turbomachine
TRANSCRIPT
1
Turbo-machinery
Pelton Wheel
Rjukan Hydroplant (Telemark)
http://exviking.net/man/Rjukan p.htm
2
Fluid Machines.
There are two basic Fluid Machine designs.
Positive Displacement Machines Positive
displacement machines force fluid into or out of
the volume of a chamber by changing the volume
of the chamber. Examples are bicycle pumps,
the lungs, the heart, and the cylinders of an
internal combustion engine (ICE). In a bicycle
pump, the device does work on the fluid. In an
ICE, the fluid does work on the piston head.
Turbomachines Turbomachines involve blades,
buckets, or passages arranged around an axis of
rotation. The rotations will either add or
subtract energy from the fluid. Window fans,
propellers, gas turbines. jet engines. The key
feature is some sort of rotary motion is involved.
It is more important to know about pumps than
turbines since there are many more pumps in the
world.
3
Turbomachinery
There are two different turbomachine functions.
Pumps These are devices to add energy to a fluid.
Usually want to direct the fluid to move to a
given places. These can be pumps, fans, blowers
or compressors.
Turbines These are devices designed to extract
energy from a fluid flow. Gas and steam
turbines.
4
Turbomachines
Turbo-machines can be characterized as radial-flow
or axial flow.
In a radial flow machine,
the fluid has a signifi-
cant velocity component
around the axis of the
machine.
In an axial flow machine,
the fluid has a signifi-
cant velocity component
along the axis of the ma-
chine.
Rotor
Inlet
Housing or casing
Outlet
(a) Radial flow fan
(b) Axial-flow fan
Housingor casing
Inlet
Rotor
Outlet
Stator
ω
There is a 3rd type of machine, the mixed-flow
machine. These different machines can be used for
different applications.
5
The centrifugal pump
(a)
Discharge
Impeller
Eye
Inflow
Blade
Hub plate
Casing, housing,or volute
(b)
Consists on an
• Impeller. The impeller is attached to the
rotating shaft.
• Housing. The impeller is enclosed in a housing,
casing or volute.
The impeller has a number of rotating blades or
vanes. As the impeller rotates, fluid is sucked in the
eye, the vanes add energy to the fluid
The vanes can be radial, forward inclined or
backward inclined.
6
The centrifugal pump
In an open impeller the blades are arranged on a hub
or backing plate.
In a enclosed impeller the blades are covered on the
hub and also by a shroud on the inlet side.
7
Head Losses
The head-gain by a centrifugal pump is
hl =(ωR)2
g−
ωR cot(β2)Q
2πRbg
• R is distance to end of vane
• ω is angular velocity of shaft
• Q is volume flow-rate
• b is impeller blade height on rim
• β2 is angle of impeller at rim
8
Head Losses
The actual pressure increase for a real pump is
slightly less than the ideal cases.
Flowrate
Head
Otherlosses
Actual head, ha
Theoretical head, hi
Friction losses
• There are shock losses at entrance when fluid
does not enter impeller smoothly. Shock losses
small near optimum flow rate.
• Friction losses increases as Q2
• Loss of fluid between impeller blades and casing.
9
Pump Performance parameters
The performance of a pump can be determined by
measuring water pressure immediately before and
after pump.
(1)
(2)
z2 – z1
The actual head rise ha = hs − hL , depends on shaft
head work hs , and head loss hL through the pipe
and valves in pump.
ha =p2 − p1
γ+ z2 − z1 +
v22− v2
1
2g
Typically, the changes in elevation and fluid velocity
are small so,
ha ≈p2 − p1
γ
10
Pump Performance parameters
The power gained by the fluid as it moves through
the pump is
Pf = γQha
A measure of the overall efficiency η of the pump is
η =Pf
Wshaft
=γQha/550
bhp
Sometimes the shaft power is given in terms of of the
brake horsepower of the pump when using BG units
(the 550 would be 746 for SI units).
The efficiency is effected by hydraulic (e.g. viscous)
losses, mechanical losses (e.g. energy loss in
bearings) and volumetric losses (e.g. loss of fluids
between end of impeller blade and casing)
11
Performance curves for pumps
The performance characteristics of a given pump are
summarized in plots of ha , η and bhp versus Q .
