Fluid Mechanics Exercices Enstab Cou SW
Fluid Mechanics Exercices Enstab Cou SW
Fluid Mechanics Exercices Enstab Cou SW
Fluid Statics
2.1 For the dam shown in Fig. 2-1, find the horizontal pressure acting at the face of the dam at 20-ft depth.
! p = yh = (62.4)(20) = 1248 lb/ft?
2.2 For the vessel containing glycerin under pressure as shown in Fig. 2-2, find the pressure at the bottom of the
tank. :
i p =50+ yh = 50+ (12.34)(2.0) =74.68kN/m? or 74.68kPa
50 kPa
y —_
Glycerin 2m
Ls Fig. 2-2
2.3 If the pressure in a tank is 50 psi, find the equivalent pressure head of (a) water, (b) mercury, and (c) heavy
fuel oil with a specific gravity of 0.92. , ,
i h=ply,
(a) h = [(50)(144)]/62.4 = 115.38 ft
(b) h = [(50)(144)]/847.3 = 8.50 ft
(c) h = [(50)(144)]/[(0.92)(62.4)} = 125.42 ft
25
26 0 CHAPTER 2
2.4 A weather report indicates the barometric pressure is 29.75 in of mercury. What is the atmospheric pressure in
pounds per square inch?
f p = yh = [(13.6)(62.4)][(29.75/12)]/144 = 14.61 lb/in? or 14.61 psi
2.5 Find the atmospheric pressure in kilopascals if a mercury barometer reads 742 mm.
2.6 A pressure gage 7.0m above the bottom of a tank containing a liquid reads 64.94 kPa; another gage at height
4.0.m reads 87.53 kPa. Compute the specific weight and mass density of the fluid.
i y = Ap/Ah = (87.53 — 64.94)/(7.0 — 4.0) =7.53kKN/m? or 7530N/m?
. p= y/g = 7530/9.81 = 786 kg/m?
2.7 A pressure gage 19.0 ft above the bottom of a tank containing a liquid reads 13.19 psi; another gage at height
14.0 ft reads 15.12 psi. Compute the specific weight, mass density, and specific gravity of the liquid.
i Ap=y(Ah) (15.12 — 13.19)(144) = (y)(19.0- 14.0) —_-y = 55.6 1b/ft®
p= y/g = 55.6/32.2 = 1.73 slug/ft? $.g. = 55.6/62.4 = 0.891
2.8 An open tank contains 5.7 m of water covered with 2.8 m of kerosene (y = 8.0 kN/m’). Find the pressure at the
interface and at the bottom of the tank.
i | Pine = yh = (8.0)(2.8) = 22.4 kPa
Pros = 22.4 + (9.79)(5.7) = 78.2 kPa
29 An open tank contains 9.4 ft of water beneath 1.8 ft of oil (s.g. = 0.85). Find the pressure at the interface and at
the bottom of the tank.
| Pin = yh = [(0.85)(62.4)](1.8)/144 = 0.663 psi
Poot = 0.663 + (62.4)(9.4)/144 = 4.74 psi
2.10 If air had a constant specific weight of 0.076 lb/ft? and were incompressible, what would be the height of the
atmosphere if sea-level pressure were 14.92 psia?
| h = p/y = (14.92)(144)/0.076 = 28 270 ft
2.11 If the weight density of mud is given by y = 65.0 + 0.2h, where y is in lb/ft® and depth h is in ft, determine the
pressure, in psi, at a depth of 17 ft.
I dp=ydh = (65.0+0.2h) dh. Integrating both sides: p = 65.0h + 0.1h”. For h = 17 ft:
p = (65.0)(17)/144 + (0.1)(17)7/144 = 7.87 psi.
2.12 If the absolute pressure in a gas is 40.0 psia and the atmospheric pressure is 846 mbar abs, find the gage pressure
in (a) lb/in?; (6) kPa; (c) bar.
2.13 ‘Ifthe atmospheric pressure is 0.900 bar abs and a gage attached toa tank reads 390 mmHg vacuum, what is the
absolute pressure within the tank?
f p=yh Patm
= 0.900 X 100
= 90.0 kPa
Page = [(13. 6)(9.79) Goes)= 51.9kPa vacuum or —51.9kPa
] A B 7;
3m
Air
5m
D
IK
F ¢c 5m
aK]
3m Water
4 | Fig. 2-3
2.16 The system in Fig. 2-4 is at 70 °F. If the pressure at point A is 2900 Ib/ft?, determine the pressures at points B,
C, and D.
