ࡱ> c#` bjbjmm (Q    $B JbJbJbPb^c$B +dfLjfjfjfnZn$ o$;hȉ p6mlnppȉ jfjftttp "jf jftptt"y r yjfd FKЃJbqy~ 0+y8sqbs(yy\s Zz4o0do"too4o4o4oȉȉtj4o4o4o+ppppB B B 4@D"B B B @B B B   INCLUDEPICTURE "http://www.csun.edu/~pubrels/NewLogoSm.JPG" \* MERGEFORMATINET  College of Engineering and Computer Science Mechanical Engineering Department Mechanical Engineering 390 Fluid Mechanics INCLUDETEXT include.doc firstPage \* MERGEFORMAT Spring 2008 Number: 11971 Instructor: Larry Caretto April 15 Homework Solutions 8.5 Carbon dioxide at 20oC 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. To ensure turbulent flow, the Reynolds number must be greater than 4000. This gives:  EMBED Equation.3  We are given the weight flow rate which is the mass flow rate times g. We can use this to get a relationship between the velocity and the area as follows:  EMBED Equation.3  Substituting this expression for velocity into the Reynolds number (and the condition for turbulent flow) gives.  EMBED Equation.3  The dynamic viscosity of carbon dioxide at 20oC and atmospheric pressure is found in Table 1.8 as m = 1.47x10-5 N(s/m2. Since the dyanmic viscosity does not have a significant dependence on pressure, except for very large pressures, we can use this value at the given pressure of 550 kPa(abs). Substituting this value and the other given data into the inequality for diameter gives  EMBED Equation.3 = 8.83 cm 8.6 The pressure distribution measured along a straight horizontal portion of a 50-mm-diameter pipe attached to a tank is shown in the table below. Approximately how long is the entrance length? In the fully developed portion of the flow, what is the wall shear stress? The data in the table below have been used to compute a finite-difference approximation to the pressure gradient. That is we compute Dp/Dx as an approximation to dp/dx according to the following equation.  EMBED Equation.3  Locating the value of the estimated gradient at the mid point of the measured pressure data gives a so-called cnetral difference approximation to the deivative which is more accurate than placing the gradient at either of the end points. The gradient data are also shown in table below along with a plot of the data. The plot also shows a straight line fitted through the last few data poitns and extrpolated back to x = 0.. x (m)p (mm H2O)Dp/Dx 0.00520mm H2O/m0.25-1860.504270.75-1521.003511.25-1261.502881.75-1042.002362.25-962.501882.75-863.001453.25-723.501093.75-724.00734.25-744.50364.75-725.000We see that the magnitude of the pressure gradient decreases with distance and eventually reaches a nearly constant value after x = 3.25 m. In the plotted data the straight line fit appears to match the data to a point slightly below x = 3.0 m. Thus a reasonable estimate of the entry length, to within the accuracy of the data is that the entry length is 3 meters. The relationship between the wall shear stress and the pressure gradient in fully-developed flow can be used to compute the wall shear stress.  EMBED Equation.3  The average pressure gradient in the region where the gradient is nearly constant is 72.5 mm H2O/m. To get this into conventional units we have to multiply by the specific weight of water. Thus, 1 mm H2O = (0.001 m)(9800 N/m3) = 9.80 N/m2. With this conversion factor we can find the wall shear stress as follows  EMBED Equation.3 8.88 N/m2 8.15 A fluid of density r = 1000 kg/m3 and viscosity m = 0.30 N(s/m2 flows steadily down a vertical 0.10-m-diameter pipe and exits as a free jet from the lower end. Determine the maximum pressure allowed in the pipe at a location 10 m above the pipe exit if the flow is to be laminar. The maximum pressure allowed in the pipe will be the pressure that produces the largest flow rate that remains laminar. Since laminar flow requires the Reynolds number to be less than 2100, we have to satisfy the following inequaltiy.  