Only two curves are
really needed since
ha , η and bhp are
closely related.
Shutoff head
Head,
ha
Bra
ke h
ors
epow
er,
bhp
Eff
icie
ncy,
Head
Efficiency
Brake horsepower
00
Normal ordesign flowrate
Flowrate, Q
η
The design flow-rate is usually the point where the
efficiency is largest (best efficiency point or BEP ).
12
Pump Performance terminology
Typical pump per-
formance curve for
centrifugal pump.
0 40 80 120 160 200 240 280 320
NPSHR
NP
SH
R,
ft
8 in. dia
7
6
50%
55
60
63
65
50
5560
6365
40 bhp
30
25
20
15
Capacity, gal/min
Head,
ft
0
100
200
300
400
500
15
10
5
0
13
Net Positive Suction Head (NPSH)
On the suction side of a pump one encounters low
pressures. Leads to possibility of cavitation
occurring at high speeds. Cavitation occurs when
pfluid < pvp pvp = pvp(T ).
Cavitation leads to loss in efficiency and structural
damage. The Net Positive Suction Head required,
NPSHR , is a plot of the pressure that must be
maintained at the pump inlet to avoid cavitation vs
Q .
Referenceplane
p1 = patm
(1)
(2)
z1
14
NPSHR
The NPSH is defined as
NPSH =ps
γ+
v2s
2g−
pvp
γ
This is the total (static + dynamic) pressure minus
the vapour pressure.
The required NPSH , or NPSHR is determined by
the pump manufacturer. The NPSHR is a function
of flow-rate.
The NPSH is defined
with the dynamic pres-
sure included since this
means NPSHR vs Q
curve given by manufac-
turers builds in flow-rate
effects.
Referenceplane
p1 = patm
(1)
(2)
z1
15
Net Positive Suction Head Available, NPSHA
The modified
Bernoulli equation
will be applied
between (1) and
(2) .
Referenceplane
p1 = patm
(1)
(2)
z1
At (1) , v1 = 0 , z1 = 0 and p1 = patm . Now apply
Bernoulli equation (z1 in diagram is really z2 ) with
hL head loss between tank and impeller inlet.
patm
γ− hL =
p2
γ+
v22
2g+ z2
p2
γ+
v22
2g=
patm
γ− hL − z2
NPSHA =patm
γ− hL − z2 −
pvp
γ
To operate the pump without cavitation requires
NPSHA > NPSHR
16
NPSHA
NPSHA =patm
γ− hL − z2 −
pvp
γ
The NPSHA decreases as z2 (height of impeller
above fluid) is increased. There is a critical value of
z2 . If z2 is too large then cavitation will occur.
0 40 80 120 160 200 240 280 320
NPSHR
NP
SH
R,
ft
8 in. dia
7
6
50%
55
60
63
65
50
5560
6365
40 bhp
30
25
20
15
Capacity, gal/min
Head,
ft
0
100
200
300
400
500
15
10
5
0
The NPSHR increases as the flow-rate increases.
Note, at zero flow-rate the NPSHR is only about 2
m ; local water speeds in pump can be much larger
than water speeds in pipes.
17
Interpretation of NPSH and NPSHA
18
Control Valve positioning
The flow rates of pump/pipe systems are often
controlled by placing control valves somewhere in the
system.
As a general rule it is best to place the valve on the
downstream side of the pump. Placing the control
valve before the pump will decrease the NPSHA and
thus make cavitation more likely to occur.
Note, using a control valve to adjust the flow rate is
a bit like controlling the speed of a car by using the
brake while maintaining constant accelerator pedal
pressure!
19
Pump Selection
(1)
Pump
(2)
z1
z 2
Typical pump selection scenario. Water in a tank at
one elevation needs to be pumped to a tank at
another elevation. With modified Bernoulli equation
p1
γ+
v21
2g+ z1 + hP − hL =
p2
γ+
v22
2g+ z2
Now v1 = v2 = 0 and p1 = p2 = 0 , so
z1 + hP − hL = z2
⇒ hP = hL + z2 − z1
The head supplied by the pump must be large
enough to overcome head losses in the pipe and
elevation changes.
