Principles of Hydraulics
Principles of Hydraulics
Principles of Hydraulics
Principles of Hydraulics
Chapter 2
Principles of Hydrostatics
𝐹𝑜𝑟𝑐𝑒, 𝐹
p= Eq. 2-1
𝐴𝑟𝑒𝑎, 𝐴
In the English system, pressure is usually measured in pounds per square inch (psi); in international usage,
in kilograms per square centimeters (kg/cm2), or in atmospheres; and in the international metric system
(S), in Newtons per square meter (Pascal). The unit atmosphere (atm) is defined as a pressure of
1.03323 kg/ cm2 (14.696 lb/ in2), which, in terms of the conventional mercury barometer, corresponds to
760 mm (29.921 n) of mercury. The unit kilopascal
(kPa) is defined as a pressure of 0.0102 kg/ cm 2 (0.145 lb/sq in).
PASCAL'S LAW
Pascal's lauw, developed by French mathematician Blaise Pascal, states that the pressure on a fluid is equal
in all directions and in all parts of the container. In Figure 2- 1, as liquid flows into the large container at
the bottom, pressure pushes the liquid equally up into the tubes above the container. The liquid rises to the
same level in all of the tubes, regardless of the shape or angle of the tube.
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The laws of fluid mechanics are observable in many everyday situations. For example, the pressure
exerted by water at the bottom of a pond will be the same as the pressure exerted by water at the bottom of
a much narrower pipe, provided depth remains constant. If a longer pipe filled with water is tilted so that it
reaches a maximum height of 15 m, its water will exert the same pressure as the other examples (left of
Figure 2-2). Fluids can flow up as well as down in devices such as siphons (right of Figure 2- 2).
Hydrostatic force causes water in the siphon to flow up and over the edge until the bucket is empty or the
suction is broken. A siphon is particularly useful for emptying containers that should not be tipped.
28
ABSOLUTE AND GAGE PRESSURES
pressure difference from local atmospheric conditions. For higher pressure differences, a Bourdon gauge,
named after the French inventor Eugène Bourdon, is used. This consists of a hollow metal tube with an
oval cross section, bent in the shape of a hook. One end of the tube is closed, the other open and connected
to the measurement region.
Absolute Pressure
Absolute pressure is the pressure above absolute zero (vacuum).
Note:
• Absolute zero is attained if all air is removed. It is the lowest possible pressure attainable.
• Absolute pressure can never be negative.
• The smallest gage pressure is equal to the negative of the ambient atmospheric pressure
29
Note: Unless otherwise specified in this book, the term pressure signifies gage pressure.
MERCURY BAROMETER
A mercury barometer is an accurate and relatively simple way to measure changes in atmospheric
pressure. At sea level, the weight of the atmosphere forces mercury 760 mm (29.9 in) up a calibrated glass
tube. Higher elevations yield lower readings because the atmosphere is less dense there, and the thinner air
exerts less pressure on the mercury.
ANEROID BAROMETER
In an aneroid barometer, a partially evacuated metal drum expands or contracts in response to changes in
air pressure. A series of levers and springs translates the up and down movement of the drum top into the
circular motion of the pointers along the aneroid barometer's face.
30
VARIATIONS IN PRESSURE
Consider any two points (1 & 2), whose difference in elevation is h, to lie in the ends of an elementary
prism having a cross-sectional area a and a length of L. Since this prism is at rest, all forces acting upon it
must be in equilibrium.
Note: Free liquid surface refers to liquid surface subject to zero gage pressure or with atmospheric pressure only.
[ƩFx = 0]
F2 – F1 = w sinɵ
p2 a- p1 a = γ(aL) sinɵ
p2 - p1 = γL sinɵ but L sinɵ = h
p2 - p1 = γh Eq. 2-3
Therefore; the difference in pressure between any two points in a homogeneous fluid at rest is equal to the
product of the unit weight of the fluid (γ) to the vertical distance (h) between the points.
31
p2 = - p1 + wh Eq. 2-4
This means that any change in pressure at point 1 would cause an equal change at point 2. Therefore; a
pressure applied at any point in a liquid at rest is transmitted equally and undiminished to every other point
in the liquid.
Let us assume that point 1 in Figure 2 - 4 lie on the free liquid surface, then the gage pressure p 1 is zero
and Eq. 2-4 becomes:
P = wh Eq. 2-5
This means that the pressure at any point “h” below a free liquid surface is equal to the product of the unit
weight of the fluid (γ) and h.
