Fluid Statics, Hydrodynamics and Hydraulic Machines · Fluid Statics, Hydrodynamics and Hydraulic...

2/8/2010Fluids Statics, Hydrodynamics and Hydraulic Machines; © B. Rauf1

Fluid Statics,

Hydrodynamics and

Hydraulic Machines

Bobby Rauf ©


Fluid Facts

1) Liquids and gases can both be categorized as fluids.

2) Liquid fluids are assumed be incompressible.

3) Gaseous fluids are compressible

4) Fluid pressure, defined as force per unit area, is measured in

pounds per square inch, psi, or pounds per square foot, psf, in

US unit system. Pressure is measured in Pascals, Pa, or

kilopascals, kPa, in the SI (metric) System.

5) Standard Atmospheric Pressure: 1.00 Atm; 14.7 psia; 407.1

inches of water gage; 29.92 in Hg; 760 mm Hg; 1.013 x 105

Pa; 101.3 kPa.

6) Pabs = Pgage + Patm


Fluid Facts

7) Fluid Density = = mass / volume = 1 / Specific Volume

8) Fluid Density, , in US units: lbm / ft3

9) Fluid Density, , in SI units: Kg / m3

10) Specific Gravity of Liquid = SGliquid = liquid / water

11) Specific Gravity of Gas = SGgas = gas / air

12) Specific Weight of Fluid = = g . ; in lbf / ft3 or N / m3

13) STP, Standard Temperature Pressure: 0 C ( 32 F); 101.3

kPa, or 1 atm, or 1 bar (14.7 psi).

14) rh = Hydraulic Radius = Area in Flow / Wetted Perimeter


FLUID STATICS - Manometer

p1

p2

h

p1 + .g.h = p2

If p1 represents atm. pressure, then:

p2 - p1 = pgage, 2 = .g.h


Manometer & Multiple Fluid Densities

p0

p

h1

p0 + 1.g. h1 + 2.g. h2 = p

If p0 represents atm. pressure, then:

p - p0 = pgage = 1.g. h1 + 2.g. h2

h2


Problem 1 – Inverted Fluid

A glass filled with a fluid is inverted. The bottom is open. See figure 1. What is the pressure at the closed end A?

The pressure, PA, at point A , plus the pressure (.g.h) exerted by the water equals the atmospheric pressure, Patm, outside the glass.


Problem 1 – Figure 1

A

h

D

.g.h = Wt. Of

the Water

Patm

PAPA +.g.h = Patm

PA = Patm - .g.h


HYDRODYNAMICS - Problem 2

Consider the holding tank shown in figure 2. Calculate the velocity of the water exiting to the atmosphere.


Problem 2; Figure 2

v2m

P1 = Patm

1 P2 = Patm

2


Problem 2, contd.

Apply Bernoulli‟s equation between the free

surface (point 1) and the exit (point 2)

g.z1 + v12/2 + p1/ = g.z2 + v2

2/2 + p2/ {SI}

p1 = p2 (both are, or are at, atmospheric

pressure)

v1 = 0 (the free surface is stationary)

g.z1 + (0) = g.z2 + v22/2 + (p2/ - p1/ )

g.z1 = g.z2 + v22/2


Problem 2, contd.

g.z1 = g.z2 + v22/2

v2 = 2.g . (z1 - z2)

= (2).(9.81).(2)

= 6.3 m/s


Problem 3 –

A pressure tank contains a fluid with weight density 81.5 lbf/ft3 ( ). The pressure in the air space is 100 psia. Fluid exits to the atmosphere from the bottom of the tank. What is the exit velocity, v?


Problem 3

p

v

10 ft

v1, z1, p

v, z2, patm


Problem 3 –

Apply Bernoulli‟s equation, for energy conservation, between the surface and the exit at the lower right end:

p/ + g.z1+v12/2 = patm/ + g.z2 + v2/2 {SI}

v1=0 (at the free surface),

z2=0 (at the exit)

p/ + g.z1= patm/ + v2/2


Problem 3, contd.

p/ + g.z1= patm/ + v2/2

v = 2.g.{(p - patm) /.g+z1)}

Since p.g = = 81.5 lbf/ft3,

v = (2).(32.2ft/sec2)[(100 psi-14.7psi).(144in2/ft2)/ 81.5lbf/ft3+10ft]

v = 101.7 ft/sec

Problem 3, contd.


Problem 3 - i

Buoyancy

A hot air balloon is to used to transport a welder across a ravine.

The balloon and the payload, including the welder and other

accessories, have a mass of 200 kg. The ambient pressure and

temperature are 102 kPa and 15.6 C. Assume wind to be calm,

aloft and at ground elevation.

(a) If density of ambient air is 1.2 kg/m3 and if air in the balloon

can be heated to a density of 1.0 kg/m3, determine the volume

and the radius of the balloon to achieve a lift off.

(b) To what temperature must the air in the balloon be heated in

order to achieve the density to 1.0 kg/m3?

Problem 3 – I Buoyancy


Problem 3 – I, contd.Buoyancy Solution 3 – i

Hot Air = 1.0 kg/m3

Amb. Air = 1.2 kg/m3

PAmb = 102 kPa = 102,000 Pa

Total Load, including the Payload & the Baloon = 200 Kg

VHA = Volume of Hot Air in the Baloon

= VAA = Volume of Ambient Air Displaced

Wpayload + Wballoon = 200 kg x g

Whot air in balloon = 1.0 x g x VHA = 1.0 x g x VAA

Wamb. air displaced = 1.2 x g x VAA = 1.2 x g x VHA



(Wpayload + Wballoon) + Whot air in balloon = Wamb. air displaced

(200 kg x g) + 1.0 x g x VHA = 1.2 x g x VHA

200 kg + 1.0 kg/m3 x VHA = 1.2 kg/m3 x VHA

VHA = VAA = 1000 m3

Volume of a Spherical Baloon = V = 4/3 x 3.14 x rb3

Radius of the Baloon = rb = {(V x 3/4) x (1/3.14)} 1/3

= 6.204 m



b) Calculate the temperature of the hot air in the balloon

for a density of 1.0 kg / m3 = THA

For ideal gases, p = x R x THA

THA = p / ( R x ) = 355.8 K

Or , THA = 82.8 C

Note: Max. Op. Temp for Nylon/Nymex Balloon: 120 C



Ancillary Note:

