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exam1_review_problems - Chapter 3 Solution 40 The gage...

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Unformatted text preview: Chapter 3, Solution 40. The gage pressure of air in a pressurized water tank is measured simultaneously by both a pressure gage and a manometer. The differential height h of the mercury column is to be determined. Assumptions The air pressure in the tank is uniform (i.e., its variation with elevation is negligible due to its low density), and thus the pressure at the air-water interface is the same as the indicated gage pressure. Properties We take the density of water to be pw =1000 kgfms. The specific gravities of oil and mercury are given to be 0.72 and 13.6, respectively. Analysis Starting with the pressure of air in the tank (point 1), and moving along the tube by adding (as we go down) or subtracting (as we go up) the pgh terms until we reach the free surface of oil where the oil tube is exposed to the atmosphere, and setting the result equal to Pam gives JP] +pwgkw _pI-Igthg _poi]ghoi] :Pafm Rearranging AIR JP] _Pann : poilghoil +pHgthg _pwghw 01‘, Water P1 ,gage : pspilhoi] + ps,thHg _hw pwg Substituting, 2 653k” 2 1000 kg “1:5 — 0.72><(0.75 m)+13.6><hHg —0.3rn (1000 kgfm )(9.81mfs ) 1kPa.-m Solving for hHg gives his 2 0.47 m. Therefore, the differential height of the mercury column must be 47 cm. Discussion Double instrumentation like this allows one to verify the measurement of one of the instruments by the measurement of another instrument. Chapter 3, Solution 43. A load on a hydraulic lift is to be raised by pouring oil from a thin tube. The height of oil in the tube required in order to raise that weight is to be determined. Assumptions 1 The cylinders of the lift are vertical. 2 There are no leaks. 3 Atmospheric pressure act on both sides, and thus it can be disregarded. Properties The density of oil is given to be p =780 kgfms. Analyst's Noting that pressure is force per unit area, the gage pressure in the fluid under the load is simply the ratio ofthe weight to the area ofthe lifi, P _ _ W A new Ir(l.20m)2/4 W mg _(500kg)(9.81n1152)[ lkN —4.34 mm2 = 4.34 kPa 1000 kg - mfs?‘ The required oil height that will cause 4.34 kPa of pressure rise is nge—pgh > h— Pm _ 4.341(101112 1000 kg-mj’sz pg (780kgfm3)(9.81m/s2) lkN/mz - 0.567 m Therefore, a 500 kg load can be raised by this hydraulic lift by simply raising the oil level in the tube by 56.7 cm. Discussion Note that large weights can be raised by little effort in hydraulic lifi by making use of Pascal’s principle. Chapter 3, Solution 12]. The air pressure in a duct is measured by an inclined manometer. For a given vertical level difference, the gage pressure in the duct and the length of the differential fluid column are to be determined. Assumptions The manometer fluid is an incompressible substance. Properties The density of the liquid is given to be p = 0.81 kgfL = 810 kga’ms. Analysis The gage pressure in the duct is determined fi‘om Pgage =Pahs _Pann=:0gh 1N lPa = 810k im3 9.3mm? 0.08m — ( g X X )[lkg-ms2][1N/m2] =636 Pa The length of the differential fluid column is L = hisinfi = (8cm) isin 35O =13.9 cm Discussion Note that the length of the differential fluid column is extended considerably by inclining the manometer arm for better readability. Chapter 3, Solution 64. A room in the lower level of a cruise ship is considered. The hydrostatic force acting on the window and the pressure center are to be determined. Assumptions The atmospheric pressure acts on both sides of the window, and thus it can be ignored in calculations for convenience. Properties The specific gravity of sea water is given to be 1.025, and thus its density is 1025 kgfmg. Analysis The average pressure on a surface is the pressure at the centroid (midpoint) of the surface, and is determined to be PM : PC : pghc : (1025 kg/m3)(9.31m/s.2)(5 m)[i] 1 kg mfsz 5 m = 50,276 N/mz Then the resultant hydrostatic force on each wall becomes FR =PmA=Pm[m‘32 {4}: (50,276 N/m2)[;r(0.3 m)2 [4]: 3554 N F“ 03“, The line of action of the force passes through the pressure center, whose vertical distance from the free surface is determined fi‘om I xR4/4 R2 0.15m 2 Yr YCl KC YCl _5+( ) 2 —5.0011 In YCA yC-JZR 45%: 4(5 111) Discussion Note that for small surfaces deep in a liquid, the pressure center nearly coincides with the centroid of the surface. Chapter 3, Solution 68E. The flow of water from a reservoir is controlled by an L-shaped gate hinged at a point A. The required weight Wfor the gate to open at a specified water height is to be determined. 