BME 516 (1)
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Transcript of BME 516 (1)
Ryerson University Department of Mechanical and Indnstrial Engineering
MEC 516 Fluid Mechanics I Midterm Exam October 26, 2009
15:10 -17:00 Instructor: Prof. W. Leong Associate Examiner: Prof. Z. Saghir
Instructions: 1. This is a closed-book exam, i.e. no aids permitted. Only a scientific calculator (non
communicable and non-programmable) is allowed.
2. Write your solutions using pen only (no pencil).
3. Read each question carefully. It is your responsibility to write clearly and legibly. Illegible solutions will not be graded.
4. Always sketch free-body diagram indicating the forces involved and system diagrams indicating relevant information, such as, control volume with dashed lines, symbols, distances, velocities, locations, etc ..
5. State all assumptions made for answering the questions.
6. Use necessary unit conversion factors to have consistent units in your solutions, and make sure that the final answers are in the required units.
7. Formula and data sheet is on page 2.
Failure to do these things will result in loss of marks, even if the correct answer is obtained.
Marks (in %) are shown on the question sheets.
Name: __ '-----
Section: O~
Marks· I
2
3
4
Total:
Page 1 of7
q I·p) II ) '2.A: .. So~
MECU
Meichaillics I MidtE'rm Exam October 26, 2009
15:10 -17:00 Instructor: W. Leong Associate Ex,aminer: Prof. Saghir
Instructions: L This is a clo,scII-boo,kexam, no aids permitted. Only a scilentitiic c:alcullltor (non-
non-programmable) is allowed.
2. Write your solutions (no pencil).
3. Read each carefully. It is your responsibility to write clearly and legibly. Illegible solutions will be graded.
4. Always sketch indicating the forces involved and system diagra:ms indicating such as, control with dashed distances, velocities, locations, etc ..
5. State assumptions made answering the questions.
6. t'"'llm's to have COIlsi:;tellt units your solutions, and the final answers are the renruil'ed
7. Fonnula sheet is on page 2.
to do these things will loss of marks, even if the correct answer is obtained.
Marks (in %) are shown on question sheets.
Marks: --_._--_.
1
2
3
4
Total:
Page lof7
Obtained from Ryerson's Mechanical Engineering Course Union (MECU)-www.ryemecu.com
MECU Page 1 of 6 www.ryemecu.com
Page 2 of7
Formula and Data Sheet
Ideal-gas law: P = pRT, Ral, = 287 J 1 kg· K
Surface tension: /';p = Y(R,-' - R~I )
Hydrostatics, constant density: p, - PI =-r(z, -zJ, P= rh+ Po' r= pg
Hydrostatic panel force:
Centroidalmomentsofinertia: A=bL,1xx =bL3 112,!xy =0; A=7rR',!xx =7rR 4 14,1xy =0
Buoyant force: F = r fI,'d \I dl.'plae,d
CVmass: O=!!.... fpdl/+ "(pAV) - "(PAV) ~ L _ L m
Yo,
Table 1.1 Primary Dimensions in Primary dimension SI unit
SI and BG Systems Mass 1M) Kilogram (kg) Lenglh IL} Meter (m) Time IT} Second (s) Temperature {®, Kelvin (K)
,'liable 1.2 Secondary Dimensions Sec.ondary dimension SI unit m Fluid Mechanics