Verification of the ambient air density assumption of 1.2

kg/m3

For T Amb = 15.6 C

= 288.6 K

Ambient Air = p / (R x TAmb)

= 102,000 / (286 x 288.6)

= 1.2 kg/m3


Problem 4

Calculate the overturning moment, per unit width, acting on the

dam shown below:

h = 8 ft &

hav = ½(0 + 8)

hav = 8/2 ft = 4 ft

hr =

2/3h

1/3h

F

Pav = P = . g . hav = . hav

= Specific weight of water = 62.4 lbf/ft3

ToeHeel


Problem 4

Typical Hydroelectric Dam Construction :

http://en.wikipedia.org/wiki/File:Hydroelectric_dam.svg


Problem 4, contd.

F = Pav . A = . g . hav . A = . hav „ A

F = (62.4 lbf/ft3 ) . (4) . ( 8 x 1)

F = 1996.8 lbf

F per foot = 1996.8 lbf /ft

M = Moment (per foot) = F per foot x 1/3h

M = Moment (per foot) = 1996.8 lbf/ft x (8/3 ft)

M = 5325 ft – lbf/ft

Ancillary questions: What force is needed to prevent the

dam from sliding and what constitutes it?


Problem 4 - i

Calculate the horizontal and vertical components of the resultant force on a 2

meter wide inclined surface in the reservoir shown below. Note that this

reservoir has just experienced a massive oil spill. This oil has a density of 900

kg/m3

30

h1 = 1 m

hi = 1 m

ho = 0.3 m

Oil Slick

Waterh2 = 2 m

30

Problem 4 – i, contd.

Given:

Density of Oil, o = 900 kg/m3

Density of Water, w = 1000 kg/m3

Angle of the Inclined Plane, = 30

Height of the Inclined Surface, hi = 1 m

Depth of Water Above Inclined Surface, h1 = 1 m

Total Height or Depth of Water, h2 = h1 + hi = 2.0 m

Depth of the Oil Slick, ho = 0.3 m

Avg. Depth of Water Above Incl. Surface = hw

= 1/2 x (h2 + h1) = 1.5 m



Length of the Incl. Surface = hi/Sin(30) = 2.0 m

Width of the Inclined Surface = 2 m (Given)

Area of the Inclined Surface Section = 2m x 2m = 4.0 m2

Pressure Contributed by the Layer of Oil

= o x g x ho = 2,649 N/m2 or Pa

Average Pressure Contributed by the Water = w x g x hw

= 14,715 N/m2 or Pa

Total Average Pressure in the Water at Inclined Surface:

17,364 N/m2 or Pa



Problem 4 – I, contd.

Resultant Force Component Analysis:

RRx

Ry


Perp. Resultant Force on any Point Along the Incl.

Surface

= Total Avg. Press x Area of Incl. Surface =

69,455 N

Ry, the y - Comp. of the Resultant Force = R x Cos ()

= 60,150 N

Rx, the x - Comp. of the Resultant Force = R x Sin ()

= 34,727 N



Problem 5

Figures 1 and 2, below, depict a pitot tube arrangement devised to

measure the flow rate of water. Calculate the height of mercury

column that would be sustained by the velocity pressure of the

water.

Water

v = 10 m/s

Mercury

http://en.wikipedia.org/wiki/File:Faa_pitot_static_system.JPG


Problem 5, contd.

2 1v2 = v = 15 m/s

v1 = 0

zHg = zwater = ?

water = Density of water is 1000 kg / m3

mercury = The density of mercury is 13,567 kg/m3

z1 = z2 and

z1 - z2 = 0


Problem 5, contd.

Apply Bernoulli‟s energy conservation equation

at points 1 and 2, for fluid dynamics portion:

p1/ + g.z1+v12/2 = p2/ + g.z2 + v2

2/2 {SI}

Since z1 - z2 = 0, v1 = 0 and

= water = 1000 kg/m3

p1/ = p2/ + v22/2

p1 - p2 = (v22/2) . =(225/2). 1000 = 112,500 pa


Problem 5, contd.

Apply fluid statics principle to the pitot tube

segment:

p2 + mercury . g . zHg = p1 + water . g . zwater

p1 - p2 = mercury . g . zHg - water . g . zwater

zHg = zwater = (p1 - p2) / g / (mercury - water )

zHg = 112,500 / (9.81) / (13,567 – 1000) = 0.913 m


Problem 6

An irrigation water channel, or canal, is to be modified. The

current and future specifications are listed below:

Change

Orig. Shape As % of Orig. New Shape

Side Elevation or Vertical Component of the Side = 10 ft 10 ft

Side Elevation Angle = 1.57 0.79

Radians,

Calct’d.

Horizontal Component of the Side Wall = 0 ft 10.00 ft

Length of Sides = 10.00 ft 14.14 ft, Calct’d

Length or Reach of the Water Channel, on Per Unit Basis: 1 30% 0.70

Fall or Drop on Per Unit Basis: 1.0 1.0

Slope, "S" (Given) = 0.05 0.0714 Calct’d

Basin Width = 40 ft 20 ft

Roughness, "n" 0.025 0.025


Problem 6, contd.

Original or Existing

Design:

Proposed or Final

Design:

40 ft

20 ft

10 ft

10 ft

1

1


Problem 6, contd.

