HYDRAULICSBy. Engr. Yuri G. Melliza

XAVIER UNIVERSITY – ATENEO DE CAGAYANMechanical Engineering Department

FLUID MECHANICS

1. Density ()

3mkg

Vm

=ρ

2.Specific Volume ()

kgm

mV

=3

υ

3.Specific Weight ()

3mKN

1000

g

1000Vmg

VW

=ρ

==γ

Properties of Fluids

4. Specific Gravity or Relative Density

For Liquids: Its specific gravity (relative density) is equal to the ratio of its density to that of water at standard temperature and pressure.

W

=γ

γ L

W

LL ρ

ρ=S

For Gases: Its specific gravity (relative density) is equal to the ratio of its density to that of either air or hydrogen at some specified temperature and pressure.

where: At standard conditionW = 1000 kg/m3

W = 9.81 KN/m3ah

G

γ

γ=

ah

GG ρ

ρ=S

5. Temperature

460FR

273CK

32+1.8°C=°F1.8

32-°F=°C

6. Pressure

KPa AF

=P

where: F - normal force, KN A - area, m2

If a force dF acts on an infinitesimal area dA, the intensity of pressure is,

KPa dAdF

=P

PASCAL’S LAW: At any point in a homogeneous fluid at rest the pressures are the same in all directions.

x

y

z

A

BC

P1 A1

P2 A2

P3 A3

Fx = 0 and Fy = 0P1A1 – P3A3 sin = 0 1P2A2 – P3A3cos = 0 2

From Figure:A1 = A3sin 3A2 = A3cos 4

Eq. 3 to Eq. 1P1 = P3

Eq. 4 to Eq. 2P2 = P3

Therefore:

P1 = P2 = P3

Atmospheric pressure: The pressure exerted by the atmosphere.

At sea level condition:Pa = 101.325 KPa = .101325 Mpa = 1.01325Bar = 760 mm Hg = 10.33 m H2O = 1.133 kg/cm2

= 14.7 psi = 29.921 in Hg = 33.878 ft H2O

Absolute and Gage PressureAbsolute Pressure: is the pressure measured referred to absolute zero and using absolute zero as the base.Gage Pressure: is the pressure measured referred to atmospheric pressure, and using atmospheric pressure as the base

Absolute Zero

Atmospheric pressure

Pgage

Pvacuum

Pabs

Pabs

Pabs = Pa+ Pgage

Pabs = Pa - Pvacuum

7. Viscosity: A property that determines the amount of its resistance to shearing stress.

x dxv+dv

v

moving plate

Fixed plate

v

S dv/dxS = (dv/dx)S = (v/x)

= S/(v/x)

where: - absolute or dynamic viscosity in Pa-secS - shearing stress in Pascalv - velocity in m/secx -distance in meters

8. Kinematic Viscosity: It is the ratio of the absolute or dynamic viscosity to mass density.

= / m2/sec

9. Elasticity: If the pressure is applied to a fluid, it contracts,if the pressure is released it expands, the elasticity of a fluid is related to the amount of deformation (contraction or expansion) for a given pressure change. Quantitatively, the degree of elasticity is equal to;

Ev = - dP/(dV/V)Where negative sign is used because dV/V is negative for a positive dP. Ev = dP/(d/) because -dV/V = d/

where: Ev - bulk modulus of elasticity, KPa dV - is the incremental volume change V - is the original volume dP - is the incremental pressure change

r h

10. Surface Tension: Capillarity

Where: - surface tension, N/m - specific weight of liquid, N/m3

r – radius, mh – capillary rise, m

C

0 0.0756

10 0.0742

20 0.0728

30 0.0712

40 0.0696

60 0.0662

80 0.0626

100 0.0589

Surface Tension of Water

rcos2

hγ

θσ

Variation of Pressure with Elevation

FREE SURFACE

1•

2•

h1

h2h

dP = - dh

Note:Negative sign is used because pressure decreases as elevation increases and pressure increases as elevation decreases.

MANOMETERSManometer is an instrument used in measuring gage pressure in length of some liquid column. Open Type Manometer : It has an atmospheric surface and is capable in measuring gage pressure. Differential Type Manometer : It has no atmospheric surface and is capable in measuring differences of pressure.

Pressure Head:

where: p - pressure in KPa - specific weight of a fluid, KN/m3

h - pressure head in meters of fluid

hP

γ

Open Type Manometer

Open

Manometer Fluid

Fluid A

Differential Type Manometer

Fluid B

Manometer Fluid

Fluid A

Determination of S using a U - Tube

xy

Open Open

Fluid A

Fluid B

SAx = SBy

Example no. 1A building in Makati is 84.5 m high above the street level. The required static pressure of the water line at the top of the building is 2.5 kg/cm2. What must be the pressure in KPa in the main water located 4.75 m below the street level. (1120.8 KPa)

Point 1: Main water line, 4.75 m below street levelPoint 2: 84.5 m above street level∆h = h2 – h1 = (84.5 + 4.75) = 89.25 mP2 = 2.5 kg/cm2 = 245.2 KPa

