Hydraulic Engineering

Home , Inch of water, Specific weight

Lecture Notes for CIVE 401

Part I

Hydrostatic Review and Closed Conduits

Pierre Y. Julien

with the assistance of Joonhak Lee

2020 Fall Semester 8-17-20 2

Table of Contents

1. Review of Hydrostatics (Week 1) 3

1.1. Dimensions and Units 3

1.2. Properties of Water 6

1.3. Specific Gravity 9

1.4. Atmospheric Pressure 10

1.5. Hydrostatic Pressure 11

1.6. Vapor Pressure 12

1.7. Piezometric Head 14

1.8. Hydrostatic Force on a Plate 15

1.9. Hydrostatic Force on a Dam 18 2. Flow in Pipes (Week 2) 24

2.1. Continuity/Conservation of Mass 24

2.2. Friction Losses in Pipes 25

2.3. Three Reservoirs 30

2.4. Pipe Networks 32

2.5. Minor Losses in Pipes 35

2.6. Negative Pressure/Suction 40

3. Hydrodynamic Forces (Week 3) 41

3.1. Momentum Force on a Plate 41

3.2. Hydrodynamic Force on a Pipe Bend 43

4. Flow Meters (Week 4) 46

4.1. Manometer 46

4.2. Venturi Meter 47

4.3. Flow Nozzle 48

4.4. Various Flow Meters 49

1. Review of Hydrostatics (Week 1)

We review hydrostatics and start to solve problems!

1.1. Dimensions and units Physical water properties are usually expressed in terms of the following fundamental dimensions: mass (M), length (L), time (T), and temperature (T°) is also considered.

In the SI system of units, the units for mass, length, time, and temperature are the kilogram (kg), the meter (m), the second (s), and degrees Celsius (°C). A Newton (N) is the force to accelerate 1 kg at 1 m/s2, or 1N = 1 kg m/s2. The gravitational acceleration at the

Earth’s surface is g = 9.81 m/s2. The weight of one kilogram is F = mass  g = 1 kg 

9.81 m/s2 = 9.81 N. The pressure is a perpendicular force per unit area given in pascals from 1 Pa = 1 N/m2. The unit of work (or energy) is the joule (J), which equals the product of 1 N  1 m. Table 1.1 gives the unit of power is a watt (W), which is 1 J/s.

In English units, the time unit is a second, the units for mass, length, time, and temperature are the slug (slug), the foot (ft), the second (s), and degrees Fahrenheit (°F).

The force to accelerate a mass of one slug at 1 ft/s2 is a pound force (lb). A pound always refers to a force, not a mass.

The temperature in degrees Celsius T°C is converted to degrees Fahrenheit T°F as T°F = 32.2 °F + 1.8 T°C. A useful conversion factor for weight is 1 lb = 4.448 N, and for discharge 1 m3/s = 35.32 ft3/s, with abbreviation 1 cms = 1m3/s and 1 cfs = 1ft3/s.

Other conversions are found in Table 1.2.

Prefixes indicate multiples or fractions of units by powers of 10:

(micro)== 10−63 ,k(kilo) 10 , m(milli)== 10−36 ,M(mega) 10 , c(centi)== 10−29 ,G(giga) 10 .

For example, one megawatt (MW) equals one million watts (1,000,000 W or 106 W). 4

Table 1.1. Geometric, kinematic, dynamic, and dimensionless variables

Variable Symbol Fundamental SI Units dimensions

Geometric (L)

Length L, x, h L m

Area A L2 m2

Volume ∀ L3 m3

Kinematic (L, T)

Velocity V LT–1 m/s

Acceleration a, g LT–2 m/s2

Kinematic viscosity  L2T–1 m2/s

Unit discharge q L2T–1 m2/s

Discharge Q L3T–1 m3/s

Dynamic (M, L, T)

Mass m M 1 kg

Force F = ma, mg MLT–2 1 kg m/s2 =1N

Pressure p = F/A ML–1T–2 1 N/m2 = 1 Pa

Work or energy E = Fd ML2T–2 1 Nm= J

Mass density ρ ML–3 kg/m3

Specific weight γ = ρ g ML–2T–2 N/m3

Dynamic viscosity μ = ρ v ML–1T–1 1 kg/m s = 1 Pa s

Dimensionless

Slope S0, Sf – –

Specific gravity G = γs /γ – –

Reynolds number Re = V h / – –

Froude number Fr = u / (g h)0.5 – – 5

Table 1.2. Useful unit conversions

Unit kg, m, s N, Pa, W

1 acre 4,047 m2

1 atmosphere (atm) 101,3 kg/m s2 101.3 kPa

1 cubic foot per second (ft3/s) 0.0283 m3/s

1 degree Fahrenheit (°F) = 32+1.8 T°C 0.5556 °K

1 foot (ft) 0.3048 m

1 gallon (U.S. gal) (1 U.S. gal = 3.785 liters) 0.003785 m3

1 horsepower (hp) = 550 lb ft/s 745.7 kg m2/s3 745.7 W

1 inch (in.) (1 ft = 12 in.) 0.0254 m

1 inch of mercury (in. Hg) 3,386 kg/m s2 3,386 Pa

1 inch of water 248.8 kg/m s2 248.8 Pa

1 mile (statute), (1mile = 5,280 ft) 1609 m

1 million gallons per day (1 mgd = 1.55 ft3/s) 0.04382 m3/s

1 fluid ounce (U.S.) 2.957  10–5 m3

1 pascal (Pa) 1 kg/m s2 1 N/m2

1 pound-force (lb) (1 lb = 1 slug  1 ft/s2) 4.448 kg m/s2 4.448 N

1 pound per square foot (lb/ft2 or psf) 47.88 kg/m s2 47.88 Pa

1 pound per square inch (lb/in.2 or psi) 6,895 kg/m s2 6,895 Pa

1 quart (U.S.) (1 qt = 2 pints) 0.0009463 m3

1 slug 14.59 kg

1 ton (S.I. metric) (1,000 kg = 1 Mg) 1,000 kg

1 ton (U.S. short) = 2,000 lb 8,900 kg m/s2 8.9 kN

1 yard (yd) (1 yd = 3 ft) 0.9144 m 6

1.2. Properties of water Mass density of water ρ. The mass of water per unit volume defines the mass density ρ. The maximum mass density of water at 4 °C is 1,000 kg/m3 and decreases slightly with temperature, as shown in Table 1.3. In comparison, the mass density of sea water is approximately 1,025 kg/m3 and the mass density of air at sea level is 1.29 kg/m3 at 0 °C. The conversion factor for mass density is 1 slug/ft3 = 515.4 kg/m3.

Specific weight of water γ. The weight per unit volume is the specific weight γ.

At 10 °C, water has a specific weight, γ = 9,810 N/m3 or 62.4 lb/ft3 (1 lb/ft3 = 157.09 N/m3).

Specific weight varies with temperature as given in Table 1.3. The specific weight γ is the product of mass density ρ and gravitational acceleration g = 32.2 ft/s2 = 9.81 m/s2:

= g

Dynamic viscosity μ. As a fluid is brought into low rates of deformation, the velocity of the fluid at any boundary equals the velocity of the boundary. The fundamental dimension of the dynamic viscosity μ is M/LT. In Table 1.3, the dynamic viscosity of water decreases with temperature. The dynamic viscosity of water at 20 °C is

0.001 kg/m s = 0.001 N s/m2 = 0.001 Pa s. The conversion factor for the dynamic viscosity is 1 lb s/ft2 = 47.88 N s/m2 = 47.88 Pa s.

