hydraulics for engineers
DESCRIPTION
TRANSCRIPT
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