Principles of Hydrostatic Pressure
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Transcript of Principles of Hydrostatic Pressure
Part II
Principle of Hydrostatic Pressure 1. Fluid Pressure, p: the force exerted on a unit area. If F represents
the total force on an area A, and dF is the elemental force on an
elemental area dA, the intensity of pressure is,
Note: If the pressure is uniform over the area A, then
Otherwise this formula gives only the average pressure
Units:
English System:
Metric System: S.I. :
dA
dFp
A
Fp
psfft
lb2 or psi
inlb
2
2cmgr
2mN
2. Pascal’s Law: states that at any point in a fluid at rest the pressure is
the same in all directions.
3. Variation of Pressure in a Fluid: The pressure at any point in a fluid
at rest is equal to its specific weight, γ multiplied by the vertical depth of the point from the surface, or
hp
Liquid surface
h
p = γ.h
Note: This relationship is only true for liquids in which the specific weight varies with small changes in elevation
3.1 Variation of Pressure of a Fluid
The theory states that “a pressure of a fluid are rest or in equilibrium condition ( or called static fluid) varies directly with the depth (or elevation) measured from the surface of that fluid”. Consider a vertical homogenous infinitesimal fluid cylindrical element column having a height of h and a cross-sectional area A with the upper end at the surface of the fluid as shown below.
A
Fatm
W
F
h
Applying equilibrium of vertical forces , we obtain: F = W + Fatm
p x A = γ (h x A) + patm x A p = γh + patm
= ρgh + patm
Thus we obtain a relation
atmphp
atmpghp
atmphp
atmpghp
equation (2) where: p = pressure at lower end (at depth h) N/m2
ρ = density of fluid kg/m3
γ = specific weight of the fluid N/m3
g = acceleration due to earth gravity m/s2
W = weight of the fluid cylindrical column N F = The total fluid force acting on the lower end area of the column element. N Fatm = The total atmospheric force acting on the upper end of the column element.
Patm = atmospheric pressure N/m2
= 1.013 bar (or 14.7 psi) If the effect of atmospheric pressure is not considered, then equation (2) is reduced to the following: p = ρgh (the pressure exerted by the fluid, which varies directly with the depth h).
3.2 Equipment for Pressure Measurement There are various types of instruments for measuring pressure. The commonly used ones are as follows: 3.2.1 Pressure gauge Elliptical cross section Metal tube
spring
Pinion gear
Dial scale joint
Axial
Pressure gauge mechanism (Bourdon Tube)
3.2.2 Piezometer
This is a very simple instrument used for measuring liquid pressure only. The Piezometer is a glass tube with an end open to the atmosphere. The lower end of the glass tube is connected to the liquid source to be measured causing the liquid inside the source of container to move freely within the tube as shown in Fig. above. The height of the liquid in the tube varies directly with the magnitude of that liquid pressure by the following relation: where: p = liquid pressure in the container N/m3
ρ = density of the liquid kg/m3
γ = specific weight of the liquid N/m3
g = acceleration due to the earth’s gravity m/s2
h = height of liquid in the piezometer tube m
h
Piezometers
p+
hghp
3.2.3 Manometer (a) A Simple manometer or open-end manometer.
p +
h
Simple manometer or open-end manometer.
The simple manometer consists of a glass U tube containing a liquid (normally mercury) with one end open to the atmospheric pressure and the other end connected to the fluid source to be measured. The pressure in the fluid source or container will push the liquid in both sides of the U tube with different levels of liquid as shown in Fig. above. The level difference of liquid in the U tube can be used to determine the pressure of the fluid in the container.
(b) Differential manometer
Differential manometer
The differential manometer consists of a U-shape glass tube for measuring the pressure difference of the fluids in two different containers as shown in Fig. above. The level difference of liquid in the U tube can be used to determine the pressure difference of the fluids in the two containers.
