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Unformatted text preview: 531 Water enters a conical diffusing passage
(see Fig. P53) with an average velocity of 10
ft/s. If the entrance cross section area is 1 ftz, how
large should the diffuser exit area be to reduce
the average velocity level to 1 ft/s‘? Section (2) lOft/s V2 3 lft/s
1&2 FIGURE 95.; H‘C
ll ll 5r Shem/y m candprﬁsﬂbk: "LT/aw éci‘wcgn gecko,” (gandﬂz) 5.2 Various types of attachments can be used with the shop
vac shown in Video vsz. Two such attachments are shown in
Fig. P5.2—a nozzle and a brush. The ﬂowrate is 1 ft3/s. (a)
Determine the average velocity through the nozzle entrance, V".
(b) Assume the air enters the brush attachment in a radial di
rection all around the brush with a velocity proﬁle that varies
linearly from O to Vb along the length of the bristles as shown in the ﬁgure. Determine the value of Vb. ' 77m) AIM:Q2 0" 14:”sz
so 47
V” = 45.9% (w%=ntmmnpvélm4@:2@wmm
lg = aye/"age Veloc/i’y ml (3) = 3L” Vb and ,43 = 7703 63 T/IVSJ 3
swwsntem =/ or
m=2a4§ ‘ 5.3 The pump shown in Fig. P53 produces El steady ﬂow of
10 gal/s through the nozzle. Determine the nozzie exit diame—
ter, 1);, if the exit velocity is to be: V; = 160 ft/s. IFIGURE P553  w I in? m3 a
For sfeady How QI—Qzl where Q, :Iaigq(zmmﬂm) = 1.337%
Thus, wﬁh vz=m§ﬁ *3
1.337%— zﬂzvz =%D:(zoo%*) 01‘
D,_ = o./3oﬁ= 2.57 in. 5.4 Air ﬂows steadily between two cross sections in a long.
straight section of 0.1m inside—diameter pipe. The static tem— Section (1) sects‘cm (2)
perature and pressure at each section are indicated in Fig. P54. If the average air velocity at section (I) is 205 m/s, determine P1: 77 “Pa (3‘35) p2 = 45 “:3 (abs) . . . T = 268 K T = 240 K
the average at veloc1ty at section (2). v1: 205 m/s 2 I FSGURE P5.4 7/7}: anaéw/x i: JIM/Var 75 #1: one a?“ éi’dmp/e 5’. 2 .
For sfead’y f/ow Ae/Weeh 'Jccr‘l'ans (I) ana’{2]
m . n
S .5: l  lee. Asfum/nj uﬂm; under ﬂie Mimi; 19m; 0/ f4}; fmé/em/ m} Aehmxe; a s an idea/ 945 we: use #16 z‘dea/ ya;
cguaﬁ'on 0f sfa/e #3) 79 967‘ = (2)
“ ’2 ’3 7, (Oman/Jay €55. / man 2. and abserw/la; Ma)! 14, =41
We. gef Vi = P; 72— V, = [feﬂms](240/<) “<5
H 5.5 The wind blows through a 7 ft X 10ft garage door
opening with a Speed of 5 ft/s as shown in Fig. P55. Deter
mine'the average speed, V, of the air through the two 3 ft X 4 ft openings in the windows. 16ft I FIGURE P5.5 For Sfead‘y 0‘7 compress/He f/atv .— anmge ' Qw/édaw + @00de
ddor
0* A V g 71 V + 4 V 7mm Mm; +v Mr.de W/nde
dddr fanatic, door so ﬁe am}; glgeed') V) m‘ fhe d/r Mmujk ﬁre *wo [5
V # Agme mm; +1, Wm w _ C714) (70 760(5 4?) mm = 4,9? ZANlrIIIOW I, L 5.6 A hydroelectric turbine passes 5“ million gal/ min
through its blades. If the average velocity of the ﬂow in the
circular cross—section conduit leading to the turbine is not to
exceed 30 ft/s, determine the minimum allowable diameter of
the cOnduit. For xhcompresﬁé/e f/ow ﬂimujh Me candor/f and furé/hc .—
 lander/'1'“ fare/me
73a;
A V .= Q .
candalf candy/f furbme
and ._
' d = :4 EMMMC ==
Corlde
= m H
dwmﬁu’f i— 55 5.7 Water ﬂows along the centerline of a SO—mmdiameter pipe
with an average velocity of 10 m/s and out radially between two
large circular disks as shown in Fig. P57. The disks are paraliel
and Spaced 10 mm apart. Determine the average velocity of the
water at a radius of 300 mm in the space between the disks. For sfeddy I}? compress/6!: flew ' (4)277 '71 _ mzdwoommom) 0” Q, g 82 fﬁ=10mls
AV, z 4sz FIGURE P5.7
Thus 1 ._.. ‘2.
