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[Pages:8]Fluid Statics and Manometers

January 24 and 29, 2008

Introduction to Fluid Statics and Manometers

Larry Caretto Mechanical Engineering 390

Fluid Mechanics

January 24 and 29, 2008

Outline

? Review course introduction ? Pressure independent of direction ? Pressure-density-distance relationship

in a static (nonmoving fluid) ? Use of manometers for pressure

measurements ? Calculations with manometers.

2

Review

? Dimensions and units

? SI, BG and EE unit systems

? Fluid density, , (mass/volume) and specific weight (weight/volume), = g, and specific gravity

? States of matter and vapor pressure ? Viscosity ? Surface tension

3

Review Typical Units

Quantity

SI units EE units BG units

Density Pressure & shear stress Velocity Viscosity

Specific weight = g

kg/m3 kPa = kN/m2

m/s

lbm/ft3

slug/ft3

1 psi = 1 lbf/in2 =

144 psf = 144 lbf/ft2

ft/s

N?s/m2 = lbf?s/ft2 =

lbf?s/ft2 =

kg/m?s 32.2 lbm/ft?s slug/ft?s

N/m3

lbf/ft3

Tabulated values at standard gravity

4

Pressure Solid

Review States of Matter

Boiling line

shows Pvapor = f(T)

? Triple point: solid, liquid and vapor coexist

Liquid

Critical ? No liquid-gas Point transition above

critical point

Triple Point Gas Temperature

? Vapor pressure for liquid-gas transition

5

Viscosity

Newtonian Fluids

have a linear variation of shearing stress with rate of shearing strain ? slope is viscosity Figure 1.4 (p. 15)

= u y

Fundamentals of Fluid Mechanics, 5/E by Bruce Munson, Donald

Young, and Theodore Okiishi,Copyright ? 2005 by John Wiley &

6

Sons, Inc. All rights reserved.

ME 390 ? Fluid Mechanics

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Fluid Statics and Manometers

January 24 and 29, 2008

Review Surface Tension

? Vertical force balance: R2h = 2Rcos

? Surface tension depends on fluid, temperature;

wetting angle, , depends on fluid and surface

h = 2 cos

R

Fundamentals of Fluid Mechanics, 5/E by Bruce Munson, Donald 7 Young, and Theodore Okiishi,Copyright ? 2005 by John Wiley &

Sons, Inc. All rights reserved.

Gravity in ?z direction

Sum forces in each direction and divide by xyz

Fz = 0 = -(xyz)g

+ p - p z xy z 2

- p + p z xy z 2

Fundamentals of Fluid Mechanics, 5/E by Bruce Munson, Donald Young, and Theodore Okiishi,Copyright ? 2005 by John Wiley & Sons,

Inc. All rights reserved.

Results

p = 0 x

p = 0 y

p + = 0

z

9

Pressure Relations

? Pressure is a scalar ? The force exerted by a pressure is the

same in all directions ? Want to see how pressure changes in a

static (nonmoving) fluid ? Look at balance of pressure force and

fluid weight over a differential volume element, xyz

8

Integrating the Result

? Since p/x =p/y = 0, p = f(z) only, and we can write p/z = ? as dp/dz = ?

? Multiply by dz and integrate between

two points (p1, z1) and (p2, z2)

p2

z2

dp = p2 - p1 = - dz

p1

z1

? To integrate dz we have to know how

depends on z

10

Incompressible Fluid

? An incompressible fluid has constant density (and specific weight)

? For an incompressible (constant density)

fluid then

z2

z2

p2 - p1 = - dz = - dz = -(z2 - z1)

z1

z1

p2 + z2 = p1 + z1

11

Incompressible Fluid II

p2 + z2 = p1 + z1

? Which pressure is higher?

z2

p2

p1 = p2 + (z2 - z1 )

h

p1 = p2 + h > p2

z1

p1

? Pressure

increases with

depth

12

ME 390 ? Fluid Mechanics

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Fluid Statics and Manometers

January 24 and 29, 2008

Problem

? If the pressure at the surface of a body of water ( = 9789 N/m3 at 20oC) is 101 kPa, what are the pressures at depths of 10 m and 100 m?