(c) An inclined manometer
An inclined manometer
In the case the pressure in the container is very low which causes the level difference in the liquid in the U tube to be very small. This makes it very difficult to read the value of h. Thus if the tube on the right hand side is inclined the linear length of the liquid in the right hand tube will increase resulting in a more accurate reading of the length H as shown in Fig. above. The ratio of h /H is sinθ or h = H.sinθ Where: sin θ can be 1/10, 1/5 or 1/2 or any other fractions. Thereby the value of H obtained from the length reading of the inclined tube can then be used to determine the fluid pressure in the container.
3.2.4 Pressure sensor and indicator (optional) The digital pressure indicator consists of sensor head capable of transmitting a pressure value and convert it into electrical ( i.e., mA or V) signals. The signal is then further transmitted and converted into digital numbers on the monitor screen.
3.2.5 Mercury Barometer (optional)
patm h
Barometer
This type of pressure measuring equipment is essentially a manometer for measuring atmospheric pressure. It consists of a glass tube closed at one end and open at the other end. The tube is initially filled with a liquid (normally mercury). It is then turned over into an upside down position with the open end submerged in a small container of the same liquid as show in Fig. above. The height of the liquid column h rising above the surface of the liquid in the container can be used to determine the atmospheric pressure.
3.2.6 Aneroid Barometer (optional)
Aneroid barometer is a device for measuring atmospheric pressure. The aneroid is an exhausted chamber with corrugated diaphragm walls, the collapsing of which is resisted by a spring. The deflections of the diaphragms against the spring are indicated or recorded by a lever mechanism. The aneroid barometer is also made in the more accurate null type in which the diaphragm is brought back to an initial position by changing the tension of the loading spring. This same motion rotates the scale to give the new reading.
In case of the atmospheric pressure, the variation is given by,
a) If γ is assumed a constant,
b) If γ is considered to vary,
c) For isothermal process, n = 1
hp
pdhdp
o 0
111
n
n
oRT
gh
n
n
po
pWhere; n = 1.2 for wet adiabatic process n = 1.4 for dry adiabatic process
oTR
hg
o
ep
p
d) Temperature decreasing linearly with elevation
Kp
T
o
o
o
o
o
hKT
T
p
p
po = absolute pressure at sea level p = absolute pressure at h above sea level R = gas constant To = absolute temperature at sea level K = -0.00356 °R/ft = -0.00650 °K/m
Case d
Case c
Atmospheric, Gage and Absolute Pressure
4.1 Atmospheric or Barometric Pressure, The pressure exerted
by the atmosphere on every surface with which it comes in
contact. This is measured by means of Mercury Barometer which
was discovered by Torricelli in 1643. Under normal conditions
at sea level the atmospheric pressure is 14.7 psi or 2116 psf or
1.01325 bar or 101.325 kPa or 30 in. (760 mm)of mercury column in the barometer.
4.2 Gage Pressure, The pressure measured by means of gages
above or below the atmospheric level. At sea level, gage
pressure is zero.
:atmP
:gp
4.3 Absolute Pressure, The pressure measured above
absolute zero. At sea level, under normal conditions,
absolute pressure is 14.7 psi or 2116 psf. Obviously, a
negative absolute pressure is impossible.
The relationship between these pressures is
:absp
gatmabs ppp
Absolute Zero Level
Atmospheric Level
x z
v
y
Atmospheric Pressure
See illustration below
y – gage pressure at point A (+) z – gage pressure at point B (-) x – absolute pressure at point A v – absolute pressure at point B
A
B Note: The term “ vacuum” means a negative gage pressure and the term “atmosphere” is used for absolute pressure. 1 atmosphere = 14.7 psi absolute = zero gage
5. Pressure Head, h: The height of a column of a homogeneous fluid of
specific weight, γ that will produce an intensity of pressure p at its
bottom, or
6. Transmission of Pressure: At any point, 1, h units below point 2, the
pressure is,
This principle is attributed also to Pascal and was applied to the
concept of hydraulic jack
ph
hpp 21
1F
2F
1
2 h
1W
2W
Liquid,γ
1
111
A
WFp
2
222
A
WFp
where: W₁ and W₂ - weights of the plungers A₁ and A₂ - areas of the plungers F₁ - applied force to raise up the weight W₂ and F₂
7. Solution of an Open Manometer Problem
i. Make an appropriate sketch of the given manometer
ii. Number in order the levels of contact of fluids of different
specific gravities starting at the atmospheric level.
iii. Starting with the atmospheric pressure head, proceed from
level to level, adding pressure heads as the elevation decreases
and subtracting pressure heads as the elevation increases,
multiplying the increase or the decrease in the elevation by the
corresponding specific gravities of the fluids.