_ 6ft
Water
— Fig. 2-4
eS
2.17 The system in Fig. 2-5 is at 20 °C. If atmospheric pressure is 101.03 kPa and the absolute pressure at the bottom
of the tank is 231.3 kPa, what is the specific gravity of olive oil?
FH 101.03 + (0.89)(9.79)(1.5) + (9.79)(2.5) + (s.g.)(9.79)(2.9) + (13.6)(9.79)(0.4) = 231.3 s.g. = 1.39
ww. -
SAE 300i | *5™
Water 25m
Mercury 0-4 m
Fig. 2-5
28 JU GHAPTER 2
Epa = (62.4)(4 + 2) = 374 lb/ft, ps = —(62.4)(2) = —125 Ib/ft®. Neglecting air, pc = pp = — 125 lb/ft’;
Pp = —125 — (62.4)(4-+ 2+ 2) = —624 lb/ft.
Fig. 2-6
2.19 The tube shown in Fig. 2-7 is filled with oil. Determine the pressure heads at A and B in meters of water.
i (hu,0)(Yx,0) = (Aon) (You) = (Aca) {(8-8-00) (1,0) 15 therefore, Ayo = (Agu)(S-8. 01). Thus, hy =
—(2.2 + 0.6)(0.85) = —2.38 m H,O and h, = (—0.6)(0.85) = —0.51 m H,0.
Fig. 2-8
2.21 Convert 9 psi to (a) inches of mercury, (6) feet of water, (c) feet of ichor (s.g. = 2.94).
2.22 Express an absolute pressure of 5 atm in meters of water gage when the barometer reads 760 mmHg.
i Pave = (5)(101.3)/9.79 = 51.74 m of water — Patm = (0.760)(13.6) = 10.34 m of water
Figure 2-9 shows one pressurized tank inside another. If the sum of the readings of Bourdan gages A and B is
34.1 psi, and an aneroid barometer reads 29.90 inHg, what is the absolute pressure at A, in inHg?
Q. oO
Fig. 2-9
- 224 Determine the heights of columns of water, kerosene (ker), and nectar (s.g. = 2.94) equivalent to 277 mmHg.
"225 In Fig. 2-10, if h = 25.5 in, determine the pressure at A. The liquid has a specific gravity of 1.85.
i p= yh = [(1.85)(62.4)][25.5/12] = 245.3 lb/ft? or 1.70 psi
Fig. 2-10
2.26 For the pressure vessel containing glycerin, with piezometer attached, as shown in Fig. 2-11, what is the
pressure at point A?
i p = yh =[(1.26)(62.4)](40.8/12) = 267 lb/ft?
Open to atmosphere
_ Glycerin
Fig. 2-11
30 OG CHAPTER 2
2.27 For the open tank, with piezometers attached on the side, containing two different immiscible liquids, as shown
in Fig. 2-12, find the (a) elevation of the liquid surface in piezometer A, (6) elevation of the liquid surface in
piezometer B, and (c) total pressure at the bottom of the tank.
I (a) Liquid A will simply rise in piezometer A to the same elevation as liquid A in the tank (i.e., to elevation
2m). (b) Liquid B will rise in piezometer B to elevation 0.3 m (as a result of the pressure exerted by liquid B)
plus an additional amount as a result of the overlying pressure of liquid A. The overlying pressure can be
determined by p = yh = [(0.72)(9.79)](2 — 0.3) = 11.98 KN/m?. The height liquid B will rise in piezometer B as a
result of the overlying pressure of liquid A can be determined by h = p/y = 11.98/[(2.36)(9.79)] = 0.519 m.
Hence, liquid B will rise in piezometer B to an elevation of 0.3 m + 0.519 m, or 0.819 m.
(€) Poottom = [(0.72)(9.79)](2 — 0.3) + [(2.36)(9.79)](0.3) = 18.9 kPa.
A B
A A
El 2m $$ 4____--—_—_ H
Liquid A
(s.g. = 0.72)
=
—)
E} O3m
Liquid B
(s.g. = 2.36)
El .Om g Fig. 2-12
2.28 The air—oil—water system shown in Fig. 2-13 is at 70°F. If gage A reads 16.1 lb/in? abs and gage B reads
2.00 Ib/in? less than gage C, compute (a) the specific weight of the oil and (6) the reading of gage C.
F(a) (16.1)(144) + (0.0750)(3) + (Yon)(2) = Pa» Pa + (You)(2) + (62.4)(3) = Dc. Since pc — pz = 2.00,
(You)(2) + (62.4)(3) = (2.00)(144), Yon = 50.4 lb/ft. (b) (16.1)(144) + (0.0750)(3) + (50.4)(2) = pp,
Pp = 2419 lb/ft? pc = 2419 + (2.00)(144) = 2707 Ib/ft2, or 18.80 Ib/in?.
