv]]-[v-vw]+ vVzw
- 96. Details of the manipulations involved in this last step are given here using the abbreviated notation of §A.9 with the Einstein summation Convention: * -[V x [V - vv]] = ~s, ¡kâ¡(â, v,vk ) = -s¡¡kâ¡(v, â,vk - vkâ, v,) but 8,12, = O i = -8¡¡k[(â¡v, Xâmk) + (v, â¡â¡vk = s¡k¡(â¡v, )(â, vk) - (v, â,(ei¡kâjvk = [ezfVv - VVH¡ - [v - V[V X VH i in Which e = E226 ô õ õ is a third-Order tensor. mnrmnr
- 97. 3D.3 Altemate form of the equation of motion Take the divergence of the equation of motion for an incompressible fluid in the form of Eq. 3.2-9, but with the stress- tensor term written in terms of the Viscosity and the Laplacian of the velocity. This gives o= »<v~1v-w1›-lv2p P OI' O = -(Vv: (Vv)I ) - àVzp Then use the definitions in Eq. 3D.3-2 to get Eq. 3D.3-1.
- 98. 4A.1 Time for attainment of steady state in tube flow. a. . ln Figure 4D.2, the centerline velocity comes within 10% of its final Value when ut/ R2 = 0.45, giving t = 114512171/ = (0.45)(0.49 x 10-4 m2)/ (3.45 x 10'** mz/ s) = 0.064 s b. If Water at (68°F=20°C) is used, with u = 1.0037 >< 10's mz/ s from Tables 1.1-l and F .3-6, the time required is t = (o.45)(o.49 x 10"* m2)/ (1.O037 x 10-6 mz/ s) = 22 s
- 99. 4A.2 Velocity near a moving sphere. From Eq. 4.2-l4 at 9 = 7r / 2, the fluid velocity relative to the approach velocity falls to 1% of 'voo at v9 = -0.99v°o relative to the sphere, giving -°›99=~[1eâ(? )›â(? )3 with R/ r < l. If R/ r « l, the cubic term will be unimportant, giving 3 R 0.01 _ Z r 300 125-? ” Clearly, the neglect of the cubic term at this distance is justifiable. OI'
- 100. 4A.3 Construction ofstreamlines for the potential flow around a cylinder. In the following drawing we show the construction of the streamline I/ = à by the method described in the problem statement.
- 101. 4A.4 Comparison of exact and aproximate profiles for ñow along a fiat plate. 3 13 zl/ ísñy' Let IL, = vI/ voo and Y = yy/ voo/ uz; then Eq. 4.4-18 gives the approximation 1 13 3/2 __ 3 2 (280) Y = 03232091* - 0005002373 II u. . and Fig. 4.4-3 shows Blasius' “exact” IT. , vs. Y. These two velocity representations will now be compared. 7 Location, Approx. 1T”, “Exact" H”, Approx. IT. , / Exact 1T. , Y Eq. 4.4-18 Fig. 4.4-3 a. 1.5 0.468 0.49 0.96 b. 3.0 0.835 0.84 0.99 c. 4.0 0.973 0.96 1.01
- 102. 4A.5 Numerical demonstration of the von Karman momentum balance. a. The integrals in Eq. 4.4-13 are I¡ = f pv, (v, ._ - v, )dy and Ig = f p(ve - v, )dy 0 o Figure 411-3 gives f' = vI/ vm as a function of a dimensionless coordinate, Y = yy/ vco/ (uz) and ve = voo in this geometry. Thus', v, = voof' and pdy = g/ ppx/ vmdY, so that these integrals take the forms - OO O0 I¡ = Mppvmzf f'(1 ~ f')dY and Ig = Jppvooxf (1 ~ f')dY o o Numerical evaluation of the integrals over Y then gives OO 00 f f'(1 - f')dY = 0.664 and f (1 - f')dY = 1.73 o o if an accurate table of the solution is used. The following calculation was made from Fig. 4.4-3: Y 0 1.0 2.0 3.0 4.0 5.0 6.0 f' 0 0.34 0.63 0.844 0.955 0.983 1.00 (1 - f') 1. 0.66 0.37 0.156 0.045 0.017 0.000 f'(1 - f') O 0.2244 0.2331 0.1309 0.0430 0.0167 0.000 Application of the trapezoidal rule gives the values I¡ = [1/2 + 0.66 + 0.37 + 0.156 + 0.045 + 0.017 + 0/2] X 1.0 = 1.74 Ig = [0/2 + 0.2244 + 0.2331 + 0.1309 + 0.043 + 0.0167 + 0/2] X 1.0 = 0.65 which agree, within their uncertainty, with the accepted values 1.73 and 0.664. b. Use of Eq. 1.1-Z and the results of a. in Eq. 4.4-18 gives dI = +012 -1/2 = ossi/ ppvgf 2 = 0.325¡/ p;1v§°/ z c. The force in the x-direction on a plate of width W and length L, wetted on both sides, according to the result in b, is L F, = ZWIO Tw| y=0dz L =2W(0.325¡/ p¡1vâ°)f z-Wds 0 = 1.3O¡/ ppvã°LW2; The recommended coefficient is 1.328.