\7=A.V, 779 V, _ (50mm)(10m/s?
z __
dz v1 = 1.04 g: 5.8 A hydraulic jump (see Video V105) is in place down" ,j
stream from a spillway as indicated in Fig. P5.8. Upstream of
the jump», the depth of the stream is 0.6 ft and the average stream
velocity is 18 ft/s. Just downstream of the jump, the average
stream velocity is 3.4 ft/s. Calculate the depth of the stream, 11, i just downstream of the jump. ft/s
ﬁflsm'sj ” " ,1. ' ' 7  m,
I FIGURE £35.60) 0r_ ﬂ
VA, = ngAz
Thus
W, = V2111
and
A2 Viv/4. = (/5300261‘7‘) _ 31mg
1 gTT—m— /‘"wmm\ 60/1790, va [u me, 5.? A water jet pump (see Fig. P53) in i
volves a jet cross section area of 0.01 ml, and a
jet velocity of 30 m/s. The jet is surrounded by entrained water. The total cross section area as sociated with the jet and entrained streams is i
0.075 m2. These two ﬂuid streams leave the pump ‘
thoroughly mixed with an average velocity of 6 ‘
. .m/sthrough a cross section area of 0.075 mg. Determine the pumping rate (i.e., the entrained I
ﬂuid ﬂowrate) involved in liters/s. Far steady in camp/essi/a/e 790w ﬁrm/(9h 7%: com‘w/ Vo/ume
a + a = Q3 a2 = Iva/43 ~ W, = [(6%)(0075 W“)  (30%>(00'm3]tm$§ 5 5? 5,10 530 An evaporative coating tower (see Fig. 135.10) is used to
cool water from 110 to 80°F. Water enters the tower at a rate of f
250,000 lbm / hr. Dry air (no water vapor) flows into the tower
at a rate of 151,000 lbm/hr. If the rate of wet air flow out of the
tower is 156,9001bm/hr, determine the rate of water evapora
tion in lbw/hr and the rate of envied water ﬂow in Ham/hr. _ We: air
m = 156.900 lbm/hr gecﬁan (2 ) Warm wars! —> ..
ti} = 250000 lbmthr Dry air —>
rh = 151.000 tbmlhr For Sfeaa’y f/ow of diy air FIGURE P5.10 M3 : m2) 0'43; ah” (I )
For Sfeady How of Wafer m/ : m2, wafer + 4 (Z)
I4/50 m2 = m2, dry air + m2, Mid/Er (3 Comb/Why E35. I and 3 we ger‘ n}; = m —— n33 ~== [Mfg of wafer empamf/mw 2/ MW ,2 50 I I6
 s /5 00 Lin . 15/ 000 [£2 =— 5900 .5:
mmwdlcr 6’ 9 5,, ’ hr Ar Hom 53. Z we 96+ m = Ii? ~ 105 = rmLe of coo/ed wafer ﬂay
q / z,wa7‘er
or [6
' g 250 000 IL» _. 59700 1%: _.= 24165 000 .32
m 1/ ’ ,5, hr hr s—q 5.11’. At cruise conditions, air ﬂows into a jet
engine at a steady rate of 65 lbm/s. Fuel enters
the engine at a steady rate of 0.60 lbm/s. The
average velocity of the exhaust gases is 1500 ft/s
' relative to the engine. If the engine exhaust ef—
fective cross section area is 3.5 ftz, estimate the
density of the exhaust gases in 1bm/ft3. 5r steady {low
I773 = I77, '7‘ /7’l‘Z 0F 5—10 secﬁbn 6') (/h/ef ) 5.1 '2 Air at standard atmospheric conditions
is drawn into a compressor at the steady rate of 30 :m3/min, The compressor pressure ratio, pm/ pink“ is 10 to 1. Through the compressor p/p" '
remains constant with n = 1.4. If the average
velocity in the compressor discharge pipe is not
to exceed 30 m/s, calculate the minimum dis charge pipe diameter required. 5.13 Two rivers merge to form a larger river as shown in
Fig, P513. At a location downstream from the junction (before the two streams completely merge), the nonuniform velocity
proﬁle is as shown» Determine the value of V. 0156 fire. Cm’ﬂ’YD/ VOA/MC 5ﬁywn wﬁﬁqn'n brake» //’)’)CJ'
I}; 71106 Ska/$51 above, We maize ﬂaf Mcloﬂv and Mn W CWSCNﬂ/ﬁ’an 0/ may: IUYI‘nc/‘p/e WC
9‘67“ ' l 2‘ '3 0.3V v 7M: my, mall/2 == {0/2832 v + [7 Av \/ and _ .