This method gives all the pressure heads in head of water.
Note: For the differential manometer problems, the method given
above can be used except that there is no atmospheric leg and
the addition and subtraction of pressure heads may be started
from any of the two points of unknown pressures.
PRESSURE
FEET METER
0 0 14.69 psi = 101.26 kPa
1000 305 14.17 psi = 97.67 kPa
2000 610 13.66 psi = 94.16 kPa
3000 915 13.16 psi = 90.71 kPa
4000 1220 12.68 psi = 87.40 kPa
5000 1525 12.21 psi = 84.16 kPa
5280 1610 12.08 psi = 83.26 kPa
6000 1829 11.76 psi = 81.06 kPa
7000 2134 11.32 psi = 78.03 kPa
8000 2439 10.89 psi = 75.06 kPa
9000 2744 10.48 psi = 72.24 kPa
10000 3049 10.09 psi = 69.55 kPa
ALTITUDE ABOVE SEA LEVEL
8. Variation in Atmospheric Pressure
3049
69.55
9. Vapor Pressure Heads of Water ( Feet of Water)
10. Vapor Pressure Heads of Water ( Meters of Water)
0 0.05 80 1.16 160 10.9
20 0.13 100 2.17 180 17.28
40 0.28 120 3.87 200 26.52
50 0.41 140 6.63 212 33.84
˚F ˚F ˚Fv
P
vP
v
P
Temperature °C
Vapor Pressure Head Pv/γ (m)
Temperature °C
Vapor Pressure Head Pv/γ (m)
0 0.06 60 2.03
10 0.12 70 3.20
20 0.25 80 4.96
30 0.44 90 7.18
40 0.76 100 10.33
50 1.26
Example 1. What height of a column of special gage liquid ( s = 2.95)
would exert the same pressure as a column of oil 4.57 m high
( s = 0.84) ?
where:
os pp ooss hh or
wss s 3981095.2
m
N
35.939,28
m
N
woo s 3981084.0
m
N
34.240,8
m
N
mm
Nh
m
Ns 57.44.240,85.939,28
33
mhs 3.1
Example 2. What height of a column of water ( s = 1.0 ) would exert the
same pressure as a column of liquid with specific gravity ?
Solution.
Ls
Lw pp or LLww hh
LwLww hsh
LLw hsh
Example 3. A water barometer reads 10 m. If the temperature is 20 °C, what is the pressure of the atmosphere in kPa? Solution: From table, the pressure head of water at 20 °C is
mp
w
v 25.0
w
v
w
a pph
The height of water in the barometer is expressed by the difference of atmospheric pressure head and the vapor pressure head.
25.010 w
ap
25.010w
ap
mm
Npa 25.109810
3
2553,100
m
Npa kPapa 553.100or
Example 4. If the pressure in a tank of oil (s = 0.80) is 414 kPa, what is the head in: (a) m of oil; (b) m of water (c) mm of mercury ( s=13.6) Assuming the specific weight of water is 9810 N/m3. Solution: (a) (b) ( c)
o
o
ph
wos
p
3
2
3
981080.0
10414
m
Nm
Nx
m752.52
w
w
ph
3
2
3
9810
10414
m
Nm
Nx
m202.42
m
m
ph
wms
p
3
2
3
98106.13
10414
m
Nm
Nx
m103.3 mm3103
Example 5. The pressure in a gas tank is 2.75 atmosphere. Compute the absolute pressure in psi , the gage pressure in psi and the pressure head in ft. of water assuming the specific weight of water is 62.4 lb/ft3. Solution: The absolute pressure in psi is,
atm
psiatmpabs
7.1475.2 psi425.40
The gage pressure in psi is,
atmabsg ppp 7.14425.40 psi725.25
The pressure head in ft of water is,
w
w
ph
3
2
4.62
725.25
ft
lbin
lb
3
2
2
2
4.62
144725.25
ft
lb
ft
inx
in
lb
ft365.59
Example 6. A gage on the suction side of a pump shows a vacuum of 250 mm of mercury. Compute (a) the pressure head in m of water; (b) The pressure in kPa; ( c) absolute pressure in kPa if the barometer reads 737 mm of mercury. The specific weight of water is 9810 N/m3. Solution: (a) The pressure head in m of water is,
w
w
ph
39810
250
m
N
mmHg
3
2
3
9810
760
10325.101
250
m
N
mmHg
m
Nx
mmHg
m4.3
(b) The pressure in kPa
mmHg
kPammHgp
760
325.101250 kPa33.33
(c) The absolute pressure in kPa is,
atmgabs ppp
mmHg
kPammHgkPapabs
760
325.10173733.33
kPapabs 928.64
Example 7. Assuming the specific weight of air is constant at 12
, what is the approximate decrease in pressure in Pascal
corresponding to a rise in elevation of 305 m?
