7/2/2017
Introduction •
Pumps produce produce flow of fluid and develop pressure pressure adding adding energy to the fluid
•
Turbines Turbines produc produce e power extra extractin cting g energy energy from fluid fluid
•
Fluid machi machines nes are are divided divided into into two catego categories: ries:
Turbomachines
– positive displacement displacement machines; machines; – turbomachines
Energy consideration •
•
•
Positive displacement machines force the fluid into or out of a chamber by changing the volume of a chamber. Thus the pressure developed (pump) or the work done (engine) are the result of essentially static forces rather then dynamic effects.
•
Turbomachines involve a collection of blades on a rotor. Rotation of a rotor produces dynamic effects that either add energy to the fluid (pumps) or remove energy from the fluid (turbines)
•
Turbomachines are classified as axial-flow, radial-flow , or mixed-flow machines depending on the predominant direction of the fluid motion relative to the rotor’s axis as the fluid passes the blades. Note
Energy consideration
Consider Consider a fan blade blade driven driven at constan constantt angular velocity, velocity, , by a motor as is shown in Fig. We denote the blade speed as U where r is the radial distance from the axis of the fan. U = ῳr Absolute velocity (V): (V): The absolute fluid velocity that seen by a person sitting stationary at the table on which the fan is denoted V.
•
Relative velocity velocity (W) : The relative velocity that seen by a person riding on the fan blade.
•
The actual actual absolut absolute e fluid velocit velocity y is the vector vector sum of the relative velocity and the blade velocity
V=W+U
1
7/2/2017
Energy consideration
Example The rotor shown in Figure (a) rotates at a constant angular velocity of = 100 rad/s. Although the fluid initially approaches the rotor in an axial direction, the f low across the blades is primarily radial. Measurement indicate that the absolute velocity at t he inlet and outlet are V 1 =12 m/s and V 2 = 25 m/s, respectively. respectively. Is this device a pump or a turbine?
Example (cntd)
Example (cntd)
2
7/2/2017
Energy consideration
Example The rotor shown in Figure (a) rotates at a constant angular velocity of = 100 rad/s. Although the fluid initially approaches the rotor in an axial direction, the f low across the blades is primarily radial. Measurement indicate that the absolute velocity at t he inlet and outlet are V 1 =12 m/s and V 2 = 25 m/s, respectively. respectively. Is this device a pump or a turbine?
Example (cntd)
Example (cntd)
2
7/2/2017
Example (cntd)
Example (cntd)
Answer: It is a pump
Pump or Turbine?
•
When blades blades move because because of the fluid fluid force, force, we have a turbine; when when blades are are forced to move fluid, we have a pump
•
When shaft shaft torque and rotation rotation are are in the same direction, direction, we we have a pump; otherwise we have a turbine
Angular momentum momentum consideration
notes
3
7/2/2017
Angular momentum consideration
Angular momentum consideration
Balance of angular momentum
Balance of angular momentum
r F
cs
r F
r V V n dA
cs
r V V n dA
Shaft torque (Euler turbomachine equation)
1 rV T shaft m 1 1 m2 r2V 2
notes
notes
Angular momentum consideration
Angular momentum consideration
Balance of angular momentum
Balance of angular momentum
r F
cs
r F
r V V n dA
cs
Shaft torque (Euler turbomachine equation)
r V V n dA
Shaft torque (Euler turbomachine equation)
1 rV T shaft m 1 1 m2 r2V 2
shaft power , is related to the shaft torque and angular velocity by
Shaft power
1 rV T shaft m 1 1 m2 r2V 2 Shaft power
1 U1V 1 m 2 U2 V 2 W shaft m
1 U1V 1 m 2 U2 V 2 W shaft m
W shaft is positive when power is supplied to the contents of the control volume (pumps) and negative otherwise turbines.