EMBED Equation.3  So we have the result that V < 6.30 m/s; we need to connect this result to the presure. The relationship for pressure drop in laminar flow in a pipe that makes an angle q with the horizontal is given by equation 8.11 on page 412 of the text, where Dp is defined as pin  pout.  EMBED Equation.3  Setting the inequality that V < 6.30 m/s and solving for the inlet pressure gives.  EMBED Equation.3  Since the exit is a free jet, pout = 0. The coordinate system for q takes q = 0 when the flow is in a horizontal direction to the right. A pipe that has a vertical upflow would have q = 90o; a pipe, like the one here, that has a vertical downflow has q =  90o. and sin(q) =  1. We are given the density of the fluid as 1000 kg/m3, so its specific weight, g = rg = (1000 kg/m3)(9.80655 m/s2) (1N(s2/kg(m) = 9807 N/m3.  EMBED Equation.3 pin <  3.759x104 Pa =  37.6 kPa 8.20 Oil flows through the horizontal pipe shown in Figure P8.20 (copied from the text at the right) under laminar conditions. All sections of the pipe are the same except one. Which section of the pipe (A, B, C, D, or E) is slightly smaller diameter than the others? Explain. For laminar flow in a horizontal pipe the pressure drop is related to the diameter by the following equation.  EMBED Equation.3  This equation tells us that the pressure gradient, Dp/l, varies as D-4 if all other terms in the equation are constant. If we assume that the temperature in the pipe is constant, the viscosity and density in the pipe will be constant, and the constant density means that the volume flow rate, Q, will be constant, even if the diameter changes, according to the continuity equation We can compute the gradient in head measured by the piezometer tubes as a measure of the pressure gradient. This is shown in the table below. SectionABCDEDistance, x (ft)1525506696Head, h (in)6056463926Dh/Dx-0.4-0.4-0.438-0.433We see that the gradient is the same (-0.4 in/ft) for the differences between section A and B and section B and C; the first change occurs in section D, where the magnitude of the gradient increases. This indicates that section D is the one with the smaller diameter. To confirm this we can compute the gradient in section E to see if it is the same as the previous gradients, -0.4 in/ft. To compute the gradient in section E, we have to start at the measurement point in section C, where the total distance, x = 50 ft. The distance to the end of section C (which the same as the start of section D) is 10 ft. (We know this because we are told that each section is 20 ft long.) Thus with the gradient of -0.4 in/ft, the head at the start of section D would be 46 in (0.4in/ft)(10 ft) = 42 in. We can then compute the gradient in D because we know that the head is 39 in over a 6 ft length of section D. This gives the gradient in section D as (39 in 42 in)/(6 ft) = -0.5 in/ft. This gradient would reduce the head over the remaining 14 ft section D to 39 in (0.5 in/ft)(14 ft) = 32 in. With this head at the end of section D, which is the same as the start of section E, we can compute the gradient in section E from the measured head of 26 in, 6 ft into section E as (26 in  32 in)/(6 ft) = -0.4 in/ft, the same as at the start of the pipe. This confirms our preliminary conclusion that section D has the smaller diameter. 8.22 Oil of SG = 0.78 and a kinematic viscosity, n = 2.QTUV     " # > ? @ A B H J ̿̿󡖎{{skd]V h;5CJ h)"5CJ h15CJ h1CJaJh6VOJQJhOJQJhzdhOJQJh}xOJQJhnih}xOJQJjhnih}xOJQJUh156CJOJQJaJh8 56CJOJQJh156CJOJQJh15CJ aJ h15OJQJh1OJQJh1jh1U!UV? 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Determine the manometer reading, h. We can write the head loss in the energy equation as Dp/g = Dp/rg =[ f (l/D) rV2/2 ] / (rg) = f(l/D)V2/2g so that the energy can be written to give the following relationship between the downstream point (2) and the upstream point (1).  