Area IL') 10'
Volume (L3) m'
Velocity ILr-') m/s Acceie1'3.1jon ILT-2
} mig' Pressure or stress (ML -IT-2 J Pa """ N/m2 AnguJa.r velocity IT- J) ,-' Energy, heat, wOIk IML'T -'I l~N'm
Power IML'T-') W ~ JI, Dengity IML -') kflm' Viscosity (ML-1T-'} kg/(m . s) Specific heat (L2
T-2e-1
) m2/(s2 . K)
rable A.I Viscosity and Density of T,oC p, kg/ml p., N . slm2
P, m2/s Water at 1 atm 0 1000 1.788 E-3 1.788 E-6
10 1000 1.307 E-3 1.307 E-6 20 998 1.003 E-3 1.005 E-6 30 996 0.799 E-3 0.802 E-6 40 992 0.657 E-3 0.662 E-6 50 988 0.548 E-3 0.555 E-6 60 983 0.467 B-3 0.475 E-6 70 978 0.405 E-3 0.414 B-6 80 972 0.355 E-3 0.365 B-6 90 965 0.316 E-3 0.327 B-6
100 958 0.283 E-3 0.295 E-6
Table A.2 Viscosity a..,d Density of T,oe p, kglml p., N . s/m'l. P, m2/s Air at 1 atm
-40 1.52 1.51 E-5 0.99 E-5 0 129 1.71 B-5 1.33 E-5
20 1.20 1.80 E-5 1.50 E-5 50 [,09 1.95 E-5 1.79 E-5
100 0.946 2.17 E-5 2.30 E-5 150 0.835 2.38 B-S 2.85 B-5 200 0.746 2.57 B-S 3.45 B-5 250 0.675 2.75 E-5 4.08 E-5 300 0.616 2.93 E-5 4.75 B-5 400 0.525 3.25 B-5 6.20 E-5 500 0.457 3.55 E-5 7.77 B-5
BG unit Conv~rsion factor
Slug 1 slug ~ 14.5939 kg Foot (ft) 1 ft ~ 0.3048 m Second (g) I s = 1 s Rankine ("R) 1 K ~ 1.8°R
BG unit Conversion factor
rt' 1 m' ~ 10.764 rt' ft' 1 ml = 35.315 ftl
ftf' I ft/s ~ 0.3048 mls ftfg' I ftfi' ~ 0.3048 mli' Ibr/tt' 1 Ibflft' = 47.88 P. s-' IS-I = 18-1
ft . Ibf 1 ft . Ibf ~ 1.3558 1 ft . Ibf/g 1 ft . Ibf/' ~ 1.3558 W ,Iugs!fl' 1 'iuglft' = 515.4 kg/m' slugs/(ft • s) I ,Iug/(ft . g) ~ 47.88 kg/(m . s) rt'/(g' . OR) 1 m'/(,' . K) = 5.980 tt'/(,' . OR)
T, OF p, slugtW /L, Ib . sift' v, £t?-/s
32 1.940 3.73 E-5 1.925 E-5 50 1.940 2.73 E-5 1.407 E-5 68 1.937 2.09 B-5 1.082 E-5 86 1.932 1.67 E-5 0.864 E-5
104 1.925 1.37 E-5 0.713 B-5 122 1.917 l.I4 B-5 0.597 E-5 140 1.908 0.9756-5 0.511 6-5 158 1.897 0.8468-5 0.4468-5 n6 1.886 0.741 E-5 0.393 E-5 194 1.873 0.660 E-5 0.352 E-5 212 1.859 0.591 E-S 0.318 E-5
T, OF p, slug/ftJ /L, Ib . sift' P, tr/s, -40 2.94 E-3 3.16 E-7 1.07 E-4
32 2.51 B-3 3.58 E-7 1.43 E-4 68 2.34 B-3 3.76 E-7 1.61 E~!
122 2.12 E-3 4.08 B-7 1.93 E-" 212 1.84 E-3 4.54 B-7 2.47 E'" 302 1.62 B-3 4.97 E-7 3.07 E-'j 392 1.45 E-3 5.37 E-7 3.71 E- . 482 1.31 B-3 5.75 B-7 439E-4 572 1.20 B-3 6.11 B-7 5.12 E-1 752 1.02 E-3 6.79 E-7 6.67 E-. 932 0.89 E-3 7.41 B-7 8.37 E-4
MECU
Page 2 of7
Data
lU\.i'"-g''' law: p = pRT, Rai, = 287 J I kg·
Surface tension: t\p = Y(Rr-r - R2~1 )
Hydrostatics, constant density: h-Pl =-r(z,-zJ, p=rh+Pa' r=pg
Hydrostatic panel force: .F=yhcc;A, Yep =
Centroidal moments of inertia: A = b L, Ixx = bL'
B " l' ~ V uoyant lorce: ; = Yflwd dt.'placed
v) -out
CV momentum: - d [ f~' J F = dt v"V p dV +
Table 1.1 Primary Dimensions in 81 and BG Systems
dimension
(PAV)in
SI unit
Mass {M} Lenglh [L)
Tim' IT} Temperature {€I)
Kilogram (kg) Meter (m) Second (s) Kelvin (K)
1.2 Dimensions dimension SI unit
Fluid Mecha,,;cs Area IL'I m' Volume (L:l1 m3
Velocity I ml' Acceleral.ion (l.T -"} m!s2 Pressure or stress {MC·IT" 1.J Pa ...." NJm2 Angu1ar veloclty IT-II ,-I
