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Unformatted text preview: The University of Michigan
Department of Mechanical Engineering ME 320  Sections 1 & 2
Homework #5 — Due October 19, 2011 Problem 1: Problems 3.46, 3.88 and 3.90 of the text. Assume quasi—steady ﬂow in all
problems. Problem 2: Consider the reservoir system shown in the ﬁgure. The reservoir is initially ﬁlled
to a height H with water of density ,0. At time t : O, a valve is Opened and the water begins
to drain through the disks at the bottom. Neglecting viscous effects and assuming a large
reservoir (so that the velocities in the reservoir are negligible), obtain a differential equation
governing the ﬂow rate Q(t) for t Z 0 in this system. Problem 3: A sanding operation injects 105 particles / sec into the air in a room as shown in
the ﬁgure. The amount of dust in the room is kept at a constant level by a ventilating fan
that draws clean air into the room at section (1) and expels dusty air at section Consider
a control volume whose surface is the interior surface of the room (excluding the sander) and
a system consisting of the material within the control volume at time t : O... (a) If N is
the number of particles, discuss the physical meaning and evaluate the terms DNSyS/Dt and
8ch/8t. (b) Use the Reynolds transport theorem to determine the concentration of particles
(particles / m3) in the exhaust air for steady state conditions. (c) Consider this problem for
the case that the ventilating fan fails to work so that V1 : V2 : 0. Evaluate and discuss the
meaning of the terms DNSyS/Dt and 8NOU/8t for this case. ‘ ‘ F ' ' ' * I v 1 * '  ' v * 1 +  I ' I r  r I v r I   r   m  a a   I     . + . I 1 I  r I r I o  r a + . I . . I I 4 a . . . . . . . r . . . . . . . . . . . . . . . . . . . 1 . . l . , . . . . , . . . . . . . . . r       I i s v a l r . . . t  u I a . . . . I . . I . , . . . . . . . . . . i . . _ _ . _ ' _ _ _ _ ' _ _ _ _ _ _ *Il:i:lllll:HJ1I'1:r:I‘I‘J‘l‘i‘l.lIi'l 15minI.Ia“IrII.I.91‘.I:..‘:hl:1:I:u:I:+::.:;:.:I:n.:.:.:.:.I:'.._:.:.Ir»:.I..:.:.:.'lII.Iar...I..i...II.u.'_.I.:..m.I“.,+,,.:.:..II:‘I;.':ﬁlialF:If?”:II_:.‘II_:.II":.I:_:I:.I'I‘:,:I':_:.I_:IIrlll“I'm”...1 3'1"..." "iii. _ /
.',4_ F I  I i   I   ' I 4 I I I I 3 I I  I  r    4 p h  I a    u     1 n  p p u   . p u  . . a . . . . a n a u . . a . . . a . . a . . . . . . . . . . . . . . . . . . . . . _ , . . . . . . . ..._._. l 2 I I I E;
n"r' ' tr. ln... .' .'. .:.: .I . .I_.I . a in.) I I I... 1:: o‘ .. my 2
ﬂ? A2=O.1 m
“M L. Problem 4: Problems 5.13 and 5.16 of the text. Problem 5: Problems 5.26 and 5.31 of the text. 3.1M Pop (with the same properties as water) ﬂows from a
4~in. diameter pop container that contains three holes as shown in
Fig. P336 (see Video 33?). The diameter of each ﬂuid stream is
0.15 in., and the distance between holes is 2 in. If viscous effects
are negligible and quasisteady conditions are assumed, determine
the time at which the pop stops draining from the tap hole.
Assume the pep surface is 2 in. above the top hole when t = 0.
Compare your results with the time you measure from the video. Q=Q,+Qz+03=h/qrg{l aFiGuaE P345
were 0‘1 M = W a and x», a a anew (£51,213) = /.227X/0—y7C/2
—_'Z.7 i 1.. 2
Th”: HT v 4 {,2 ﬂ) .. o. 0873 {J
J +10; 4%] .747" 36%.; when: htsth2=h+LJh3:h+2/' and L =2 I27.