'V'IEES Assumptions 1 The atmospheric pressure acts on both sides of the gate, and thus it can be ignored in calculations for convenience. 2 The weight of the gate is negligible. Properties We take the density of water to be 62.4 lbmffl3 throughout. Analysis The average pressure on a surface is the pressure at the centroid (midpoint) of the surface, and is determined to be PM = Pghc = pawl?) = (62.41bmm3 )(32.2 as2 )(12/ 2 fi)[i] 32.21bm-fifs2 _ 2 — 374.4lbfi’ft , , B Then the resultant hydrostatic force acting on the dam becomes FR 2 PMA 2 (374.41bf/ft2)(12ft><5 ft) = 22,4641bf The line of action of the force passes through the pressure center, which is 2M3 from the free surface, _%_ 2x02fi) 3 3 y? :83 Taking the moment about point A and setting it equal to zero gives ZMA =0 4) FR(s+yP)=WE Solving for Wand substituting, the required weight is determined to be W = 52’” FR = (3 “mt (22,464lb1) = 30,900 lbf AB Sfi Discussion Note that the requin weight is inversely proportional to the distance of the weight from the hinge. Chapter 3, Solution 65. The cross-section of a dam is a quarter-circle. The hydrostatic force on the dam and its line of action are to be determined. Assumptions The atmospheric pressure acts on both sides of the dam, and thus it can be ignored in calculations for convenience. Properties We take the density of water to be 1000 kgz’m3 throughout. Analysis We consider the fi‘ee body diagram of the liquid block enclosed by the circular surface of the dam and its vertical and horizontal projections. The hydrostatic forces acting on the vertical and horizontal plane surfaces as well as the weight of the liquid block are: Horizontal force on vertical surface: FH =12r =PMA=pghCA=pg(R/'2)A = (1000 kg1m3)(9.81 ms2 )(10/ 2m)(10 mX100 mifi] :4.905><10T N Vertical force on horizontal surface is zero since it coincides with the free surface of water. The weight of fluid block per in length is E, :W:,ogv:pg[w><;z;s2 l4] IN = (1000 kg/m3)(9.8l 1an2 )[(100 m)rr(10m)2 l4}[m] =7.705><10T N Then the magnitude and direction of the hydrostatic force acting on the surface of the darn become FR 21/1913, +13,2 = ‘{(4.905><10T N)2 +(7.705><10T N32 =9.134><107 N Fr _ 7n705><10T N 7 —1.571 > 9—575" FH 4.905><10 N taut?— Therefore, the line of action of the hydrostatic force passes through the center of the curvature of the dam, making 575° downwards from the horizontal. Chapter 3, Solution 85. The height of the portion of a cubic ice block that extends above the water surface is measured. The height of the ice block below the surface is to be determined. Assumptions 1 The buoyancy force in air is negligible. 2 The top surface of the ice block is parallel to the surface of the sea. Properties The specific gravities of ice and seawater are given to be 0.92 and 1.025, respectively, and thus the corresponding densities are 920 kg/rn3 and 1025 kglmg. Anaivsis The weight of a body floating in a fluid is equal to the buoyant force acting on it (a consequence of vertical force balance fi'om static equilibrium). Therefore, in this case the average density of the body must be equal to the density of the fluid since W = F3 Pbodygvml = pfluid gvsubmerged 10 cm Vsubmerged : Pbody i. _- View Pnuid The cross-sectional of a cube is constant, and thus the “volume ratio” can be replaced by “height ratio”. Then, I: 7 0.92 hsubmerged 7 pbody > ’3 pica ) h+0.10 1.025 ”total pfluid h T 0-10 pwater where h is the height of the ice block below the surface. Solving fork gives it = 0.876 m = 87.6 cm Discussion Note that the (192/1025 = 90% of the volume of an ice block remains under water. For symmetrical ice blocks this also represents the fiaction of height that remains under water. Chapter 4, Solution 36. 