? Given: p1 = 101 kPa at z1 = 0 ? Find: p at z2 = ?10 m and z3 = ?100 m ? Equation:

p2 + z2 = p1 + z1 p2 = p1 + (z1 - z2 )

13

Solution

Depth = 10 m

p2 = p1 + (z1 - z2 ) = 101 kPa +

9789 N [0 m - (-10 m)] kPa m2 = 198.9 kPa

m3

1000 N

Depth = 100 m

p3 = p1 + (z1 - z3 ) = 101 kPa +

9789 m3

N

[0 m - (-100 m)] kPa m2

1000 N

= 1080 kPa

14

Pressure Head

p1 = p2 + (z2 + z1)

p1 = p2 + h > p2

?

Fluid height equivalent to h =

p2 - p1

z2

p2

a pressure difference

h

? h is called pressure head

z1

p1 ? For p2 ? p1 = 14.696 psia =

101.325 kPa, h = 0.76 m =

29.92 in for Hg

? What is h for water at this

p?

15

P = 1 Atm for Water at 20oC

? At 20oC, water = 9789 N/m3 = 62.32 lbf/ft3 (p 761, text) 14.696 lb f 144 in2

h = p2 - p1 =

in2

ft 2 = 33.96 ft

62.32 lb f

ft 3

h=

p2 -

p1

=

101.325

kPa

1000 N kPa m2

= 10.35 m

9789 N

m3

16

Free Surface

? Surface of liquid open to atmosphere is called a "free surface"

? Pressure, p0, is atmospheric pressure, p0 ? Height, z0 = 0 ? In the liquid, p + z = p0 + z0, where z < 0 ? Depth h = z0 ? z > 0 ? p = p0 + (z0 ? z) = p0 + h

? Pressure, p, at depth, h, not influenced by size or shape of container

17

ME 390 ? Fluid Mechanics

Free Surface II

? Pressure, p, at depth, h, not influenced by size or shape of container

Figure 2.4, Fundamentals of Fluid Mechanics, 5/E by Bruce Munson, Donald Young, and Theodore Okiishi Copyright ?

18

2005 by John Wiley & Sons, Inc. All rights reserved.

3

Fluid Statics and Manometers

January 24 and 29, 2008

Reference Pressure

? Free surface equation: p = p0 + (z0 ? z) ? Apply this to two different pressures

? p1 = p0 + (z0 ? z1) ? p2 = p0 + (z0 ? z2)

? Find p2 ? p1 from these equations

? Result: p2 ? p1 = p0 + (z0 ? z2) ? [p0 + (z0 ? z1)] = (z1 ? z2) independent of p0 or z0

? Reference pressure cancels in taking pressure differences

19

Gage Pressure

? For taking pressure differences, we can use any reference pressure

? Many pressure measurement methods measure the difference between actual and atmospheric pressure

? We can used this measured pressure difference, called gage pressure, directly in p calculations

? pabsolute = pgage + patmosphere

20

Gage and Absolute Pressure

Figure 2.7, Fundamentals of Fluid Mechanics, 5/E by Bruce Munson, Donald Young, and Theodore Okiishi Copyright ?

21

2005 by John Wiley & Sons, Inc. All rights reserved.