Kinematic viscosity of water . The kinematic viscosity is obtained when the dynamic viscosity of a fluid  is divided by its mass density  . The kinematic viscosity v of water in L2/T in Table 1.3 decreases with temperature. The viscosity of clear water at

20 °C is 0.01 cm2/s = 1  10-6 m2/s. The conversion factor is 1 ft2/s = 0.0929 m2/s. The kinematic viscosity of water v depends on temperature T° as

=

  = =[1.14 − 0.031(TT − 15) + 0.00068( − 15)2 ]  10− 6 m 2 / s  CC 7

Table 1.3. Physical properties of water in SI Units at atmospheric pressure

Temperature Vapor Density Specific Dynamic Kinematic Surface Modulus of Pressure Weight Viscosity Viscosity Tension Elasticity

푝 γ μ  σ 퐸 T 푣 ρ 푣 (◦C) (kN/m2) (kg/m3 ) (kN/m3 ) (N-s/m2) (m2/s) (N/m) (kN/m2) × 106 absolute × 10−3 × 10−6

0 0.61 999.8 9.805 1.781 1.785 0.0756 2.02

5 0.87 1000.0 9.807 1.518 1.519 0.0749 2.06

10 1.23 999.7 9.804 1.307 1.306 0.0742 2.10

15 1.70 999.1 9.798 1.139 1.139 0.0735 2.14

20 2.34 998.2 9.789 1.002 1.003 0.0728 2.18

25 3.17 997.0 9.777 0.890 0.893 0.0720 2.22

30 4.24 995.7 9.764 0.798 0.800 0.0712 2.25

40 7.38 992.2 9.730 0.653 0.658 0.0696 2.28

50 12.33 988.0 9.689 0.547 0.553 0.0679 2.29

60 19.92 983.2 9.642 0.466 0.474 0.0662 2.28

70 31.16 977.8 9.589 0.404 0.413 0.0644 2.25

80 47.34 971.8 9.530 0.354 0.364 0.0626 2.20

90 70.10 965.3 9.466 0.315 0.326 0.0608 2.14

100 101.33 958.4 9.399 0.282 0.294 0.0589 2.07

Commonly used values of specific weight and viscosity are:

Weight mass ퟏ, ퟎퟎퟎ 퐤퐠 9.81 m ퟗ. ퟖퟏ 퐤퐍 γ = = × g = ρ × g = ퟑ × 2 = ퟑ Vol Vol 퐦 s 퐦

1,000 kg ퟏ × ퟏퟎ−ퟔ 퐦ퟐ ퟏ × ퟏퟎ−ퟑ 퐤퐠 1 × 10−3 N. s µ = ρ × υ = × = = m3 퐬 퐦. 퐬 m2

Table 1.4. Physical properties of water in English units at atmospheric pressure

Temperature Vapor Density Specific Dynamic Kinematic Surface Modulus Pressure Weight Viscosity Viscosity Tension of

Elasticity 푝 γ μ ν σ T 푣 ρ 퐸푣 (◦F) (psia) (slug/ft3 ) (lb/ft3 ) (lb-s/ft2) (ft2/s) (lb/ft) (psi) × 10−5 × 10−5 × 10−3 × 103

32 0.09 1.940 62.42 3.746 1.931 5.18 293

40 0.12 1.940 62.43 3.229 1.664 5.14 294

50 0.18 1.940 62.41 2.735 1.410 5.09 305

60 0.26 1.938 62.37 2.359 1.217 5.04 311

70 0.36 1.936 62.30 2.050 1.059 4.98 320

80 0.51 1.934 62.22 1.799 0.930 4.92 322

90 0.70 1.931 62.11 1.595 0.826 4.86 323

100 0.95 1.927 62.00 1.424 0.739 4.80 327

120 1.69 1.918 61.71 1.168 0.609 4.67 333

140 2.89 1.908 61.38 0.981 0.514 4.54 330

160 4.74 1.896 61.00 0.838 0.442 4.41 326

180 7.51 1.883 60.58 0.726 0.385 4.27 318

200 11.52 1.868 60.12 0.637 0.341 4.13 308

212 14.70 1.860 59.83 0.593 0.319 4.04 300

Commonly used values of specific weight and viscosity are:

Weight mass ퟏ. ퟗퟒ 퐬퐥퐮퐠 ퟑퟐ. ퟐ 퐟퐭 ퟔퟐ. ퟒ 퐥퐛 γ = = × g = ρ × g = ퟑ × ퟐ = ퟑ Vol Vol 퐟퐭 퐬 퐟퐭

1.94 slug ퟏ. ퟎ × ퟏퟎ−ퟓ 퐟퐭ퟐ ퟐ. ퟎ × ퟏퟎ−ퟓ 퐬퐥퐮퐠 2.0 × 10−5lb. s µ = ρ × υ = × = = ft3 퐬 퐟퐭. 퐬 ft2 9

1.3. Specific Gravity The specific gravity, S.G. or G, of a fluid defined as the ratio of the density of the fluid to the density of water at some specified temperature. Normally the temperature is taken as 4°C

(39.2°F). At this temperature, the density of water is 1.94 slugs/ft3 or 1,000kg/m3. One can also define the specific gravity as the weight of the liquid divided by the specific weight of water, or

휌 훾 G = 푓푙푢푖푑 = 푓푙푢푖푑 (1.1) 휌푤 훾푤

The specific gravity is a dimensionless number. The specific weight of water, 훾푤, at 4°C and under atmosphere pressure is 9,810N/m3.

Example 1: If 2 m3 of oil has a mass of 1,720 kg, calculate its specific weight, density, and specific gravity.

푊 1,720푘푔 γ = = × 9.81푚/푠2 = 8,437 N/m3 푉표푙 2푚3

훾 8,437 ρ = = = 860 kg/m3 푔 9.81

훾 8,437 G = 표푖푙 = = 0.86 훾푤 9810

Example 2: A container weighs 5,000 lb when filled with 60ft3 of a certain liquid. The empty weight of the container is 500 lb. What is the density of the liquid?

푊 5,000−500 γ = = = 75 lb/ft3 푉표푙 60

훾 75 ρ = = = 2.33 slugs/ft3 푔 32.2

1.4. Atmospheric Pressure

At sea level, the atmospheric pressure is patm = 101.3 kPa = 14.7 psi = 2,116 psf

Table 1.5. Atmospheric pressure decrease with altitude (1 psi = 6.895 kPa)

Altitude Above Sea Level ||Absolute Barometer ||Absolute Atmospheric Pressure Feet meters inches Hg mm Hg psia psfa kPa 01) 0 29.9 760 14.7 2,116 101.3 1,000 305 28.9 733 14.2 2,045 97.7 2,000 610 27.8 707 13.7 1,973 94.2 3,000 914 26.8 681 13.2 1,901 90.8 4,000 1,219 25.8 656 12.7 1,829 87.5 5,000 1,524 24.9 632 12.2 1,757 84.3 6,000 1,829 24.0 609 11.8 1,700 81.2 7,000 2,134 23.1 586 11.3 1,627 78.2 8,000 2,438 22.2 564 10.9 1,570 75.3 9,000 2,743 21.4 543 10.5 1,512 72.4 10,000 3,048 20.6 523 10.1 1,454 69.7 15,000 4,572 16.9 429 8.29 1,194 57.2 20,000 6,096 13.8 349 6.75 972 46.6 25,000 7,620 11.1 282 5.45 785 37.6 30,000 9,144 8.89 226 4.36 628 30.1