20
Pump Selection
To a first approximation, the head losses are
proportional to Q2 (e.g. constant friction factor).
hP = z2 − z1 + hL
hP = z2 − z1 + KQ2
This equation is called the system equation.
The operating (duty)
point of the pump is de-
termined from the in-
tersection of the system
and pump curves. So-
lution of two simulta-
neous non-linear equa-
tions. Graphical solu-
tion may be easiest.
Change insystem equation System
curve
Efficiencycurve
Operatingpoint
Pump performancecurve
Pum
phead,
hp
Eff
icie
ncy
(A)(B)
Flowrate, Q
Elevation (static) head= z2 – z1
The system curve can change over time, e.g.
build-up of deposits in pipe walls.
21
Pumps in series and parallel
Two pumps
(B)
(A)
Head,
ha
Head,
ha
Flowrate, Q Flowrate, Q
Two pumps
One pump
One pumpP
P
PP
(A)
(B)
Systemcurve
Systemcurve
(a) (b)
Connecting two pumps in series gives a larger
pressure head. The fluid can be raised to a higher
elevation.
Connecting two pumps in parallel results in a higher
flow-rate.
22
Example: Pump performance Curve
Water is to be pumped from one large open tank to
another. The pipe diameter is 0.50 ft and the pipe
length is 200 ft . There are minor losses at the
entrance, exit and throughout the pipe. The friction
factor will be taken as 0.02 . What is the flow-rate
and shaft-power needed?
KL = 0.5
(1)
KL = 1.5
Pump
KL = 1.0
(2)
Water
Diameter of pipe = 6 in.
Total pipe length = 200 ft
0 400 800 1200 1600 2000 24000
20
40
60
80
100
Flowrate, gal/min
Head,
ftE
ffic
iency,
%
Head
10 ft
Efficiency
(b)
(a)
23
Example: Pump performance Curve
Apply the modified Bernoulli equation
p1
γ+
v21
2g+ z1 + hP − hL =
p2
γ+
v22
2g+ z2
Now v1 = v2 = 0 , z1 = 0 , z2 = 10 ft , p1 = p2 = 0 .
The head loss is
hL =flv2
2dg+∑
KLv2
2g
where v = Q/A is fluid velocity in pipe. Using
numbers
hL =0.02 × 200Q2
2 × 0.5 × 32.2A2+ (0.5+1.0+1.5)
Q2
2 × 32.2A2
=20Q2
32.2 × 0.196352+ 3.0
Q2
64.4 × 0.196352
= 4.43Q2
So the modified Bernoulli equation is
hP = 10 + 4.43Q2
24
Example: Pump performance Curve
Since the pump curve is given in gal/min , need to
do a conversion
hP = 10 + 4.43Q2
hP = 10 + 2.2 × 10−5Q2
What has to be done
is to solve two simul-
taneous non-linear
equations. Resort to
graphical solution.
(c)
0 1600 24000
66.5
100
Flowrate, gal/min
Head,
ftE
ffic
iency,
%
Head
Efficiency
Operatingpoint
System curve (Eq. 4)
The point of interaction is the operating point for
the pump and system. One finds a flow-rate of
Q = 1600 gal/min = 3.56 ft3/s
25
Example: Pump performance Curve
Finally, the power required is
Wshaft =γQha
ηgal/min
Wshaft =62.4 × 3.56 × 66.5
0.84= 17000 ft lb/s
26
Dimensional analysis: centrifugal pump
The aim is to determine the groupings of variable
that describe the important pump parameters.
There are three parameters of interest
• The pump shaft-power, Wshaft
• The actual head rise, ha
• The efficiency of the pump, η
These parameters will depend on the following
• The characteristic diameter, D
• The surface roughness, ǫ
• The pump volumetric flow rate, Q
• The pump rotation speed, ω
• The fluid viscosity, µ
• The fluid density, ρ
• Some characteristic lengths describing the pump
geometry, li
So Parameter = F (D, li, ǫ, Q, ω, µ, ρ)
27
Dimensional analysis: centrifugal pump
Parameter = F (D, li, ǫ, Q, ω, µ, ρ)
Will choose ρ , ω and D as the repeating variables.