Determine the following water channel performance

parameters based on the data provided above:

a) The slope of the modified water channel is:

Slope “S” = Rise / Run

Let Rise = R , and let Run = Length = L and Lo = Orig. L

Therefore, So = Original Slope = Ro / Lo = 0.05 {Given}, & Ro= 0.05

Lo

Since the total Rise or Fall stays constant, Ro = Rf = R = 0.05 Lo

Also, the length of the modified channel = Lf = (0.7) Lo

New Slope or Slope of the Modified Channel = R / Lf

= R / (0.7) Lo = 0.05 Lo / (0.7) Lo = 0.0714


Problem 6, contd.

b) The volumetric flow rate in the original, or existing, design is :

Manning’s Equation :

Volumetric Flow Rate = Q = (1.486/n).(A).(R2/3).(S1/2)

Where,n = Orig. Roughness = Modified Channel Roughness = 0.025 {Given}

Ao = Original Area in Flow = 40 ft x 10 ft = 400 sq. ft

Wetted Perimeter = 10 + 40 + 10 = 60

Ro = Original Hydraulic Radius = Area in Flow / Wetted Perimeter= 400 / 60 = 6.7

So = 0.05, {Given}

Therefore,

Volumetric Flow Rateo = Qo

= (1.486/0.025).(400).(6.72/3).(0.051/2)

= 18,832 cu-ft./sec


Problem 6, contd.

c) The velocity of water in the original, or existing, design is :

Volumetric Flow Rate = Q = V.A

Where,

V = Fluid Velocity, in ft/sec

A = Area of cross-section or Area in Flow, in sq. ft.

Therefore,

Velocity of water in the original, or existing, design

is = Q / A = 18,832 cu-ft./sec / 400 sq. ft.

= 47.08 ft. / sec.


Problem 6, contd.

d) Which of the following factors influence the volumetric flow

rate, Q, in Cu-ft/Sec?

__ Intensity of storm, distance or total “run” of the water

channel, size of the basin.

__ Area, energy, type of terrain..

__ Area, type of terrain, slope, wetted perimeter..

__ Velocity, amount of rain, size of the basin, shape of

water channel.


Problem 6, contd.e) The volumetric flow rate in the proposed or final design is :

Manning‟s Equation :

Volumetric Flow Rate = Q = (1.486/n).(A).(R2/3).(S1/2)

Where,

n = Orig. Roughness = Modified Channel Roughness = 0.025 {Given}

Af = Modified or Prop. Area in Flow = (20 ft x 10 ft) + 2.(1/2 x 10 ft x 10 ft)

= 300 sq. ft

Wetted Perimeter, modified = 2.(102 + 102 ) + 20 = 48.28 ft

Rf = Original Hydraulic Radius = Area in Flow / Wetted Perimeter

= 300 / 48.28 = 6.2

Sf = 0.0714, as calculated in part (a)

Therefore,

Final or proposed Volumetric Flow Rate = Qf

= (1.486/0.025).(300).(6.22/3).(0.07141/2)

= 16,117 cu-ft./sec


Problem 6, contd.

f) The velocity of water in the modified or final design is :

Volumetric Flow Rate = Q = V.A

Where,


A = Area of cross-section or Area in Flow, in sq. ft.

Therefore,

Velocity of water in the final design

is = Q / A = 16,117 cu-ft./sec / 300 sq. ft.

= 53.72 ft. / sec.


Fluid Energy Flow Analysis Concepts;

1) Equivalent Diameter, De, or Hydraulic

Diameter, dh :

The Equivalent Diameter, De, or Hydraulic

Diameter, dh are used to calculate the dimensionless

Reynolds Number, Re, to determine if a flow is

turbulent or laminar.

De = dh = 4. rh = 4 x Hydraulic Radius


Fluid Energy Flow Analysis Concepts;

2) Reynolds Number, Re:

Reynolds Number, Re, is a dimensionless number

representing the ratio between inertial forces and

viscous forces in the fluid

Re = De. v /

Where,

v = Fluid Velocity

= Kinematic Viscosity


Fluid Flow Energy Analysis :

Vmax

V=0

Parabolic Velocity

Distribution

3) Laminar and Turbulent Velocity Distributions in Circular

Pipes:

Laminar Flow, where,

Re < 2300 (2100)

Turbulent Flow,

where, Re > 4000

(10,000)Transient Flow

for 2300 < Re < 4000


Fluid Flow Energy Analysis, Contd. :

Energy Loss, or Head Loss due to friction, in turbulent flow

situations is calculated using the Darcy Equation Method,

based on the following formula:

Hf = (f . L . V2 ) / (2 . D. g)

Where,

f = Fluid Friction Factor read from Moody‟s Chart, using calculated

Reynolds Number, Re. See the chart on the next page.

L = Length of pipe, in ft.


D = Diameter of circular pipe or hydraulic diameter, of a non-circular

vessel

g = 32.2 ft/sec2


Problem 7, contd. – Moody’s Graph



Notes :

Darcy’s Eq. can be applied for turbulent and laminar flow

conditions.

Challenge to accurately interpret the value of “f” from the

Moody’s Chart makes Darcy’s method for calculation of frictional

head loss less favorable. It is because of this reason that the

Hazen-Williams equation and method are preferred choice

under common situations, e.g. for Civil Engineering

applications.



Energy Loss, or Head Loss calculation, in most Civil

Engineering analysis is conducted using the Hazen-Williams

Method, or equation. The Hazen-Williams equation is as

follows:

HL = (4.73 . Q1.85 L ) / (C1.85 . D4.87)

Where,Q = Volumetric Flow Rate in cu-ft./sec

L = Length of pipe in ft.

C = Hazen-Williams Roughness Coefficient, available for various pipes

constructed from different types of material. See the chart in the next page .

D = Dia. of circular pipe or hydraulic diameter, of a non-circular vessel, in ft.

Alternate Form of H-W Equation, with Q in gpm and D in inches:

HL = (10.44 . Q1.85 L ) / (C1.85 . D4.87)



MaterialHazen-Williams Coefficient

- c -

Asbestos Cement 140

Brass 130 - 140

Brick sewer 90 - 100

Cast-Iron - new unlined (CIP) 130

Cast-Iron 10 years old 107 – 113; Mean Value = 110

Cast-Iron 20 years old 89 – 100

Cast-Iron 30 years old 75 - 90

Cast-Iron 40 years old 64-83; Mean Value = 73.5

Cast-Iron, asphalt coated 100

Cast-Iron, cement lined 140

Cast-Iron, bituminous lined 140

Cast-Iron, wrought plain 100

Concrete 100 - 140

Copper or Brass 130 - 140

Corrugated Metal 60

Ductile Iron Pipe (DIP) 140

Fiber 140

Fiber Glass Pipe - FRP 150

Galvanized iron 120

Glass 130

Lead 130 - 140

Metal Pipes - Very to extremely smooth 130 - 140


Problem 7

A pump delivers water from a reservoir to a closed conduit

system and an elevated water tank as shown in the figure below.