1.5. Hydrostatic Pressure

Z 퐖 = 훒퐠퐕퐨퐥 = 후퐕퐨퐥 = 후∆풙∆풚∆풛 p +∆p ∑ 푭풛 = 풑∆풙∆풚 − (풑 + ∆풑)∆풙∆풚 − 휸 ∆풙∆풚∆풛 = ퟎ

∆z ∆퐩 = −후∆퐳 W=γV Y 풅풑 − = 휸 풅풛

∆x ∆y 퐏 퐙 X ∫ 퐝퐩 = ∫ −후퐝퐳 퐏퐨 퐙퐨 p

퐩 − 퐩퐎 = −후(퐳 − 퐳퐨) = 후퐡

퐩 = 퐩퐎 + 후퐡

Absolute and Relative Pressure

P = 0 Patm o

zo Relative Absolute pabs = prel + patm

z pressure pressure

Example 3: At an altitude of 10,000 ft and water temperature of 5C, determine the

absolute and relative pressure 10 ft below the surface of a reservoir. (1 psf = 47.88 Pa)

푙푏 2 prel = 62.4 × 10 푓푡 = 624 lb/ft = 29.9 kPa 푓푡3

푝푠푓 pabs = 10.1 푝푠𝑖푎 × 144 + 624 푝푠푓 = 2,078 푝푠푓 = 99.5 kPa 푝푠𝑖 12

1.6. Vapor Pressure Values of vapor pressure for various water temperatures are given in Tables 1.1 and 1.2.

There is a wide variation in vapor pressure among liquids. Mercury has the lowest vapor pressure and was therefore used in traditional barometers.

Table 1.4. Vapor pressure of mercury and water at 20C (68F)

Liquid Pressure (psia) Pressure (N/m2 abs) Pressure (mbar, abs)

Mercury 0.000025 0.17 0.0017

Water 0.34 2,340 23.4

A Pv

Pv hv Patm B

Water Hg

Example 4: Vapor pressure at 20C, pv=0.339 psi = 2.34 kPa in water, find hv =?

2 pB = pA + γhv , patm = pv + γhv , (1kPa = 1kN/m )

3 (Patm = 101.36 kPa, γ = 9.81 kN/m )

P −P 101.3 kPa−2.34kPa 101.3 kN/m2−2.34kN/m2 h = atm v = = = 10.1m  33.1 ft v γ 9.81kN/m3 9.81kN/m3

Example 5: Hg : Mercury, γHg=13.6γ

Vapor pressure at 20C, pv= 0 kPa for mercury

patm − pv 101.36 kPa − 0 101.36 kN/m2 hvHg = = = = 760 mm of Hg γHg 13.6 × 9.81kN/m3 13.6 × 9.81kN/m3 13

Relative vapor pressure

푝푟푒푙 푝푎푏푠 > 0 Sea Level 푝 푎푡푚 10, 000 ft 푝푟푒푙 < 0 푝 푣푎푝표푟 푟푒푙 < 0 푝푎푡푚 ◦F 푝푎푏푠 > 0 70 32 ◦F 푝 = 0 푝 푣푎푝표푟 푎푏푠 > 0

Table 1.5. Atmospheric and vapor pressure

Temperature, T Absolute Vapor Pressure Altitude Atmosphere Pressure (◦F) (always positive) (ft) (always absolute)

푝푣 (psia) 푝푎푡푚 (psia)

32 0.09 0 14.7

40 0.12 5,000 12.2

50 0.18 10,000 10.1

60 0.26 15,000 8.3

70 0.36 20,000 6.8

80 0.51

풑풂풃풔 = 풑풂풕풎 + 풑풓풆풍

풑풓풆풍 = 풑풂풃풔 − 풑풂풕풎

Example 6: Find the relative vapor pressure of water at sea level and T = 70F

푝푣 푟푒푙 = 푝푣 푎푏푠 − 푝푎푡푚

푝푣 푟푒푙 = 0.36 psia − 14.7 psi = −14.34 psi = −2,065 psf

Water will turn to vapor when the relative pressure decreases below 푝푣 푟푒푙

1.7. Piezometric Head The piezometer is used to measure pressure. The piezometric head is given by

퐩 퐩퐢퐞퐳퐨퐦퐞퐭퐫퐢퐜 퐡퐞퐚퐝 = 퐳 + 후

Within the same static fluid, the piezometric head is constant.

Example 7: Absolute and relative pressure at B Patm A pC = patm + γ2h2 , pC = pB + γ1h1 B h pB = pC − γ1h1 = patm + γ2h2 − γ1h1(absolute) 1 2 h1 pB rel = pB abs − patm = γ2h2 − γ1h1(relative) C

 2 Notice that the piezometric line at both A and C is at the free surface.

Example 8: Pressure and piezometric head

2 Oil Given S.G.oil = 0.8, find the pressure pA when pB = 200lb/ft pC = pB + γ ∙1ft + γHg ∙ 3ft = pA + γoil ∙ 6ft pC = pB + γ ∙1ft + 13.6 γ ∙ 3ft = pA + 0.8 γ ∙ 6ft

3 (γ = γwater = 62.4 lb/ft )

2 3 3 pC = 200 lb/ft + (62.4 lb/ft ) ∙1ft + (13.6×62.4 lb/ft ) ∙3ft

3 = pA + (0.8×62.4 lb/ft ) ∙6ft

2 3 3 3 pA = 200 lb/ft + (62.4 lb/ft ) ∙1ft + (13.6×62.4 lb/ft ) ∙3ft - (0.8×62.4 lb/ft ) ∙6ft

= 200 + 62.4 + 2,546 – 299.5 = 2,509 lb/ft2 = 2,509 psf

1 psi = 144 psf, thus, pA = 17.4 psi; and 1kPa =0.145038 psi, thus, pA = 120 kPa

The reader can check that the piezometric head at point A and point C are identical

(note that γoil is used). 15

1.8. Hydrostatic Force on a Plate 1.8.1. Area Moment of Inertia

2 The moment of inertia of a planar surface is given as Io = ∫ y dA

For instance, for a rectangular surface, the moment of inertia about the center of gravity is:

O L a/2 dy dA = bdy a CG -a/2 b a 3 2 b a 3 −a 3 ba I̅ = ∫ y2dA = ∫ y2bdy = [( ) − ( ) ] = a A − 3 2 2 12 2

2 From the parallel-axis theorem, we obtain Io = I̅ + AL

1.8.2. Force magnitude F on a plate

θ =

ℎ̅=

p = γh = γysinθ

dF = pdA = γysinθdA

̅ F = ∫F dF = ∫A γysinθdA = γsinθ ∫퐴 y dA= γsinθAy̅ = γAh

퐅 = 후퐀퐡̅

1.8.3. Center of pressure ycp

ycp

Fyc

y dA=bdy 2 dy b a

The center of pressure ycp defines the point of application of the resultant hydrostatic force F 17

Moment about O

dMO = ydF = ypdA = y(γysinθ)dA

y2 2 2 MO = ∫ dMO = ∫ y(γysinθ)dA = ∫ (γsinθ)y dA = (γsinθ) ∫ y bdy = F ycp M A A y1

γsinθ F ycp I = ∫ y2dA = ∫ y2dA = o γsinθ γsinθ

γsinθ I γsinθ (I + Ay̅2) I + Ay̅2 I y = o = xc = xc = y̅ + xc cp F γy̅sinθA y̅A y̅A

퐈 퐲 = 퐲̅ + 퐱퐜 퐜퐩 퐲̅퐀

Example 9: Force on a vertical plate

Find force F on plate and point of application of the force ycp on a vertical (y=h) rectangular

ba3 1ft(24 m)3 plate of height = 24 m per unit width, i.e. area A = 24m2, I̅ = = = 1,152 m4 12 12

y̅ = 12m

ycp = 16 m 24 m F

1 m N F = γh̅A = 9,810 × 12 m × 24m2 = 2,825 kN m3

I̅ 1,152 m4 y = y̅ + = 12m + = 16 m of depth, or 8 m from bottom cp Ay̅ 24m2(12 m)

1.9. Hydrostatic Force on a Dam The basic concept of gravity dam analysis involves the hydrostatic force and the weight of concrete. The resultant force should pass through the central third of the base.