16.1 Ib /in? abs
) Z A
Air 3ft
¥
2ft
Oil —_—
2ft B
Avs
Water 3ft
Cc Fig. 2-13
2.29 For a gage reading at A of —2.50 psi, determine the (a) elevations of the liquids in the open piezometer
columns E, F, and G and (8) deflection of the mercury in the U-tube gage in Fig. 2-14. Neglect the weight of
the air. 7
f(a) The liquid between the air and the water would rise to elevation 49.00 ft in piezometer column E as a
result of its weight. The actual liquid level in the piezometer will be lower, however, because of the vacuum in
the air above the liquid. The amount the liquid level will be lowered (h in Fig. 2-14) can be determined by
FLUID STATICS JZ 31
(—2.50)(144) + [(0.700)(62.4)](h) = 0, h = 8.24 ft. Elevation at L = 49.00— 8.24 = 40.76 ft; (—2.50)(144) +
[(0.700)(62.4)][49.00 — 38.00) = py, Pa = 120.5 Ib/ft?. Hence, pressure head at M = 120.5/62.4 = 1.93 ft of
water. Elevation at N = 38.00 + 1.93 = 39.93 ft; 120.5 + (62.4)(38.00 — 26.00) = po, Po = 869.3 lb/ft”. Hence,
pressure head at O.= 869.3/[(1.600)(62.4)] = 8.71 ft (of the liquid with s.g. = 1.600). Elevation at Q =
26.00 + 8.71 = 34.71 ft. (b) 869.3 + (62.4)(26.00 — 14.00) — [(13.6)(62.4)](h,) = 0, h, = 1.91 ft.
El. 65.0057) 4 EF G
Air
El. 49.00 ¢¢ © H
El. 14.00 ¢t
Fig. 2-14
2.30 A vessel containing oil under pressure is shown in Fig. 2-15. Find the elevation of the oil surface in the attached
piezometer.
I Elevation of oil surface in piezometer = 2 + 35/[(0.83)(9.79)] = 6.31 m
. f
Air pressure = 35 kPa
Elev. 2m Vv.
Oil
(s.g. = 0.83)
2.31 The reading of an automobile fuel gage is proportional to the gage pressure at the bottom of the tank
(Fig. 2-16). If the tank is 32 cm deep and is contaminated with 3 cm of water, how many centimeters of air
remains at the top when the gage indicates “full”? Use Y,asciine = 6670 N/m? and y,;, = 11.8 N/m’.
I When full of gasoline, Pease = (6670)(0.32) = 2134 Pa. With water added, 2134 = (9790)(0.03) +
(6670)[(0.32 — 0.03) — h] + (11.8)(A), h = 0:0141 m, or 1.41.cm:
32 0 CHAPTER 2 ~
(ve
Z Air m
i
& ; 32cm . Gasoline ,
ne TE = Water 3cm
2.32 The hydraulic jack shown in Fig. 2-17 is filled with oil at 55 lb/ft’. Neglecting the weight of the two pistons, what
force F on the handle is required to support the 2200-Ib weight?
Lob I The pressure against the large and the small piston is the same. p = W/Aiarge = 2200/[2(33)"/4] =
“ 44 818 lb/ft”. Let P be the force from the small piston onto the handle. P = pA,inan = (44 818)[(45)"/4] = 244 Ib.
For the handle, £M, = 0 = (16 + 1)(F) — (1)(244), F = 14.4 1b.
ROIS
WY
oe
SER
WK
\S
F=?
Cylinder ~
Weight = 10,000 Ib
‘Cross-sectional 15 ft
area = 500 in?
2.34 For the vertical pipe with manometer attached, as shown in Fig. 2-19, find the pressure in the oil at point A.
i Pa + [(0.91)(62.4)](7.22) ~ [(13.6)(62.4)](1.00) =0 pp, = 438.71b/ft? or 3.05 Ib/in?
FLUID STATICS 2 33
(“<TD
|
Oi... |
(s.g.
= 0.91)
om
.
Mercury
(sg. = 13.6) Fig. 2-19
A monometer is attached to a tank containing three different fluids, as shown in Fig. 2-20. What will be the
‘difference in elevation of the mercury column in the manometer (i.e., y in Fig. 2-20)?