The elevations are MSL, or with respect to the Mean Sea Level.

The pump operates, continuously, at 1,000 gpm, gallons per

minute. The pump efficiency is 65 % and the efficiency of the

pump motor is 90%.

The water level in the elevated water tank is varies between a 300

ft and 330 ft.

The user demand in the closed conduit varies between 1,500 gpm

and 500 gpm.

The closed conduit system consists of 20 year old 10 “ cast iron

pipe.


Problem 7, contd.

Elevated Water Tank, EWT

330‟ MSL

300‟ MSL

Pump

Reservoir

User Demand

System

200‟ MSL

110‟ MSL

100‟ MSL

650‟

1,200‟

2,100‟

P2 = Patm

Z2=330 ft

V2= 0

P1 = Patm

Z1=110 ft

V1= 0

2

1A

B

C

D

PD

ZD=330 ft

VD= 2.04 ft/sec

@ 500 gpm;

Conversion

factor: 0.13368

cu-ft/gal

Note: This slide is ref. to as

38 in the Audio.


Problem 7, contd.

a) Identify the most appropriate value for the 20 year old, 10”

diameter, cast iron pipe in the system.

From the given Hazen-William Roughness Coefficient table:

C, for 20 year old, 10” diameter, cast iron pipe is 100.

Alternate, interpolation method for C calculation:

C, for 40 year old, cast iron pipe is, approx. 80

C, for 10 year old, cast iron pipe is, approx. 110

C / Age = (110 – 80) / (40 – 10) = 1/1 = 1

C, for 20 year old, cast iron pipe is,

approx. = 80 + (1/1) . 20 = 100


Problem 7, contd.

b) Using Hazen-Williams Equation for determination of Friction

Head Loss, calculate the maximum pump horsepower

developed. Assume a C, Hazen-William Roughness

Coefficient, value of 105.

The maximum Req. pump horsepower would be required under the following

conditions:

• The pump is operating at a constant flow rate of 1,000 gpm.

• The flow to the user water distribution would be at a minimum flow rate of

500 gpm.

• The flow rate from point “B” to “C” is 500 gpm.

• The water tower is full or at maximum level, producing maximum head of

330 ft., for the pump to work against and sustain.


Problem 7, contd.

• Power Capacity of the Motor for Maximum Required Pump Head, in hp :

hp = ( . Q . H ) / ( 550 . )

Where,

= Sp. Weight of Water, in lbf/cu-ft. = 62.4 lbf/Cu-ft

Q = Flow Rate in Cu-ft. / sec.: 1000 gpm => 2.228 Cu-ft. / sec.

= Hydraulic Pump Efficiency = 0.65 {Given}

H = Pump Head in ft. = ?

- Calculate maximum H


Problem 7, contd.

Apply Bernoulli‟s equation between the free surface (point 1)

and the exit (point 2)

z1 + v12/2gc + p1/ + H = z2 + v2

2/2gc + p2/ + HL

{US, standard version divided by g }

Note: • p2/ - p1/ = 0, since p1 & p2, both, are at atmospheric pressure

• & v1 = v2 = 0; the free surface is stationary

• The Pump Head contrib. to cover the Vel. Hd.

developed at Pt. D is negligible, at 0.06 ft, and therefore, left out

of the Eq.

• The Pump Head contrib. to cover the Potential Hd. developed at

Pt. D, is within the 330 ft. Elev. Tower Hd., and therefore out of the

Eq.

H = z2 - z1 + HL , or

H = z2 - z1 + (HL-AB + HL-BC + HL-BD )

H = 330 – 110 + (HL-AB + HL-BC + HL-BD )


Problem 7, contd.

Head Loss Calculation:

HL = (4.73 . Q1.85 L ) / (C1.85 . D4.87)

HL-AB = (4.73 x 2.2281.85 x 2,100 ) / (1051.85 x 0.8334.87)

= 19.37 ft.

HL-BC = (4.73 x 1.1141.85 x 1,200 ) / (1051.85 x 0.8334.87)

= 3.07 ft

HL-BD = (4.73 x 1.1141.85 x 650 ) / (1051.85 x 0.8334.87)

= 1.67 ft.

Therefore,

H = 330 – 110 + (HLAB + HLBC + HLBD ), or

Max. Head Req. to be Delivered by

Pump = H = 330 – 110 + (19.37+3.07+1.67) = 244.15 ft.

And,

Motor hp = ( . Q . H ) / ( 550 . )

= (62.4 lbf/ft3 x 2.228ft3/s x 244.15ft.) / (550 lbf-ft/s/hp x 0.65)

= 94.9 hp.

Select a 95 hp motor.


Problem 7, contd.

c) Calculate the maximum water pressure, in psig, available to

the user.

Maximum water pressure condition at user distribution point “D”

purports the following:

• The pump is operating at a constant flow rate of 1,000 gpm.

• The flow to the user water distribution would be at a minimum flow

rate of 500 gpm.

• The flow rate from point “B” to “C” is 500 gpm.

• The water tower is full or at maximum level, producing maximum

head of 330 ft., for the pump to work against and sustain.

Apply Bernoulli’s Energy Equation at points “D” and (2).”

z2 + v22/2gc + p2/ = zD + vD

2/2gc + pD/ + HL


Problem 7, contd.

z2 + v22/2gc + p2/ = zD + vD

2/2gc + pD/ + HL

Note:

Since v2 = 0, v22/2gc = 0

QD = 500 gpm => 1.114 ft3/s => vD = (1.114 ft3/s)/A = 2.04 ft/s

Vel. Hd. at “D” = vD2/2gc => 0.065 ft

HL = HL-BC + HL-BD = 3.07 + 1.66 {As calc. in part (b) }

pD/ - p2/ = z2 - zD - vD2/2gc – HL = 330 – 200 – 0.065 – 3.07 –

1.66 = 125.2 ft

pD-gage = (pD/ - p2/ ) . = pD - p2 = 125.2 ft x 62.4 lbf / ft3

= 7813 lbf/ ft2 => 54.3 psig


Problem 7, contd.

d) Assume that the pump motor has just tripped. Calculate the

maximum volumetric flow rate at user water distribution

point “D.”