HBHCGJ —(V¥A,/Ar)jfn =f0 A II dh . . .  wﬁere I! I: 71/79 Ill/979 7’
(W7 +Vh+z WW) {an fez/"1% {mm/trace 1‘0 recto/7 7% V/a/arﬁa/e or L (lyea ’
f = A7 a“? 1 Mr 0(V/7+Vh_+2+ mu = 0. as 73 H’ L d1;
(/.227XIO+H2)[(1){32'2“3:81;ng +1417 W14 +2.1. )
771m; 1.
of}; 2
)1: b ‘F‘H50J667
a (VFW/97PM) W m ’2 {’1
Mlle: lat/x76 if» 7426’] 1%}; email/'0» 9/1/63 15 I}? sew/talc. (600’!) 7720 numerical valve of {he 1711'69m/ is ebb/ﬂed [11/03/779 7/90
{rapezoidal rule since #70 closed form ana/yﬁcaf 30/1175an
is 17011 given in 1771’199/‘4/ fab/es. 7Z6 EXCEL sprang/35.99% wed {or #1711 is 911/00 lye/ow. L I
7 = 01.7 5 10140 were 101= O 4. 00.7%; (1‘, +£1,)(h..,—h.)] =(6’87 ﬁﬂoIwVﬁk I075 n, in. h. 11 KM. 11ft"2 (172)1410010“  h1). ﬂ” 7
0.0 0.0000 1.015 0.00004 1
0.1 0.0003 0.914 0.00743 2
0.2 0.0107 0.070 0.00711 3
0.3 0.0250 0.037 0.00000 4
0.4 0.0333 0.010 0.00005 5
0.5 0.0417 0.700 0.00040 5
0.0 0.0500 0.704 0.00029 7
0.7 0.0503 0.745 0.00014 0
0.0 0.0007 0.720 0.00000 9
0.9 0.0750 0.712 0.00507 10
1.0 0.0033 0.097 0.00575 11
1.1 0.0917 0.004 0.00504 12
1.2 0.1000 0.071 0.00554 13
1.3 0.1003 0.059 0.00544 14
1.4 0.1107 0.047 0.00535 15
1.5 0.1250 0.037 0.00520 10
1.0 0.1333 0.027 0.00510 17
1.7 0.1417 0.017 0.00510 10
1.0 0.1500 0.000 0.00503 19
1.9 0.1503 0.599 0.00490 20
2.0 0.1007 0.591 21 Sum of column = integral = 0.12011 Thus. t = 88.7*0.12011 =10] S 392 3.83 A long water trough of triangular cross
section is formed from two planks as is shown in
Fig. P338 A gap of 0.1 in. remains at the junc
tion of the two planks. If the water depth initially was 2 ft, how long a time does it take for the
water depth to reduce to 1 ft.? FIGURE P388 :—%+V22 2.
ﬁluVL'FZ '5’?“ +32 (1) (I) J” 2; ’
:0 J 2, =5, and Ez=0 h
ﬂ/so =lé/42 01‘ since I>>ﬂr [7‘ d fallow“;
follows ﬁai‘ 1/, << 14; J Wéere Z = " a?
777st £310} git/85 lé Lil/2; So #7677! _ '1g.§l=,qz 23/, Wff/g 4=5Z=266 dﬂd 142:5:4/ w/aere b is Me fame? ﬂangf/i or_I
{h a“! = — HIng bub/d? can be fnfegmfed 1‘0 9/1/19
hf] 3’96 3.90 When the drain plug is pulled, water flows from a hole in the
bottom of a large, open cylindrical tank. Show that if viscous ef—
fects are negligible and if the ﬂow is assumed to be quasisteady,
then it takes 3.41 times longer to empty the entire tank than it does
to empty the ﬁrst half of the tank. Explain why this is so. a=av =a‘aﬂv =4“ (—3?)
where Infeyrmle from 5“” at {=0 7‘0 /2 ate: hath d1;
HEW "We—Ma” or ’1 1
21W "@651!
H 0f 2 2
a We) [WTWT]
77211.; flee/Mm? the fan/g
/=  (3W1
[3:0
and fa lmlf empty 2%? hulk) 26/ =43”
2. (0): free ratio“: uh. +1? (warn'7‘ (i)! 384%) Jim: , rifLJ’GJ‘ 9118. Brio!
I a} W‘IVE' (04 r: r.) A4" mil SECJfOA. (5.4 rzg) 'u‘ we ‘9 ﬂ 2 ' z 5‘ 2
zmxwt .. £+A¢3y+53145 3" 2 g 2 3+ EAL R, E m awrflen‘ao V950 m H» NJQIVNV.  '3.
3(2a“%;)7— iii—.9 g E‘nafzH(+}
o l 'é‘rko tév “La 1
3.! a! ‘3 7:0” <f __ Jr .. __ §3V
E“ i ' ([34 ' SM Jr ,3?» J" {L‘ﬁ'Jf MIWVoJV s q / Xx
. '3 X
,__ as QR \
M Maw + Mm Mao ;/ a
. I /’
c s 5 I K I I3 C V \ \ v. N {I}
\ﬂ’fi\v/‘u._,~,,ﬂwn ) \ a 'j\ Ex?»
I a C \xx» ' ’(LOEF W 4 ' VI . a ‘2’ ( VI ‘\\\—v'/W‘_,r’
Chats?! L M) .