KPa 743.120,1P)25.89(81.92.245P

)hh(PP)hh(PP

1

1

1221

1212

Example No. 2A mercury barometer at the ground floor of a high rise hotel in Makati reads 735 mm Hg. At the same time another barometer at the top of the hotel reads 590 mmHg. Assuming air density to be constant at 1.22 kg/m3, what is the approximate height of the hotel. (1608 m)

Point 1: Ground floorh1 = 0 mP1 = 735 mm Hg = 98 Kpa

Point 2: Roof Toph2 = h (height)P2 = 590 mm Hg = 78.7 KPa

meters 1608.33 h sec

m 81.9g

h h-h :gminassu-

)P-(Ph-h

mKN

1000

g

)h-(h-P-P mkg

1.22

mKN

012.01000

1.22(9.81) :air For

2

12

12123

12123

3

Example No. 3The reading on a pressure gage is 1.65 MPa, and the local barometer reading is 94 KPa. Calculate the absolute pressure that is being mea-sured in kg/cm2. (17.78 kg/cm2)

Example No. 4A storage tank contains oil with a specific gravity of 0.88 and depth of 20 m. What is the hydrostatic pressure at the bottom of the tank in kg/cm2. (1.76 kg/cm2)

Example No. 5A cylindrical tank 2 m diameter, 3 m high is full of oil. If the specific gravity of oil is 0.9, what is the mass of oil in the tank?

Forces Acting on Plane Surfaces

Free Surface

•C.G.•C.P.

hhp

S S

•C.G.•C.P.

S

M

N

M

N

y

yp

e

F

F - total hydrostatic force exerted by the fluid on any plane surface MNC.G. - center of gravityC.P. - center of pressure

where:Ig - moment of inertia of any plane surface MN with respect to the axis at its centroidsSs - statical moment of inertia of any plane surface MN with respect to the axis SS not lying on its planee - perpendicular distance between CG and CP

Forces Acting on Curved Surfaces

Free Surface

C.G.

C.P.

h phA

B

C

D E

C’

B’

C

B

F

Fh

FV

Vertical Projection of AB

L

Ah=Fh γA = BC x LA - area of the vertical projection of AB, m2

L - length of AB perpendicular to the screen, m

V=FV γ

V = AABCDEA x L, m3

2

v

2

h FF=F +

Hoop Tension

1 m

D

P = h

h

T

T

F

F = 02T = FT = F/2 1S = T/AA = 1t 2

T

T

F

1 m

D

t

S = F/2(1t) 3From figure, on the vertical projection the pressure P;P = F/AA = 1DF = P(1D) 4substituting eq, 4 to eq. 3S = P(1D)/2(1t)

where:S - Bursting Stress KPaP - pressure, KPaD -inside diameter, mt - thickness, m

KPa t2

PDS

Laws of BuoyancyAny body partly or wholly submerged in a liquid is subjected to a buoyant or upward force which is equal to the weight of the liquid displaced.

1.

Vs

W

BF

W = BFW = BVB KNBF = LVs KN

W = BFW = BVB

BF = LVs

where:W - weight of body, kg, KNBF - buoyant force, kg, KN - specific weight, KN/m3

- density, kg/m3

V - volume, m3

Subscript:B - refers to the bodyL - refers to the liquids - submerged portion

W2.

Vs

BFT

W = BF - TW = BVB KNBF = LVs KN

W = BF - TW = BVB

BF = LVs

where:W - weight of body, kg, KNBF - buoyant force, kg, KNT - external force T, kg, KN - specific weight, KN/m3

- density, kg/m3

V - volume, m3

Subscript:B - refers to the bodyL - refers to the liquids - submerged portion

Vs

W

BF

T3.

W = BF + TW = BVB KNBF = LVs KN

W = BF + TW = BVB

BF = LVs

where:W - weight of body, kg, KNBF - buoyant force, kg, KNT - external force T, kg, KN - specific weight, KN/m3

- density, kg/m3

V - volume, m3

Subscript:B - refers to the bodyL - refers to the liquids - submerged portion

W = BF + TW = BVB KNBF = LVs KN

W = BF + TW = BVB

BF = LVs

Vs

W

BF

T4.

VB = Vs

W = BF - TW = BVB KNBF = LVs KN

W = BF - TW = BVB

BF = LVs

Vs

W

BFT

5.