Following conditions or parameters apply in this situation:

• The water tower is full and elevation head is 330 ft.

• Since the pump is de-energized, there is no water flow from A to B

and QBC = QBD = Q = ?

• To achieve maximum flow at D, the line must be open to

atmospheric pressure; therefore, ( pD/ - p2/ ) = 0

• v22/2g = 0; and calculated in previous parts, vD

2/2g = 0.06 0

Apply Bernoulli’s Energy Equation at points “D” and 2.

z2 + v22/2g + p2/ = zD + vD

2/2g + pD/ + HL

0 0

Note: This slide is ref. to as 46 in the

Audio.


Problem 7, contd.

z2 - zD = HBC + HBD

HL = (HBC + HBD ) = ( 4.73 . Q1.85 L ) / (C1.85 . D4.87)

= { 4.73 . Q1.85 . (1200 + 650) } / (1051.85 . 0.8334.87)

= 3.88 Q1.85

Therefore,

3.88 Q1.85 = z2 - zD = 330 – 200

Q = {(330 – 200)/3.88} 1/1.85

Q = 6.68 cfs = 2997 gpm { Practice conversion in class}

Note : This Q = 6.68 cfs => 12.25 ft/sec => vD2/2g = 2.33 ft; which is still <<

Other heads.

Hydraulic Machines


Hydraulic Machines

Two basic types of hydraulic machines, that convert energies between fluid and mechanical forms are:

– Pumps:

Pumps convert mechanical energy into fluid energy;

e.g. a hydraulic pump converts mechanical energy

into pressure energy of the fluid which can then be

used to perform work through hydraulic cylinders.

– Turbines:

Turbines convert fluid energy into mechanical

energy; e.g. turbines in hydroelectric power

plants.



Hydraulic Machines

Hydraulic Machines

Major Types of Pumps:1) Positive Displacement (PD) Reciprocating Pumps. These pumps

are suitable for viscous fluids and slurries; fluids sensitive to shear. These pumps provide fixed-displacement volume per cycle by entrapping a constant amount of fluid in each cycle.

a) Power Pumps: It is a cylinder-operated pump. Power pumps

can be Single Acting and Double Acting

b) Direct Acting Pumps: Direct acting pumps , sometimes

referred to as steam pumps , are double acting

c) Diaphragm Pump: A diaphragm pump push s fluid

mechanically through movement of a membrane diaphragm;

check valves ensure flow of fluid in one direction. While

diaphragm pumps are less efficient than other types of pumps

they are suitable for pumping fluids that are sensitive to

shearing.


Diaphragm Pump


Hydraulic Machines

Major Types of Pumps, contd:

2) Rotary Pumps: Rotary pumps are positive displacement pumps that move fluid by means of screws, progressing cavities, gears, lobes, or vanes turning within a fixed casting. Rotary pumps produce smooth discharge. However, rotary pumps do experience “slip,” or leakage of some rotational volume back to the suction or intake side. This slip reduces the pump capacity.



Circumferential Piston

Vane Type Rotary Pump

Rotary Pump Types:

Lobe Type

Hydraulic Machines; Major Types of Pumps contd:

3) Centrifugal Pumps: Centrifugal pumps are classified according to the way their impellers impart energy to the fluid. Speed range and specific application determine the type of centrifugal pump that should be selected. Centrifugal pumps are commonly used to move liquids through a piping system. The fluid enters the pump impeller along or near to the rotating axis and is accelerated by the impeller, flowing radially outward into a diffuser chamber, from where it exits into the downstream piping system.

A centrifugal works on the principle of conversion of the kinetic energy of a flowing fluid (velocity pressure) into static pressure.

Centrifugal pumps and reciprocating pumps are used extensively for pumping sludge.



Centrifugal Pump:

Hydraulic Machines, Major Types of Pumps,

contd.

Some of the centrifugal pump types are listed below:

a) Radial Flow: Suitable for adding high pressure at low flow

rates. Radial flow pumps can be single or double –suction type.

b) Axial Flow: Axial flow impellers impart energy to the fluid in

form of pressure energy through compression. Suitable for

adding low pressures at high fluid flow rates.


Hydraulic Machines

Hydraulic Machine Formulae & Terminology :

Circular Blade Pitch: Circular blade pitch is the

impeller’s circumference divided by the number of

impeller vanes.

Impeller Tip Speed, vtip: Impeller tip speed is the

tangential velocity at the periphery of the impeller

and is a function of impeller diameter and rotational

speed.

Vtip = . D. n / 60 sec/min = D. ω / 2

Where,

D = Diameter of the impeller

n = Rotational speed in rpm

ω = Angular speed in rads/sec


Hydraulic Machines

Hydraulic Machine Formulae & Terminology, contd:

Suction: Suction is the inlet point of a pump. Suction

end parameters are subscripted with “s.”

Discharge: Discharge is the outlet point of a pump.

Discharge end parameters are subscripted with “d.”

Friction Head, hf: Friction Head is the head required to

overcome the resistance offered to flow of fluid in

pipes, fittings, elbows, valves, entrances and exits.

Hf = (f . L . V2 ) / (2 . D. g) { Darcy Equation }


Hydraulic Machines


Velocity Head, hv: The specific kinetic energy of the

fluid. The velocity head is also referred to as dynamic

head.

hv = v2/2g

Static Suction Head, hz(s): Static Suction Head is

defined as the vertical distance above the centerline

of the pump inlet to the level of fluid in the reservoir.