(30% m, c vim 7» v1: Jim» W; mg): 'L
$31] "5 “ L z \X/Lefﬁ 3—K 7 L fig
, 3* ' anLI /: i9; (55,, \ “Q'Rmumg
4+ 2,214 2 r M22223 /. (ﬂ. 2 i9 + ,9 (“Wag H0») ) HIM/x}!sz 3:” 5m“
ZQL‘ 0H“ ERZK'LL’Z 0 +20 Pleem 5, _
A sanding operation into the air in a room as shown in Fig. jamounx of damn; the mom 'ismainwimd gt a" can?
by: {mm draws dammit into mermam
In (I) and expel: dusty air a! section (2). Consider a
contra! veiume whose surface is 111: interior surface aidla room
(«angling the mast) and a system consisting of the material
within the come! volume at time I In. 0. (a) If N is'lhe number
of particles. discuss 1h: physical meaning of and avaium the
W M'me and an/at. (1:) Us: the Reynolds transport
themcm to determine the concentration of panic!” (pani
cles/m’) in the alums: air for Steady state caudiﬁqns._ mm: 5 >: . m‘ pm‘liu’trh 'mJ.‘ nu... _ 2 .DM 5 d)’ D =' him rafe of change cf Ha: number of purities I}? 1%: V22mls a Azio‘l WI2 .sysfem. ﬂf ﬂare #0 Hie sysfem 5mm: of Npm‘icles. In fact H73 sysfem is +53“ paffgbles for all Hm ism. ' ﬁssummy Hid {be par{ides do 1101‘ gef "gleeffaydbor, IV remam: (:0me
Thus ,V 31%“ *0  ' ‘3'?th x gm “"9 0*. 5119099 of H78 maiber of parh‘c/es 131' 1%
confra/ pro/ma; Depend/395: an 1%: Mia af win“ {he Sandw crearespari‘ic/as and ejem‘: Men Ma Ma mm
compared 1‘0 fhe raft: m‘ whiz/I {be fan draw: ﬂam {rm 1%: mm, we could have 95%! % 0. . N I
5) gal/jig = 3553 +179} rd: of Hoard parh'cles am‘ of Mm} volume
or {or ﬁend); sfale éﬁﬁw Sg‘Maf: Haw mt parficles info awry] valvme (f/m sander, none enfer I41)
 Flow of farm/e: am‘ of mnfral viz/am ( filmy/a I’m aha/:1" :91 7720.3, “’05 ﬁffc/es_ 3 I ﬂam J Mm»: ni= parfiqle iconrzmlraﬁon (matski) Hence, rib, ‘ .
. muff“J = 5m; gar£15k: // CC) DMZ; : O L ( 9mg we wumbgy of particle; Yeww‘ws (ow/472;?
’ 9* prov‘de paﬂft‘cle; d0 WC bewwe #5516164" @Wev . 0*, 1%va “We COMJCVQI Volume
WW4 \A =szo 7 We only How 0(CV099 we cox/Jim! garage (‘9 ﬁrm We gander 61W! F9 eapmf “to ,, [05’ lam/tides
'9  Thu; 7 856/: U) become; I BNGV. _. 9 E’ffﬁdes) (BNGV \ 5 fl
0 __ . 0 Y ————4\ YlCEf
2+ H I 9 ' 0 M “9 ﬂ; Wet air
m = 156,900 Ibmihr
53.13 An evaporative cooling tower (see Fig. P513) is used to cool water from 110 to 80°F. Water enters the tower at a. rate ’of is 156,900 lbm/hr. determine the rate of water evaporation in
lbm /hr and the rate of cooled water ﬂow in lbm/hr. ' Dryajr —)
rh =151.0001er For sfeady flow of 00y air “F‘IQURE 95.13 If] :m (I) 3 2.) d1? airr'
Far H2009 F/0w of Wafer ‘z' (I l 2’ wafer
’1/50 = ' + ’1'?
m2 m2, dry 0,) 2, wafer (3) Coméimhj E135. l and 3 we yef ' = m « bl? = I’d/’6 0/ wafer euaporab‘m—r
mzzwaier 2 3 50
s 156 900 “’m _ 15/2000 L4: 0 5900 L5: .._— m30W4‘lo" [7,. hr Ar 50m £3. 2 we 96+  M —— rid = I‘m‘e 0F coo/ed wafer ﬂow?