VB = Vs

Energy and HeadBernoullis Energy equation:

Reference Datum (Datum Line)

1

2

z1

Z2

HL = U - Q

1. Without Energy head added or given up by the fluid (No work done bythe system or on the system:

L2

222

t1

211 H+Z+

2g

v+

γ

P=h +Z+

2g

v+

γ

P

L2

222

1

211 H+Z+

2g

v+

γ

P=Z+

2g

v+

γ

P

h+H+Z+2g

v+

γ

P= +Z+

2g

v+

γ

PL2

222

1

211

2. With Energy head added to the Fluid: (Work done on the system

3. With Energy head added given up by the Fluid: (Work done by the system)

Where:P – pressure, KPa - specific weight, KN/m3v – velocity in m/sec g – gravitational accelerationZ – elevation, meters m/sec2+ if above datum H – head loss, meters- if below datum

L2

222

1

211 HZ

g2

vPZ

g2

vP

g2v

1C1

H2

22

v

L

APPLICATION OF THE BERNOULLI'S ENERGY THEOREM

where: Cv - velocity coefficient

NozzleBase

Tip

Jet

Q

/secm AvQ 3

Venturi Meter

A. Without considering Head loss

flow ltheoretica QvAvAQ

Zg2

vPZ

g2vP

2211

2

2

221

2

11

γγ

inlet

throat exit

Manometer

1

2

B. Considering Head loss

flow actual 'QvAvA'Q

HZg2

vPZ

g2vP

2211

L2

2

221

2

11

γγ

Meter Coefficient

Q'Q

C

PUMPS: It is a steady-state, steady-flow machine in which mechanical work is added to the fluid in order to transport the liquid from one point to another point of higher pressure.

LowerReservoir

Upper Reservoir

Suction GaugeDischarge Gauge

Gate Valve

Gate Valve

1

2

FUNDAMENTAL EQUATIONS

1. TOTAL DYNAMIC HEAD

meters HZZ2g

vvPPH L12

2

1

2

212t

γ

2. DISCHARGE or CAPACITY Q = Asvs = Advd m3/sec

3. WATER POWER or FLUID POWER WP = QHt KW

4. BRAKE or SHAFT POWER

KW 60,000

TN2BP

π

5. PUMP EFFICIENCY

100% xBPWP

P η

6. MOTOR EFFICIENCY

100% xMPBP

mη

7. COMBINED PUMP-MOTOR EFFICIENCY

mPC

C

ηηη

η

100% xMPWP

8. MOTOR POWER

KW 1000

)(cosEMP

θI

For Single Phase Motor

For 3 Phase Motor

KW 1000

)(cosE 3MP

θI

where: P - pressure in KPa T - brake torque, N-m v - velocity, m/sec N - no. of RPM - specific weight of liquid, KN/m3 WP - fluid power, KW Z - elevation, meters BP - brake power, KW g - gravitational acceleration, m/sec2 MP - power input to HL - total head loss, meters motor, KW E - energy, Volts I - current, amperes (cos) - power factor

HYDRO ELECTRIC POWER PLANT

A. Impulse Type turbine (Pelton Type)

Headrace

Tailrace

Y – Gross HeadPenstock turbine

1

2

B. Reaction Type turbine (Francis Type)

Headrace

Tailrace

Y – Gross Head

Penstock

ZB

1

2Draft Tube

B

Generator

B – turbine inlet

Fundamental Equations

1. Net Effective Head

A. Impulse Typeh = Y – HL

Y = Z1 – Z2

Y – Gross Head, metersWhere:

Z1 – head water elevation, mZ2 – tail water elevation, m

B. Reaction Typeh = Y – HL

Y = Z1 –Z2

meters Zg2

vPh B

2BB

Where:PB – Pressure at turbine inlet, KPavB – velocity at inlet, m/secZB – turbine setting, m - specific weight of water, KN/m3

2. Water Power (Fluid Power)FP = Qh KW

Where:Q – discharge, m3/sec

3. Brake or Shaft Power

KW 000,60

TN2BP

Where:T – Brake torque, N-mN – number of RPM

4. Turbine Efficiency

mvh eeee

100% x FP

BPe

Where:eh – hydraulic efficiencyev – volumetric efficiencyem – mechanical efficiency

5. Generator Efficency

100% x BP

GP

100% x powerShaft or Brake

Output Generator

g

g

6. Generator Speed

RPM n

f120N

Where:N – speed, RPMf – frequency in cps or Hertzn – no. of generator poles (usually divisible by four)

Pump-Storage Hydroelectric power plant: During power generation the turbine-pump acts as a turbine and during off-peak period it acts as a pump, pumping water from the lower pool (tailrace) back to the upper pool (headrace).

Turbine-Pump

A 300 mm pipe is connected by a reducer to a 100 mm pipe. Points 1 and 2 are at the same elevation. The pressure at point 1 is 200 KPa. Q = 30 L/sec flowing from 1 to 2, and the energy lost between 1 and 2 is equivalent to 20 KPa. Compute the pressure at 2 if the liquid is oil with S = 0.80. (174.2 KPa)

300 mm100 mm

1 2

A venturi meter having a diameter of 150 mm at the throat is installed in a 300 mm water main. In a differential gage partly filled with mercury (the remainder of the tube being filled with water) and connected with the meter at the inlet and at the throat, what would be the difference in level of the mercury columns if the discharge is 150 L/sec? Neglect loss of head. (h=273 mm)

The liquid in the figure has a specific gravity of 1.5. The gas pressure PA is 35 KPa and PB is -15 KPa. The orifice is 100 mm in diameter with Cd = Cv = 0.95. Determine the velocity in the jet and the discharge when h = 1.2. (9.025 m/sec; 0.071 m3/sec)

1.2 m

PA

PB