Shaft work per unit mass
w shaft U1V 1 U2V 2
notes
notes
4
7/2/2017
Angular momentum consideration
or (alternative) w shaft
Balance of angular momentum
r F
cs
V22
V12 U22 U12 W22 W12 2
r V V n dA
Shaft torque (Euler turbomachine equation)
1 rV T shaft m 1 1 m2 r2V 2 Shaft power
1 U1V 1 m 2 U2 V 2 W shaft m Shaft work per unit mass
w shaft U1V 1 U 2V 2
it is an important concept equation because it shows how the work transfer is related to absolute, relative, and blade velocity changes. Because of the general nature of the velocity triangle in above Fig and Eq. 12.8 is applicable for axial-, radial-, and mixed-flow rotors.
Centrifugal Pump
Pump Theory Flow through the pump is unsteady and three-dimensional
Pump arrangements: Volute or diffuser casing Open or shrouded impeller Single- or double suction Single- or multistage Work is done on the fluid by r otating blades (centrifugal action and tangential blade force acting on the fluid over a distance) creating a large increase in kinetic energy of the fluid through the impeller. This kinetic energy is converted into an increase in pressure as t he fluid flows from the impeller into casing
notes
5
7/2/2017
Pump Theory
Pump Theory
Flow through the pump is unsteady and thr ee-dimensional
Flow through the pump is unsteady and three-dimensional
Assume the average one-dimensional fluid flow from inlet to outlet of the impeller as blades rotate
Assume the average one-dimensional fluid flow from inlet to outlet of the impeller as blades rotate Draw velocity diagrams at inlet and outlet
notes
Pump Theory
notes
Ideal Head
Flow through the pump is unsteady and thr ee-dimensional Assume the average one-dimensional fluid flow from inlet to outlet of the impeller as blades rotate Draw velocity diagrams at inlet and outlet Apply the angular momentum equation and obtain: shaft torque
m r2V 2 rV 1 1
or
Tshaft Q r2V 2 rV 1 1
Q r2V 2 rV 1 1
or
W shaft Q U 2V 2 U 1V 1
T shaft
shaft power
W shaft
shaft power per unit mass of flowing fluid
w shaft U 2V 2 U 1V 1
notes
notes
6
7/2/2017
Ideal Head
Ideal Head vs. Flowrate
Energy equation
If fluid has no tangential component of velocity at the inlet then
pout
2 Vout
zout
2 g
pin
V in2 2g
zin
W shaft in mg
hL hi
1
= 90°. In this case
U 2V 2 g
maximum (ideal) head rise
hi
W shaft
From velocity diagram
gQ
or
cot 2
U 2V 2 U1V 1
hi or
hi
1
U 2 V 2 V r 2
1
g
Then
V22 V12 U 22 U12 W12 W22
2 g
K.E rise Pressure head rise due to centrifugal effect
hi
Diffusion of relative flow in blade notes
U 22 g
U 2V r 2 cot 2 g
notes
Ideal Head
Ideal Head
The flow rate Q is related to the radial component of the absolute velocity through the equation
Q 2 r2 b2Vr 2
hi
U 22 g
U 2 cot 2 2 r2 b2 g
Q
where b2 is the impeller blade height at the radius r 2. Then ideal head rise:
hi
U 22 g
U 2 cot 2 2 r2 b2 g
Q
Ideal head rise for a centrifugal pump varies linearly with Q for a given blade geometry and angular velocity Thus, ideal head rise for a centrifugal pum p varies linearly with Q for a given blade geometry and angular velocity
Blades with 2 < 90° are called backward curved. Blades with 2 > 90° are called forward curved Pumps are not usually designed with forward curved blades since such pumps tend to suffer unstable flow conditions.
notes
For actual pump, the blade angles have a normal range: 20° < 2 < 25°, 15° < 1 < 50°
notes
7
7/2/2017
Example
Example
Water is pumped at the rate of 1400 gpm through a centrifugal pump operating at a speed of 1750 rpm. The impeller has uniform blade height, b, of 2 in. with r 1 = 1.9 in. and r 2 = 7.0 in., and the exit blade angle 2 is 23°. Assume ideal flow conditions and that the tangential velocity component, V 1, of the water entering the blade is zero ( 1 = 90°). Determine (a) the tangential velocity component, V 2, at the exit, (b) the ideal head rise, hi , and (c) the power transferred to the fluid.