Consider that points 1 and 2 in Figure 2-4 lie on the same elevation, such that h=0; then Eq. 2-4 becomes:
p1 = p2 Eq. 2-6
This means that pressure along the same horizontal plane in a homogeneous fluid at rest are equal.
32
Consider the tank shown to be filled with liquids of different densities and with air at the top under a gage
pressure of Pa, the pressure at the bottom of the tank is:
PRESSURE HEAD
Pressure head is the height "h" of a column of homogeneous liquid of unit weight γ that will produce an
intensity of pressure p.
𝑝
h= Eq. 2-8
𝛾
𝑆𝐴 𝑃𝐴 𝛾𝐴
hB = hA or hB = hA or hB = hA Eq. 2-9
𝑆𝐵 𝑃𝐵 𝛾𝐵
To convert pressure head (height) of any liquid to water, just multiply its height by its specific
gravity
33
MANOMETER
A manometer is a tube, usually bent in a form of a U, containing a liquid of known specific gravity, the
surface of which moves proportionally to changes of pressure. It is used to measure pressure.
Types of Manometer
Open Type - has an atmospheric surface in one leg and is capable of measuring gage pressures.
Differential Type - without an atmospheric surface and capable of measuring only differences of pressure
Piezometer - The simplest form of open manometer. It is a tube tapped into a wall of a container or
conduit for the purpose of measuring pressure. The fluid in the container or conduit rises in this tube to
form a free surface.
Limitations of Piezometer:
34
Steps in Solving Manometer Problems:
1. Decide on the fluid in feet or meter, of which the heads are to be expressed, (water is most
advisable).
2. Starting from an end point, number in order, the interface of different fluids
3. Identify points of equal pressure (taking into account that for a homogeneous fluid at rest, the pressure
along the same horizontal plane are equal). Label these points with the same number.
4. Proceed from level to level, adding (if going down) or subtracting (t going up) pressure heads as the
elevation decreases or increases, respectively with due regard for the specific gravity of the fluids.
Solved Problems
Problem 2-1
If a depth of liquid of 1 m causes a pressure of 7 kPa, what is the specific gravity of the liquid?
Solution
Pressure, p = γh
7 = (9.81 x s) (1)
S = 0.714 → 𝑆𝑝𝑒𝑐𝑖𝑓𝑖𝑐 𝐺𝑟𝑎𝑣𝑖𝑡𝑦
Problem 2-2
What is the pressure 12.5 m below the ocean? Use sp. gr. = 1.03 for salt water.
Solution
p = γh
p = (9.81 x1.03)(12.5)
p = 126.3 kPa
35
Problem 2-3
If the pressure 23 meter below a liquid is 338.445 kPa, determine its unit weight γ, mass density p, and
specific gravity s.
Solution
1500
s=
1000
s = 1.5
Problem 2-4
If the pressure at a point in the ocean is 60 kPa, what is the pressure 27 meters below this point?
Solution
p2 = p1 + γh
= 60 + (9.81x1.03) (27)
p2 = 332.82 kPa
36
Problem 2-5
If the pressure in the air space above an oil (s= 0.75) surface in a closed tank is 115 kPa absolute, what is
the gage pressure 2 m below the surface?
Solution
p = psurface + γ h
psurface = 115 - 101.325 note: patm = 101.325 kPa
psurface = 13.675 kPa gage
p = 13.675+ (9.81x0.75)(2)
p = 28.39 kPa
Problem 2-6
Find the absolute pressure in kPa at a depth of 10 m below the free surface of oil of sp. gr. 0.75 if the
barometric reading is 752 mmHg.
Solution
Problem 2-7
A pressure gage 6 m above the bottom of the tank containing a liquid reads 90
kPa. Another gage height 4 m reads 103 kPa. Determine the specific weight of the liquid.
Solution
p2 - p1 = γh
103-90 = γ(2)
γ = 6.5 kN/m3
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Problem 2-8
An open tank contains 5.8 m of water covered with 3.2 m of kerosene (γ=8
kN/m3). Find the pressure at the interface and at the bottom of the tank.
Solution
Problem 2-9
If atmospheric pressure is 95.7 kPa and the gage attached to the tank reads 188
mmHg vacuum, fnd the absolute pressure within the tank.