\/ = 42”; + ’42 l/z. _(50H)(3ﬂ)(31g +{gﬁ’ﬁ{04ﬂ ,4 (m + ,4 (yofﬁéﬂjﬂmgﬂ (705%2‘7‘) 0.9V V V s 963751
.5 —_
————— 5"42 5.1% I ' 5.14 Oil having a speciﬁc gravity of 0.9 is pumped as Section (1)
illustrated in Fig. P5.i4 with a water jet pump (see Video '\"3.6).
The water volume flowrate is 2 m3/s. The water and oil mix
ture has an average speciﬁc gravity of 0.95. Calculate the rate. Section (3) in m3/s, at which the pump moves oil. _>_
water Water
Q} = an?
« "Lm3/s “‘3'”
Section (2) (8G = 0'95)
Oil (80 = 0,9)
For “eddy {my FIGURE P514
. . z ml + ﬂ’l2~ 3
Of“
ﬂqﬂk/fQLz/gqa (I)
14/50 Slhce, #12 Waller and 0// may be. cow/'45er ,,‘, Com/Wx/é/ﬂ
/
Q, + €22 = a} (2) COMé/n/nj 5?; /ama’z we. gei‘
pIQ/ + 542 = gamma) 0f” , Q; + SGZQz = 563(6)! + Q2)
and Q n a, (1— 563) z _
5G3  SGZ Thus 3 \ Q = (1%)(/”01%_/ .—, 2001": 0.625 r 0.70 5 O
5. 15‘ l WW“— Air at standard conditions enters the compressor
shown in Fig. P5.15 at a rate of 10' ft3/s. It leaves the tank through a 1.2—in.—diameter pipe with a density of 0.0035
slugs/ft3 and a uniform speed of 700 ft/s. (a) Determine the
rate (slugs/s) at which the mass of air in the tank is increasing or decreasing. (b) Determine the average time rate of change of
air density within the tank. m 2:»: r U56 the va/umc Wif1”I;’), M6 low/Zen Air/165.
(a) Fray/s ‘Hm’, (macaw/4790M all WSI pnhclk/t We wy‘:M.—Vh asia’aiaoﬂ V pf . m amL I}: In “7' 0“" “ff 1
D MWI — @0023? f/uj (a f+j_ @1925 353.9 77 ["2 "0/700 353‘)
W ‘ ‘"‘ m 1H 44 if
pf .‘ff'3 5 ‘ / ff>)
24457; = 0.00156 {9f 18' crew/£37
Dz‘ "' 5
(b) 9.44:7: WP '57: D: ‘4” DE _ 0. 00456 5;"!
at 5,: Dz‘
/ J}
50 D 0.00 ‘HK If,” __ 0.00954 {try ._. 2.29xm ﬁg
4 z: ' 3 pg— V 2.0 7"?"
I}: O
p 5" ” 5.16 An appropriate turbulent pipe flow velocity proﬁle is R... r‘ l/nA
V: i
””( R where u, =Acente1'line velocity, r = local radius, R = pipe
radius, and i = unit vector along pipe cen‘terline. Determine
the ratio of average velocity. E, to centerline velocity, uc, for
(an = 4, (b)n I 6, 02):: = 8, ((1)12: = 10. cross Sec ﬁzm
area For any Cross sad/On area. h'v=/0Au = {/2 Mme/A ﬂ/so .1 _.3 ’1 we R~F
. z o = M .__.......