Water is pumped at the rate of 1400 gpm through a centrifugal pump operating at a speed of 1750 rpm. The impeller has uniform blade height, b, of 2 in. with r 1 = 1.9 in. and r 2 = 7.0 in., and the exit blade angle 2 is 23°. Assume ideal flow conditions and that the tangential velocity component, V 1, of the water entering the blade is zero ( 1 = 90°). Determine (a) the tangential velocity component, V 2, at the exit, (b) the ideal head rise, hi , and (c) the power transferred to the fluid.
Solution
Solution (a)
U2
r 107 ft/s 2
V r 2
cot 2
V 2
Q 2 r2 b2
5.11 ft/s
U 2 V 2 V r 2
U 2 V r 2 cot 2 95.0 ft/s
(b)
hi
U 2V 2 g
316 ft
(c)
W shaft QU 2V 2 61500 ft lb/s = 112 hp notes
Actual Head
Effect of losses on the pump head-flowrate curve (backward curved blades)
notes
Pump Performance Characteristics
notes
notes
8
7/2/2017
Pump Performance Characteristics
Pump Performance Characteristics Actual head is determined experimentally
Power gained by the fluid
ha
P f
is made up of hydraulic, mechanical and volumetric efficiencies
Qha
Pump overall efficiency
p2 p1
power gained by the fluid shaft power driving the pump
P
f W shaft
hmv
Experimental setup for determining the head rise gained by a fluid
Experimental setup for determining the head rise gained by a fluid
notes
Pump Performance Characteristics
•
notes
Pump Performance Characteristics
Rising head curve
Head curve continuously rise as flow rate decreases •
Shuttoff Head
Head develop by the pump at zero discharge. It represent the rise in pressure head across the pump with discharge valve closed. At this point the efficiency of pump is zero. Power supplied to pump is dissipated as heat. •
Falling head curve
Break horse power (BHP). As the discharge increases from, the BHP increases, until the maximum discharge is achieved i.e. Best efficiency points (BEP)
Typical performance characteristics for a centrifugal pump of a given size operating at a constant impeller speed
Performance curves for a two-stage centrifugal pum p operating at 3500 rpm. Data given for three different impeller diameters NPSHR – required net positive suction head
notes
back
9
7/2/2017
Net Positive Suction Head (NPSH)
Net Positive Suction Head (NPSH)
Low suction pressure can cause cavitation NPSH is the difference between total head on the suction side and vapor pressure head
NPSH
p s
Vs2 2 g
pv
NPSHR is the required NPSH that must be maintained, or exceeded, so that cavitation will not occur. NPSHR is determine experimentally NPSH A is the available NPSH, which represent the head that actually occurs for the particular flow system. It may be determine experimentally, or calculated if the system parameters are known.
notes
notes
NPSH A for Typical Flow System
NPSH A for Typical Flow System Energy equation
patm
z1
ps
2
V s
2 g
h L
Available head
p s
Vs2 2 g
patm
z1 h L
NPSH available
NPSH A
patm
z1 h L
pv
For pump to operate properly
NPSH A
NPSHR
As the height of the pump above the fluid surface is increased, the NPSH A is decreased. There is some critical value of z 1 above which the pump cannot operate without cavitation. If reservoir is above the pump, NPSH A will increase as the height is increased
notes
notes
10
7/2/2017
Example
Example
A centrifugal pump is to be placed above a large, open water tank, as show in Figure, and is to pump water at a rate of 0.5 ft3/s. At this flowrate the required net positive suction head, NPSHR, is 15 ft, as specified by the pump manufacturer. If the water temperature is 80°F and atmospheric pressure is 14.7 psi, determine the maximum height, z 1, that the pump can be located above the water surface without cavitation. Assume that the major head loss between the t ank and the pump inlet is due to a filter at t he pipe inlet having a minor loss coefficient K L = 20. Other losses can be neglected. The pipe on the suction side of the pump has a diameter of 4 in.