The central third condition maintains compression inside the concrete at the base of the dam.

When the force is outside the central third, there is tension at the base of the dam.

Example 10: Dam Stability

Will tension develop at the base of 16m the following dam configuration? 24

F = 2,825kN, and h = F H cp 16m, as shown in Example 9. O 10m 20m F 24 kN 30m × 24m F = γ WA = (1m) ( ) = 8,640kN C C C m3 2

The sum of moments about O is

∑ MO = FC lC + FH lH = 8,640kN × 10m + 2,825kN × 8m = 109,000kNm

∑ Mo 109,000kNm At the base, ∑ MO = FC x + FH × 0, from which, x = = = 12.6m Fc 8,640kN

The resultant force passes through the central third, i.e. 10 < x < 20 m in this case.

Example 11: For the dam below, determine the following per unit width:

2 Y = kx

24 m Concrete

12 m dx 4 m 8 m O A

(a) the constant k of the parabolic equation y = kx2 y = kx2, 24 = k(12)2, 푘 = 1/6 (b) the horizontal hydrostatic force 9810 N 퐹 = 훾ℎ퐴 = × 12 m × 24 m2 = 2,825kN 퐻 m3 (c) the weight of water above the dam 12 12 12 1 푥3 A = ∫ 푦푑푥 = ∫ 푘푥2푑푥 = [ ] = 96 m2 6 3 0 0 0 A′ = 12 × 24 − 96 = 192 m2 9,810 N 퐹 = 훾V = A × 192 m3 = 1,883 kN 푤 m3

(d) divide the concrete part into three segments

24 m Concrete W2

O 12 m 4 m 8 m A

(e) determine the weight and CG for each segment

24푘N 푊 = 훾 푉 = × 96 m3 = 2,304 kN 1 푐 m3 24푘N 푊 = 훾 푉 = × 4 × 24 × 1 m3 = 2,304 kN 2 푐 m3 24푘N 푊 = 훾 푉 = × 8 × 24 × 0.5 × 1 m3 = 2,304 kN 3 푐 m3

24 m Concrete (14,12)

(18.66,8) (9,

O 12 m 4 m 8 m A

(f) calculate the sum of horizontal and vertical forces

∑ 퐹퐻 = 2,825kN

∑ 퐹푣 = 1,883+ 3ⅹ2,304 = 8,795 kN

(g) determine the resultant force

2 2 2 2 |퐹⃗⃗⃗⃗푅 | = √(∑ 퐹퐻) +(∑ 퐹푉) = √(2,825) +(8,795) = 9,237 kN

(h) calculate the sum of moments about O dA = (24 − y)dx dA = (24 − kx2)dx 12 푥 1 12 1 푥2 푥4 푥̅ = ∫ 푑퐴 = ∫ 푥(24 − 푘푥2) 푑푥 = [24 − ] 퐴 192 192 2 24 0 0 12 1 122 1 124 1,728 − 864 [24 − ] = = 4.5m 192 2 6 4 192 0

∑ MO = FH × 8 + 퐹푊 × 4.5 + 2,304 × (9 + 14 + 18.66) = 127,073 kNm

(i) is the resultant passing through the central third of the base?

∑ Mo 127,073kNm x푅 = = = 14.45푚, Yes F푉 8,795kN

Answers: k = 1/6;  FH = 2,825 kN;  Mo = 127,000 kNm; xR = 14.5 m 21

1.9.1. Gravity Dams

The detailed analysis of gravity dams involves more force components: (1) wave and ice forces near the surface; (2) uplift pore pressure at the base of the dam; (3) active sediment force at the base of the dam; and (4) seismic loading and inertia of the dam. Dam design requires knowledge of hydraulics, structures, geotechnical and geological engineering.

In bedrock canyons, it is possible to anchor the dam on the side rather than the base. In wide open areas, buttresses and multiple arch dams can also be favorably designed.

-> Can you search on the web: Photos Manic 5 22

1.9.2. Rock and Earth-fill Dams Rock and earth-fill dams are pervious and it is important to control the seepage. On impervious surfaces, this can be done with a rock fill toe or a drainage blanket. On pervious surfaces, cutoff walls or cutoff impervious trenches can be designed.

Options to reduce seepage include a rock-fill toe and drainage blankets as illustrated below

Other options include cutoff walls upstream of the dam and a clay core at the dam center.

Permeable dams are also sensitive to failure from sudden drawdown in reservoir levels.

2. Flow in Pipes (Week 2)

We develop programming skills and solve more difficult problems!

2.1. Continuity/Conservation of Mass D 2 A = Conservation of mass implies that mass cannot be 4 created or destroyed. D A P = D Internal mass change Mass flux in – mass flux out = Time Wetted Incompressible fluids, ρ constant, require that a fixed fluid volume contains a constant mass. Perimeter (P)

푨 흅푫ퟐ 푫 Hydraulic Radius: 푹 = = = 풉 푷 ퟒ흅푫 ퟒ 푸 ퟒ푸 퐐 = 푨푽, 풐풓 푽 = = 푨 흅푫ퟐ

Note, 1 m3/s = 35.32 ft2/s, and 1 ft3/s = 450 gpm (gallons per minute)

Continuity for pipe contraction /expansion

2 2 2 4 퐴2 퐷2 푉1 퐷2 푉1 퐷2 = ( ) , 퐴1푉1 = 퐴2푉2 , = ( ) , ( ) = ( ) 퐴1 퐷1 푉2 퐷1 푉2 퐷1

Q 1 = Q 2 푉2 4푄 2 1 D D V V 1 2 2 1 = ( ) 2푔 휋퐷2 2푔

Continuity for pipe branches

2.2. Friction Losses in Pipes

The Hydraulic Grade Line (HGL) is the sum of the elevation at the pipe centerline and the pressure head p/γ. The HGL is the same as the piezometric line from Section 2.1. Adding the velocity head V2 /2g to the piezometric line defines the Energy Grade Line (EGL). The friction losses H represent the decrease in EGL in the flow direction.