What a Barometer Measures

? It actually measures the local pressure ? A barometer in a undersea submersible,

pressurized to 4 times atmospheric pressure would measure this level ? Weather barometer readings are corrected to mean sea level ? Standard atmosphere: 760 mm Hg, 760 torr, 29.921 in Hg, 101.325 kPa, 14.696 psia, 2116.2 psfa, 1013.25 mbar

23

ME 390 ? Fluid Mechanics

Barometric Pressure

? Mercury barometer used to measure atmospheric pressure

? Top is evacuated and fills with mercury vapor

? Patm = h + pvapor ? pvapor = 0.000023 psia =

0.1586 Pa at 68oF (20oC)

? h = 760 mm = 29.92 in for standard atmosphere

Figure 2.8, Fundamentals of Fluid Mechanics, 5/E by Bruce Munson, Donald Young, and Theodore Okiishi Copyright ?

22

2005 by John Wiley & Sons, Inc. All rights reserved.

Gage/Absolute Notation

? For pressure differences a specification of gage or absolute is not required

? Traditional notation is psig (or psfg) and psia (or psfg) for gage and absolute pressure, respectively

? Can also use kPa(abs) or kPa(gage) ? Munson uses psi or kPa for gage

pressures and psi(abs) or kPa(abs) for absolute pressures

24

4

Fluid Statics and Manometers

Variable Density

? Problem: integrate dp/dz = ?z when density (and hence ) is not constant

? Simple solution: for gases is small so that p does not change much with z

? E. g. air at atmospheric pressure and T = 20oC has = 11.81 N/m3

? If = 11.81 N/m3 were constant an elevation change of 10 m gives p = (11.81 N/m3)(10 m) = 118.1 N/m2 = 0.1181 kPa

25

January 24 and 29, 2008

Variable Density II

? Result: the pressure change of 0.1181 kPa is only 0.12% of patm = 101.325 kPa

? Simple solution: for gases with small elevation changes we can assume that the specific weight is constant!

? This is not valid for changes of several kilometers as in the atmosphere

? Standard atmosphere defined and used for aerospace designs

26

Constant specific weight is a good assumption for gases except for large elevation changes

Figure E2.2, Fundamentals of Fluid Mechanics, 5/E by Bruce

27

Munson, Donald Young, and Theodore Okiishi Copyright ?

2005 by John Wiley & Sons, Inc. All rights reserved.

Piezometer

Open means p = patm

? A passage between a container (such as the pipe at A) and the atmosphere is called a piezometer tube

? For a piezometer tube the pressure in the fluid, pA = p1 = patm + 1h1

? patm(gage) = 0

Figure 2.9, Fundamentals of Fluid Mechanics, 5/E by Bruce Munson, Donald Young, and Theodore Okiishi Copyright ? 2005 29

by John Wiley & Sons, Inc. All rights reserved.

ME 390 ? Fluid Mechanics

U. S. Standard Atmosphere

(See appendix C for more data)

Figure 2.6, Fundamentals of Fluid Mechanics, 5/E by Bruce Munson, Donald Young, and Theodore Okiishi Copyright ? 2005 by John Wiley & Sons,

Inc. All rights reserved.

28

Problem

? Find the pressure at point A if the fluid is water at 20oC, h1 = 0.2 m, and patm = 101 kPa

? From table B.2, 1 = water = 9.789 kN/m3 at 20oC

? pA = p1 = patm + 1h1

Figure 2.9, Fundamentals of Fluid Mechanics, 5/E by Bruce Munson, Donald Young, and Theodore Okiishi Copyright ? 2005 30

by John Wiley & Sons, Inc. All rights reserved.

5

Fluid Statics and Manometers

January 24 and 29, 2008

Solution

? h1 = 0.2 m, patm = 101 kPa, and 1 =water = 9.789 kN/m3

? pA = patm + 1h1

PA

= 101

kPa

+

9.789 kN m3

(0.2

m)

kPa m2 1 kN

? pA = 103 kPa (absolute)

? pA = 1.96 kPa (gage)

Figure 2.9, Fundamentals of Fluid Mechanics, 5/E by Bruce Munson, Donald Young, a3n1d Theodore Okiishi Copyright ? 2005 by John Wiley & Sons, Inc. All rights reserved.