Bnergy, heat, work {ML2r -2} J=N'rn Power {ML'1"-' I W = J/, Density [ML -3J kglm3
Viscosity {ML--'T- 1 J kgl(m . ,) Specific heat {L 2T -28:- 1 J m1/(s2 . K)
fable A.I Viscosity and Density of T~ <Ie
Water at 1 atm p, /L, N . slm2 v, m2/s
0 1000 1.788 E-3 1.788 E6 10 1000 1.307 B-3 1.307 E-6 20 998 1.003 E-3 1.005 E-6 30 996 O~799 Ii-) 0.802 E-6 40 992 O~657 E-3 0.662 ij-6 50 988 0.548 £-3 D~555 8-6 60 983 OA67 1l-3 0.475 1i-6 70 978 DADS E-3 0.4t4 E-6 80 972 0.355 E-3 0.365 E-6 90 965 0.316 E-3 0.3276-6
100 958 0.283 E-3 0.295 E-6
Toole A.2 Viscosity ailld Density of 1', .,c p., N . slm2
P, m2/s Air at 1 atm p,
--40 1.52 1.5t E-5 0.99 8-5 (} 129 1.71 E-5 1.33 1l-5
20 1.20 1.80 Il- 5 1.50 B-5 50 L09 1.95 E-5 1.79 E~5
100 0.946 2.17 E-5 2.30 B-5 t50 0.835 2.38 E-5 2.85 E-5 200 0.746 2.57 1l-5 3.45 E-5 250 0.675 2.75 E-5 4.0B 8-5 300 0.616 2.93 E-~5 4.75 B-5 4CO 0.525 3~25 E-5 6.20 £-5 500 0.457 3.55 B-5 7.77 ,E-""j
=7rR4
14,/xy =0
BGunH ConversIuu factor
Slug' Foot (ft) Soomd (s) Rankine ("R)
I slug ~ 14~5939 kg I ft = 0.3048 m ) s = 1. s I K ~ 1.8'R
BG unif Conversion factor
ri' I m' ~ 10.764 It' it' 1 m3 = 35.315 fit!!
ft/S I rtl, ~- 0.3048 ml' ft/s2 I ft/s7. = 0.3048 m/..r
I IbOO' = 47.88 Pa s 18-1 = ls-1
it . lof 1 ft • Ibf ~ 1.3558 J f! . Ibf/s I ft . lbf/' = 1.3558 W slugs1ft' 1 'l"glft' = 515.4 kg/m' slugsl(ft • 5) I ,lng/(ft . ,) = 47.88 kJlI(m . 5) fi'/(,' . OR) I m'I(,' . K) ~ 5~980 f?-I(s' ~ 'R)
T,"'F 1', /" Ib . ,Iff JI'~ rtt/s 32 1.940 3.73 B-5 1.925 B-5 50 1.940 2.73 E-5 1.407 B-5 68 1.937 2.09 6-5 J.[i!l2 E-5 86 1.932 1.67 1l-5 0.864 E-5
104 1.925 137 6-5 0.713 8-5 122 1.917 1.14 85 0.597 8-5 140 1.908 0.975 1l~5 {l.51! E-5 158 1.897 0.8468-5 OM68-5 1-76 1.886 0.741 8-5 0.393 E-5 194 1.873 0.660 E-5 0.352 E-5 212 1.859 0.591 8-5 0318 E-5
T,"'fi' p, slug/nJ p., Ib . ,1ft' p~ rrls, -40 2.94 E-3 3.161l-7 1.07
32 2.51 B-3 3.58 E-7 1.43 68 2.34 B-3 3.76 E-7 1.61
122 2.12 B-3 4.08 E-7 1.93 212 t.84 1l-3 4.54 B-7 2.47 3()2 1.62 E-3 4.97 il-7 3~1l1
392 1.45 B-3 5.37 B-7 3.71 482 1.31E~3 5,75 B-7 4.39 572 1.20 E-3 6.11 E-7 5.12 752 1.02 B-3 6.19 E-7 6.67
932 0.89 E-3 7.41 E-7 B.37
Obtained from Ryerson's Mechanical Engineering Course Union (MECU)-www.ryemecu.com
MECU Page 2 of 6 www.ryemecu.com
Page 3 of7
Problem 1 (20% Marks) On a calm day with air temperature of 200 e and barometric pressure of 1 atm, a skydiver goes for skydiving. Before opening parachute, the drag force on the skydiver is approximated by the
following relation: FD = e J.L pV 2
where e is a constant, J.L is the air viscosity, p is the air density, and V is the velocity of falling skydiver. (a) Determine the dimensions of the constant e, in terms of the primary dimensions. (b) If the skydiver has a total weight of 580 N and a body volume of 0.3 m3
, using e = 1.528x 1 04
with the basic SI units for its dimensions obtained from part (a), find the terminal velocity V, in mis, of the skydiver?