2.2.1. Energy equation with friction losses

The Bernoulli Equation with friction losses is ퟐ ퟐ 푷ퟏ 푽ퟏ 푷ퟐ 푽ퟐ + 풛 + = + 풛 + + ∆퐇 휸 ퟏ ퟐ품 휸 ퟐ ퟐ품

The Bernoulli sum defines the energy level of the fluid in terms of the Energy Grade Line EGL

H 2 V 1 2 E .G .L 2g V 2 2g H .G .L P 2  P1  Q

Z 2

Z 1

Datum

L 26

2.2.2. Pipe types and roughness

Draw tube, commercial steel, cast iron, asphalted, concrete, and riveted pipes

The pipe roughness ks measures the size of the roughness of the pipe inner surface

The relative roughness ks /D or /d of a pipe is calculated from roughness and

pipe diameter

The pipe roughness increases with the age of the pipe (the real diameter may also decrease!) 27

2.2.3. Darcy-Weisbach Equation

The Darcy-Weisbach equation is a relationship between the energy loss in a pipe and the mean flow velocity (or discharge) It also defines the Darcy- Weisbach friction factor f

풇푳 푽ퟐ 풇푳 ퟒ푸 ퟐ ퟏ ퟖ풇푳 ∆퐇 = 풉 = = ( ) = [ ] 푸ퟐ = 푲푸ퟐ 풇 푫 ퟐ품 푫 흅푫ퟐ ퟐ품 품흅ퟐ푫ퟓ

ퟖ풇푳 푳푻ퟐ 푻ퟐ 푲 = , [ = ] 품흅ퟐ푫ퟓ 푳푳ퟓ 푳ퟓ

2.2.4. Moody Diagram

A reasonable approximation for the Darcy-Weisbach friction factor f from Swamee and Jain with Re = VD/υ = 4Q/πDυ is 휺 ퟓ. ퟕퟒ −ퟐ 휺 ퟓ. ퟕퟒ −ퟐ 풇 = ퟎ. ퟐퟓ [풍풐품 ( + )] = ퟎ. ퟐퟓ [풍풐품 ( + )] ퟑ. ퟕ푫 푹풆ퟎ.ퟗ ퟑ. ퟕ푫 (ퟒ퐐/훑퐃훖) ퟎ.ퟗ

Typical fixed values of the Darcy-Weisbach friction factor f will be given for most problem calculations in this class. It is always desirable to double check that the assumed f value is appropriate. 28

Table 1.6. Pipe characteristics

Example 12: Consider pipe losses and find Q? (hf = 25ft)

(1) (2)

 = 12"= 1ft 1 = 6" 2 L = 10 00 ft L1 = 10 00 ft 2 f = 0.02 f1 = 0.02 2

In this case, the minor losses from expansion and entrance/exit are neglected. 푓퐿 푉2 푓퐿 4푄 2 1 8푓퐿 ℎ = = ( ) = [ ] 푄2 = 퐾푄2 푓 퐷 2푔 퐷 휋퐷2 2푔 푔휋2퐷5

푓 퐿 푉 2 푓 퐿 푉 2 1 1 1 2 2 2 2 2 ( ) 2 ℎ푓 = ℎ푓1 + ℎ푓2 = + = 퐾1푄 + 퐾2푄 = 퐾1 + 퐾2 푄 퐷1 2푔 퐷2 2푔 8푓 퐿 8 × 0.02 × 1,000푓푡. 푠2 푠2 퐾 = 2 2 = = 0.5 2 2 5 2 5 5 푔휋 퐷2 32.2푓푡 × 휋 × 1푓푡 푓푡

5 Note that 퐾1 = 2 퐾2 = 32 퐾2 = 16.1 since the diameter ratio is 2 other parameters being unchanged. 푠2 ℎ = (퐾 + 퐾 )푄2 = (16.1 + 0.5) 푄2 = 25푓푡 푓 1 2 푓푡5 25 푓푡5 푓푡3 Q = √푄2 = √ ft = 1.23 16.6 푠2 푠 2 2 ℎ푓1 = 퐾1푄 = 16.1 × 1.23 = 24.2푓푡 2 2 ℎ푓2 = 퐾2푄 = 0.5 × 1.23 = 0.76푓푡

Note that the head loss in the small pipe is 32 times the head losses in the larger pipe.

2 4푄 4 × 1.23 2 2 ( 2) 푉 휋퐷 ( 2) 1 = 1 = 휋 × 0.5 = 0.61푓푡 2푔 2푔 2 × 32.2

The velocity head in the small pipe is 24 = 16 times the head losses in the larger pipe. Can you sketch the HGL and the EGL on the above figure?

2.3. Three Reservoirs

Three reservoir problems involve the calculation of the head ℎ표 at the junction of the three pipes and the flow discharge in each pipe.

푓퐿 푉2 8푓퐿 ℎ = = ( ) Q5 = 퐾푄2 푓 퐷 2푔 푔휋2퐷5 2 ℎ1 − ℎ표 = 퐾1푄1 2 ℎ2 − ℎ표 = 퐾2푄2 2 ℎ0 − ℎ3 = 퐾3푄3

There are four steps to solve this type of problem

i) Calculate 퐾1, 퐾2, 퐾3 ii) Assume ℎ0, Calculate 푄1, 푄2, 푄3 – (it is useful to first assume ℎ0 = ℎ1) iii) Check continuity 푄3 = 푄2 ± 푄1 If ℎ표 < ℎ1, 푄1 + 푄2 = 푄3 , and if ℎ표 > ℎ1, 푄2 = 푄1 + 푄3 iv) Change ℎ0 and iterate until 푄3 = 푄2 ± 푄1

Example 13: Three-reservoir problem 31

i) Calculate 퐾1, 퐾2, 퐾3 8푓퐿 8 × 0.02 × 750 퐾 = = = 2.867 1 2 5 8 푔휋 퐷 32.2 × 휋2( )5 12 8푓퐿 8 × 0.02 × 1000 퐾 = = = 0.503 2 푔휋2퐷5 32.2 × 휋2(1)5 8푓퐿 8 × 0.02 × 500 퐾 = = = 1.911 3 2 5 8 푔휋 퐷 32.2 × 휋2( )5 12 ii) Assume ℎ0, Calculate 푄1, 푄2, 푄3 1/2 2 ℎ1−ℎ표 First, assume ℎ표 = ℎ1, ℎ1 − ℎ표 = 퐾1푄1 , 푄1 = ( ) = 0 퐾1 1 1 2 2 ℎ2 − ℎ표 120 − 110 3 푄2 = ( ) = ( ) = 4.46 ft /s 퐾2 0.503 1 1/2 2 ℎ표 − ℎ3 110 − 70 3 푄3 = ( ) = ( ) = 4.58 ft /s 퐾3 1.911 iii) Check continuity 푄3 = 푄2 ± 푄1, here 푄3 > 푄2, so we need to lower ℎ0 iv) Iterate until 푄3 = 푄2 ± 푄1

Second, assume ℎ표 = 109 ft 1 1/2 ℎ1−ℎ표 110−109 2 3 푄1 = ( ) = ( ) = 0.59 ft /s 퐾1 2.867 1 1 2 2 ℎ2 − ℎ표 120 − 109 3 푄2 = ( ) = ( ) = 4.68 ft /s 퐾2 0.503 1 1/2 2 ℎ표 − ℎ3 109 − 70 3 푄3 = ( ) = ( ) = 4.52 ft /s 퐾3 1.911 Check continuity, here 푄3 < 푄2 + 푄1, so we need to raise ℎ0

Third, Assume ℎ표 = 109.9 ft 1 1/2 ℎ1−ℎ표 110−109.9 2 3 푄1 = ( ) = ( ) = 0.19 ft /s 퐾1 2.867 1 1 3 ℎ2 − ℎ표 2 120 − 109.9 2 ft 푄2 = ( ) = ( ) = 4.48 퐾2 0.503 s 1 1 3 ℎ표 − ℎ3 2 109.9 − 70 2 ft 푄3 = ( ) = ( ) = 4.57 퐾3 1.911 s

Check continuity, in this case, we find 푄3~푄2, so we are close to the answer. This process can be repeated/programmed to obtain more accurate answers.