Simple U-Tube Manometer

? Manometers measure pressure by measuring height differences

? Point A is fluid ( = 1) in a pipe

? h1 and h2 are measured ? Gage fluid has = 2 ? What is pressure at A?

32 Figure 2.10, Fundamentals of Fluid Mechanics, 5/E by Bruce Munson, Donald Young, and Theodore Okiishi Copyright ? 2005 by John Wiley & Sons, Inc. All rights reserved.

Simple U-Tube Manometer II

? Right side: p3 = patm + 2h2 ? Left side: p2 = pA + 1h1

patm

? p3 = p2 gives patm + 2h2 = pA + 1h1

p2 = p3 = p0 ? 2h0

? Conclusion: pA = patm + 2h2 ? 1h1

h0

33p0

Figure 2.10, Fundamentals of Fluid Mechanics, 5/E by Bruce Munson, Donald Young,

and Theodore Okiishi Copyright ? 2005 by John Wiley & Sons, Inc. All rights reserved.

Simple U-Tube Manometer III

? Result for absolute pressure: pA = patm + 2h2 ? 1h1

? Result for gage pressure: pA = 2h2 ? 1h1

34 Figure 2.10, Fundamentals of Fluid Mechanics, 5/E by Bruce Munson, Donald Young, and Theodore Okiishi Copyright ? 2005 by John Wiley & Sons, Inc. All rights reserved.

? Given: Known specific weights and measured heights shown above

? Find: pA ? pB Equation: p + z = p + z

35 FigureE2.5, Fundamentals of Fluid Mechanics, 5/E by Bruce Munson, Donald Young, and

Theodore Okiishi Copyright ? 2005 by John Wiley & Sons, Inc. All rights reserved.

pA = p1 + 1h1 p1 = p2 = p3 p3 = p4 + 2h2

p5 = p4 pB = p5 + 1(h1 + h2 )

36 FigureE2.5, Fundamentals of Fluid Mechanics, 5/E by Bruce Munson, Donald Young, and

Theodore Okiishi Copyright ? 2005 by John Wiley & Sons, Inc. All rights reserved.

ME 390 ? Fluid Mechanics

6

Fluid Statics and Manometers

January 24 and 29, 2008

FigureE2.5, Fundamentals of Fluid

Mechanics, 5/E by Bruce Munson, Donald

Young, and Theodore Okiishi Copyright ?

2005 by John Wiley & Sons, Inc. All rights reserved.

Combine first set of equations

pA = p1 + 1h1 p1 = p2 = p3 p3 = p4 + 2h2

pA = p1 + 1h1 = p3 + 1h1 = p4 + 2h2 + 1h1

37 FigureE2.5, Fundamentals of Fluid Mechanics, 5/E by Bruce Munson, Donald Young, and

Theodore Okiishi Copyright ? 2005 by John Wiley & Sons, Inc. All rights reserved.

( ) Combine result from previous page with

pA = p4 + 2h2 + 1h1

second set of equations p5 = p4 pB = p5 + 1 h1 + h2

pA = p5 + 2h2 + 1h1 = pB - 1(h1 - h2 )+ 2h2 + 1h1

pA - pB = (2 - 1)h2

38

Incline Manometer Problem

? Incline used to increase accuracy for small pressure differences

? Want to find pB ? pA when we know 1, 2, 3, h1, l2, h3, and

Equation: p + z = p + z

Figure 2.9, Fundamentals of Fluid Mechanics, 5/E by Bruce Munson, Donald Young, a3n9d Theodore Okiishi Copyright ? 2005 by John Wiley & Sons, Inc. All rights reserved.