4 H 1 2,w0+er 10,,
M», __ 5700 [£0 ._= 244/, 000 __
g: [Ir hr“ 5!6 5.]6 Freshwater ﬂows steadily into an Open
SSgal drum initially ﬁlled with seawater. The
freshwater mixes thoroughly with the seawater
and the mixture overﬂows out of the drum. If the
freshwater ﬁowrate is 10 gal/min, estimate the
time in seconds required to decrease the differ
ence between the density of the mixture and the
density of fresh water by 50%. A {raced , nan deformriy conﬁm/ Vo/uma 741ml cmfmhs 1%:
(h the 55  94/ drum 1': used. Fresh wafer
confra/ I’d/um: wafer mixfure
errhrs MC wifh densffy, ’6’? J and volume
flowml'e’ ’  The Ml'x'f‘urc i5 dimmed 7’» 15¢ homageneous
+hroughou+ ﬁle. General Volume and leave: Me Camﬁ/ Vd/umc
wiH: density, 3 I and volume flown/rach az. App/,‘caﬁ'an
of ﬁve. Cmatcrvaﬁan 9/ mass eqaaﬁ'an 55) 2‘0 7‘68. f/ow f/trauy/q 1%"; Camt‘m/ Ila/um: yields
_ = {/J
346” t 4% M 0
v 7,“: ,. are, IhCOMp/GSS/é/C’ Q: = Q; :. Q . WM} Volume. is msv‘ztnf 771115 slhce 'H’IC. {/m'ds' (Java/wed
Mia, fhe. volume of M £35.! lead: 1'29
t: 1" g Q = Q
di .
or
4’?) + (L) g ., 21 (2)
CH ,4 47ch {73/
7776 Joluﬁan of Q53 2 i5
6: = C 5‘wa + 10 (3)
/.9
' 5/117:
M f: 0) = freame , {’99 4'3): [026
I fﬁesémhr (/ 94 mfg/#3)
Hour
C = 0.02é
(can't) " 'fm'a/ ml‘KﬁH’e densi‘fy im'w‘ia/ miﬁwe density fubsz‘h‘uf/n'j ﬂu‘: m/ue (fggl/ngcrff/Vm‘ "3"? 533 W3 3”
(55y!)(6ar§§ag)  [.013 = 0,026 e + [0 and
1E :— 22? s 5172 5.26 Estimate the time required to ﬁll with water a cone—
shaped container (see Fig. P526) 5 ft high and 5 ft across at the
top if the ﬁlling rate is 20 gal/min. I FIG U RE P526 513m Willa/7'0!) af‘ #4: (mien/47245;: 6% pics: pr/hcxjo/e 74? {£5 can 'fml volume. Slum/n 1h ﬂit £39qu W6 lam/a +f/aﬁ3dﬁ = 0 at cs For I}? compressive flow éi“_a =0
9t 0r 1‘ f
[w = Q/dt‘
0 0 Th 2 1':
"5 Jo; 72m A , 7f (gawk; f+)('728 a") 5 f g a 1261 ’ (121(20w)(23/§_’3 ' 3 Min ‘7‘!
f = [2 2. Min 53! Storm sewer backup causes your basement to flood at 60%”, I Volume ﬁnd
the steady rate of 1 in. of depth per hour. The basement ﬂoor f canfﬁfns Wafer
area is 1500 ftz. What capacity (gal/min) pump would you rent —— ——  — — — — — — I
I0 (a) keep the water accumulated in your basement at a constant //I x I : h level until the storm sewer is blocked off, (b) reduce the water / i / / ]
accumulation in your basement at a rate of 3 in./hr even while /  . _ / ’4 “ ‘1'
the backup problem ex1sts?  [‘br 57 661610??an Conﬁ/o/ Vii/«me ﬂmz‘ can/67x37: J‘Ac wa/or
over fhe Ansemenf ﬂoor (566’. 51$;th agave) J fee Conserva/a‘m 09‘ mas: pram/we, (.53 52/7) Ami; f0 *fﬂlfﬁa/ﬂ :0
at CV C; or 761» Cam’fanf f/m'd dens/7 and area (A) dé “ (4) 5r Pan‘ 6?. J / 4344576)
Qﬂue = Q5?
729 eva/uml: 4;,“ we use, 53'. I wi/t, QMso. 771m, _ = A (111 z. {1500 f¥‘){/ 31. /
at»; df hr)//Z )= 125 19‘? . 1‘1. 1”
and ﬂ
= (/25 fig/29485627 _/_ __ ,56 34/
Q0101" ;,,. {:3 “ﬂ " _'_ I”;
In (b) For powf" 6) 53/ yie/d:
Q0114 = Q"; F A Q o’t
=15.6!3_’ _/ 0029‘ 3/_11 (Tﬁﬁﬂ)/J__
Q0“. mi” X hr)125.) {1,3 60,???
Q + = 62.4 22!
0“ _— Md}, 525 ...
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