10 mfs Scale: —)- =IGURE 1 f'elocity vectors in the upper right quadrant 'or the given velocity field. Solution For the given velocity field, the locationls) of stagnation point(s) are to be determined. Several velocity vectors are to be sketched and the velocity field is to be described. Assumptions 1 The flow is steady and incompressible. 2 The flow is two- dimensional, implying no Z—component of velocity and no variation of u or v with 2. Analysis (at) The velocity field is V =(u,v)=(1+2.5x+y)f+(—0.5—1.5x—2.5y); (1) Since V is a vector, all its components must equal zero in order for I7 itself to be zero. Setting each component of Eq. 1 to zero, a: l+2.5x+ y:0 Simultaneous equations: v :41571.5x72.5y :0 We can easily solve this set of two equations and two unknowns simultaneously. Yes, there is one stagnation point, and it is located at Stagnation point: x : -0.421 m y : 0.0526 m (b) The x and y components of velocity are calculated from Eq. 1 for several (Jay) locations in the specified range. For example, at the point (x = 2 m, y = 3 m), u = 9.00 mJ's and v = -ll mJ's. The magnitude of velocity (the speed) at that point is 14.21 mfs. At this and at an array of other locations, the velocity vector is constructed from its two components, the results of which are shown in Fig. l. The flow can be described as a counterclockwise turning, accelerating flow from the upper lefi to the lower right. The stagnation point of Part (a) does not lie in the upper right quadrant, and therefore does not appear on the sketch. Discussion The stagnation point location is given to three significant digits. It will be verified in Chap. 9 that this flow field is physically valid because it satisfies the differential equation for conservation of mass. Chapter 4, Solution 37. Scale: 10 ms” FIGURE 1 Acceleration vectors in the upper right quadrant for the given velocity field. Chapter 5, Solution 13. Solution For the given velocity field, the material acceleration is to be calculated at a particular point and plotted at several locations in the upper right quadrant. Assumptions 1 The flow is steady and incompressible. 2 The flow is two- dimensional, implying no z-component of velocity and no variation of u or v with 2. Analysis (1:) The velocity field is V :(u,v):(l+2.5x+y)i'+(70.571.5x72.5y)} (1) Using the velocity field of Eq. 1 and the equation for material acceleration in Cartesian coordinates, we write expressions for the two non-zero components of the acceleration vector: Bu Bu Bu Bu ax=—+u— +v— +w— a! 8): By 32 o (1 25x1yx25):(0515x 25yXl)1 and 3v 3v 3v 3v :2 =—+u— +v— +w— ’ at 3.): By 32 =0+(1+2.5x+y)(71.5)+(7o.571.5x72.5y)(7 2.5)+o At ()5: 2m,y: 3 m), ax:11.5 nu'szanday:14.fl mlsz. (b) The above equations are applied to an array ofx and y values in the upper right quadrant, and the acceleration vectors are plotted in Fig. 1. Discussion The acceleration vectors plotted in Fig. 1 point to the upper right, increasing in magnitude away from the origin. This agrees qualitatively with the velocity vectors of Fig. 1 of Problem 4-36; namely, fluid particles are accelerated to the right and are tun-red in the counterclockwise direction due to centripetal acceleration towards the upper right. Note that the acceleration field is non-zero, even though the flow is steady. A smoking lounge that can accommodate 15 smokers is considered. The required minimum flow rate of air that needs to be supplied to the lounge and the diameter of the duct are to be determined. Assumptions Infiltration of air into the smoking lounge is negligible. Properties The minimum fresh air requirements for a smoking lounge is given to be 30 US per person. Analysis The required minimum flow rate of air that needs to be supplied to the lounge is determined directly from V31: = an perm (No.0f persons) = (30 Lls- person)(15 persons) = 450 Lls = 0.45 mals The volume flow rate of flesh air can be expressed as V: VA: VUzDZ {4) Solving for the diameter D and substituting, 132F— 4—(0. 