2.4. Pipe Networks (this is the most challenging section of Part I) Pipe networks are common in urban water distribution systems. As opposed to branching networks, the flow can reach all nodes even when some pipes are closed for maintenance.

20 50

K=5

35 K=2 K=1 K=1 70 35

K=4

100 30

푓퐿 푉2 8푓퐿 ℎ = = ( ) 푄2 = 퐾푄2 when minor losses are neglected 푓 퐷 2푔 푔휋2퐷5 푓퐿 42 1 퐾 = [( + 퐾 ) ] with the minor losses 퐾 . The next section will show how 퐷 푒 휋2퐷4 2푔 푒 to convert 퐾푒into an equivalent pipe length to be added to the actual pipe length.

Find flow discharge in each pipe and head at each node

Solve flow networks in nine steps: 1) Determine K values for each pipe 2) Assume a flow distribution (these steps are shown on the above figure) 3) Calculate ℎ푓 in each pipe 4) Sum 퐾푄2 in each loop 5) Sum |2퐾푄| in each loop ∑ 퐾푄2 6) Correction ∆푄 = for each loop ∑|2퐾푄| If ∆푄 > 0, add |∆푄| 퐶퐶푊 If ∆푄 < 0, add |∆푄| 퐶푊 7) Compute new discharge 푄new = 푄old ± ∆푄 (each pipe & each loop) 8) Repeat steps 3~7 until convergence 2 9) The head at each node is calculated from ℎ0 − ℎ1 = 퐾1푄1 flow from 0 to 1 33

Example 14: Solved pipe network – In this study example, the lower loop calculations are presented on the left-hand side, and the upper loop on the right side. Pay attention to all details and keep track of the signs. Use a calculator to double check the calculations. For instance, notice the double (two loops) discharge correction in the diagonal pipe. This iterative procedure converges rapidly.

Q2×K |2×Q×K| Q2×K |2×Q×K|

It is important for engineers to develop computer programming skills. In any programming language of preference, the following tips will be helpful in developing a robust program.

1) Head losses are calculated from the highest point down in the flow direction.

Because of possible flow reversal, ∆H = kQ2 is replaced with

2) Inside a loop, there is a sign for the flow direction in each

pipe. For instance, the sign from A to B is positive as it is

clockwise around the loop (SAB = +1, and SBA = -1). Now

consider that CA is clockwise (CW) and SCA = +1, but AC

is counter-clockwise (CCW), and SAC = -1.

3) The sign is valid for a single loop. When a pipe is

adjacent to two loops, the signs for the two loops are

opposite. For instance, SBC1 = +1 for CW flow in loop 1,

but SBC2 = -SBC1 = -1 for CCW flow in loop 2.

4) The sum of head losses is

calculated as follows

5) The correction ∆Q for this loop

(calculated CW) is in the opposite

(CCW) direction. For instance,

QAB new = QAB old - ∆Q and QAC new = QAC old + ∆Q. This can be programmed as

QXY new = QXY old – SXY ∆Q. Pipes adjacent to two loops will have two corrections. 35

2.5. Minor Losses in Pipes Table 1.7. Minor losses in pipes (this table is sufficient to solve the problems)

2.5.1. Additional information about minor pipe losses

푽ퟐ 푲 푸 ퟐ 푲 ퟒ ퟐ ퟏퟔ 푲 ퟏ ퟖ푲 ∆퐡 = 푲 = 풆 ( ) = 풆 ( ) 푸ퟐ = 풆 푸ퟐ, ∆퐡 = ( 풆 ) 푸ퟐ 풆 ퟐ품 ퟐ품 푨 ퟐ품 흅푫ퟐ ퟐ 품흅ퟐ 푫ퟒ 품흅ퟐ푫ퟒ

Entrance losses

Contraction/expansion losses

Expansion losses

Elbows

Can you search photos of flanged and threaded pipe fittings to tell the difference? 37

2.5.2. Equivalent pipe length

Minor losses in pipes are usually treated as an equivalent pipe length Le 푉2 푓퐿 푲풆푫 ∆h = 퐾 , 퐾 = 푒 and the equivalent pipe length is 푳 = 푒 2푔 푒 퐷 풆 풇 And this equivalent length is added to the physical pipe length in pipe networks.

Table 1.8. Equivalent pipe length for minor losses

2.5.3. Valves (you can also learn from your own web search)

Globe and needle valves

Gate and swing valves

Butterfly, ball and plug valves

Angle valves

Foot valve and strainer

Example 15: Find the equivalent length, and V, Q with entrance/exit losses

A hf = 25ft

(1) (2) (3) B V K 1.0 K = 1.0 e = e f = 0.02 L = 200ft D = 6'' 퐿 푉2 200푓푡 푉2 h = h + h + h = (퐾 + 푓 + 퐾 ) = (1 + 0.02 + 1) 푓 푓1 푓2 푓3 푒 퐷 푒 2푔 0.5푓푡 2푔 푉2 h = 10 = 25ft 푓 2푔 푉2 = 2.5푓푡 , V = √5 × 32.2 = 12.6ft/s 2푔 휋 퐷2 ft 휋 푄 = 푉퐴 = V = 12.6 (0.5ft)2 = 2.49ft3/s 4 s 4

Alternatively, it may be simpler to solve directly with the equivalent length. The equivalent lengths of the entrance and exit losses are both 퐾 퐷 1×0.5푓푡 퐿 = 푒 = = 25 푓푡 푒 푓 0.02 The total equivalent pipe length is 퐿푡 = 퐿 + 2퐿푒 = 200 + 50 = 250 ft 2 8푓퐿푡 8 × 0.02 × 250푓푡 푠 퐾 = = = 4.03 푡 2 5 5 푔휋 퐷 32.2푓푡 × 휋20.55푓푡 25 푄 = √ = 2.49ft3/s 퐾푡 The head loss at the pipe entrance and exit are equal to 10% of the total losses, or 2.5 ft each. The friction loss in the pipe is 20 ft. Can you plot the HGL and EGL?

2.6. Negative Pressure/Suction Example 16: With reference to Example 15, the central part of the pipe is lifted above the water level in the reservoir. How far can you raise the high point C before water turns to vapor? Assume that the reservoir is at sea level and T = 70F.