- 103. 4A.6 Use of boundary-layer formulas. The data for this problem are: W=10ft L=3ft v°o=20ft/ s From Table 1.1-2 and Appendix F: u = (01505 cm2/s)/ (12 X 2.54 cm/ ft)2 = 1.62 >< 10'** ft” / s p = (0.01813 mPa~s)(10'3 Pa/ mPa)(6.7197 x 10'* lbm/ ft-s/ Pa) = 1.218 x 10v" lbm/ ft-s p 2 11/1; = 1.218 x 10"5/1.62 x 10** = 7.5 x 10-2 lbm/ ft3 a. The local Reynolds number at the trailing edge (z = L = 20 ft) is: Re = Loco/ u = (3ft)(20 ft/ s/(1.62 x 10'** ftz/ s) = 3.7 x 105 b. According to Eq. 4.4-17, the boundary layer thickness at the trailing edge is 6(L) = 4.64,/1'-Ii voo _ (1.62 x 10-4 ft2/s)(3 ft) _ 4.64, / --_-_----20 WS = 2.4 x 10"** ft c. . According to Eq. 4.4-30, the total drag force of the fluid on both sides of the plate is F, = 1.328y/ ppLW2v§° = 1.328¡/ (0.075 lbm/ ft3)(1.22 X 10“5 lbm/ ft-s)(3 ft)(10 ft)7(20 ft/ s)3 = 0.62 lbm-ft/ sz = 0.019 lb¡
- 104. 4A.7 Entrance fiow in conduits. a. With the indicated substitutions, Eq. 4.4-17 gives UL. ; vmax 1 -D=4. 4 2 6 Setting uma¡ = 2(v) at the end of the entrance region, we obtain the following estimate of Le: D 2 9.28 __ 2 D2 (v) ” (92292 1/ = 0.023DRe which is similar to the expression given in §2.3, except that the coefficient is about 2/ 3 as large. VL, __ 2(v) T 01' Lc b. At the typical transition locus muco /1/ z 3 >< 105 for flow along a fiat plate, Eq. 4.4-17 gives 6/1: = 4.64, / L uma: = 4.64(3.5 x 1050-1” = 0.00847 and the transition Reynolds number based on the characteristic length 6 is 0006 _ voor 6 V V Z = (3 x 105)(0.00s47) = 2542 v For flow in tubes, with transition occurring when 6 = D/ 2 and with vo. , = 2(v), the latter result gives D(v)/ 1/:2542 as the minimum transition Reynolds number, in fair agreement with the reported value of 2100. c. For laminar flow between parallel planes, the method in Problem 4.C gives 6 = B and uma¡ = (3/2)(v, ) at the end of the entrance region. Insertion of these results into Eq. 4.4-17 gives VL, B = 4.64 1.5(v, ) whence 1.5 B2 (1), ) (4.64)2 11 = 0.070B7(1›, ) / u = 0.070BRe With Re= B(v, )/u Le:
- 105. 4B.1 Flow of a fluid with a suddenly applied constant wall stress a. Differentiation of Eq. 4.1-1 with respect to y gives 32v, z viam Or 3p” 80x): Vip” 8o, ] âyât 81 912 ât ôy 612 ay Then using Newton's law of viscosity, we get 2 dry, = V8 ty, at âyz b. This equation is to be solved with the initial condition that ty, = 0 for t5 O, and the boundary conditions that ty, = T0 at y = 0, = 0 at t: 0°. c. The solution is exactly as in Example 4.1-1 with appropriate Changes of notation, and the solution is given in Eq. 4B.1-1. d. TO get the velocity profile, we integrate Newton's law of Viscosity: 1 0 _ ____ : E ao _ ; V _ fvxdvx- H yryxdy or v, #fil erfx/ Êjdy Changing variables we get 'Z' eo v, = ;2«/4vt_[y/ Jm(1-erfu)du g_ _Lu/ ix_ 17g y ixMvtL/ Eey 1 erf The velocity at y = 0 is then j4vt I t vx(0,t)= í : :Zico -_ IJ 7¡ "HP 4~3
- 106. 4B.2 Flow near a wall Suddenly set in motion (approximate solution) a. Integration of Eq. 4.1-1 over y gives «do 232v d s au t” 4d = -J-d h - d = x plo at z plo ay, z w ence dtjo pv, z u ay F0 Since the velocity gradient is zero at infinite distance from the plate, we end up with Eq. 4B.2-2. b. We introduce the variable ' n = y/ ô(t). Then when Eqs. 4B.2-3 and 4 are substituted in Eq. 4B.2-2, We find âõ(t)lãpvs(lrãn+àn3)dn= -uvm(-%+-â-n2)In=0g:7) - Then after dividing by pve, and evaluating of the integral, we get 3 d 3 1 - -V-à 2 6a) (see Eq. 4B.2-S) c. Eq. 4B.2-5 when integrated gives [fada = 445111 Or a0) = «sw Then the velocity profile is given by v, 3 y 1 y r E --=1-- --- - -- f _ _V8 v_ ztvsvtlüix/ zavt °r 0<y< vt 34:1 for y. >. 8vt