Vn \/c c ) 73a: for a um'zérm 4: cr’llrz‘r/éufed densify/ ﬂ Over area A
I? [K n
4 ac (Rf) Zvrrdr
Z TI: 77"? 5.17 Water flows steadily thruugh the control volume shown
in Fig. P5.17. The volumetric ﬂowrate acmas section (3) is
2 fig/s and the mass ﬂowrate across section (2) is 3 slugs/s. IFIGURE P517 (a) f 6‘ VI? 51/! ==~weigl1f flawrafe across area (I) Lyn},
m (/Vofe 5 ‘5‘ am < 0 since ‘7'»? <0 for Mei/917W area (1))
m B y conserv‘aﬁon of maSsJ for arfeacly How)
"a, r ”;2. +Ih3 1: "52 + (33 Q3 = 35/1/73/3 S/VyJ/fijZ 0!"
0"), =3 6385/0951; Thus; from 137.02,
[3' (M :(“32.2 H/s‘)(6.ee Slugs/s) 5‘ " ZZZ/dug ' H/s")/S 3’ :222 MA: (I) (b) I V9 V513 M = Momenfdm f/w: amss area (I)
(I) L ' 0/7 (01 ﬁ=+c:053005'+sm30f and
V,‘ \{("603300? ’Jsiﬂ30?) =‘Wl31, where 9/], 5/75, r6W 35% (from 5712))
Thus , V: m, v 6.33% __ ’ 6’7" (I9*if%?)(a.mf)
Hence,
WWW/H Wemﬁ.)~h‘,/I, =€\4M=»'v,14 “may (I)
= (6.88 %£)(8.87§)(59330?4 91130;“)
.2 (52.82+ 30.51) = 8.8795 : 52.8;‘+3o.5; lb .5 “I6 (I) O
5. m ]——"—’”———__' 7N As shown in Fig. P5119, at the entrance to a 3—f‘t—wide
channel the velocity distribution is uniform with a velocity V.
Further downstream the velocity proﬁle is given by u 2 4y —
2y2, where u is in ft/s and y is in ft. Determine the value of V. we f’he wm’n/ Vo/ume Mal/cake! éy m6 [am/<04 Am"; pa
ﬂit Skefolq above
BM #11; cdm‘erwné'on 0/ max! FYI/7690A? A2 Jar 2 “If H3
— 3’ 47,2 7b: 9% __
vMeng Z? g}: o ? j 5. 2 o _
5.20 Flow of a viscous ﬂuid over a ﬂat plate seam" (2)
surface results in the development of a region of Section (ll \ U Outer d
e ge reducedvelocity adjacent to the wetted surface \
as depicted in Fig. P520. This region of reduced
ﬂow is called a boundary layer. At the leading boundary
layer edge of the plate, the velocity proﬁle may be r___,,f
considered uniformly distributed with a value U. ‘ »;,, ,,,,,,”,,,,,,,, ,1” I,” III/III“
All along the outer edge of the boundary layer, ‘ ' ' ‘ , x the ﬂuid velocity component parallel to the plate 
surface is also U. If the x direction velocity proﬁle FIGURE P520 at section (2) is
1/7
E ___; X
U (5) develop an expression for the volume ﬂowrate through the edge of the boundary layer from the leading edge to a location downstream at x where
a the boundary layer thickness is (5. From #13 comm/a 1%“ of mass pr/hc/‘p/c app/fed 7b 7%: {/0W 7%,???” 7V“: canﬁv/ Volume Shoot/n I): Me f/jm’é
we have. ' s m = \7’. "a’A
2
1‘5 r Iiawmpressib/el “ﬂaw
I
=ﬂVIy g: 7 d
/0 QM 0(5) 6%) where
I :: Laid/ﬂ 0/ 726 p/al‘é and #1445 = JUN
634/ 2? 5.22 Estimate the time required to ﬁll with water a cone
shaped container (see Fig. P522) 5 ft high and 5 ft across at the
top if the ﬁlling rate is 20 gal/min. IF‘IGUHE P532 50m aff/iczﬁ'an 02‘ ﬂc (anscryaﬁbn 0/ was: pr/‘nc/jo/e 25 ﬁt carav‘ml valume, :Aown M Ma {ya/c
we have
_3 7t. ﬂVqu/l =0
31‘
0/ CS éi‘  a s 0
at
0r 1* +
[M = 42/414
0 0
Th 2 l 2:3
a: g V g WDA g 7(5;+)C5{+)(/I728 “2)
Q. ’2 Q (/2) (20 537/ 23/ [if
and mm>( if] 1f 5 mil?) 5.24 I 5.2 L!“ Storm sewer backup causes your basement to ﬂood at defymm‘a 7% 7L
' the steady rate of l in. {of depth per hour. The basement ﬂoor f 6500;] Vie/“9‘”: a
area is 1500 ftz. What capacity (gal/min) pump would you rem —— — —— w — — ~   a, r
to (a) keep the water accumulated in your basement at a constant / A ~ / I
level unlnl the storm sewer is blocked off, (b) reduce the water / l / I ll
accumulation in your basement at a rate of 3 in. / hr even while / l l l
the backup problem exists? /. _ 7.1.“: _: : : t. :l" ‘ ‘7
V , _ , , l /
/ n n
:/ a” ‘1 :éW/%wmm
'.I _ _ J7. __ .I .. :‘l/ _ {low ml ﬂaw in
E?!” a ﬂeﬁrmmj Con/ra/ Via/«me 7‘th Came/37.; 7%6 Wﬂ/er (mar 7%8 éasemcnf f/aor (see “614% awe} / Me
Conserua/fon 09‘ Mas; Mme/we, (53 5.17) /66245 +0 f" :0
at CV cs P
or 7%)» Conf/‘an/ {I’M/d dens/)7 and area, (A)
A 513  62,." + QM ==0 I m (a) '5)’ par‘l‘ 6!, J / brads7‘0.