I <1 . ~- <. "
~/ v-
b) rD" C,£.{;\j2
MECU
Pagc3 of7
On a calm day ""1th air temperature of 200 e and barometric pressure of 1 atm, a skydiver goes for skydiving. Before parachute, the drag force on the skydiver is approximated by the following relation: = e JI pV 2
where e is a constant, JI is the viscosity, p is density, and V is the velocity falling skydiver. (a) Detennine the constant e, in tenns of the m:i!!!!,[Y~:rl!e~ill!li. (b) If the skydiver has a total weight of 580 N a body volume of 0.3 , using e = l.528x 104
with the basic SI units for its dimensions obtained from part (a), find the velocity V, mis, skydiver?
~ltI'{X3M )
/ ftP18j
Obtained from Ryerson's Mechanical Engineering Course Union (MECU)-www.ryemecu.com
MECU Page 3 of 6 www.ryemecu.com
Problem 2 (300/, . ' A . • Ma""j ,~'"" gate III the diagram b 1 ' at 0 d e ow IS 30 h' an prevented to 0 e cm Igh, 60 cm wide into
gate to open to the right! n to left by a block B. What wate~~:~e~, ;Vhich is. hinged at its top . _ p, III m, will first cause the
~--.-----L-~ - I J
water at I 200t - - - --I
I a I AiratlOkPa I (gage) I
£ M 0 '" 0 ~ (0,., (1"i',;t~,r) - A" ("r ) O~J<:J61(hJ')[.l~A (X-":~ ] - 180~
V =Q'1 b I h - 'Jl.I..i.1 S )[,,51' »>.10.31
-7...':::10 y.._.16 J
o ~ ::l~4.IE>h -i i3:j}.~~~ _ ;,G\.6Z25 -I,q8_ -2QO - h-. 15 -))-.1'0
(h-, 15)2'" .i5h. ,'.2'",), -(h-.I~)N ,<226 -I ,q 8-2'l0'0
::: ,,- 04 "15~ ')iiO·
-52.8,3
wJJ ~tL
MECU
Page 4 of7
A gate in the below is 30 em high, 60 em wide into the paper, which is hinged at top at 0 prevented to to left by a block B. What water depth In, cause the gate to open to the right?
,,"- /Ole
r:. "" 6o~1 /1,<-
7_.I'C"''' I
h C[)~ ~';"b'l)\ h ~ h-
, I
·hc<J.. q)::h
8 ('h
. h ;,<L
.rB)
0:::: l ~ (./. I h..j ~..:.:.:::-;:-;;:--
C ) - ( .l':') .~
h-.15 Z(,t.1./5),-+
.ISL- i I
4.1 ht.-,----h-
, .
AiratlOkPa I (gage) I
"" ,l{l(
:::::1
(~
I Oy5) v
.x
0::: 0
1 .
.;j r"
Obtained from Ryerson's Mechanical Engineering Course Union (MECU)-www.ryemecu.com
MECU Page 4 of 6 www.ryemecu.com
Page 5 of7
Problem 3 (20% Marks) A mercury manometer is connected at two points to a horizontal pipe with flowing water at 20ce for measuring pressure drop due to pipe friction. If the manometer reading is h = 3.5 em, what is the pressure difference, in Pa, between points I and 2? Is the water flowing to the left or to the right? Explain why, based on your result, the water is flowing in that direction.
, .A "'.)1'" ~ }-1) .\k <.J
() I \ '. , CD CV water ....,_~3i-ed ~,.,.,.V~ I.b '\ 1'01\'\ 'f'-e.l.'T - 5fW4 CAY- "~ \ '/, MA -eq-<:.",h ~{' S,,"'~o,. ... .{, ~Y\>lcf\
~ (YI£J.\"O~1<'\' ""~5~',-J \ ----r;- ec..,).,<.~I- {uj ~ ~ loss ~v(.., 'Tc '" . "of' ~ ::;r
r«-:7,,-,, ~ ( I
f' _f<V.='J5'z,-J,/<"j,'h '~) Jl V'" ~ < ~ W~ f,,::' ~-i,=- ~ -t h:o~A-~
g, r:- (k~ j;) -==- -f <Y) <j
/0 <-0@0/& ·-{JL,}
CVf'~@~ ,~ ~
r;;t~ 0:: 11$:lI3,&b)
m' V
MECU
Page50f7
A mercury manometer is connected at two points to a pipe with flowing water at 20°C meastrrin.g preSSUlre drop due to pipe friction. manometer reading is h = em, what is
dif'r",·en,,,,. in Pa, points 1 and 27 Is the water flowing to or to right? "lilly, based on your the water is flowing in that dil·ecitiolil .
, :0(\ ~
) ( water
I mercmy
'~~~'!JiPSGm ~ 13.55
...,.A~>"""<:.. }-o
Obtained from Ryerson's Mechanical Engineering Course Union (MECU)-www.ryemecu.com
MECU Page 5 of 6 www.ryemecu.com