A centrifugal pump is to be placed above a large, open water tank, as show in Figure, and is to pump water at a rate of 0.5 ft3/s. At this flowrate the required net positive suction head, NPSHR, is 15 ft, as specified by the pump manufacturer. If the water temperature is 80°F and atmospheric pressure is 14.7 psi, determine the maximum height, z 1, that the pump can be located above the water surf ace without cavitation. Assume that the major head loss between the t ank and the pump inlet is due to a filter at the pipe inlet having a minor loss coefficient K L = 20. Other losses can be neglected. The pipe on the suction side of the pum p has a diameter of 4 in. Solution
NPSH A
z1 max
h
L
pv
patm
z1 h L
patm
KL
h L
V2 2 g
K L
0.5069 psi,
Answer:
pv
pv
NPSH R 2
Q 10.2 ft 2g A 1
62.22 lb/ft
z 1 max 7.65 ft
2
Where to install the valve?
notes
System Characteristics and Pump Selection
notes
System Characteristics and Pump Selection From energy equation between (1) and (2) ha
z2 z1 hL
With the lost head proportional to Q 2 , the system equation ha
z2 z1 KQ 2
System equation shows how the actual head gained by fluid f rom the pump relates to system parameters Each system has its own specific system equation.
notes
notes
11
7/2/2017
Example
System Characteristics and Pump Selection
Water is to be pumped from one large open tank to a second large open tank as shown in Figure (a). The pipe diameter throughout is 6 in. and the total length of the pipe between the pipe entrance and exit is 200 ft. Minor loss coefficients for the entrance, exit, and the elbow are shown on the figure, and the friction factor for the pipe can be assumed constant and equal to 0.02. A certain centrifugal pump having the performance characteristics shown in Figure (b) is suggested as a good pump for this flow system. With this pump, what would be the flowrate between the tanks? Do you think t his pump would be a good choice?
To select a pump for particular application, it is necessary to utilize system and pump characteristics
notes
notes
Example
Example
Water is to be pumped from one large open tank to a second large open tank as shown inFigure (a). The pipe diameter throughout is 6 in. and t he total length of the pipe between t he pipe entrance and exit is 200 ft. Minor loss coefficients for the entrance, exit, and the elbow are shown on the figure, and the friction factor for the pipe can be assumed constant and equal to 0.02. A certain centrifugal pump having the performance characteristics shown in Figure (b) is suggested as a good pump for t his flow system. With this pump, what would be the flowrate between the tanks? Do you think this pump would be a good choice?
Water is to be pumped from one large open tank to a second large open tank as shown inFigure (a). The pipe diameter throughout is 6 in. and t he total length of the pipe between the pipe entrance and exit is 200 ft . Minor loss coefficients for the entrance, exit, and the elbow are shown on the figure, and the friction factor for the pipe can be assumed constant and equal to 0.02. A certain centrifugal pump having the performance characteristics shown in Figure (b) is suggested as a good pump for t his flow system. With this pump, what would be the flowrate between the tanks? Do you think this pump would be a good choice?
Solution
Solution Energy equation between (1) and (2)
With
z2 f
l V2 D 2 g
K L
V 2 2 g
Q A
10 22 10 5Q 2
system equation
ha
Flowrate
Q 1600 gal/min
Power
notes
V
z1 h p
W shaft
Qha
32.0 hp notes
12
7/2/2017
Pumps Arrangement
Dimensionless Parameters and Similarity Laws Dimensional analysis is used in the study and documentation of pump characteristics
Effect of operating pumps in (a) series and (b) in parallel
notes
Dimensionless Parameters and Similarity Laws
notes
Dimensionless Parameters and Similarity Laws Neglecting Reynolds number and relative roughness effects, for geometrically similar pumps (all pertinent dimensions, l i , scaled by a common length scale), dependent pi terms are functions of only Q/ D 3:
Dimensional analysis is used in the study and documentation of pump characteristics
Principal dependent pump variables are actual head rise, shaft power and efficiency
gha
C h
C P
Dependent variable f D , li , , Q , , ,
2
D
2
Q 1 3 D
W shaft 3 D5
li Q D2 , , , 3 D D D
Dependent pi term
Dimensionless parameter
notes
C Q
Q D 3
Q 2 3 D Q 3 3 D
is called the flow coefficient
notes
13
7/2/2017
Pump Scaling Laws
Use of Pump Scaling Laws
For two pumps from the family operating at the same v alue of flow coefficient:
gha gha 2 D 2 2 D 2 1 2 W shaft W shaft 3 5 3 5 D 1 D 2 1 2 These equations are called pump scaling laws
Pump scaling laws are used to predict the performance of different-sized, geometrically similar pumps.