Solution
Problem 2-10
The weight density of a mud is given by y = 10 + 0.5h, where y is in kN/m 3 and h is in meters. Determine
the pressure, in kPa, at a depth or m.
38
Solution
Since the density of the mud varies with depth, the pressure
should be solved by integration
dp = γ dh
dp = (10+ 0.5 h)dh
𝑝 5
∫0 𝑑𝑝 = ∫0 (10 + 0.5h)𝑑h
P = 10h +0.25h2]50
= [10(5)+ 0.25(5)2] - 0
p = 56.25 kPa
Problem 2-11
In the figure shown, if the 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?
Solution
[p = Ʃγh]
P = γm hm + γo ho +γw hw + γoil hoil
130.27=(9.81x13.6)(0.4)+(9.81xs)(2.9)+9.81(2.5)+(9.81x0.89)(1.5)
s = 1.38
39
Problem 2-12
lf air had a constant specific weight of 12.2 N/m 3 and were incompressible, what would be the height of
the atmosphere if the atmospheric pressure (sea level) is 102 kPa?
Solution
𝑝
Height of atmosphere, h =
𝛾
102 𝑥103
=
12.2
Solution
Pbot - Ptop = γh
(γm hm)bottom-(γm hm)top(γ h)air
9,810 x 13.6)(0.654) - (9,810 x 13.6) (0.48) = 12 h
h =1,934.53 m
40
Problem 2-14
Compute the barometric pressure in kPa at an altitude of 1,200 m if the pressure at sea level is 101.3 kPa.
Assume isothermal conditions a 21°C. Use
R= 287 Joule /kg-°K.
Solution
For gases:
dp = -pg dh
𝑝
p=
𝑅𝑇
𝑝
=
287(21+273)
p = 0.00001185 p
dp = -(0.00001185 p)(9.81) dh
dp = 0.0001163 dh
𝑝 1200
𝑑𝑝
∫ = 0.0001163 h ∫ 𝑑h
𝑝
101.3 𝑥 103 0
𝑝
ln p]101.3 𝑥103 = -0.0001163 ]1200
0
In p – ln(101.3 x 103) = -0.0001163(1200-0)
ln p = 11.386
p = 𝑒 11.386
p = 88,080 Pa
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Problem 2-15
Convert 760 mm of mercury to (a) oil of sp. gr. 0.82 and (b) water.
Solution
𝑆𝑚𝑒𝑟𝑐𝑢𝑟𝑦
(a) hoil = hmercury
𝑆𝑜𝑖𝑙
13.6
= 0.76
0.82
= 0.76(13.6)
Solution
Problem 2-17
A hydraulic press is used to raise an 80-kN cargo truck. If oil of sp. gr. 0.82 acts on the piston under a
pressure of 10 MPa, what diameter of piston is required?
Solution
Since the pressure under the piston is uniform:
Force = pressure x Area
𝜋
80,000 = (10 x 103) D2
4
D = 0.1 m = 100 mm
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Problem 2- 18 (CE November 1998)
Piston A has a cross-section of 1,200 sq. cm while that of piston B is 950 sq. cm with the latter higher than
piston A by 1.75 m. If the intervening passages are filled with oil whose specific gravity is 0.8, what is the
difference in pressure between A and B.
Solution
PA -PB = γo ho
= (9,810 x 0.8) (1.75)
PA-PB =13,734 Pa
Problem 2-19
Solution
Since points 1 and 2 lie on the
same elevation, p1 = p2
1.5 𝑊
𝜋 =𝜋
(0.03)2 (0.3)2
4 4
W = 150 kN
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Problem 2- 20
A drum 700 mm in diameter and filled with water has a vertical pipe, 20 mm in diameter, attached to the
top. How many Newton’s of water must be poured into the pipe to exert a force of 6500 N on the top of
the drum?
Solution
Force on the top:
F = p x Area
𝜋
6500 – p x (7002 -202)
4
p = 0.016904 MPa
p = 16,904 Pa
[p = γh]
16,904 = 9810 h
H =1.723 m
Weight = γ x Volume
𝜋
= 9810 x (0.02)2(1.723)
4
Weight = 5.31 N
Problem 2-21
The figure shown shows a setup with a vessel containing a plunger and a cylinder. What force F is
required to balance the weight of the cylinder if the weight of the plunger is negligible?
