MEN 321 Problems For Chapter 8 with Solution, Fall 2009

&.1 8.1 Rainwater runoff from a parking lot flows through a 3-ft-diameter pipe, completely filling it. Whether flow. in ’ a pipe is laminar or turbulent depends on the value of the Reynolds number. (See Video V8.1.) Would you expect the flow to be laminar or turbulent? Support your answer.with appropriate calculations. Re = oye = VD If Re>#000 the flow is turbulent, The corresponding velocity iso y= Rev. (zoo0)(1.2Ix1o > £ ) ~ DO” 3 fF Most likely the velocity will be greater than this, ¢., turbulent how. = 0.0161 # 3.2. enter a pip P8.2. Woul problem if 8.2 Blue/and yellow streams of paint at 60 °F (each with a den- sity of 1.6 slugs/ft? and a viscosity 1000 times greater than water) Yellow Separate streams of blue and yellow paint? Explain. Repeat the | Green? 2 in. e with an average velocity of 4 ft/s as shown in Fig. Spli id you expect the paint to exit the pipe as green paintor =| the paint were “thinned” so that it is only 10 times caren 25 tt more viscous than water. Assume the density remains the same. ALS j FIGURE P8.2 Tf the flow is laminar the paint would exit as separate blve and ellow streams. ’ evo _ _eVD _ 16 HE (48) (#A) Re= He ~ “7000 fky,g ~ 1000 (2.3#xi0°* Bs) Thus, laminar flow so bive and yellow streams, If use fp =/04y, 9 obtain Re = #560>4000 so have forbulent flow with natural mixing and green paint. Note + Check to determine if the 25 f! length js greater than the entrance length, J,. For laminar flow 2 =0.06 Re, or 4,=0.06 (45.6)(% fl) =0.456 f<25H For turbulent flow 42 =44 Rel, op by = %4(¥560)4(7B i) = 2.99 ft <25H = $5,6 <2/00 0 20 40 60 80 8.3 | 83 The flow of water in a 3-mm-diameter pipe is to remain laminar. Plot a graph of the maximum flowrate allowed as a function of temperature for 0 < T < 100 °C. For laminar flow Re= vp $2100 , where ye$= zo Thus, the maximum Q is given by Re _ (os)D = #8 _=32/00, or Q= 21007 VD V 7wvbd v7 or ( , ; 00 7 (0.003 Q= 2/0 7 oc my, = ¥95 V , where yr and Qr~ With values of ¥ from Table B.2 we obtain T,degC v,m%s Q,miis 1.79E-06 8.86E-06 Flowrate vs Temperature 1.00E-06 4.95E-06 6.58E-07 3.26E-06 1.E-05 4.75E-07 2.35E-06 3.65E-07 1.81E-06 8.E-06 + 100 2.90E-07 1.44E-06 2 6.E-06 £ Gg 4-06 | 2.E-06 0.E+00 | ) 20 40 60 80 100 T, deg C 8-3 8.4 8,4 Air flows through a 0.08-mm-diameter passageway in a person’s lungs at a rate of 10°” m*/s. Is the flow laminar or turbulent? Support your answer with appropriate calculations. If Re= vp < 2100 the flow is laminar. Ve = 40 _ #10) _ A 7 D® ” F (.08%10%m)y Thus, Re= (0.199 #)(0.08x10 m) e 1. ¥6 x/0* = =0./992 = 1,09 <2100, £.e. laminar flow [Ee [a5 J 8.5 Carbon dioxide at 20 °C and a pressure of 550 kPa (abs) flows in a pipe at a rate of 0.04 N/s. Determine the maximum diameter allowed if the flow is to be turbulent. For turbulent flow, Re = evo > 4000 | where Q=VA -fp’y or Re Oe = £6. = $000 Thus -5 Ni ‘D= oar , where geQ= 0.048 and w=L4 x10”. (Table 8) Hence, y (0, o4ty( ES) D= 4000 7 (1,47 xjo* HS) = 0.0688 im [8-6] x (m) (+0.01 m) p (mm H,0) (+5 mm) 5.0 (pipe exit) 8,6 The pressure distribution measured along 0 (tank exit) 520 a straight, horizontal portion of a 50-mm-diam- 0.5 427 eter pipe attached to a tank is shown in the table 1.0 351 below. Approximately how long is the entrance 15 288 length? In the fully developed portion of the flow. 2.0 236 what is the value of the wall shear stress? 