In a Buckingham Π analysis one find
Π(Parameter) = F
(
liD
,ǫ
D,
Q
ωD3,ρωD2
µ
)
The dependent Π terms are
Head rise coefficient. This is the actual head rise per
unit mass (i.e. gha ).
CH =gha
ω2D2= F1
(
liD
,ǫ
D,
Q
ωD3,ρωD2
µ
)
Power coefficient.
CP =Wshaft
ρω3D5= F2
(
liD
,ǫ
D,
Q
ωD3,ρωD2
µ
)
The efficiency, η is already dimensionless
η =ρgQha
Wshaft
= F3
(
liD
,ǫ
D,
Q
ωD3,ρωD2
µ
)
28
Dimensional analysis: Centrifugal pump
F3
(
liD
,ǫ
D,
Q
ωD3,ρωD2
µ
)
The last argument ρωD2
µ is a type of Reynolds
number. Pumps are usually operated at high
rotation speeds, so relative impact of viscous effects
is not important. This term has little impact.
The surface roughness term, ǫD is unimportant since
interiors of pumps are irregular with sharp bends.
For geometrically similar pumps, liD does not
change. So
CH =gha
ω2D2= = F1
(
Q
ωD3
)
CP =Wshaft
ρω3D5= F2
(
Q
ωD3
)
η =ρgQha
Wshaft
= F3
(
Q
ωD3
)
CQ =Q
ωD3
29
Pump Scaling laws
For equal flow coefficients,(
Q
ωD3
)
1
=
(
Q
ωD3
)
2
One finds equality between other coefficients(
gha
ω2D2
)
1
=
(
gha
ω2D2
)
2(
Wshaft
ρω3D5
)
1
=
(
Wshaft
ρω3D5
)
2
η1 = η2
100%
80
60
40
20
0
100
80
60
40
20
0
Bra
ke h
ors
epow
er
Eff
icie
ncy
0
10
20
30
40
50
60
70
80
0 1000 2000 3000 4000 5000
Capacity, gal/min
(a)
Head,
ft
Horsepower
Head
Efficiency
0
0.05
0.10
0.15
CH
0.20
0.25
0 0.025 0.050 0.075 0.100
CQ
(b)
100%
80
60
40
20
0
0.016
0.012
0.008
0.004
0
C
η
C (pump)
CH
η
P
P
30
Special Pump Scaling laws
There are a number of special cases where the
similitude relations collapse to give some very useful
principles
Same CQ and D1 = D2 (same pump)
One finds
Q1
Q2
=ω1
ω2
ha1
ha2
=ω2
1
ω22
Wshaft1
Wshaft2
=ω3
1
ω32
So at a fixed flow coefficient, the flow-rate is
proportional to speed, while the head varies as
the square of the speed and the power varies as
the cube of the speed,
31
Same CQ and ω1 = ω2 (same speed)
Q1
Q2
=D3
1
D32
ha1
ha2
=D2
1
D22
Wshaft1
Wshaft2
=D5
1
D52
The flow rate is proportional to the diameter
cubed, the head generated is proportional to the
square of the diameter and the amount of shaft
work is proportional to the diameter to the fifth
power.
Pump manufacturers often put different sized
impellers in identical casings, so exact geometric
similarity is not maintained. OK to using these
scaling relations (pump affinity laws) if impeller size
does not change by more than 20% .
32
Specific speed
A dimensionless parameter that is useful in pump
section is the specific speed. It is a combination of
two Π terms, namely C1/2
Q /C3/4
H .
Ns =C
1/2
Q
C3/4
H
=ω√
Q
(gha)3/4
The specific speed at the flow coefficient
corresponding to peak efficiency are listed for given
pumps. It gives an indication of what type pump
works most efficiently for a given combination of Q
and ha . Centrifugal pumps often have low-capacity
and high-heads, so they have low specific speeds.
50
0
60
0
70
0
80
0
90
0
10
00
15
00
20
00
30
00
40
00
50
00
60
00
70
00
80
00
90
00
10
00
0
15
00
0
20
00
0
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
2.0
3.0
4.0
5.0
6.0
7.0
8.0
Radial flow
Specific speed, Ns
Specific speed, Nsd
Mixed flow Axial flow
Axis of
rotation
Vanes
Hub
Impellershrouds
Vanes
Hub
Vanes
Hub
Vanes
Hub
Impellershrouds
Vanes
Impellershrouds
Impellerhub
33
Impulse Turbines
Impulse turbines use momentum transfer from a
water jet to spin a turbine. One of the easiest to
understand is the Pelton wheel (invented in 19th
century by Lester Pelton, a mining engineer in
California).