Static Suction Head & Static Suction Lift


Static Suction Head, hz(s):

Static Suction

Lift, - hz(s

Static Discharge Head


Static Discharge Head, hz(d)

Hydraulic Machines - Formulae & Terminology, contd:

Hydraulic machine parameters such as power factor, motor efficiency, motor input horsepower (EHp), brake horsepower (BHp), hydraulic horsepower (WHp), total efficiency, and pump efficiency are typically calculated or retrieved through reference tables.

Additional Definitions:

Head - a measure of the pressure or force exerted by the fluid.

Power input - the electrical input to the motor expressed in kilowatts (kW). A measure of the rate at which work is done.

Power factor - the ratio of the true power to the volt-amperes in an alternation current (ac) circuit.

Motor efficiency - a measure of how effectively the motor turns electrical energy into mechanical energy. It is the ratio of power input to power output.


Hydraulic Machines - Formulae & Terminology, contd:

Motor input horsepower (EHp) - the power input to the motor expressed in horsepower.

Brake horsepower (BHp) - the power delivered to the pump shaft expressed in horsepower.

Hydraulic horsepower (WHp) - the pump output or the liquid horsepower delivered by the pump.

Total efficiency - the ratio of the energy delivered by the pump to the energy supplied to the input side of the motor. Sometimes refered to as the ‘wire to water efficiency’.

Pump efficiency - the ratio of the energy delivered by the pump to the energy supplied to the pump shaft.


Efficiency & Power Delivery Stages


Pump

Efficiency,

pump

Motor

Eff.

motor

WHP, Delivered

by the Pump to

the Fluid

BHP, Delivered by the

Motor to the Pump

BHP = WHP / pump

HP, Delivered by

the Utility to the

Motor

HP Del. By Utility

= BHP / motor

Electrical

Power

from

Utility

Note: For a steam or hydro based electric power generating system, the

power flow would be the reverse of the sequence depicted above. See

Problem 9.

EHP > BHP > WHP

Machine Formulas & Terminology, contd:

Pumping Power Formulas, contd:

2. Pumping Horsepower as a function of the head added, hA, volumetric flow rate, Q, and the Specific Gravity, SG :

• When hA is given in ft, Q, in gal/min, Pumping

Power Pp , in hp, can be calculated using the

following equation:

Hydraulic Horsepower = WHP, Water

Horsepower :

Pp= hA . Q . SG / 3956 ……….…Eq. 2(a)


Hydraulic Machine Formulas & Terminology, contd:


3. Pumping Horsepower as a function of the Differential Pressure, P, and the volumetric flow rate, V, in ft3 / sec :

• Hydraulic Horsepower = WHP, Water

Horsepower :

Pp= P . V / 550, if P is in psf ..……Eq. 3(a)

Pp= P . V / 3.819, if P is in psi..……Eq. 3(b)




4. Pumping Horsepower as a function of W, Work or Energy (Specific) in ft-lbf /lbm and the volumetric flow rate:

• Hydraulic Horsepower = WHP, Water Horsepower :

Pp= W . Q. (SG) / 3956; where Q is in gal/min

……Eq. 4(a)

Pp= W . ṁ. / 550; where ṁ is in lbm/sec.

……Eq. 4(b)

Pp= W . V. (SG) / 8.814; where V is in ft3 / sec

……Eq. 4(c)

Pp= hA . V. (SG) / 8.814; where V is in ft3 / sec

…………..Eq. 4(d)



Specific Speed:

The capacity and efficiency of a centrifugal pump are, partially, a function of the impeller design. Each impeller design will yield a certain flow rate and added head. The quantitative index used to optimize the impeller design is

known as specific speed, ns.

ns = n . V / (hA)0.75 ….. SI Units, …Eq. SS-1

ns = n . Q / (hA)0.75 ….. US Units, …Eq. SS-2

Note: n and ns are in rpm



Affinity Laws:If the impeller diameter is held constant and the speed is varied, the following

ratios hold true:

Q2 / Q1 = n2 /n1 …………………….Eq. AF(1)

h2/h1 = (n2/n1)2 = (Q2/Q1)

2 …..…………….Eq. AF(2)

P2/P1 = (n2/n1)3 = (Q2/Q1)

3 …….…………………..Eq. AF(3)

If the impeller speed is held constant and the impeller size is varied, the following ratios hold true:

Q2 / Q1 = D2 /D1 …..........……………Eq. AF(4)

h2/h1 = (D2 /D1)2 ………………………….……Eq. AF(5)

P2/P1 = (D2 /D1)3 ……………………………………..…Eq. AF(6)


BLANK SLIDE



Problem 8Case study of an elevated water tower, service pump and a fire pump. The

service pump operates at constant 80 gpm. The inline fire pump can draw

water, in case of fire, at the rate of 40 gpm. The local fire code requires

that the fire pump must be able to sustain the required supply of water for

a minimum period of 3 – hours. The service demand per hour, for a 24

hour period, is given on the next slide. Calculate or determine the

following based on the data provide:

a) Maximum flow of water INTO the water tower.

b) Average or uniform demand rate for the 24 hour period.

c) Maximum flow OUT of the water tower in gpm.

d) The hour when maximum water flows OUT of the water tower.

e) The hour when the maximum amount of water flows INTO the

water tower.

f) Required , minimum, water tower volume or storage capacity.

Problem 8


Service Pump, 80 gpm

Fire Pump, 40 gpm

Elevated Water Tower, EWT


Problem 8, contd.

a) Maximum flow of water into the water tower occurs

when the service demand is the least and the no fire

emergency conditions exist.

The calculations required for this part and other are

premised on conversion of service volumes demanded

each hour to respective volumetric flow rates in gpm.

This data and the demand to supply deficit for each

hour are calculated and tabulated on Table 8.1.

The net flow of water into the EWT, under these

conditions would be

= 80 gpm – 23.33 gpm = 56.67 gpm { 1400 gph / 60m/h = 23.33

gpm}


Problem 8, contd.