- 107. 4B.3 Creeping flow around a spherical bubble a. According to Eq. B.1-18, the vanishing of the shear stress is rí(v_9)+-1-av*=0 orri 1 + âr r r86 dr r rsin6âr In the second form, we have inserted the expressions for the velocity components in terms of the stream function from the last line in Table 4.2-1. Next we insert 1;/ = f(r)sin2 6 and obtain Eq. 4B.3-l. b. Equations 4.2-7 through 10 are still valid for this problem, as well as the values of C3 = -àvm and C4 = O. Hence We have to determine the remaining constants in f (r) = C11** + C21' by requiring that Eq. 4B.3-1 be valid at r = R, as well as f = O at r = R (Eq. 42-3). These boundary conditions lead to C¡ = O and C2 = ívmR. Then Eqs. 4B.3-2 and 3 follow directly. c. When the velocity distributions in Eq. 4B.3-Z and 3 are substituted into the equations of motion (Eqs. B.6-7 and 8), we get 3 2 ? E = Cos 6 and i0: = sin 6 86 R r dr R2 r Integration of these two equations gives ? P = cos6 + F(6) and 3° = cos6 + G(r) -. .h 1 . .ai, n96 r2sin6 86 In order for the solution to be unique, P (6) and G(r) must be equal to a constant. If we require that the modified pressure be equal to po at z = O infinitely far from the sphere, we then get oo R 2 p= p0-pgz-(Hã-)(-) cos6 f' d. The z-component of the force acting on the sphere is p, = 45'10" (s, 15,-gas + TH)L= RR2 sin 6d6dcp
- 108. -27rR2 _Íoqpcos 6 + 1,, cos 9 - 1,9 sin 0)| R sin @dB where Eqs. A.6-28 and 29 have been used for getting the dot products of the unit Vectors. Then we use the result of (c) and Eqs. B.1-15 and 18 for the components of the stress tensor (along with Eqs. 4B.3-2 and 3) to get the three contributions to the force: _ n yum R 2 . P”, - -27IR2IO (po _pgz cos B] r= R cos @sm 6:19 = §7rR3pg + ânuRvm 2 II aU, . ” Fm” = -2rtR [O [eu ar -cos e) sm ado r= R = +47mR2v°° fãêmosz O] r sin 6d9= â ? If/ JRU” r= R (-sin 9)sin6d9=O r= R rã? râü = -ZNRZÍÃY-IO( a [v9] + 1%) When these are added together, we get Eq. 4B.3-S.
- 109. 4B.4 Use of the vorticity equation a. For the postulate vz = vz(x) the vorticity w = [Vxv] has but one component w, =-dvz/ dx. Then at steady state, the y- component of Eq. 3D.2-l is (vvíígcz) = VVZÍÊÍ) +(w-V)vy But vy is postulated to be zero so that the last term drops out. Also the first term drops out, because v, and v, are zero, and vz is postulated to have no z-dependence. Consequently the equation simplifies to 113v, /dx3 = O. Integration gives _ 2 _ 2 vz-àClx +C2x+C3 or vz/ vzlmax-Dlê +D2§+D3 where we have redefined the integration constants. From the three boundary conditions it is found that D¡ = -1, D2 = O, and D3 = 1, Eq. 4B.4-1 results. Then the z-component of the equation of motion becomes at steady state d? d? dzv, p(v - V)vz = -íz- + uVzvz or - dz dx¡ Knowing the velocity distribution we can evaluate the second derivative of the velocity and get the pressure distribution thus d? 2 O = -E + yvzlmaxpí) and 2 P0-'ñ°(z)= -B-Z$íz For z = L, this gives vzlmax = (P0 - 'tPL)B2 /2_uL in agreement with Eq. 2B.3-Z. b. Since it is postulated that v¡ = v, (r), the only nonzero component of of the vorticity vector is we = ~dvz/ dr. Then the 9- component of the steady-state vorticity equation is: 442.