notes
notes
Example
Example
An 8-in.-diameter centrifugal pump operating at 1200 rpm is geometrically similar to the 12-in.-diameter pump having the performance characteristics of Figs. (a) and (b) while operating at 1000 rpm. For peak efficiency, predict the discharge, actual head rise, and shaft horsepower for this smaller pump. The working fluid is water at 60°F
An 8-in.-diameter centrifugal pump operating at 1200 rpm is geometrically similar to the 12-in.-diameter pump having the performance characteristics of Figs. (a) and (b) while operating at 1000 rpm. For peak efficiency, predict the discharge, actual head rise, and shaft horsepower for this smaller pump. The working fluid is water at 60°F Solution For a given efficiency the flow coefficient has the same value for a given family of pumps From Fig (b) at peak efficiency C Q = 0.0625, C H = 0.19 and C = 0.014 . Thus, for 8-in. pump: P
Q CQ D3
1046 gpm
ha
2
C H D g
2
41.4 ft
Wshaft
CP 3 D5 12.9 hp
Power gained by the fluid
P f
Qha 6020 ft lb/s
Thus, efficiency
P f
shaft W
85%
which checks with the efficiency curve of Fig. (b)
notes
notes
14
7/2/2017
Special Pump Scaling Laws (Pump Affinity Laws)
Special Pump Scaling Laws (Pump Affinity Laws) Q Q D3 D 3 1 2
gha gha 2 D 2 2D 2 1 2
W shaft W shaft 3D 5 3D 5 1 2
1. For the same flow coefficient with D1 = D2 (the same pump operating at different ):
Q1 Q2
1 2
ha1
,
ha 2
12 22
shaft W 1
,
W shaft 2
13 23
1.For the same flow coefficient with 1 = 2 (pumps from the family operating at given )
Q1 Q2
D13 D23
,
notes
Example
ha1 ha 2
D12 D22
,
W shaft 1 Wshaft 2
D15 D25
notes
Example
Answer: 6.9 kL/min; 12.5 m; 17.7 kW
Answer: 0.0328 m3/s; 8 m
15
7/2/2017
Example
Specific Speed Specific speed is obtained by eliminating diameter D between the flow coefficient and the head rise coefficient:
Answer: 4 kL/min; 240 m
N s
Q 34
gha
For any pump, the value of specific speed is specified at the flow coefficient corresponding to peak efficiency
Specific speed is used to select the m ost efficient pump for particular application
Centrifugal pumps are low-capacity, high-head pumps, they have low specific speed.
Axial pumps are high-capacity, low-head pumps, they have high specific speed.
Mixed-flow pumps lie in between
Specific Speed
notes
Suction Specific Speed Suction specific speed , defined as
S s
Q
g NPSH R
34
has a fixed value for a family of geometrically similar pumps. If this value is known, then the NPSH R can be estimated for other pumps w ithin the same family operating at different speed and flow rate
notes
notes
16
7/2/2017
Example
Axial-Flow and Mixed-Flow Pumps
notes
Axial-Flow and Mixed-Flow Pumps
Pumps Comparison
notes
notes
17
7/2/2017
Turbines
End of Lecture
Impulse Turbines
Water enters and leaves the control volume surrounding the wheel as free jets
Hydraulic turbines
•
Turbin es generate power extracting energy from a flowing fluid
•
In turbines, fluid exerts a torque on the rotor in the direction of its rotation
•
In hydraulic turbines working fluid is a water
•
In compressible flow turbomachines working fluid is a gas or steam
•
Two basic types of hydrauli c turbines areimpulse turbines and reaction turbines
•
For hydraulic impulse turbines, the pressure drop across t he rotor is zero; all of the pressure drop across the turbine stage occurs in the nozzles
•
For reaction turbines, part of the pressure drop occurs across the guide vanes and part occurs across the rotor. Reaction is related to the ratio of static pressure drop across the rotor to static pressure drop across the turbine stage
•
In general, impulse turbines are high-head, low-flowrate devices, while reacton turbines are lowhead, high-flowrate devices
•
Which turbines are used in Tarbela power station?