(b) Average or uniform demand rate for the 24 hour period is

determined by summing the total demand over the 24 hour

period and dividing by 24, see Table 8.1:

= 86,700 gal / 24 hrs / 60 min / hr = 60.21 gpm

(c) The conditions for maximum flow out of the EWT, in gpm,

would be as follows, see Table 8.1:

Maximum Service Demand , at 1700 hrs, of 8,000 gal,

translating into = 8,000 gph / 60 min/hr = 133.33 gpm

Full Demand by the Fire Pump = 40 gpm

Therefore, the Max. Flow Out of EWT = 133.33 + 40 = 173.33 gpm

{This is assuming that the supply pump is off line}

2/8/2010Fluids Statics, Hydrodynamics and Hydraulic Machines; © B. Rauf 93

I II Vol. Flow Rate in Pump Demand Minus Deficit to be

Hour Peak DemandGPM Supply Rate Supply Offset by Tank

in GallonsGallons Ea. Hr. Stg. Capacity

0 1,400 23.33 4800 3,400

100 1,400 23.33 4800 3,400

200 1,400 23.33 4800 3,400

300 1,600 26.67 4800 3,200

400 2,100 35.00 4800 2,700

500 2,600 43.33 4800 2,200

600 2,900 48.33 4800 1,900

700 3,500 58.33 4800 1,300

800 3,900 65.00 4800 900

900 4,100 68.33 4800 700

1000 4,900 81.67 4800 (100) 100

1100 5,000 83.33 4800 (200) 200

1200 6,000 100.00 4800 (1,200) 1,200

1300 5,500 91.67 4800 (700) 700

1400 6,500 108.33 4800 (1,700) 1,700

1500 6,600 110.00 4800 (1,800) 1,800

1600 6,700 111.67 4800 (1,900) 1,900

1700 8,000 133.33 4800 (3,200) 3,200

1800 3,100 51.67 4800 1,700

1900 2,500 41.67 4800 2,300

2000 2,100 35.00 4800 2,700

2100 1,900 31.67 4800 2,900

2200 1,600 26.67 4800 3,200

2300 1,400 23.33 4800 3,400

86,700 10,800

Gallons

Problem 8, contd.Table 8.1


Problem 8, contd.

(d) The hour when maximum water flows OUT of the water

tower is 1700 hrs, see Table 8.1, at the rate of 133.33 gpm.

(e) The hour when the maximum amount of water flows INTO

the water tower are 0000, 0100, 0200 & 2300 hrs, see Table

8.1, at the rate of 23.33 gpm.

(f) Required EWT storage capacity:

The minimum required EWT storage capacity would have to

comply with 3 – hour fire pump supply code and must

meet the service demand during the supply – demand

deficit periods; therefore, the EWT capacity

components/constituents would be as follows:

Volume of water required to operate the fire pump for 3 -

hours: = 3 hrs x 40 gpm x 60 min/hr = 7200 gallons


Problem 8, part (f), contd.

The EWT must have the capacity to meet the short fall

during the periods when demand exceeds pump capacity.

See Table 8.1. The deficit periods span from 1000 hrs

through 1700 hours. The last column in the table lists the

specific deficits associated with each of the eight (8)

deficit hours.

The total deficit from the last column in table 8.1 is

= 10,800 gallons

http://www.watertowers.com/photos_2493_Örebro_Night_The_water_tower.html


Problem 8, part (f), contd.

There fore the minimum required capacity of the EWT

would be:

= 10,800 gallons + 7,200 gallons

= 18,000 gallons or

= 2406.24 cu-ft.

In other words, you would specify a water tank with the

following radius:

Since the EWT Capacity or volume is

= 2,406.24 cu-ft

And the volume of a sphere is = 4/3 x ᴨ x r3

Radius of the EWT would be = r = 8.31 ft


Problem 9Pressures on the intake and exit ends of a turbine are measured to be 50 psia

and 10 psia, respectively. The volumetric flow rate for the fluid (water) is

120 ft3/sec. The turbine is driving a electric generator with nameplate

efficiency of 90%. Calculate the following:

a) Water Horsepower delivered by the water to the turbine.

b) Brake Horsepower delivered by the turbine to the shaft driving the

generator. The efficiency of the turbine is 65%.

c) Maximum power generated, in KW’s, by this hydroelectric power

generating system .

Problem 9

Given:Pin = 50 psia

Pout = 10 psia

P = 40 psia

P Conv. to psf = 40 lb/in2 x 144 in2 / ft2 = 5760 psf

V = Volumetric Flow Rate = 120 cu-ft / sec

Turbine Efficiency = 65 %

Generator Efficiency = 90 %

a) WHP = Water Horsepower delivered by the water to the

turbine:

WHP = Pp= P . V / 550, if P is in psf ……Eq. 3(a)

Pp= 5760 lbf/ft2 . 120 ft3/sec / 550 = 1256.73 hp


Problem 9, contd.

b) Brake Horsepower delivered by the turbine to the shaft driving

the generator; the turbine efficiency is 65%:

BHP = WHP x Turbine Efficiency

= 1256.73 hp x 0.65

= 816.87 hp

c) Maximum power generated, in kW‟s, by this hydroelectric

power generating system :

Power Generated By The Hydro Elect. Power System, in

kW's = BHP x Eff. x 0.746 kW / hp

= 816.87 hp x 0.9 x 0.746

= 548.45 kW


Problem 10Consider a hydroelectric reservoir where water is flowing through the

turbine at the rate of 1100 ft3 / sec. The turbine exit point is 700 ft lower than the elevation of the reservoir water surface. The turbine efficiency is 90% and the total frictional head loss through the shaft (Draft Tube) and turbine system is 52 ft. a) Calculate the power output of the turbine in MW's. b) If the efficiency of the Electric Power Generator is 92%, what would the electric power output be for this hydroelectric power generating system?

Given: specific weight = 62.4 lbf / ft3

hf = 52 ft

hz = 700 ft

hA = hz - hf = 648 ft

V = 1100 cu-ft / sec

Turbine Efficiency 90 %

Generator Efficiency 92 %


Problem 10, contd.

a) Calculate the power output of the turbine in MW's.