풇 = ퟎ. ퟎퟐ, 푳 = ퟐퟎퟎ′(퐟퐭), 푫 = ퟔ′′ = ퟎ. ퟓ′(퐟퐭)

Bernoulli Equation 푃 푉2 푃 푉 2 + 푍 + = 표 + 푍 + 표 훾 2푔 훾 표 2푔

푃 푉 2 푐 + 푍 + 푐 + ∆ℎ = 푍 훾 푐 2푔 푐 표 푃 푉 2 푐 = 푍 − 푍 − 푐 − ∆ℎ = −푥 − 퐻 훾 표 푐 2푔 푐 푐 푃 Water will turn to vapor when 푃 ≤ 푃 , 표푟 푥 = 푋 , when 푣푟푒푙 = −푋 − 퐻 푐 푣푟푒푙 푚푎푥 훾 푚푎푥 푐 푃 Therefore, 푋 = − 푣푟푒푙 − 퐻 푚푎푥 훾 푐

◦ ◦ 2 From Example 6, at sea level and T = 70 F, 푃푣 = −2,065 lb/ft , and (−2,065) 푋 = − − 15 = 33.1 ft − 15 ft = 18.1 ft 푚푎푥 62.4

Can you repeat the calculations at an altitude of 10,000 ft with T◦ = 40 ◦F? (Answer: 푋푚푎푥 = 8.0 ft) – note the large difference with altitude. 41

3. Hydrodynamic Forces (Week 3) We apply force and momentum concepts to hydraulic problems!

The direction of the force vectors is towards the control volume

3.1. Momentum Force on a Plate (constant atmospheric pressure) Flow diagram in a horizontal plane or planview

Ɵ V = V A = A 2 1 2 1

= V2 A 1 푉⃗ = 푉푥푖 +푉푦푗 = V2 V = 푉cos휃푖 + 푉sin휃푗 1

∆풎풂풔풔 흆 = ∆풗풐풍풖풎풆

∆풎풂풔풔 = 흆∆풗풐풍풖풎풆 = 흆푨∆푺 = 흆푨푽∆풕 = 흆푸∆풕

Force that the plate applies onto water to change its linear momentum

∆풎푽⃗⃗ ퟏ + ∑ 푭⃗⃗ ∆풕 = ∆풎푽⃗⃗ ퟐ ∆풎 흆푸∆풕 푭⃗⃗ = (푽⃗⃗ − 푽⃗⃗ ) = (푽⃗⃗ − 푽⃗⃗ ) = 흆푸(푽⃗⃗ − 푽⃗⃗ ) ∆풕 ퟐ ퟏ ∆풕 ퟐ ퟏ ퟐ ퟏ 푭⃗⃗ = 흆푸[(푽풄풐풔휽풊 + 푽풔풊풏휽풋 ) − (푽풊 + ퟎ풋 )] 푭⃗⃗ = 흆푸푽[(풄풐풔휽 − ퟏ) × 풊 + 풔풊풏휽 ⃗⃗풋 )] = 푭풙풊 + 푭풚풋 푭풙 = 흆푸푽(풄풐풔휽 − ퟏ), 푭풚 = 흆푸푽(풔풊풏휽) Force diagram (Free Body Diagram)

ρQV Ɵ

퐹 = 휌푄푉(1-cos휃) ρQV 푥

퐹푦 = 휌푄푉(sin휃) 42

Example 17: Force and power on a moving plate

A jet is impacting the inclined plate below, neglect the weight of water and find the following per unit width 3 when  = 60, Q = 2 ft /s, V = V0 = 85 ft/s, and A = Q/V:

(1) the force F required to hold the plate in place (u = 0);

(2) define the expression for the work done on the plate per unit time when the plate is moving to the right at 푢푝 = 10 ft/s; and

(3) at what constant plate velocity up max (푢푝 to the right) will the work done be maximum?

(1) The force tangential to the plate is zero because there is no friction on the plate. The force normal to the plate is now coming at an angle with the horizontal.

퐹 = 휌퐴푉표푉표 sin 60° = 1.94 × 2 × 85 sin60° = 285 lb

퐹퐻 = 퐹 sin 60° = 285 sin 60° = 247 lb

(2) Note here that when the plate is moving to the right at 푢푝, not all the fluid will reach the plate because it accumulates in front of the plate. Therefore, the flow rate entering the control volume moving with the plate is such that the area is the same, but the velocities both entering and leaving the control volume are

reduced to 푉 − 푢푝;

푊표푟푘 2 푃표푤푒푟 P = = 퐹 . 푢 = 휌퐴(푉 − 푢 ) sin260° 푢 푡𝑖푚푒 퐻 푝 푝 푝

2 = 1.94 × × 752 × 0.8662 × 10 = 1,925 lb ∙ ft/s 85

2 2 2 2 2 3 (3) P = ρ퐴(푉 − 푢푝) sin 60° 푢푝 = ρ퐴sin 60°(푉 푢푝 − 2푉푢푝 + 푢푝 )

휕P 2 2 2 = ρ퐴sin 60°(푉 − 4푉푢푝 + 3푢푝 ) = 0 휕푢푝

(푉 − 푢푝)(푉 − 3푢푝) = 0

푢푝 푚𝑖푛 = 푉 , because P = 0, and

푉 4 3 2 푢 = , and Pmax = ρ퐴V sin 60° = 3,115 lb-ft/s when 푢 = 28.3 ft/s 푝 푚푎푥 3 27 푝 43

3.2. Hydrodynamic Force on a Pipe Bend (variable pipe pressure)

For flows in pipes, we usually proceed as follows: (1) solve continuity and calculate areas, flow velocities, discharges and hydrodynamic forces 휌QV for each pipe; (2) apply Bernoulli to determine the pressure and pressure force pA in each pipe; and (3) sketch a free-body diagram with all forces (hydrostatic pA and hydrodynamic 휌QV) to determine the anchor force R required to hold the system in place. As a general case, the flow diagram below is in elevation such that it also includes the weight component of the fluid W inside the control volume.

Example 18: The horizontal pipe below (z2 = z1) discharges water into the atmosphere on the right side ( p2

= 0), what is the pressure p1?

푃 푉 2 푃 푉 2 1 + 푍 + 1 = 2 + 푍 + 2 훾 1 2푔 훾 2 2푔 푉 2 − 푉 2 푃 = 훾 ( 2 1 ) 1 2푔 휋(2)2 푄 = 푉 퐴 = 10 × = 31.4 ft3/s 1 1 1 4 휋(1)2 퐴 = =0.785 ft2 2 4

402 − 102 lb 푃 = 62.4 ( ) = 1,453 = 1,453 psf 1 2 × 32.2 ft2 44

Example 19: What is the force required to hold this pipe bend in place?

Note that the direction of the force vectors is towards the control volume

∑ 퐹푥 = 푅푥 + (푝1퐴1 + 휌푄푉1)푐표푠 70° − 휌푄푉2=0

푅푥 = 휌푄푉2 − (푝1퐴1 + 휌푄푉1)푐표푠 70° 1.94 푠푙푢푔 31.4 푓푡3 40 푓푡 1,453 푙푏 푅 = × × − ( × 3.14 푓푡2) 푐표푠 70° 푥 푓푡3 푠 푠 푓푡2 1.94 푠푙푢푔 31.4 푓푡3 10 푓푡 − × × 푐표푠 70° = 668 lb 푓푡3 푠 푠

∑ 퐹 = (푃 퐴 + 휌푄푉 )푠𝑖푛 70° − 푅 = 0 – note that Ry is opposite to the y direction 푦푥 1 1 1 푦 1,453 푙푏 1.94 푠푙푢푔 31.4 푓푡3 10 푓푡 푅 = ( × 3.14 푓푡2 + × × ) 푠𝑖푛 70° 푦 푓푡2 푓푡3 푠 푠 = 4,860 lb

2 2 2 2 |퐹 | = √(푅푥 + 푅푦 ) = √(668 + 4,860 ) = 4,906 푙푏

Example 20: From the pipe junction in a horizontal plane shown below, what is the velocity of section 3 if the diameter is 0.15m? If the pressure head at (2) is 10 m, determine the pressure head at (3) and find the net force required to hold this junction when  = 30.