Quad = Q5?
779 evalualc aim we use, {5%, WW, Q’W=0. .ﬂws
. s A W» ; (/500 f+"){/ m. /  2
‘2’ "' h7)/ ‘ )= /25 H ” dt /2 31. h:
“"4 3 Ff
' ; 125 ﬁ)/Zygyal) _/_ W 56 4/
Q” ( hr ,4? w ’ ___i'—.— it:
hr
(b) Far parf AJ 53/ yie/d:
Q = Q,~ — A «.111
aul‘ ’1 dt
.2 [5,6 in! __ 00,9“ (.3151. (796731 J,
QM 55? ﬁg ll hr)lzf') #3 6031.5
f r
Qawf = 6—————2"4 7""! ml}, 5.25 Two 8ftwirde rectangular crates are wheeled into the k 6 “:1 trailer portion of a semitruck a; a speed of V = 2 ﬂt/s as shown ii DA;
in Fig. 1F5.25.Attime 23 == 0 the ﬁnal of the ﬁlm cram is 2 ft from
the open back of thc trailer. Plot a graph of the air ﬂowrate across
the open and of the trailer as a function of time for 0 5. t s 10 s. Consider ‘Hve deform/M emira/ volume
shown in {he skeia/I. From 5715/7) 3%]941/ '1‘“) W’gdﬂ‘ﬂ
CV CS or sinus Pﬁcamhm‘
ﬁde+fWIﬁmzo or $1,? {maﬁa/xi: v,// “cam/m Q, = airﬂow rafe across and of frank/t 19/50, 1
£5 = r‘mlg ml which #73 w/me mpg/1‘ in 7%? I‘M/Var Manges MUM 79m =V/1m;e J where Harm = cram—sem‘iml area of fl): amfc Md
[.5 crass/177 ﬁle and of 7%6' I‘M/7W. 777W,
— Vﬂcrdfc 5 ~ 01' 0" Q, 3‘ Vﬂcmfe J W/Iel'e V‘ 2 6%“ ‘ (I) From 1%: 911/3” Jain z
ﬂora}: : 0 7%" 0$£<lsj ﬂank ‘1 {3H)(3ﬂ) 5664f)! far/351R 645]. Amalie :0 {W [7‘35 g <55; IZ 5 (5H)(<9#) 5549;!” far 5:5 I! 57.: 1% ft? 7778 Correspona/Ma f/au/I'm‘es {297m 59, 0)) ﬂ/‘c‘ Q, = 01 Q, = 54791722044) 5 I28 WA) and 4), s Wf/‘(Zf/M = WW. 124‘
a: :ﬁown [Ire/3W. I
I Ii
ll
% 5.27 1:! takas you 1 min to ﬁll your car’s fuel tank with 8.3 gal
lons 0f gasoline. What is the approximate averqgc velocity of
the gasoline licxaving the: G.i:n.—diameter nozzle at this pump? V ,4 I .z Q = (8. 8 yd/ ) _
“033/6 '70)? C , 7. $4725! 6’0 3: I and Amy/C = 72—62.. "0331c = 7770.651? 2
2 W222
sa V s M '9‘"
mjﬂa (Wﬁﬁfﬁwjﬂn
V a 62.49 gﬁ‘ nayjle ...
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 Spring '08
 HARTMANN

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