44
Solution
[p2 – p1 = γoh]
𝐹 𝐹
p1 = =
𝑎 0.323
p1 = 309.6 F (kPa)
𝑊 44
p2 = =
𝐴 0.323
p2 = 136.22 kPa
Problem 2-22
The hydraulic press shown is filled with oil with sp. gr. 0.82. Neglecting the weight of the two pistons,
what force F on the handle is required to support the 10.kN weight?
45
Solution
[ƩMo = 0]
F(0.425) = F2(0.025)
F(0.425) = 1.11(00025)
F = 0.0654 kN
F = 65.4 N
Problem 2- 23
The fuel gage for a gasoline (sp. gr. = 0.68) tank in a car reads proportional to its bottom gage. If the tank
is 30 cm deep an accidentally contaminated with 2
cm of water, how many centimeters of gasoline does the tank actually contain when the gage erroneously
reads "FULL"?
Solution
46
Since the gage reads "FULL then the reading is equivalent to 30 cm of gasoline.
Reading (pressure head) when the tank contain
1
water =( y+2 ) cm of gasoline
0.68
1
Then; y+2 = 30
0.68
y = 27.06 cm
Solution
Summing-up pressure head
from 1 to 3 in meters of water
𝑃1 𝑃3
+ h2(0.84) – x =
𝛾 𝛾
0 + 0.84 h2 – (4 – 3) = 0
h2 = 1.19 m
47
Solution
The pressure in the air space
p1 = 0
p2 = γwhw
= 9.81(2)
p2 = 19.62 kPa
p2 – p3 = γo ho
19.62 – p3 = (9.81 x 0.80)(4)
p3 = -11.77 kPa
Another solution:
Sum-up pressure head from 1 to 3 in meters of water:
𝑃1 𝑃3
+ 2 – 4(0.80) =
𝛾 𝛾
𝑃3
0 + 2 – 3.2 =
9.81
p3 = -11.77 kPa
48
Problem 2-26
Solution
Sum-up pressure head from
1 to 3 in meters of water:
𝑃1 𝑃3
+1(13.55) +1.5(0.8) =
𝛾 𝛾
𝑃3
0 +14.75 =
𝛾
𝑃3
=14.75 m of water
𝛾
p3 = 14.75(9.81)
p3 = 144.7 kPa
49
Problem 2-27 (CE Board November 2001)
Determine the value of y in the manometer shown in the Figure.
Solution
Summing-up pressure head from
A to B in meters of water:
𝑃𝐴 𝑃𝐵
+ 3(0.8)+ 1.5 - y(13.6) =
𝛾 𝛾
5 𝑃𝐵
+ 3.9 -13.6y =
9.81 𝛾
where pB = 0
y = 0.324 m
50
Problem 2- 28(CE May 1993)
In the figure shown, when the funnel is empty the water surface is at point A and the mercury of sp. gr.
13.55 shows a deflection of 15 cm. Determine the new deflection of mercury when the funnel is filled with
water to B.
Solution
51
In Figure (b):
When the funnel is filled with water to B, point 1 will move down to 1
with the same value as point 2 moving up to 2’
Problem 2-29
The pressure at point m in the figure shown was increased from 70 kPa to 105 kPa. This causes the top
level of mercury to move 20 mm in the sloping tube. What is the inclination, ɵ?
Solution
52
Sum-up pressure head from 2 to m in meters of water:
𝑃2 𝑃𝑚
+ y(13.6) - x =
𝛾 𝛾
70
13.6y – x = Eq. (1)
9.81
In Figure (b):
Sum-up pressure head from 2' to m’ in meters of water:
𝑃2′ 𝑃𝑚′
+ (0.2 sin ɵ + y + 0.2)(13.6) - (x +0.2) =
𝛾 𝛾
105
0+2.72 sin ɵ + 13.6y + 2.72 – x - 0.2 =
9.81
[13.6y- x = 13.6y – x]
70
8.183 - 2.72 sin ɵ =
9.81
sin ɵ = 0.3852
ɵ = 22.66 ̊
Problem 2- 30
A closed cylindrical tank contains 2 m of water, 3 m of oil (s =0.82) and the air above oil has a pressure of
30 kPa. If an open mercury manometer at the bottom of the tank has 1 m of water, determine the deflection
of mercury.