(1) 휋푑 2 휋(0.1)2 푑 = 0.1 m, 퐴 = 1 = = 0.00785 푚2 1 1 4 4 3 푄1 = 퐴1푉1 = 0.00785 × 4 = 0.0314 m /s 푉 2 푉 2 62 42 푃 = 푃 + 휌 ( 2 − 1 ) = 98,100 + 1000 × ( − ) = 108,100 Pa 1 2 2 2 2 2 푚3 푃 퐴 = 848.6 푁, 휌푄 푉 = 1000 × 0.0314 × 4 = 125.6 N 1 1 1 1 푠 푃1퐴1 + 휌푄푉1 = 974.2 푁 (2) 휋푑 2 휋(0.12)2 푑 = 0.12 m, 퐴 = 2 = = 0.0113 푚2 2 2 4 4 3 푄2 = 퐴2푉2 = 0.0113 × 6 = 0.0678 m /s 푃 2 = 10, 푃 = 98,100 Pa 훾 2 푚3 푃 퐴 = 1,108 푁, 휌푄 푉 = 1000 × 0.0678 × 6 = 406.8 N 2 2 2 2 푠 푃2퐴2 + 휌푄2푉2 = 1,515 푁 (3) 휋푑 2 휋(0.15)2 푑 = 0.15 m, 퐴 = 2 = = 0.01767 푚2 3 3 4 4 3 푄3 = 푄1 + 푄2 = 0.0314 + 0.0678 = 0.0992 = 0.1 m /s 푄3 0.1 푉3 = = = 5.66 m/s 퐴3 0.01767 푉 2 푉 2 62 5.662 푃 = 푃 + 휌 ( 2 − 3 ) = 98,100 + 1000 × ( − ) 3 2 2 2 2 2 = 100,082 Pa 푚3 푃 퐴 = 1,768 푁, 휌푄 푉 = 1000 × 0.1 × 5.66 = 566 N 3 3 3 3 푠 푃3퐴3 + 휌푄3푉3 = 2,334 푁 푅푥 = −1,515 + 2,334 cos 30° = 506 N (to the right) 푅푦 = −974.2 + 2,334.5 sin 30° = 194 N up Ans: 푽ퟑ = ퟓ. ퟔퟔ m/s , 푷ퟑ = ퟏퟎퟎ 퐤퐏퐚 , Rx = 506 N right , Ry = 194 N up 46

4. Flow Meters (Week 4) Applications to flow meters – are you ready for the first test?

A flow meter is a device to determine the flow discharge in a pipe.

4.1. Manometer Manometers typically measure the pressure difference in pipe contractions.

2 V 1 2g 2 E .G .L V 2 2g

H .G .L

P 2  P1 D 2 = 0.5ft 

Water 4ft

3ft

D 1 = 1ft A B 4ft 3ft Mercury, Hg Da tum

Example 21: Find the discharge Q in this pipe assuming no friction loss?

푝 푝 푝 훾 푝 퐴 = 1 + 3푓푡 = 2 + 4푓푡 + 퐻퐺 (4 − 3)푓푡 = 퐵 훾 훾 훾 훾 훾 푝 푝 1 − 2 = 13.6푓푡 + 1푓푡 = 14.6푓푡 훾 훾 푝 푝 푝 푝 ∆H = (푧 + 1) − (푧 + 2) = (6푓푡 + 1) − (8푓푡 + 2) = 14.6 − 2 = 12.6푓푡 1 훾 2 훾 훾 훾 Can you demonstrate here that the difference in piezometric head ∆H = 푅′(퐺 − 1) 훾 where 푅′ is the manometer reading (4-3) ft and 퐺 = 퐻퐺 = 13.6? 훾 2 2 2 4 2 2 푉2 푉1 푉2 퐷2 8푄 1 1 8푄 1 ∆H = ( − ) = [1 − 4] = 2 [ 4 − 4] = 2 [16 − 1] 4 2푔 2푔 2푔 퐷1 푔휋 퐷2 퐷1 푔휋 푓푡

∆퐻푔휋2 12.6푓푡 32.2푓푡휋2 Q = √푄2 = √ = √ 푓푡4 = 5.77푓푡3/푠 8 × 15 120 푠2

Can you find the velocities in the two pipes? (Answer 푉2= 29.4 ft/s, 푉1= ?) 47

4.2. Venturi Meter

The analysis of a Venturi is very similar. We demonstrate here that the flow results are independent of the angle of the pipe with the horizontal.

퐴 퐷 2 2 = ( 2) 퐴1 퐷1

2 2 4 푉 퐷 푉 퐷 2 = ( 1) , ( 2) = ( 1) 푉1 퐷2 푉1 퐷2

푃퐴 = 푃푐 , 푃퐵 = 푃퐴 − 훾푅

휌푚 푃 = 푃 − 훾푅 퐷 푐 휌

B D 푃 푃 ∆푃 휌 Or 퐵 − 퐷 = = ( 푚 − 1) 푅 A C 훾 훾 훾 휌

Note here that at the same elevation D (or B), the term ∆푃/훾 is the difference in piezometric head, which is independent of the elevations z1 and z2. ∆푃 ∆푉2 We find the flow velocity from the measured pressure difference = 훾 2푔

휌 푉 2 푉 2 푉 2 푉 2 푉 2 퐷 4 ( 푚 − 1) 푅 = 2 − 1 = 1 [( 2) − 1] = 1 [( 1) − 1] 휌 2푔 2푔 2푔 푉1 2푔 퐷2 휌 1 푉 = √푉 2 = 2푔푅 ( 푚 − 1) 1 1 √ 휌 퐷 4 [( 1) − 1] 퐷2 휋퐷 2 휌 1 푎푛푑 푄 = 퐴 푉 = 1 2푔푅 ( 푚 − 1) 1 1 4 √ 휌 퐷 4 [( 1) − 1] 퐷2 At very high flows, the pressure may become low enough to cause the formation of vapor bubbles and cavitation.

4.3. Flow Nozzle

A flow nozzle is a very short contraction in a pipe. The difference in piezometric head simply becomes ∆H = 푅′(퐺 − 1)where 푅′ is the manometer reading and G is the ratio of the densities of the two fluids in the manometer. A discharge coefficient C is pre-calibrated for the instrument as shown below, such that the discharge relationship becomes

휌푚 푄 = 퐶 퐴 √2푔푅 ( − 1) 2 휌

Example 22: Flow in a nozzle Let’s determine the flow discharge in a 6” line with a 4” flow nozzle. The mercury- water manometer measures a gage difference of 10” at a water temperature of 60˚F. 2 2 From this information, the area 퐴2 = 0.25 휋0.333 = 0.0873 ft , the ratio of areas is 2 -5 2 퐴2/퐴1 = (4/6) = 0.444, the dynamic viscosity υ = 1.22 × 10 ft /s, R= 10/12ft = 0.833 ft, and 휌푚/휌 = 13.6. Assuming C = 1.06 on the right hand side of the above figure gives the following discharge

푄 = 1.06 × 0.0873 √64.4 × 0.833(13.6 − 1) = 2.4 ft3/s,

2 To check the coefficient, V1 = 4Q/ 휋0.5 = 12.2 ft/s and

Re = VD/ υ =12.2× 0.5/1.22 × 10-5 ft2/s = 5 ×10 5, and C = 1.06 and therefore, the value of C was correctly estimated.

4.4. Various Flow Meters

4.4.1. Venturi flow meter

4.4.2. Nozzle flow meter

4.4.3. Ultrasonic flow meter

4.4.4. Coriolis flow meter

4.4.5. Magnetic flowmeter

4.4.6. Laser Doppler Velocimeter

4.4.7. Turbine flow meter

4.4.8. Propeller flow meter

4.4.9. Price current meter

4.4.10. Acoustic Doppler Current Profiler (ADCP)