Unformatted text preview: MAE 101A: Introductory Fluid Mechanics Homework 2 Due Friday January 22, 5:00 PM
Problem 1 The manometer shown in ﬁgure 1 contains water and kerosene. With both tubes open to the atmosphere, the freesurface elevations diﬀer by H0 = 20.0 mm. Determine the elevation diﬀerence when a pressure of 98.0 P a (gage) is applied to the right tube. Figure 1: Twoliquid manometer Problem 2 Water ﬂows downward along a pipe that is inclined at 30◦ below the horizontal, as shown. Pressure diﬀerence pA − pB is due partly to gravity and partly to friction. Derive an algebraic expression for the pressure diﬀerence. Evaluate the pressure diﬀerence if L = 5 f t and h = 6 in. Figure 2: Inclined pipe Problem 3 The inclinedtube manometer shown has D = 96 mm and d = 8 mm. Determine the angle, θ, required to provide a 5:1 increase in liquid deﬂection, L, compared 1 with the total deﬂection in a regular Utube manometer. Evaluate the sensitivity of this inclinedtube manometer with respect to the angle θ. Figure 3: Inclinedtube manometer Problem 3.51 !"# Problem 4 Semicircular plane gate AB is hinged along B and held by horizontal force FA applied at A. The liquid to the left of the gate is water. Calculate the force FA required for equilibrium. ,./)01$.($23)4.05$46$(.0$789:9;098/ Figure 4: Semicircular plane gate
h H$%$&'$()$ $ Problem 5 A plane gate of minimum thickness FA holds back a depth of water as shown in A ﬁgure 5. Find the minimum weight needed to keep the gate closed. R$%$*+$()$ y y
B z x 4? $ " @ A6 # 783)9.>= Problem 6 A gate, in the shape of a quartercylinder, hinged at A and sealed at B, is 3 m wide. The bottom of the gate is 4.5 m below the water surface. DEtermine the force on the !stop at B if [email protected] gate is made of concrete; R = 3 m. the "
AB $ ρ% 2 ΣCD $ + Problem 7 The gate shown is 1.5 mGHH wide and pivoted at O; a = 1.0 m−2 , D = 1.20 m, and IB0$1$I.8:A$;$/3=80A 4? $ @5% 6 1F $ 15 & H = 1.40 m. Determine (a) the magnitude and moment about O of the vertical component 6 % 15 (0./$)B$(0$=80(352
M,[email protected]==80E$=9>[email protected]$% @3)/$.>$.)B0$=9AN 5$(:89AK$!$%$5.>=)3>)[email protected])/$.>$.)B0$=9AK$A..0$9=$9>$789:9;098/ )B0$.($)[email protected]@0.35B=E$I$>.)$)B$(.::.I9>2E$8=9>2$1$3=$9>$)B$=L)5B ΣCD $ + 46 % ? $ ! " 1% @ A6 " # I9)B @ $ ρ% 2% B Problem 3.53 !"# 42,* $:**0$6,$;1+<*= $ Problem 3.76
h y Figure 5: Plane gate
L$%$"$& !"# dF W w$%$($& L'( 0: ! " $ " 0 =B # 3$< $
=0 $ ρ% 4 =9 Σ5C $ D Figure 6: A quartercyliner gate
G 3 # ρ" * Problem 3.71 <,23,J$02,&$+3$+,9*$<6=*J$=++$6<$63$*?7616K67& + HH of the force on the gate&due to the water, (b)Rthe horizontal !force applied at A needed for @ A $ 0; % B D .F $ .; 6& " R67 B % .; equilibrium. x y’ FV A W% y x [3] Part 1/2 [email protected] # A R F 0+2;9*<E$8*$3+,*$,9*$/+11+8634E$7<634$.$2<$63$,9*$<:*,;9 H WGate FB W& ! M " L% % ;+< N θO $ " . [email protected] ( # 86,9 0 $ ρ% 4% 9 $ ρ% 4% .% <63N θO D$!$E$1/5,95,D$39,2$/5$/,+(0$)>( % F% '()*+,$./0$1/234,)5* FV N)24*$0*<<7*E$<63;*$0$%$02,&$+3$+,9*$<6=*O ! ! ( " " .% 0 =B $ % .% ρ% 4% .% <63N θO % 8 =. " G+(0($3$)$*9*($30(40($95>$+$)$2(940(>$>/G5G90>3"# ρ" *" + N θO # M% ;+< N θO # Figure 7: Gate ($H)51C4>)5*$C/19,)/5I$>4($,/$G9,(0$/5$140J(>$40.91($95>$45>(05(9,+K$$;/0$140J(>$40.91($G($1/4C>$)5,(*09,( M $,+($1/51(3,$,+9,$;M$H(($N(,1+I$)$(=4)J9C(5,$,/$,+($G()*+,$/.$.C4)>$9F/J(L$95>$;O$)$(=4)J9C(5,$,/$,+($./01($/5 ! (% ρ% 4% 8% ,23N θO ! ( ( ( " % .