Impulse Turbines: Torque and Power
Hydraulic turbines
Pelton wheel
Magnitude of the relative velocity of the water across the buckets does not change, but its direction does Change in direction of the velocity of the fluid jet causes a torque on the rotor
Pelton wheel turbines operate most efficiently with a larger head and lower florates
Radial component of velocity is negligible.
Fluid leaves bucket with axial component
Design considerations: - head loss in the penstock; - design of the nozzle; - design of the buckets
notes
18
7/2/2017
Hydraulic turbines
Impulse Turbines: Torque and Power
Impulse Turbines: Torque and Power
V 1 V1 W1 U
T shaft
m U V1 mr 1 cos
W2 cos U
W shaft
U V1 1 cos Tshaft mU
V 2
Hydraulic turbines
W 2
with
W1
V 2 V 1
U V1 1 cos
Power is a function of Typical value of
is 165°
Power is maximum at U = V 1/2
Shaft torque
T shaft
m U V1 1 cos mr
Shaft power
W shaft
U V1 1 cos Tshaft mU
Maximum speed occurs at zero torque, then U = V 1 and there is no force from fluid on bucket
Typical theoretical and experimental power and torque for a Pelton wheel turbine as a function of bucket speed
Example: Water to drive a Pelton wheel is supplied through a pipe from a lake as indicated in figure. The head loss due to friction in the pipe is important, but minor losses can be neglected. Determine: (a) the nozzle diameter, D1, that will give the maximum power output ; (b) the maximum power and the angular velocity of the rotor ant the conditions found in part (a) Answer: 0.07 m; −4.2 ×10 4 N∙m at 302 rpm
Notes
Example: Water flows through the Pelton wheel turbine shown in figure. For simplicity assume that the water is turned 180º by the blade. Show, based on the energy equation, that the maximum power output occurs when the absolute velocity of the fluid exiting the tu rbine is zero
Notes
19
7/2/2017
Example: Water flows through the Pelton wheel turbine shown in figure. For simplicity assume that the water is turned 180º by the blade. Show, based on the energy equation, that the maximum power output occurs when the absolute velocity of the fluid exiting th e turbine is zero
Impulse Turbines
Multinozzle, non-Pelton wheel impulse turbine commonly used with air as the working f luid
An air turbine used to drive the high-speed drill used by your dentist is shown in figure. Air exiting from the upstream nozzle holes forces the turbine blades to move in the direction shown. The turbine rotor speed is 300,000 rpm, the tangential component of velocity out of the nozzle is twice the blade speed, and the tangential component of the absolute velocity out of the rotor is zero. Estimate the shaft energy per unit mass of air flowing through the turbine (-29 kN∙m/kg)
Reaction Turbines
Hydraulic turbines
Radial flow Francis turbine
20
7/2/2017
Reaction Turbines
Hydraulic turbines
Hydraulic turbines
N S
Axial flow Kaplan turbine
shaft W
ghT
54
Turbine cross sections and maximum efficiencies as a function of specific speed
The single-stage, axial-flow turbomachine shown in figure involves water flow at a volumetric flowrate of 9 m3/s. the rotor revolves at 600 rpm. The inner and outer radii of the annular flow path through the stage are 0.46 and 0.61 m, and β 2 = 60º. The flow entering the rotor row and leaving the stator row is axial when viewed from the stationary casing. Is this device a turbine of a pump? Estimate the amount of power transferred to or from the fluid. (816 kW)
Compressible flow Turbomachines
21
7/2/2017
Compressible flow Turbomachines
Compressors Pressure ratio
PR
p02 p01
For multistage compressor with the same PR of each stage n
p02 p01
PR
Two-stage centrifugal compressor with an intercooler Rotor from automobile turbocharger
Axial-flow Compressors
Enthalpy, velocity, and pressure distribution in an axial-flow compressor
Compressors
Performance characteristics of an axial-flow compressor
22
7/2/2017
Compressible Flow Turbines
Compressible Flow Turbines
Enthalpy, velocity, and pressure distribution in a three-stage reaction turbine
Enthalpy, velocity, and pressure distribution in two-stage impulse turbine
Compressible Flow Turbines
Wind Turbines
Typical compressible flow turbine performance map
23
7/2/2017
Wind Turbines
Ideallized Wind Turbines Theory
•
Horizontal-axis win d turbines (HAVT). Dutch m ill, American farm mill, propeller turbine.