Solution
Sum-up pressure head from
to 4 in meters of water:
𝑃𝑎𝑖𝑟 𝑃4
+ 3(0.82) + 2 + 1 – y (13.6) =
𝛾 𝛾
30
+ 2.46 +3 - 13.6y = 0
9.81
y = 0.626 m
53
Problem 2-31
The U-tube shown is 10 mm in diameter and contains mercury. If 12 ml of water is poured into the right-
hand leg, what are the ultimate heights in the two legs?
Solution
Solving for h, (see figure b):
𝜋 10
Volume of water = ( )2 h = 12 cm3 Note: 1 ml = 1 cm3
4 10
h = 15.28 cm = 152.8 mm
Since the quantity of mercury before and after water is poured remain the same, then;
120(3) = R + x + 120 +x
R+2x 240 →Eq. (1)
54
In Figure (b):
Summing-up pressure head from I to 3 in mm of water:
𝑃1 𝑃2
+ 152.8 - R(13.6) =
𝛾 𝛾
R 11.24 mm
In Eq. (2):
l1.24 + 2x = 240
x = 114.38 mm
Left-hand leg, hL = R + x
= 11.24+ 114.38
Left-hand leg, hL= 125.62 mm
Problem 2-32
For a gage reading of -17.1 kPa,
determine the (a) elevations of the liquids in the open piezometer columns E, F, and G and (b) the
deflection of the mercury in the U-tube manometer neglecting the weight of air.
55
Solution
Column E
Sum-up pressure head from l to e in metes of water;
𝑃1 𝑃𝑒
+ h1(0.7) =
𝛾 𝛾
−17.1
+ h1(0.7) = 0
9.81
h1 = 2.5 m
Surface elevation = 15 - h1
Surface elevation = 15 - 2.5 = 12.5 m
Column F
Sum-up pressure head from1 tofin meters of water;
𝑃1 𝑃𝑓
+ 3(0.7) – h2(1) =
𝛾 𝛾
−17.1
+ 2.1 - h2 = 0
9.81
h2 = 0.357m
Surface elevation = 12 + h2
Surface elevation = 12 +0.357 = 12.357 m
56
Column G
Sum-up pressure head romi to g in meters ot water;
𝑃1 𝑃𝑔
+ 3(0.7) + 4(1) – h3(1.6) =
𝛾 𝛾
−17.1
+ 2.1+ 4 - 1.6h3 = 0
9.81
h3 = 2.72 mm
Surface elevation = 8 + h3
Surface elevation = 8 + 2.72 = 10.72 m
Deflection of mercury
Sum-up pressure head from 1l to 5 in meters of water;
𝑃1 𝑃5
+3 (0.7) + 4 + 4 - h4(13.6) =
𝛾 𝛾
−17.1
+ 10.1-13.6h4
9.81
h4 = 0.614 m
Problem 2-33
An open manometer attached to a pipe shows a deflection of 150 mmHg with the lower level of mercury
450 mm below the centerline of the pipe carrying water. Calculate the pressure at the centerline of the
pipe.
Solution
𝑃1 𝑃3
+ 0.45 - 0.15(13.6) =
𝛾 𝛾
𝑃1
+ 0.45 – 2.04 = 0
9.81
p1 = 15.6 kPa 57
Problem 2-34
For the configuration shown, calculate the weight of the piston if the pressure gage reading is 70 kPa.
Solution
Weight = FA
= pA x Area
= 78.44 x (1)2
Weight = 61.61 kN
Problem 2-35
Two vessels are connected to a differential manometer using mercury, the connecting tubing being filled
with water. The higher pressure vessel is 1.5 m lower in elevation than the other. (a) If the mercury
reading is 100 mm, what
15 the pressure head difference in meters of water? (6) If carbon tetrachloride
s = 1.59) were used instead of mercury, what would be the manometer reading for the same pressure
difference?
58
Solution
(a) Gage liquid = mercury, h = 0.1 m
Sum-up pressure head from
1 to 4 in meters of water;
𝑃1 𝑃4
+ x + h - h(13.6) - X - 1.5 =
𝛾 𝛾
𝑃1 𝑃4
- = 1.5 – 0.1 + 0.1(13.6)
𝛾 𝛾
𝑃1 𝑃4
- = 2.76 m of water
𝛾 𝛾
𝑃1 𝑃4
- = 1.5 + 0.59h
𝛾 𝛾
𝑃1 𝑃4
where - = 2.76 m → from(a)
𝛾 𝛾
Problem 2-36
In the figure shown, determine
the height h of water and the gage reading at A when the absolute pressure at B is 290 kPa.