45 1113/5) memo) Smoking Lounge 15 smokers 30 Us person - —0.268 m Therefore, the diameter of the flesh air duct should be at least 26.3 cm if the velocity of air is not to exceed 8 m/s. Chapter 5, Solution 73. A fan is to ventilate a bathroom by replacing the entire volume of air once every 10 minutes while air velocity remains below a specified value. The wattage of the fan-motor unit, the diameter of the fan casing, and the pressure difference across the fan are to be determined. wt Assumptions ] The flow is steady and incompressible. 2 Frictional losses along the flow (other than those due to the fan-motor inefficiency) are negligible. 3 The fan unit is horizontal so that z = constant along the flow (or, the elevation effects are negligible because of the low density of air). 4 The effect of the kinetic energy correction factors is negligible, at: 1. Properties The density of air is given to be 1.25 kgjmg. Analysis (9) The volume ofair in the bathroom is V: 2 m x 3 mx 3 m=13 m3. - —> Then the volume and mass flow rates of air through the casing must be All' - 3 0 8 mfs D . V 13 —) V=—=—m=0.03m3/s At 10x605 m = p? = (1.25 kgjm3)(0.03 m3is) : 0.0375 1:ng We take points 1 and 2 on the inlet and exit sides of the fan, respectively. Point 1 is sufficiently far from the fan so that P 1 = Pm and the flow velocity is negligible (VI = 0). Also, P2 =Pm. Then the energy equation for this control volume between the points 1 and 2 reduces to 2 2 2 . V - _ V _ _ _ _ V m fl+a'1;+8il +mezm i+c;t12_2+gz2 +mehie+EmechJoss —> Wfanmzmazi :9 2 p 2 2 since Emmik'ss =Emh lDS-‘wumv in this case and Wmm = WP!“ _ -mech [055.me ' Substituting, - - V22 (8111193 1N 1w Wan u = m;—= (0-0375ksf8)(1-0) —, =12 w ’ 2 2 1kg -mfs 1N -rnfs - W and Wi'amelect=fl=fl=z.4w Therefore, the electric power rating of the fanfmotor unit must be 2.4 W. (b) For air mean velocity to remain below the specified value, the diameter of the fan casing should be ' 3 v : A213 :(aD22/4W2 > D2 , 4V , 4(0'03m [SJ 70.069m76.9 cm EVE ir(8m/s) (c) To determine the pressure difference across the fan unit, we take points 3 and 4 to be on the two sides of the fan on a horizontal line. Noting that z; = 24 and V; = P2. since the fan is a narrow cross-section and neglecting flow loses (other than the loses of the fan unit, which is accounted for by the efficiency), the energy equation for the fan section reduces to - P3 . P m 4 + me) u m P3 Wifanm Winn, u m! p F ) a . . 1.2w lN-rru' Substitutmg, P4 137 [ s 0.03 mafs 1 W Therefore, the fan will raise the pressure of air by 40 Pa before discharging it. J740Nl‘m274flPa Discussion Note that only half of the electric energy consumed by the fan-motor unit is converted to the mechanical energy of air while the remaining half is converted to heat because of imperfections. Chapter 5, Solution 77. Water flows at a specified rate in a horizontal pipe whose diameter is decreased by a reducer. The pressures are measured before and after the reducer. The head loss in the reducer is to be determined. Ni Assumptions 1 The flow is steady and incompressible. 2 The pipe is horizontal. 3 The kinetic energy correction factors are given to be 0.1 = a2 = a: 1.05. Properties We take the density of water to be p = 1000 kgi'ms. Analysis We take points 1 and 2 along the centerline of the pipe before and after the reducer, respectively. Noting that zl = 22, the energy equation for steady incompressible flow through a control volume between these two points reduces to P V2 P V2 P —P era/2 4/2) [‘0'] ] izlihpmp,u——2:a2 2 :zflhmwmemL _> 5L2] 2+4] 2 pg 23 pg 2g pg 2g where ' ' 3 VI :1: l: : 0.035m2/s :19ng A] EDI x4 :r(0.l§m) l4 ' ' 3 V2=1= 1: =me=696mrs A2 7.92/4 12'(0.03m) x4 Reducer Substituting, the head loss in the reducer is determined to be h 7 (470-440)kPa [lkN/m2][1000kg-m/s2]l