Pp(in hp) = ( hA . . V ) / 550 ……Eq. 1(a)

Pp= WHP = Hydraulic Horsepower

= (648 x 62.4 x 1100 ) / 550

= 80,870 hp

Pout = Power Output of the Turbine

= (0.90 x 80,870)

= 72,783 hp

Pout = Power Output of the Turbine in MW

= ( 72,783 x 0.746 ) / 1000 kW per MW

= 54 MW


Problem 10, contd.

b) If the efficiency of the Electric Power Generator is 92%, what would the electric power output be for this hydroelectric power generating system?

P Gen-out = Power Output of the Generator in MW

= ( 54 MW x 0.92 )

= 50 MW


Problem 11

Consider a hydraulic machine application scenario where the system performance requirement is governed by the mathematical expression: H = 50 + Q2 . Where H represents the required system head in ft. and Q is the volumetric flow rate in cu-ft / sec. The scenario considers and compares two (2) pumps of different size. The pump performance curves for these two pumps are shown in figure 11(a). These performance curves represent operation at 1400 rpm.

a) Determine the flow rate Q for Pump - 1 when it is operating at 1400 rpm, to meet the system head requirement.

b) What would the system head be at the performance point in part (a)?

c) Determine the flow rate Q for Pump - 2 when it is operating at 1400 rpm.


Problem 11

d) What would the system head be, with Pump - 2, at the performance point in part (c)?

e) Calculate the Brake Horse Power, in hp, required to drive Pump - 1, under part (a) conditions.

f) If Pump -1 is powered by a VFD and the shaft speed is reduced to 1200 rpm, what would the new flow rate be in the system?

g) Which pump would you select for your application if a head of approx. 220 ft is required at a volumetric flow rate of 10 cu-ft/sec.?


Problem 11


a) Determine the flow rate Q for Pump - 1 when it is operating at 1400 rpm, to

meet the system head requirement.

Given: Conversion Factor: 448.8 gpm per cu-ft/sec

System Head Requirement Eq.:

H (Q) = 50 + Q2

Q: Flow Rate System Head, Hs

0 50

2 54

5 75

7 99

10 150

15 275

20 450

25 675

30 950

35 1275

0

50

100

150

200

250

0 10 20 30 40

Hea

d, H

, in

ft.

Q, Flow Rate in cu-ft./sec

System & Pump Curve

Pump #1 Perf. Curve

Pump #2 Perf. Curve

11 cu-

ft/sec

The Pump

Performance

Curves are Based

on 1400 rpm

Operation

13 cu-

ft/sec

(g)

Fig. 11(a), Q vs. hA

Problem 11, contd.

a) Determine the flow rate for Pump - 1, at the required system head, when it is operating at 1400 rpm

Identify the point on the graph at which Pump-1's performance curve intersects with the system head requirement curve. Extrapolate a line (dashed line on the graph) to the horizontal axis representing Q, in Cu-ft./sec. The projected line intersects the Q-axis at 11 cu-ft/sec.

Therefore, the required flow rate Q:

= Q, in gals/min, or, gpm

= 11 cu-ft/sec


Problem 11, contd.b) System Head at the performance point in part (a), at Q of 11 cu-

ft./sec

= H (Q) = 50 + Q2 = 171 ft

c) Determine the flow rate Q for Pump - 2 when it is operating at 1400 rpm.

Identify the point on the graph at which Pump-2's performance curve intersects with the system head requirement curve. Extrapolate a line (dashed line on the graph) to the horizontal axis representing Q, in cu-ft./sec. The projected line intersects the Q-axis at 13 cu-ft/sec.

Therefore, the required flow rate, Q = 13 cu-ft/sec

Q, in gals/min, or, gpm = 13 cu-ft/sec x 448.8 gpm/(cu-ft/sec)

= 5,834 gpm


Problem 11, contd.d) System Head at the performance point in part (a), at Q of 13 cu-

ft/sec

= H (Q) = 50 + Q2

= 219 ft

e) Calculate the Brake Horse Power, in hp, required to drive Pump -1, under conditions stated in part (a).

Approach: Calculate the WHP, Water Horse Power, first, using Eq. 4 (d); then derive the BHP using the pump efficiency. The pump efficiency is not given and must be retrieved from the given pump curve.

Since,

WHP = Pp = hA . V. (SG) / 8.814; the power that needs to be

delivered to the fluid.

hA , as calculated in part (b) = 171 ft.

V , as determined in part (a) = 11 cu-ft/sec ,or = 4937 gpm

SG for water = 1


Problem 11, contd.Part (e) contd.

n = 1400 rpm

Therefore,

WHP = Pp = hA . V. (SG) / 8.814 WHP = Pp = 213 hp

Now, BHP = WHP / Eff.pThe pump efficiency Eff.p can be obtained through the Specific Speed

Pump Efficiency curve "E."

From equation SS-2, the Specific Speed is:

ns = n . Q / (hA)0.75

Where, n and ns are in "rpm", Q is in "gpm" and hA is in "ft“

Therefore,

ns = 1400 . 4,937/ (171)0.75

or, ns = 2080 rpm

Then, from Pump Efficiency curve "E," the pump efficiency = 87.5% Hence,

BHP = WHP / Eff.p = 213 / 0.875 = 244 hp



Problem 11, contd.

Specific Speed vs. Pump Efficiency Curve

Problem 11, contd.f) If Pump -1 is powered by a VFD and the shaft speed is reduced to

1200 rpm, what would the new flow rate be in the system?

From Affinity Laws; Equation AF (1):

Q2/Q1 = n2/n1 and

Q2 = (n2/n1) . Q1

Given and as calculated earlier:

Q1 = 4937

n1 = 1400

n2 = 1200

Therefore,

Q2 = (n2/n1) . Q1

= (1200/1400) x 4937

= 4232 gpm


Problem 11, contd.g) Which pump would you select for your application if a head of

approx. 220 ft is required at a volumetric flow rate of 10 cu-ft/sec.?

The question is premised on the assumption that the system head curve has changed such that it passes through a point on the graph where Q = 10 cu-ft/sec and the head hA is approximately 220 ft.

By inspection of the Q vs. hA graph, figure 11 (a), we can see that Pump #1 is capable of providing only 185 ft at Q of 10 cu-ft/sec.

Therefore, Pump #2 would be more suitable for meeting the specified requirement.


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