Elev.6m —
Elev. 5m Air pressure = 30kPa y
c Oil
(s.g. = 0.82)
Elev..2 m-_
Water
Elev.0m
Mercury
(s.g. = 13.6)
_ Fig. 2-20
Oil of specific gravity 0.750 flows through the nozzle shown in Fig. 2-21 and deflects the mercury in the U-tube
gage. Determine the value of h if the Pressure at A.is 20. 0psi.
Fig. 2-21
2.37 Determine the reading h in Fig. 2-22 for p, = 39 kPa vacuum if the liquid is kerosene
(s.g. = 0.83).
Fig. 2-22
2.38 In Fig. 2-22, the liquid is water. If h = 9 in and the barometer reading is 29.8
inHg, find p, in feet of water
absolute.
i Pa + (0.84)(96) — (1.0)(159) =0
Fig. 2-23
2.40 At 20°C, gage A in Fig. 2-24 reads 290 kPa abs. What is the height h of water?
What does gage B read?
7OcM] Mercury
2.41 The U-tube shown in Fig. 2-25a is 10 mm in diameter and contains mercury. If 12.0 mL of water is poured into
the right-hand leg, what are the ultimate heights in the two legs?
I After the water is poured, the orientation of the liquids will be as shown in Fig. 2-25b; h =
(12.0 x 10° mm?)/ (5 mm)? = 152.8 mm, (13.6)(240 — L) = 13.6L + 152.8, L = 114.4 mm. Left leg height above
bottom of U-tube = 240 — 114.4 = 125.6 mm; right leg height above bottom of U-tube = 114.4 + 152.8 =
267.2 mm.
Mercury
120mm 120mm
e—
L
at
120mm Fig. 2-25(a) 120mm Fig. 2-25(b)
2.42 Assuming sea water to have a constant specific weight of 10.05 kN/m’, what is the absolute pressure at a depth
of 10 km?
i p =1+ (10.05)(10 000)/101.3 = 993 atm
2.43 In Fig. 2-26, fluid 2 is carbon tetrachloride and fluid 1 is benzene. If p,,,, is 101.5 kPa, determine the absolute
pressure at point A.
i 101.5 + (15.57)(0.35) — (8.62)(0.12) =p, Pa= 105.9 kPa
Patm
fluid 1
35¢em
TC
126m
2.44 In Fig. 2-27a, the manometer reads 4 in when atmospheric pressure is 14.7 psia. If the absolute pressure at A is
doubled, what is the new manometer reading?
Ep, + (62.4)(3.5) — [(13.6)(62.4)}() = (14.7)(144), p, = 2181 Ib/ft?. If p, is doubled to 4362 lb/ft”, the
mercury level will fall x inches on the left side of the manometer and will rise by that amount on the right side of
the manometer (see Fig. 2-27b). Hence, 4362 + (62.4)(3.5 + x/12) — [(13.6)(62.4)][(4 + 2x)/12] = (14.7)(144),
x = 16.0 in. New manometer reading = 4 + (2)(16.0) = 36.0 in.
36 2 CHAPTER 2
Water
3.5 ft
Wk
|
3°
Mercury Fig, 2-27(a) Fig. 2-27(b)
2.45 In Fig. 2-28a, A contains water, and the manometer fluid has density 2900 kg/m*. When the left meniscus is at
zero on the scale, p, = 100 mm of water. Find the reading of the right meniscus for p, = 10 kPa with no
adjustment of the U-tube or scale.
Il First, determine the reading of the right meniscus for p, = 100 mm of water (see Fig. 2-285):
100 + 500 — 2.90h = 0, A = 206.9 mm. When p, = 10 kPa, the mercury level will fall some amount, d, on the left
side of the manometer and will rise by that amount on the right side of the manometer (see Fig. 2-28b). Hence,
10/9.79 + (500 + d)/1000 — [(206.9 + 2d)/1000](2.90) = 0, d = 192.0 mm. Scale reading for p, = 10 kPa is
206.9 + 192.0, or 398.9 mm.
2.46 A manometer is attached to a conduit, as shown in Fig. 2-29. Calculate the pressure at point A.
i Pa + (62.4)[(5 + 15)/12] — [(13.6)(62.4)]4)=0 p, = 957 1b/ft?
4 . . Water
cabana
Fig. 2-29
it 2.47 A manometer is attached to a pipe containing oil, as shown in Fig. 2-30. Calculate the pressure at point A.
l pa +[(0.85)(9.79)](0.2) — (9.79)(1.5)=0 © p,=13.02 kKN/m?