Solution.
3m
N
hp
p
dhdp
o 0
0305123
mm
Nppp o
mp
p
dhm
Ndp
o
305
0
312
Pap 660,3 2660,3
m
N
Example 8. On a certain day the barometric pressure at sea level is
765 mm Hg and the temperature is 20 ˚C. The pressure gage on an
airplane flying overhead indicates that the atmospheric pressure at
that point is and that the air temperature is 8 ˚C. Calculate
as accurately the height of the airplane above sea level. Assume linear
decrease of temperature with elevation.
Solution.
absm
kN,73
2
Kp
T
o
o
o
o
oo
hKT
T
p
p
absm
Nxp ,1073
2
3
mmhm 765 Hg
mwmmmo hshp
mm
mmm
m
N
1000
1765981066.13
3 2514,102
m
N
KTo 293
mKK /00649.0
301.12
m
No
continuation
Kp
T
o
o
o
o
oo
hKT
T
p
p
00649.0514,102
29301.123
00649.0293
293
514,102
1073
h
x
h
x
00649.0293
293ln
00649.0514,102
29301.12
514,102
1073ln
3
h
x
00649.0293
293ln
514,102
1073ln
29301.12
00649.0514,102 3
514,102
1073ln
29301.12
00649.0514,102 3
00649.0293
293x
eh
continuation
514,102
1073ln
29301.12
00649.0514,102 3
00649.0293
293x
eh
514,102
1073ln
29301.12
00649.0514,102 3
29300649.0293
x
e
h
514,102
1073ln
29301.12
00649.0514,102 3
29329300649.0
x
e
h
514,102
1073ln
29301.12
00649.0514,102 3
293293
00649.0
1x
e
h
mh 15.2807
Example 9. Given the figure below, determine the pressure at m if x = 760 mm and y = 760 mm.
x
y
CaCl4 ( s = 1.60)
Oil ( s = 0.856)
m
0
1
2
Air
Using addition and subtraction of pressure heads, expressed in head of water. The addition and subtraction maybe started at atmospheric level (level 0) and proceeding from level to level up to to obtain the algebraic sum of the pressure heads at the level in consideration ( In this case, it is point m).
Solution:
x = 760 mm
y = 760 mm
CaCl4 ( s = 1.60)
Oil ( s = 0.856)
0
1
2
Air
m
Starting from level 0, then to level 1, then to level 2 and lastly to level m
w
mpxy
856.06.10
39810
856.0760.06.1760.00
m
N
pmm m
2311,18
m
Npm
2311.18
m
kN
or
kPapm 311.18
Note: In meters of oil, the head is
Example 10. Given the figure below, determine (a) Pm if x = 0.30 m and y = 0.50 m; (b) how many millimeters in the 12 mm tube will the fluid rises if the pressure at m is increased by 7 kPa.
m
x
y
12 mm dia. tube
s = 2.95
4 mm dia. tube
125 mm dia.
s = 0.915
0
1
Solution: (a) Starting at the 0 level and proceeding up to the level m.