- 110. This simplifies to d 1 d dv d: = O or vz = &Clrg + C2 lnr + C3 This is the same solution found in the soution to Problem 3B.6. When the two boundary conditions and the conservation-of-mass condition are used, we finally get the solution in Eq. 3B.6-2. To get the pressure distribution we use the z-component of the equation of motion, which is dr Hrdr dr When the velocity distribution of Eq. 3B.6-Z is inserted, we get This may then be integrated to get the pressure distribution.
- 111. 4B.5 Steady potential flow around a stationary sphere a. The boundary conditions are: (i) as r -› oo, v -› vwõz, or by using Eqs. A.6-28 and 29 as r-›oo, v, :vw cos9 and v, = -vmsin9 (ii) at r= R, v, =O (iii) at z=0 as r-›oo, 'tl°-›? O b. Since v = -Vçb, we have as r -› oo -âv/ âr = v, cos 9 and -(1/r)âq>/ â9 = -v_, sin9 When these equations are integrated we find that cp = -vmr cos 9, Thus we feel that v = f (r) cos 9 may be an appropriate trial function. c. We next write the 3-dimensional Laplace equation in spherical coordinates (for a system with Symmetry about the z-axis) 1 3 â 1 . 3 ? gr-bz + rz sin9(Sm6ã%) = O Into this equation we substitute the trial velocity potential and get 1 d d . . _ ; íãpz-àj-: J-Zriz-: O whichasthesolution f= CIr+C2r 2 since this is an "equidimensional" equation of the form of Eq. C.1-14. d. Application of the boundary conditions then gives f = -vwRíâ + and q) = -U, ,R[Ê + Jcos 9 e. Then from the components of v = -Vdn we get Eqs. 4B.5-2 and 3 by differentiation. f. Then from the equation of motion for steady potential flow p(V §- vz) = -VÍP (see Eq. 4.3-2) By integrating the components of this equation we get 4-14
- 112. _ v = pví[[1-(§)3]cos2 e +[1+â(§)3]sin2 9J+C3 The integration constant is then obtained from the boundary condition given in (a): C3 = -'0°0 -pví Then when the modified pressure is evaluated at the surface of the sphere, Eq. 4B.5-Al is obtained directly.
- 113. 4B.6 Potential flow near a stagnation point a. At the origin of coordinates (z = O) the complex velocity dw/ dz = -2voz is zero, which is a stagnation point. b. By taking the real and imaginary parts of the complex velocity, we get from Eq. 43-12: v, = 2v0x and v, = -2v0y. c. When v0 is positive, the fluid is flowing toward the surface y = 0 in the upper half plane. The magnitude of v0 specifies the speed with which the fluid is flowing: v = Jvf + v; = 2vor.
- 114. 4B.7 Vortex flow a. By using Eq. 43-12, we find that dw_zT 1 _iF í _zT x-iy _ . É-E É _2-7; : É _ã x2+y2 --vxHvy in which the overbar indicates the complex conjugate. Equating the real and imaginary parts gives T' (sin9) F x F (cos9)- r 27: x +y 275 r The components in cylindrical coordinates are G H | | I IN) al” /17 H N) "l-Q § N¡ k. ; | | N) : a v, =vxcos9+v, sin9=O; v9 = -v, sin9+v, cos9=L(l) 27¡ r b. The forced Vortex is given by v9 = S21* in Eq. 3.6-37.