EMBED Equation.3  We have to compute the Reynolds number as the first step in determining the friction factotr. For this pipe D = 20 mm = 0.02 m and the area = pD2/4 = p(0.02 m)2= 0.001257 m2. The velocity is found from the volume flow rate by the usual equation and then used to compute the Reynolds number.  EMBED Equation.3   EMBED Equation.3  Since Re < 2,100, the flow is laminar and we find the friction factor as 64/Re. With this friction factor, we can find the pressure difference from the energy equation after cancelling the equal velocity terms.  EMBED Equation.3  The manometer reading can be analyzed by starting with the lower level on the right of the manometer. At this point, the pressures on both sides of the mamometer are the same and can be equated to pressures in the pipe by the manometer formula.  EMBED Equation.3  From the diagram we see that the sum h1 + h2 is the length of 4 m between the pressure taps. Furthermore, the ratio of the specific weights of the oil and manometer fluid can also be written as the ratio of the specific gravity of each fluid. Thus we can solve for the unknown height, h, as follows.  EMBED Equation.3 =18.5 m 8.29 Air at standard conditions flows through an 8-in-diameter, 14.6-ft-long, straight duct with the velocity versus pressure drop data indicated in the table below. Determine the average friction factor over this range of data. The usual equation for pressure drop in terms of friction factor can be solved for the friction factor as follows.  EMBED Equation.3  The data shown in the table below have the velocity in ft/min and pressure drop in inches of water. To get the pressure drop in lbf/ft2 we have to multiply the pressure drop in inches of water by the specific weight of water. Doing this and inserting the necessary conversion factors gives the following calculation formula for the friction factor. (The density of air at standard conditions, 0.00238 slug/ft3, is taken from Table 1.7.)  EMBED Equation.3  Applying this computational equation to all the data in the table below gives a set of data for friction factors from which we can compute the average value. faverage = 0.0162 V ft/min҅ԅ;Uer쿶춿vkWGhhWh|5OJQJ\^J'jh!oh|5EHOJQJU\jI h|UVh|5OJQJ\jh|5OJQJU\%jhj5OJQJUmHnHu%jh$Mb5OJQJUmHnHuhhWOJQJ\h0Ch|OJQJ\h0Ch|OJQJ\h|OJQJ\h0Ch|OJQJ\h|OJQJ\h|H*OJQJ\"$*>@Ј*,.0VXZ\^`ujQu:-jhjOJQJU\^JaJmHnHu0jEhhWhjEHOJQJU^JmHnHujI hjUV&jhhWOJQJU^JmHnHuhHOJQJ^JmHnHuhjOJQJ^JmHnHu hhWH*OJQJ^JmHnHuhhWOJQJ^JmHnHu&hhWhhWH*OJQJ^JmHnHu#hhWhhWOJQJ^JmHnHu#hhWhhWOJQJ^JmHnHu`Ư{g[G8-j8I h[UVh$MbOJQJ^JmHnHu&jh$MbOJQJU^JmHnHuh$Mbh$MbOJQJ\'jhjh$Mb5EHOJQJU\j I h$MbUVhj5OJQJ\jhj5OJQJU\hjOJQJ^JmHnHu-jhjOJQJU\^JaJmHnHu7j\h_Qh;EHOJQJU\^JaJmHnHujcMK h;UV$hjOJQJ\^JaJmHnHug7O~~s  !xgd|$ & p@ P D%xa$gdn & p@ P D%xgd[ $ & p@ P D%0x`0gd[$ !ha$gd[  !dgd<$ !da$gd<  ghijijĨӀ^ULUC4jhn5OJQJU\hnOJQJ\h-@g5OJQJh[5OJQJBh[h[56CJOJQJ^JaJeh  mHnHr  uh[OJQJ^JmHnHu0j} h[h[EHOJQJU^JmHnHujI h[UV h$MbH*OJQJ^JmHnHuh$MbOJQJ^JmHnHu&jh$MbOJQJU^JmHnHu0jh$Mbh[EHOJQJU^JmHnHuh{|}~678KLMNOyeVGVhOJQJ^JmHnHuh*qOJQJ^JmHnHu'j'hnh*q5EHOJQJU\j~I h*qUVhnh[OJQJ\hnH*OJQJ\hnH*OJQJ\h0Ch[OJQJ\hnOJQJ\hnhnOJQJ\jhn5OJQJU\'j@$hnhn5EHOJQJU\jI hnUVhn5OJQJ\vx>@BFHΫΒq^PP>626h Sjh SU#h[h*qOJQJ^JmHnHuhTB*OJQJ^Jph$h*qh*qB*H*OJQJ^Jph!h*qh*qB*OJQJ^JphUh*qB*OJQJ^Jph1hh*q56CJOJQJ^JaJmHnHuEhhT56CJH*OJQJ^JaJeh mHnHr uBhhT56CJOJQJ^JaJeh mHnHr uhHOJQJ^JmHnHu(2@ykkk $$7$8$H$Ifa$fkd,$$IfTlF|     4 laytT$$7$8$H$Ifa$gd*q$ !xa$gdDp (in H2O)f39500.350.016137300.320.016536100.300.016534300.270.016532800.240.016030000.200.016027000.160.0158Average f0.0162     Jacaranda (Engineering) 3333 Mail Code Phone: 818.677.6448 E-mail:  HYPERLINK mailto:lcaretto@csun.edu lcaretto@csun.edu 8348 Fax: 818.677.7062  REF date \h  \* MERGEFORMAT April 15 homework solutions  INCLUDETEXT "include.doc" headerCenter \* MERGEFORMAT ME 390, L. 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