2.5 188 3.0 145 3.5 109 4.0 73 4.5 36 0 The. entrance length extends to the fully “en, portion in which %€ =canstant. Approximate %~ $2 to obtain the following: From X= | fox=l)m 5p, mm H.0 a an sie Hao 4) 05 -93 OSs -/86 OS 10 ~76 0.5 ~/52 Lo LS -63 0.8 ~/26 1S 2.0 -52 0.5 -/04 2.0 2.8 - 48 0.5 -96 25 3.0 -43 0.S - 86 3.0 3.5 -36 0.5 -72 3.5 HO ~36 0S -72 40 4S ~37 Os ~74¥ 45S 5.0 -36 O05 -72 Within the error on &p, the pressure gradient is constant for X23,m Thus, 43m. For X>3m , aa 72 lm HO rom Hal fo VA0 im 7800/%) Since | mm t,o x ¥, =9008, then Ap _ mmth® f 9,80 m2 f= 72 m mm K, set) 706 i Since ap = Yiuk if follows that 2 “f = " o.0g0n (706!) = 8.93é 8.6 8.7 8,7 A soft drink with the properties of 10 °C water is sucked through a 4-mm-diameter, 0.25- m-long straw at a rate of 4 cm?/s. Is the flow at the outlet of the straw laminar? Is it fully devel- oped? Explain. 3,,72m 3 _ 4 10 Gm . _ -6 mt v=-£ = pee =0.3/8% Thus, with V=1.31 x10 3 (see Table B2) 318 & (4x10, a, . Re = VD. era m) = 471 <2/00; it is laminar flow. Hence , be Dp =0.06 Re , or L, = (0.06)(971)(4x10°m) = 0,233 m < 0.25m or we have fully developed flow at the exit. 8.3 The wall shear stress in a fully developed flow portion of a 12-in.-diameter pipe carrying water is 1.85 Ib/ft?. Determine the pressure gra- dient, dp/dx, where x is in the flow direction, if the pipe is (a) horizontal, (b) vertical with flow up, or (c) vertical with flow down. Ln general, 2p =H sind = a Thos, vith T=% at r=2 and 3h =o this becomes ag = 7 - S sinO a) For a fursontl p pipe O=O ag —_ Hla 48S fy) | Q D lft b) For vertical flow up 9 = 90° ie. Hy _y =~ 4.055) ~ 62.4 2 lb ~ 69.8 16. itt = Ib - 7.40 3 rye) fF and c) For vertical flow down @=-90° 1.85 + eee eee 8.9 The pressure drop needed to force water through a horizontal 1-in.-diameter pipe is 0.60 psi for every 12-ft length of pipe. Determine the ‘shear stress on the pipe wall. Determine the shear Stress at distances 0.3 and 0.5 in. away from the pipe wall. For a horizontal pipe iia or Tet of r 0.6 X44 fim = =r Oe ff) d. 3.6r2 | where reff and with r= (0.5-0.3)in. = 0.2in, T= 3.6 (92) =0,06 0.06 sf Finally, with Pr =(0.5-0,5)in. = O in, T=0 8.10 Repeat Problem 8.9 if the pipe is on a 20° hill. Is the flow up or down the hill? Explain. For a pipe on a hill G = Ft e sin@, where O= Assume the flow is uphill : get Thus, T= EF - ¥ sind] or - =¢ ore -62. wh, sinad] a = - 0,295 1, Hs Since we must have % 70, the tlw must not be uphill. Assume the tlow js down hill: B- -20° m b Thus, rs] -dsin6] or r=_] +62 #1, sinzd’| ft =/£3r $ , Where r~ ft. The flow is downhill Hence, with r= % =/43( 28) = 0.596 ih r= (05-03)in. = 0.2 in., T= 14,3 (92 92) = 0,238 With r=(05-0.5)in. =O , T=O 8/1 8.11 Water flows in a constant diameter pipe with the following conditions measured: At sec- tion (a) p, = 32.4 psi and z, = 56.8 ft; at section = (b) (b) p, = 29.7 psi and z, = 68.2 ft. Is the flow from (a) to (b) or from (b) to (a)? Explain. - (a) Assume the flow is ophill, Thos, a Me +Z, = Bit +Z, th or with Va= Ve , 32.4 psi- is h,= & +z, -& “4 = f pai-24 7psi) 14 ) +56.8 ft -68.24 8 *h, =-5,17 11 <0, which is impossible. Thus, the flow is downhill, from (b) to(a). —EEEEe——————— 8-10