l.05[(l.98m/s)2—(6.96m/s)2] 1. . (1000 kgxm3)(9.31 m2) lkPa ]kN 2(9.81mis2) = 3.06—2.38 = 0.68 m Discussion Note that the 0.79 m of the head loss is due to fi‘ictional effects and 2.27 m is due to the increase in velocity. This head loss corresponds to a power potential loss of . _ . _ 3 3 2 IN 1w _ Emhmmg—pvgh —(1000kg/m )(0.035m ls)(9.81m/s )(0.79m)[W INNS _271w Chapter 5, Solution 86. A pump is pumping oil at a Specified rate. The pressure rise of oil in the pump is measured, and the motor efficiency is specified. The mechanical efficiency of the pump is to be determined. Assumptions 1 The flow is steady and incompressible. 2 The elevation difference across the pump is negligible. 3 All the losses in the pump are accounted for by the pump efficiency and thus {2,5 = 0. 4 The kinetic energy correction factors are given to be on = a; = a= 1.05. Properties The density of oil is given to be p = 860 kg/mg. Analysis We take points 1 and 2 at the inlet and the exit of the pump, respectively. Noting that 21 = 22, the energy equation for the pump reduces to P V2 P V2 P —P aU/Z—VZ) p; 10:] 2'3 +2] mm!” — p; +012 2; +22 +1’1mrhm’e +kL —> Jimmy,“ = ng '+# where ' ' 3 Vi — V — l: — 0'1”“: 49.9sz A] m9. [4 :r(0.08m) /4 ' ' 3 V2 — V V — 0.1m ’15 -8.84mfs A2 111322 /4 :z(0.12m)2 /4 Substituting, the useful pump head and the corresponding useful pumping power are determined to be h _ ”m“ (360kg/m3j)(9.81n1152j) — 47.4 17.0 — 30.4 m 400,000 N/mz 11%.sz2 . 1.05[('8.84 m/s_)2 —(19.9 mJ’sJZ] 1N 2931111152) , . lkN IkW W = VI: = 860k! 3 0.1 3/ 9.81m/2 30.4 — — =25.6kW pumpp p g pmnp,u (. gm X. m SK 5 )( m)1000kg-mfsz lkN-mfs Then the shafi pumping power and the mechanical efficiency of the pump become Wpumpfimfi : ”motor Welerxdc = (0'90)(35 kW) : 31-5 kw Wm, n 25.6 kW ”PM” _ WPW’M _ 31.5 kW — 0.813 — 81.3% Discussion The overall efilciency of this pump/motor unit is the product of the mechanical and motor efficiencies, which is 0.9><0.813 = 0.73. Chapter 5, Solution 59. The water height in an airtight pressurized tank is given. A hose pointing straight up is connected to the bottom of the tank. The maximum height to which the water stream could rise is to be determined. \l Assumptions 1 The flow is steady, incompressible, and irrotational with negligible frictional effects (so that the Bernoulli equation is applicable). 2 The friction between the water and air is negligible. Properties We take the density of water to be 1000 kg/ms. Analysis We take point 1 at the free surface of water in the tank, and point 2 at the top of the water trajectory. Also, we take the reference level at the bottom of the tank. At the top of the water trajectory V; = 0, and atmospheric pressure pertains. Noting that 21 = 20 m, PLsnse = 2 atm, P2 = Pm“, and that the fluid velocity at the fi‘ee surface of the tank is very low (V; E 0), the Bernoulli equation between these two points simplifies to P V2 P V2 P P P —P P 1+ 1 +21 — 2 + 2 +22 > l+2] — am +22 ) 22 ——l “m +2} 2 [£385 +2] pg 2g pg 28 pg 93 90 pg Substituting, 2 2 z] _ 2e:tm 2 101,325 me 1kg mfs +20 _ 40-7". (1000 kg/m )(9.31sz ) latm 1N Therefore, the water jet can rise as high as 40.7 m into the sky from the ground. Discussion The result obtained by the Bernoulli equation represents the upper limit, and should be interpreted accordingly. It tells us that the water cannot possibly rise more than 40.7 m (giving us an upper limit), and in all likelihood, the rise will be much less because of frictional losses. Chapter 5, Solution 48. Water enters an empty tank steadily at a specified rate. An orifice at the bottom allows water to escape. The maximum water level in the tank is to be determined, and a relation for water heightz as a fimction of time is to be obtained. Assumptions ] The orifice has a smooth entrance, and thus the frictional losses are negligible. 2 The flow through the orifice is steady, incompressible, and irrotational with negligible frictional effects (so that the Bernoulli equati...
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