•
Vertical-axis wind wind turbines (VAWT). Darrieus rotor, Savonius rotor.
Idealyzed actuator-disk and streamline analysis of flow through a windmill (A. Betz, 1920)
notes
Ideallized Wind Turbines Theory Power extracted by the disk P FV
1
A V12 V22 V1 V 2 4
Maximum possible power Pmax
8 27
AV13
at
V2
1
V 1 3
Power coefficient C p
Idealyzed actuator-disk and streamline analysis of flow through a windmill (A. Betz, 1920)
P 1 AV 13 2
Maximum efficiency (Betz number) C p ,max
0.593 Estimated performance of various wind turbine designs as a function of blade-tip speed ratio. notes
24
7/2/2017
End of Lecture
World availabilit y of land-based wind energy: estimated annual electric output in kWh/kW of a wind turbine rated at 11.2 m/s
back
References
Schematic design of positive-displacement pumps: (a) reciprocating piston or plunger, (b) external gear pump, (c ) double-screw pump, (d ) sliding vane, (e) threelobe pump, (f) double circumferential piston, (g ) flexible-tube squeegee.
25
7/2/2017
Pump Theory
Velocity diagram at the inlet and exit of a centrifugal pump
back
Example 12.4
back
Example 12.4
back
back
26
7/2/2017
Hydraulic turbines
Impulse Turbines
Example 12.5
Pelton Wheel Turbine
back
•
Total head of incoming flu id is converted into velocity head;
•
both the pressure drop and change in relative speed of the fluid across the bucket (blade) are negligible
•
space surrounding the rotor is not completely filled with fluid
•
torque is generated by the impulse of jets striking the buckets
back
Hydraulic turbines
Reaction Turbines
Francis Turbine •
Rotor is surrounded by a casing completely filled with fluid;
•
there is both a pressure drop and a fluid relative speed change across the rotor;
•
guid e vanes accelerate flow and turn it in the appropriate direction as fluid enters the rotor;
•
part of pressure drop occurs across the guide vanes and part occurs across the rotor
back
27
7/2/2017
HAWT
HAWT
Dutch mill American multiblade farm mills back
back
HAWT
VAWT
Modern propeller mills Savonius rotor back
Darrieus rotor back
28
7/2/2017
VAWT
VAWT
Darrieus turbine
Savonius rotor back
back
VAWT
Ideallized Wind Turbines Theory F
F m V2 V 1
F
F pb
x
x
F
Va pa A m
V b 0
pb pa A m V1 V 2
1 1 p V12 pb V 2 2 2 1 1 pa V 2 p V 22 2 2 pb pa P
Savonius rotor + Darrieus rotor
1 FV AV V1 V2 AV12 V22 V1 V 2 4
1
V12 V22 V V1 V 2 2
2
Pmax
8 AV13 27
at
V2
V 1
1 3
Pmax
8 AV13 27
at
V2
V 1
V
1
V1 V 2 2
1 3
back
29
