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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 airwater 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 _pIIgthg _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 kgmj’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 )[lkgms2][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 yCJZR 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 Lshaped 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]
32.21bmﬁfs2 _ 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 crosssection of a dam is a quartercircle. 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 crosssectional 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 010 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 zcomponent 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 nonzero 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 436; namely, ﬂuid particles are accelerated to
the right and are tunred in the counterclockwise direction due to centripetal
acceleration towards the upper right. Note that the acceleration ﬁeld is nonzero, 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 fanmotor 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
fanmotor 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;—= (00375ksf8)(10) —, =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 crosssection 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 lNrru'
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 fanmotor 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 (470440)kPa [lkN/m2][1000kgm/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)1000kgmfsz lkNmfs Then the shaﬁ pumping power and the mechanical efficiency of the pump become Wpumpﬁmﬁ : ”motor Welerxdc = (0'90)(35 kW) : 315 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 _ 407".
(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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 Spring '08
 Bohl

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