w
mpyx
915.095.20
m
X = 0.30 m
y = 0.50 m
12 mm dia. tube
s = 2.95
4 mm dia. tube
125 mm dia.
s = 0.915
0
1
w
mpyx
915.095.20
39810
915.050.095.230.00
m
N
pmm m
2170,13
m
Npm or kPa170.13
( b ) Let d = drop in the 125 mm dia. Tube. r = rise in the 12 mm dia. Tube Note: Volume that drops = volume that rises
rd22
124
1254
dr
2
12
125
or rd
2
125
12
Consider the new levels
m
X = 0.30 m
y = 0.50 m
12 mm dia. tube
s = 2.95
4 mm dia. tube
125 mm dia.
s = 0.915
0
1 d
New level
Original level
New level
r
w
mpdydxr
915.095.20
w
mpddr
915.050.095.230.00
Note when the pressure at m is increased, the fluid inside tends to push outward.
m
X = 0.30 m
y = 0.50 m
12 mm dia. tube
s = 2.95
4 mm dia. tube
125 mm dia.
s = 0.915
0
1 d
New level
Original level
New level
r
81.9
717.13915.050.095.230.00
ddr
81.9
17.20915.0
125
1250.095.2
125
1230.00
22
rrr
mr 24.0
Example 11. In figure, fluid A is water while fluid B is oil (s = 0.85). If x = 1500 mm and y = 750 mm, find (pm – pn).
m
n y = 0.75 m
x = 1.50 m
Level n
Level m
Level 1
Level 2
z
w
Fluid B
Fluid A Fluid A
Solution: Using addition and subtraction of pressures Starting from the end pressure pm and proceeding to pn.
nwowm pzywp
nm pzwp 981075.0981085.09810
Note : x + z = y + w or x – y = w – z 1.50 – 0.75 = w – z w – z = 0.75
m
n y = 0.75 m Level n
Level m
Level 1
Level 2
z
w
Fluid B
Fluid A Fluid A x = 1.50 m
nm pzwp 981075.0981085.09810
zwpp nm 981075.0981085.0
Note: w – z = 0.75
75.0981075.0981085.0 nm pp
2611,13
m
Npp nm kPapp nm 611.13or
Exercise Problems: 1. If the pressure 4 m below the free surface is 150 kPa, calculate
its specific weight and specific gravity. 2. If the pressure at a point in the ocean is 1500 kPa, what is the pressure 20 m below this point? The specific gravity of salt water is 1.03. 3. An open vessel contains carbon tetrachloride ( s = 1.50 ) to a depth of 2.5 m and water above this liquid to a depth of 1.5 m. What is the pressure at the bottom? 4. How many meters of water are equivalent to a pressure of 120 kPa? How many cm of mercury? 5. What is the equivalent pressure in kPa corresponding to one meter of air at 15 °C under standard atmospheric conditions? 6. At sea level a mercury barometer reads 750 mm and at the same time on the top of the mountain another mercury barometer reads 745 mm. The temperature of the air is assumed constant at 15 °C and its specific weight assumed uniform at 12 N/m3. Determine the height of the mountain.
7. At ground level the atmospheric pressure is 101.325 kPa at 15 °C. Calculate the pressure at a point 6500 m above the ground, assuming (a) no density variation, (b) an isothermal variation of density with pressure. 8. If the barometer reads 750 mm of mercury, what absolute pressure corresponds to a gage pressure of 130 kPa? 9. Determine the absolute pressure corresponding to a vacuum of 20 cm of mercury when the barometer reads 745 mm of mercury. 10. If the pressure in a gas tank is 2.50 atmosphers, find the pressure in kPa and the pressure head in meter of water. 11. The gage pressure at the suction side of a pump shows a vacuum of 25 cm of mercury. Compute (a) pressure head in meter of water, (b) pressure in kPa, (c) absolute pressure in kPa if the barometer reads 750 mm of mercury. 12. The pressure of the air inside a tank containing air and water is 20 kPa absolute. Determine the gage pressure at a point 1.5 m below the water surface. Assume standard atmospheric pressure.