- 115. 4B.8 The flow field around a line source a. For this purely radial flow v, = v, (r) and the other two components are zero. Then Vzq) = O simplifies at once to Eq. 4B.8-1. b. Integration of Eq. 4B.8-1 gives q) = C1 lnr + C2. Then, since v, = -dqõ/ dr, we find v, = -Cl/ r. Next we calculate the volumetric flow rate per unit length thus I' = _fãnv, rd9 = -27rCl whence v, = L- 2m The pressure distribution is then obtained from the radial component of the equation of motion o, pqry- r par: 27rr 27172 dr Integration then gives m_ r 2m 1 p r 2 f? dñ°= p(2_n) j; [-r-3)dr or Fg-3°= -2-(_-) dv, _ di? dr _d_r Pvr
- 116. 4B.9 Checking solutions to unsteady flow problems a. Substituting the solution of Eq. 4.144 (or 15) into Eq. ,4.1-5, we have to verify that 8 2 _-2 _ 82 _-2 _ all? " M): “iv-PÉ " d") We have to use the chain rule of partial differentiation along with the Leibniz formula ãÊ(Io"e*ñ2dñ)'Ê? =“§(ío”e'ñzdñllâlz OI' visita): v<ei~w»»rsr When the definition of 77 is used, the above is found to be an identity.
- 117. 4C.1 Laminar entrance flow in a duct. a. . Calculation of the mass How rate according to Eqs. 4C.1-l and 4C.1-2 gives w = 72(vz)WB = ,wewf [2 - (%)21dy+ pveW / &B[1]dy = puew [a - É] + pveWIB - a] = pveVV [à + B] Equating the first and last expressions for w, we get (with õ < B and A = 6 / B), ve(z)= (v1›BÍ, ,3 = <v1)3ÍA The following related equations will prove useful in part b: dve = (vI)__3___5J__A_ da: (3 - 4)? dx = (avâíjgíâ $935) = ,g [(1,9, 3,9291] “b, The boundary layer in this system lies between y = 0 and y = 6, so those limits suffice for the integrals in Eq. 4.443. Evaluation of the terms in Eq. 4.4-13 (divided by p) according to the results in a. then gives: 617, _ 277v, _ 61/(113) "ã eo - . s - M375) â: '/06 17,079 - v, )dy = dj1ívÍõ)/ o1[(2u - u3)(1~ 211 + u2)]du with u = y/ õ = [211 - 4112 + 2113 - 112 + 2u3 - u4]du = ;(12111 - (473) + (274) ~ (173) + (274) - (175)) = Em? É 127151 4 -lo
- 118. d», 5 d , , 1 dz [O (ve-vJdy-; veõ d: f [1-2114-112]du 0 dve = 1196 dx [1 - 1 + (1/3)]d11 Z 9<U¡)2B A E : :- 1 3 (3 - A)3 dae l / ] With these substitutions, Eq. 4.4-13 gives 27+9A]dA A : lá = B(vz)2 ÉIÍQÍÍ5l+9WÀ2BW d, l1/3l _ B(v, )2 dA 54+18A+45A “m-_Arãl 15 l _ Em? dA 54+63A Wa-maãl 15 l <U¡)B2 Multiplication of both members by (5/3)A(3 - A)/ (B(1›, )2 gives 10 ” )_É_+_7ÉÍÊÉ (v, )B7 _ (3)-AV da: in agreement with Eq. 4C.1-ll. c. Integration of the last equation with the initial condition that A = 0 at : c z: O gives 11:1: 1 4 63 + 752 = _ d (M32 1o o (3-s)2 5 From integral tables we get the formulas 8 s 1 a / › d$- b? [ln(a+bs)+a+bs] a2 a+bx 2 [ ds= í13- [a+ba: -2aln(a+bx)- which yield the definite integrals / tiA (s -5 sv” = ll” + [i (3 : zsyds = [(3-A)-61n - à] l+c 4,21
- 119. and the solution iowffB, _(3 [ln (334) + 33,] +7 [(3-A)-6ln - à] 3-A ôA 63 _7A+4s1n(›-3 )+--3_A+21_-_3_A 3-.4 6A~21(3~A)-63 _7A+4s1n<-3 j+--í 3-A 2721 : A 1 _-_ É 7 +481( 3 )+3_A in agreement with Eq. 4C.1-õ. d. Setting A = l and x 2: Le in the last equation gives e z T6 = (0.1) [7 - 19462325 + 13.5] = 0.104 L 1 7+48ln(2/3)+Ê2z] in agreement with answer e. Application of Eq. 4.3-5 to the region 6(: c) < y < B, with v9 neglected so that 17 : v9 there, gives 1 -ipvf + 'P = constant Insertion of the result of part a gives lp(v )2 i- 2+'P= constant 2 ” 3-4 Evaluating the constant at : c = O, where A = 0 and 'P = P), We get 1 3 2 1 5907:? (íÁ) + 7° = 9mm* + 73o p = ;,112 [12 (9.3-An OI' 4-22

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