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Unformatted text preview: Exam 2 CE 3700 Fluid Mechanics Fall 2005 Show all work. Show all units. NAME: 3 ‘4me (20 pts) 1. For the open tank below, ﬁnd the value of h that will produce a steady water surface elevation
(that is. a water surface elevation that does not change in time). CV CS ® 1ft diameter —Ql + Wﬁﬂdz> :O Vout:@ﬁ/$
4— L
2% = f
j 3 (1m
Q 7.
t 2 2L w L
3 £74”)
I Z6 “3/5 L
_ _,_._— f
4321?) [141005;]
PW
0.188341 (20 pts) 2. Water ﬂows in a 0.5—mdiameter pipeline at 1.0 m3/s. The pipeline is 4 km long and is
connected to a reservoir at the upstream end. a) Calculate the critical closure time.
b) Find the pressure rise that would occur if the valve were completely closed before the end of the critical closure time. s
Q 1 M/s b) A 'm
: A : —— 2 .0 __
V A ':%:(OSM)L b 3 S
1
0.1%“?
E‘V "' 2.7. (EVA 2 2.7.xlo'1 Va. (Fun—Wide M ”Wale AS)
\_.____‘.\
Q,  ti)” s /Z.Z)<Io1 N/M‘L 4 «m
Q F5 ‘ﬁv’lAiL l 83 s looc x __._r
J< MS ““3 \
Qrﬂq’coi closure el—nMe 2 4mm»; {Sew pressure waw +9 MW rme valve lo reservu'w +3 wilt/‘8. med ‘Hme Mm valve h VQRNMY Crel’l‘ca,‘ Unsure l—me : 1(23 5) : Q4 5 = 7,563,000 FL (20 pts) 2. Water ﬂows in a 0.6mdiameter pipeline at 1.2 m3/s. The pipeline is 5 km long and has a
valve at the downstream end. The pipeline is connected to a reservoir at the upstream end. a) Calculate the critical valve closure time.
b) Find the pressure rise that would occur if the valve were completely closed before the end of the critical closure time. If; "‘73
V 2 £3: 1: 2 4.14, "4/:
TT‘ 7.
7; (OW)
L—V—F“
0.293%— Ev = 2.; C779. = 2‘?— Ktbq la. Ofﬂine“ M T034“ A‘s) E 
t, = ,2;— ' : l483 "Y:
Q) Cﬂ'h’uJ Closure Hue '; Hm! Rm TressuvQ wouelo Havel ”9mm v«\ve 4o Hitch/off h valve. Travel time. (EX/WA valve in vetervsu‘»  L goQQM
C ”(53 ”(t : 3.375 [HA{Lad doswe lime : 1(337 s): (2.75 = c.1825, 000 7"; (20 pts) 3. The siphon discharges into the atmosphere at point (2). The surface of the reservoir at point
(1) is open to the atmosphere. Ignore energy losses in the siphon. 21. Find the discharge Q in the siphon.
b. Find the pressure at point A. Reservoir 5 ft 12. 2 6H
13
i p,
V; ~ (WXZWLL 9)
2 {7.1 "21“
ngx 2 m %)[% ”ﬂ
W
0.18”»?
= 14—.\ 333
E) f 5. ° ‘ .
I A n
4;
w
9
3“:
/‘\
L9
3,
\f/ (20 pts) 4. Determine the forces in the x and y directions exerted by the oil jet on the vane. The vane is moving
to the right at 5 m/s. The vane turns the oil jet in a horizontal plane. V. 2 V3 : 16 m/s. Q = 0.13 m"/s. Ignore energy losses. y
l Wul’ln wovmg camlwl volume: on (s = 0.90) V” =  4.63 “/5 WYSL:  SSOWYS
30°
N "Y; —————
__.x .
'3
v“/
A z: {3‘ 2 (1:11,: = 0 003‘ m
Qlﬂiﬂ) — v A = \\W/,Xo.oorlm1)
Vol 3
= 0.0%‘t4 “/5
EFX : (Q(T¥v’ W1)
ELSL 3 w
h = @‘l x “300%: x L“; mostlng’iSle  n 4)
X W‘ . l Kj : — \USL N Vane ctr/1mg on gel > lLaSL (\l 3d chlm3 m \rqwe CESZN—ez 2F : (Q (martial)  U
N’S'L w}
F = Q51 X(DOo [513 x LI>(0.0314"§>(”5573M6 ~ O> : " 4'43 N vane M/HVL] W 32A» 7 443'N gel (LL{rm} w Vanna £44’3N H (20 pts) 4. Determine the forces in the x and y directions exerted by the oil jeton the vane. The vane is moving
to the right at 7 m/s. The vane turns the oil jet in a horizontal plane. V] = V; = 22 m/s. Q] : 0.18 m"/s. lgnore energy losses. ly Oil (S = 0.90) 30° l N—SL k3 lit/{W WDVYVK? oowlvm Volmmt‘. vii: 42.4% ml: law—7.50%
\s'vg ““““
,  g _ mew/s = 7,
A—_ V  22% oooxlim
QGﬁ‘i’l‘) VA : i5ﬂX0.00?l8mL)
VanMC "
= 012?) ““7; 3 m
1/0423 m§)(~ll.q1’§ — 15 £35 ‘30‘13 {\l \(vae £1:th on Set 3018 H ‘Jel (LeiM3 on vawe, (BOTBN a, M
R X ZFY : g Q (my
: K
Q3 x loco f; 2n, = M (TarVa.)
F3 : (0“! x [000 —830 N 8'50 N l
l 39* Adams on wmc N—sL
'YVI I: M ho \féme “.ij on Eel— 82w T (5 pts) 5. Start with the general control volume equation shown below and derive the general continuity
equation. stys (if J
_ _ + v I
it it bpdV bp dA Lei l5,“ : mas: a? lib» Sysle/m b a mus ” ‘
— ms: g4}; (MASSoi: “HM. TLLS'ie’w‘) :c 0 Hum Jai‘nM\i‘l:au a? ‘1 “(Sin)"; Wm.“ ”p
' Sys—lem is MSW (8 pts) 6. Write the type of ﬂow represented by each of the equations below. 5km! ‘3 yum/mum ( «(mm/M uuerUMH Wﬂom ...
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 Fall '05
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