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.