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Transcript of Fluid Mechanics Pp
Hydraulics
By: Engr. Yuri G. Melliza
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 DensityFor 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.
ah
G
γ
γ=
ah
GG ρ
ρ=S
where: At standard condition
W = 1000 kg/m3
W = 9.81 KN/m3
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 deformat-ion (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, KPadV - 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
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
Ah=F γ
yA
yA+I=y
2
g
p
s
gg
p S
I
yA
I=y =
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 γ
A = 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 = B VB g
BF = LVs g
W = BF + TW = BVB g
BF = LVs g
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 gBF = LVs g
W = BF + TW = BVB
BF = LVs
Vs
W
BF
T4.
VB = Vs
W = BF - TW = BVB gBF = LVs g
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 by
the 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
Orifice: An orifice is an any opening with a closed perimeterWithout considering Head Loss
1
2
a
a
Vena Contractah
By applying Bernoulli's Energy theorem:
2
222
1
211 Z
g2
vPZ
g2
vP
But P1 = P2 = Pa and v1is negligible, then
21
22 ZZg2
v
and from figure: Z1 - Z2 = h, therefore
hg2
v 22
gh2v2
Let v2 = vt
gh2vt
where:vt - theoretical velocity, m/sech - head producing the flow, metersg - gravitational acceleration, m/sec2
COEFFICIENT OF VELOCITY (Cv)
velocity ltheoreticavelocity actual
vC
tvv'
Cv
COEFFICIENT OF CONTRACTION (Cc)
orifice theof areacontracta vena @ jetof areaCc
A
aCc
COEFFICIENT OF DISCHARGE(Cd)
discharge ltheoreticadischarge actual
vC
QQ'
Cd
vcd CCC
where: v' - actual velocityvt - theoretical velocity
a - area of jet at vena contractaA - area of orificeQ' - actual flowQ - theoretical flowCv - coefficient of velocityCc - coefficient of contractionCd - coefficient of discharge
Jet Trajctory
v
v sin
v cos1
2
3R = v cos (2t)
d
gθsinv
R22
y
x
v = vx
y2g
xv
vx
t
If the jet is flowing from a vertical orifice and the jet is initially horizontal where vx = v.
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
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