Rotor
BucketNozzle
(a)
The energy of the water stream is partly converted
to energy to drive the turbine. There is a nozzle to
increase the velocity of the water stream.
34
Pelton Wheel
(b)
The water stream is split into two channels when it
leaves the turbine. Sending the water backward gives
a bigger momentum transfer and splitting it to either
side redirects recoiling stream away from incoming
stream.
35
Pelton Wheel
Water with a velocity of v1 strikes the Pelton wheel
bucket. The Pelton wheel bucket is moving at a
speed of U = ωrm .
Tangential
V1
V2
Ub
ω
Radial
rm
a
The redirected streams leaves the bucket in two
equal sized streams moving at a velocity v2 .
36
Pelton Wheel
The streams are di-
rected to each side at
an angle of β to in-
coming direction.
a
a
b
b
W1 = V1 – U
Blade cross section
W2 = W1 = V1 – U
β
Tangential
Axial
The velocity components in the axial direction do
not contribute to the torque generated by the wheel.
The relative velocity of the incoming stream in
tangential direction is
w1 = v1 − U
The relative velocity of the outgoing stream
(tangential direction) is
w2 cos β = v2 − U
37
Pelton Wheel
Force applied by the bucket to the water stream is
Fjet = m(w2 cos β − w1)
Assuming w2 ≈ w1 , (elastic collision in bucket ref.
frame)
F = m(w1 cos β − w1) = mw1(cos β − 1)
Force of water on bucket is equal and opposite so
Fbucket = mw1(1 − cos β)
Fbucket = m(v1 − U)(1 − cos β)
The torque applied to the shaft is
τ = Fbucketrm
= mrm(v1 − U)(1 − cos β)
The rate of shaft work being done (on the fluid, note
sign change) is
Wshaft = mU(U − v1)(1 − cos β)
38
Pelton Wheel
The rate of shaft work being done is
Wshaft = mU(U − v1)(1 − cos β)
Since v1 > U , shaft work being done is negative.
The Pelton wheel extracts energy from the fluid.
At what speed should the Pelton wheel rotate to
extract the maximum shaft power out of the water
stream?
Want β as large as possible. Typically β ≈ 165o so
cos(165o)= −0.966 . The (1− cos β) factor is 1.966 .
The torque is a maximum when U = 0 , but now
work is being done when wheel is not turning. The
maximum power out occurs when U(U − v1) is a
maximum.
Umaxpower =v1
2
39
Pelton Wheel
Efficiency of a Pelton wheel as a function of the rim
rotation speed.
Actualpower
β
Tshaft
Actualtorque
0 0.2 V1 0.4 V1 0.6 V1 0.8 V1 1.0 V1
U = rm
shaft
⋅–W –Tshaft
Tshaftmax= mrmV1(1– cos )⋅
max= 0.25mV1
2(1– cos )⋅ β
U = 0.5V1max power
shaft
⋅ W
shaft
⋅ W
ω
You should note that U = v1/2 corresponds to
v2 ≈ 0 (need v2 > 0 to get water out of way) . Most
of the kinetic energy of the incoming water stream
has been converted to the task of spinning the Pelton
wheel.
40
Reaction Turbines
In a reaction turbine, water completely fills are the
passageways in the turbine. Most useful for higher
flow rates and low pressure heads.
Rotor blades
ω
RotorAdjustableguide vanes
Draft tube
ω
Adjustableguide vane
Plan view of guide vanes
ω
(a) (b)
ω
The Francis turbine is a radial (or mixed) flow
machine. At lowest flow rates the axial-flow or
Kaplan turbine is most efficient.
41
Turbine efficiencies
10 20 40 60 80 100
10 20 40 60 80 100
′Nsd
100
90
80
70
%η
Francis
Impulse
Kaplan
Impulse turbines
Radial-flow Mixed-flow Axial flow
Reaction turbines
′Nsd