59
Solution
Sum-up absolute pressure head
from B to2 in meters of water;
𝑃𝐵 𝑃2
- 0.7(13.6) + 0.7 =
𝛾 𝛾
290 175
– 9.52 – h =
9.81 9.81
h = 2.203 m
290 𝑃𝐴
- 9.52+0.7 =
9.81 9.81
pA = 203.5 kPa abs
Problem 2-37
In the figure shown, the atmospheric pressure 1s 101 kPa, the gage reading at A is 40 kPa, and the vapor
pressure of alcohol is 12 kPa absolute. Compute x + y.
Solution
Sum-up absolute pressure head from
1 to 2 in meters of water;
𝑃1 𝑃2
– y(0.9) =
𝛾 𝛾
40+101 12
– 0.9y =
9.81 9.81
y=14.61 m
60
Sum-up absolute pressure (gage) head from 1 to 4 in meters of water;
𝑃1 𝑃4
+ x (0.9) + 1.3(0.9) -1.3(13.6) =
𝛾 𝛾
40
+ 0.9x - 16.51= 0
9.81
x=13.8l m
Then, x + y = 28.42 m
Problem 2- 38
For the manometer setup shown, determine the difference in pressure between A and B.
Solution
x + 0.68 = y+1.7
x – y = 1.02 m →Eq. (1)
in meters of water;
𝑃𝐴 𝑃𝐵
-x - 0.68(0.85) + y =
𝛾 𝛾
𝑃𝐴 𝑃𝐵
- = x –y + 0.578 →Eq. (2)
𝛾 𝛾
𝑃𝐴−𝑃𝐵
=1.598
9.81
pA - pB = 15.68 kPa
61
Problem 2-39
A differential manometer is attached to a pipe as shown. Calculate the pressure difference between points
A and B.
Solution
62
Problem 2- 40
In the figure shown, the deflection of mercury is initially 250 mm. If the pressure at A is increased by 40
kPa, while maintaining the pressure at B constant, what will be the new mercury deflection?
Solution
𝑃𝐴 𝑃𝐵
- 0.6 - 0.25(13.6) + 0.25 + 2.1 =
𝛾 𝛾
𝑃𝐴 𝑃𝐵
- = 1.65 m of water
𝛾 𝛾
63
In Figure b, pA’ = pA + 40
Sum-up pressure head from A’ to B in meters of water;
𝑃𝐴′ 𝑃𝐵
– (0.6 – x) – (0.25 + 2x) 13.6 + (2.35 + x) =
𝛾 𝛾
𝑃𝐴+40 𝑃𝐵
- 0.6 + x - 3.4 - 27.2x + 2.35 + x =
𝛾 𝛾
𝑃𝐴 40 𝑃𝐵
+ -1.65 - 25.2x =
𝛾 9.81 𝛾
𝑃𝐴 𝑃𝐵 𝑃𝐴 𝑃𝐵
- = 25.2 x -2.423 But - = 1.65
𝛾 𝛾 𝛾 𝛾
Problem 2-41
In the figure shown, determine the difference in pressure between points A and B.
64
Solution
𝑃𝐴 𝑃𝐵
+ 0.2(0.88) - 0.09(13.6) - 0.31(0.82) +0.25-0.1(0.0012) =
𝛾 𝛾
𝑃𝐴 𝑃𝐵
- = 1.0523 m of water
𝛾 𝛾
Solution
Applying Boyle's Law
(assuming isothermal condition).
Solution
[p V1= p2 V2]
101.3 (3 A) = 121.31 [(3-y) A]
3 – y = 2.505
Y = 0.495m
Problem 2-44
A bottle consisting of a cylinder 15 cm in diameter and 25 cm high, has a neck which is 5 cm diameter and
25 cm long. The bottle is inserted vertically in water, with the open end down, such that the neck is
completely filled with water. Find the depth to which the open end is submerged. Assume normal
barometric pressure and neglect vapor pressure.
Solution
h =1.15 cm
x = h + 25 = 26.15 cm