Incline Manometer Problem II

p1 = pA + 1h1 p1 = p2 + 2 (z2 - z1) = p2 + 2 l2 sin

p2 = pB + 3h3 pB = p2 - 3h3 pB = p2 - 3h3 = p1 - 2 l2 sin - 3h3 pB = pA + 1h1 - 2 l2 sin - 3h3

pA - pB = 3h3 + 2 l2 sin - 1h1

40

Solving Manometer Problems

? Basic equation: pressures at two depths open

in same fluid: p2 = p3 + (z3 ? z2) = p3 + h

? "Open" means p = patm

3

? patm = 101.325 kPa = 14.696 psia

? For gage pressure, patm = 0

? Same pressures at same level on two sides of a manometer

? p2 = p3

41

Solving Manometer Problems II

? Write equations for (1) pressures at two open depths in same fluid and (2) equal pressures at same level (with same fluid) 3 at all branches in manometer.

? Eliminate intermediate pressures from equations to get desired P

? Watch units for length, psi or

psf, N or kN

? For gases z 0

42

ME 390 ? Fluid Mechanics

7

Fluid Statics and Manometers

January 24 and 29, 2008

Problem

? Given: air at 0.50 psig, pA = 2 psig, and

oil = 54.0 lbf/ft3 and other data shown

on diagram. Find: z and h

pA - pair = oil z

z = 4 ft

z = pA - pair oil

=

2 lb f in2

-

54.0 lb f

0.5 lb f

in2 ft 2

ft3 144 in2

Figure P2.27, Fundamentals of Fluid Mechanics, 5/E by Bruce

43

Munson, Donald Young, and Theodore Okiishi Copyright ? 2005

by John Wiley & Sons, Inc. All rights reserved.

Problem Continued

? Given: air at 0.50 psig, pA = 2 psig, and

oil = 54.0 lbf/ft3 and other data shown

on diagram. Find: h

pA + oil (2 ft ) =

popen = 0 psig

popen + manoh

mano = SGmano H2O

=

3.05

62.4 ft

lb

3

f

=

190.3 lb ft 3

f

Figure P2.27, Fundamentals of Fluid Mechanics, 5/E by Bruce

44

Munson, Donald Young, and Theodore Okiishi Copyright ? 2005

by John Wiley & Sons, Inc. All rights reserved.

Problem Concluded

? popen = 0 psig, mano = 190.3 lbf/ft3, h = ?

pA + oil (2 ft ) = popen + manoh

h = pA + oil (2 ft )- popen

mano

0

h = 2.08 ft

h=

2 lb f in2

144 in2 ft 2

+

54.0 lb f ft 3

190.3 lb f

(2

ft )

ft 3

Figure P2.27, Fundamentals of Fluid Mechanics, 5/E by Bruce

45

Munson, Donald Young, and Theodore Okiishi Copyright ? 2005

by John Wiley & Sons, Inc. All rights reserved.

Problem 2.38

A hemispherical shell on the ocean floor has an internal barometric pressure of 765 mm Hg. A mercury manometer measures the differential pressure between the sea outside and the shell interior as shown in the diagram. What is the pressure at the ocean surface? 46

Problem 2.38 Part 2

Pshell

PoceanPleft = Pshell + Hg (0.735 m) = Pright = Pocean +

Equal pressure line

sea (10 m + 0.36 m)

water

Use specific weight data from Table 1-6

(ignore difference in

Pshell = Hghbaro

hbaro = 765 mm Hg

temperature)

Hg = 133 kN/m3 sea water = 10.1 kN/m3

47

Problem 2.38 Part 3

? Rearrange equations from previous

slide and substitute given data

Pleft = Pshell + Hg (0.735 m) = Pright = Pocean + sea (10.36 m)

water

Pocean = Pshell + Hg (0.735 m) - sea (10.36 m)

water

Pshell

=

Hg hbaro

=

133 kN m3

(0.765

m)

=

101.745 m2

kN

Pocean

=

101.745 m3

kN

+

133 kN m3

(0.735

m)

-

10.1 kN m3

(10.36

m)

Pocean

=

94.9 kN m2

= 94.9 kPa

48

ME 390 ? Fluid Mechanics

8

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