% 0 =B $ % " . =. $ % ρ% 4% 8% M % ,23N θO $,+9,$,+($N(,1+$/5C:$+/G$./01($,+9,$G)CC$F($4(>$,/$1/234,($,+($2/2(5,$9,[email protected] #D N θO " M " # Problem 8 The crosssectional shape of a canoe is modeled by a curve y = ax2 , where a = 3.89 m−1 and the coordinates are in meters. Assume the width of the canoe is constant ( :4 W %< # '% $ '& & ( DD% ' TUVS% ' at&WN "= O0.6,23N "D% =*4Oits entire length L $= %5.25 m. Set up a general algebraic expression (% ' % & ' m over ' L XV :W " ( :4 & & < relating the total mass of %the canoe and its contents to distance d between the water surface and the gunwale of the ﬂoating P" & canoe. Calculate the maximum total mass allowable without N* 2 # ρ" *" G" Q" R # %AAA" % SKT%" % 7" 2 % "KU" 2 % 7" 2 % '% # 7SV" NP 7 & canoe. swamping the N*" 2 2 # ρ " *" G " π" R "
& # %AAA" N* 2
7 % SKT%" 2 & % 7" 2 % π & P" % H 7" 2I % " N*" 2 & '& # &AT" NP 3 # '% $ '& ;M # %TS" NP ionary. The width b into the paper m. Neglect the weight of the gate. tion: The horizontal component of Problem 3.83 er force is
Fig. P2.86
[3] FH # & h CG A # (9790 N/m3 )(1 m)[(2 m)(3 m)] # 58,740 N s force acts 2/3 of the way down or 1.333 m down from the surface (0.667 m from C). The vertical force is the wei ght of the quartercircle of water above BC:
FV # & (Vol)water # (9790 N/m3 )[(" /4)(2 m)2 (3 m)] # 92,270 N Figure 8: Canoe acts down at (4R/3") # 0.849 m to the left of C. Sum moments clockwise about t C:
Problem 9 The bottle – champagne (SG # 0. $ 117480 ' MC # 0 # (2 m)P $ (58740 N)(0.667 m) of(92270 N)(0.849 m)= 2P96) is under pressure as shown by the mercury manometer reading. 58( 7)kN Ans. net force on the 2inradius hemispherical end Solve for P # 58,700 N # Compute the cap at the bottom of the bottle.
Solutions Manual The bottle of champagne (SG! Fluid Mechanics, Fifth Edition # ) in Fig. P2.87 is under pressure as wtion: theirst, how high is the coreading. Well, 1 fluid oz. " 1.805 in3, hence 12 fl. oz. " u n by F mercury manometer ntainer? p in " (1.5 in)2 or h 3.06 in—It 6 ute3 the#net forceh,on the$2inradius is a fat, nearly square little glass. Second, ispherical end cap at thetoward the the r of the merrygoround, noting that the rmine the acceleration bottom of cente l lar ue. velocity is % " (12 rev/min)(1 min/60 s)(2# rad/rev) " 1.26 rad/s. Then, for r " 4 ft, tion: First, from thexmanometer, com 2 (4 ft) " 6.32 ft/s2 a " %2 r " (1.26 rad/s) the gage pressure at section AA in the n, for steady rotation, the water surface in the glass will slope at the angle
Fig. P2.87 tan & " Figure 9: Bottle of champagne ax 6.32 " " 0.196, or: 'h left to center " (0.196)(1.5 in) " 0.294 in g ( a z 32.2 ( 0 s the glass should be filled to no more than 3.06 ) 0.294 $ 2.77 inches s amount of liquid is * " # 10 in)2(2.77 in) liquid in 3 $ 1ﬁgure accelerates to the right with the ﬂuid in rigidProblem (1.5 The tank of " 19.6 in the 0.8 fluid oz. Ans. body motion. (a) Compute ax in m/s2 . (b) Why doesn’t the solution to part (a) depend upon ﬂuid density? (c) Compute gage pressure at point A if the ﬂuid is glycerin at 20◦ C . 9 The tank of liquid in the figure P2.139 elerates to the right with the fluid in dbody motion. (a) Compute ax in m/s2. Why doesn’t the solution to part (a) end upon fluid density? (c) Compute e pressure at point A if the fluid is erin at 20+C. Figure 10: Liquid tank Fig. P2.139 ution: (a) The slope of the liquid gives us the acceleration: tan& " a x 28 ) 15 cm " " 0.13, or: & " 7.4+ g 100 cm Ans. (a) 4 Clearly, the solution to (a) is purely geometric and does not involve fluid density. Ans. (b) From Table A3 for glycerin, , " 1260 kg/m3. There are many ways to compute pA. example, we can go straight down on the left side, using only gravity: thus a x " 0.13g " 0.13(9.81) " 1.28 m/s 2 ...
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