model simlitude [compatibility mode]
DESCRIPTION
mast charasTRANSCRIPT
-
Lecture outline Dimensional analysis Model similitude Dimensionless numbers for turbo-
machinesmachines Physical significance of such
dimensionless numbers Specific speed for turbo-machines examples
-
Example: Drag on a Sphere
Nature of Dimensional Analysis
BITS Pilani, K K Birla Goa Campus
Drag depends on FOUR parameters:sphere size (D); speed (V); fluid density (); fluid viscosity ()
Difficult to know how to set up experiments to determine dependencies
Difficult to know how to present results (four graphs?)
-
Example: Drag on a Sphere
Nature of Dimensional Analysis
BITS Pilani, K K Birla Goa Campus
Only one dependent and one independent variable Easy to set up experiments to determine dependency Easy to present results (one graph)
-
Nature of Dimensional Analysis
BITS Pilani, K K Birla Goa Campus
-
Step 1:List all the dimensional parameters involved
Let n be the number of parameters
Example: For drag on a sphere, F, V, D, , , and n = 5
Buckingham Pi Theorem
BITS Pilani, K K Birla Goa Campus
-
Step 2Select a set of fundamental (primary) dimensions
For example MLt, or FLt
Example: For drag on a sphere choose MLt
Buckingham Pi Theorem
BITS Pilani, K K Birla Goa Campus
-
Step 3List the dimensions of all parameters in terms of primary dimensions
Let r be the number of primary dimensions
Example: For drag on a sphere r = 3
Buckingham Pi Theorem
BITS Pilani, K K Birla Goa Campus
-
Step 4Select a set of r dimensional parameters that includes all the primary dimensions
Example: For drag on a sphere (m = r = 3) select , V, D
Buckingham Pi Theorem
BITS Pilani, K K Birla Goa Campus
-
Step 5Set up dimensional equations, combining the parameters selected in Step 4 witheach of the other parameters in turn, to form dimensionless groups
There will be n m equations
Example: For drag on a sphere
Buckingham Pi Theorem
BITS Pilani, K K Birla Goa Campus
-
Step 5 (Continued)Example: For drag on a sphere
Buckingham Pi Theorem
BITS Pilani, K K Birla Goa Campus
-
Step 6Check to see that each group obtained is dimensionless
Example: For drag on a sphere
Buckingham Pi Theorem
BITS Pilani, K K Birla Goa Campus
-
Principles of model similitude By making use of the principle of similarity, the performance of one
machine from the results of tests on a geometrically similar machine can be predicted. Why is this necessary?
Even the performance of the same machine under conditions different from the test conditions can be predicted.
Necessary conditions for this principle are geometrical similarity, kinematic similarity and dynamic similarity to all significant geometrical similarity, kinematic similarity and dynamic similarity to all significant
parts of the system viz., the rotor, the entrance and discharge passages and so on.
Machines which are geometrically similar form a homologous series. Therefore, the member of such a series, having a common shape are simply enlargements or reductions of each other.
If two machines are kinematically similar, the velocity vector diagrams at inlet and outlet of the rotor of one machine must be similar to those of the other. It is generally a necessary condition for dynamic similarity.
-
Example: Drag on a Sphere
For dynamic similarity
Flow Similarity and Model Studies
BITS Pilani, K K Birla Goa Campus
then
-
Incomplete Similarity
Sometimes (e.g., in aerodynamics) complete similarity cannot be
Flow Similarity and Model Studies
BITS Pilani, K K Birla Goa Campus
complete similarity cannot be obtained, but phenomena may still be successfully modelled
-
Variable physical parameters Dimensional formula
D = any physical dimension of the machine as a measure of the machine's size, usually the rotor diameter
L
Q = volume flow rate through the machine L3 T -1N = rotational speed (rev/min.) T -1H = difference in head (energy per unit weight) across the machine. This may be either gained or given by the fluid L either gained or given by the fluid depending upon whether the machine is a pump or a turbine respectively
L
=density of fluid ML-3
= viscosity of fluid ML-1 T -1
k = coefficient of elasticity of fluid ML-1 T-2g = acceleration due to gravity LT -2
P = power transferred between fluid and rotor
ML2 T-3
-
Formation of dimensionless groups
it is more logical to consider the energy per unit mass gH as the variable rather than H or g alone.
Therefore, the number of separate variables becomes eight: D, Q, N, gH, , , k and P .
Since the number of fundamental dimensions required to express these variable are three, the number of independent terms these variable are three, the number of independent terms (dimensionless terms), becomes five. Using Buckingham's theorem with D, N and as the repeating variables, the expression for the terms are obtained as,
2
1 2 3 4 53 2 2 3 5 2 2; ; ; ;kQ gH ND P
ND N D N D N D
pi pi pi pi pi
= = = = =
1
2
( , , , , , )( , , , , , )
P f D Q N kgH f D Q N k
=
=
-
Physical significance of first pi term
All lengths of the machine are proportional to D , and all areas to D2. Therefore, the average flow velocity at any section in the machine is proportional to Q/D2 .
Again, the peripheral velocity of the rotor is proportional to the product ND .
The first term can be expressed as The first term can be expressed as
This term thus represents the condition for kinematic similarity, and is known as capacity coefficient or discharge coefficient
2
1 3fluid velocity Vrotor velocity U
QQ D
ND NDpi = =
-
Physical significance of second pi term
The second term is known as the head coefficient since it expresses the head H in dimensionless form. Considering the fact that ND rotor velocity, the term becomes , and can be interpreted as the ratio of fluid head to kinetic energy of the rotor.
2 2 2gH
N Dpi =
-
Physical significance of third pi term
The term can be expressed as and thus represents the Reynolds number with rotor velocity as the characteristic velocity.
-
The term expresses the power P in dimensionless form and is therefore known as power coefficient . Combination of and in the form of gives .
The term QgH' represents the rate of total energy given up by the fluid, in case of turbine, and gained by the fluid in case of pump or compressor. Since P is the power transferred to or from the rotor.
Physical significance of fourth pi term
compressor. Since P is the power transferred to or from the rotor. Therefore the combination becomes the hydraulic efficiency for a
turbine and reciprocal of it for a pump or a compressor.
-
From the fifth term, we get
Multiplying with , we get
Physical significance of fifth pi term
5
1 NDkpi
=
It represents the well known Mach number
21
5
fluid velocitylocal acoustic velocity
QDk
pi
pi
=
-
Overall comments For a fluid machine, handling incompressible fluid, the term can
be dropped. The effect of liquid viscosity on the performance of fluid machines is
neglected or regarded as secondary, (which is often sufficiently true for certain cases or over a limited range).Therefore the term can also be droped.also be droped.
One set of relationship or curves of the pi terms would be sufficient to describe the performance of all the members of one series.
If data obtained from tests on model machine, are plotted so as to show the variation of dimensionless parameters with one another, then the graphs are applicable to any machine in the same homologous series. The curves for other homologous series would naturally be different.
-
Example: Centrifugal Pump
Flow Similarity and Model Studies
BITS Pilani, K K Birla Goa Campus
Pump Head
Pump Power
-
Example: Centrifugal Pump
Flow Similarity and Model Studies
BITS Pilani, K K Birla Goa Campus
Head Coefficient
Power Coefficient
-
(Negligible Viscous Effects)
Flow Similarity and Model Studies
BITS Pilani, K K Birla Goa Campus
If then
-
Unit Quantities (used in turbine design)
111/2 2 1/2 1/2
2
Unit Discharge
Formula derived by thus =
( ignored due to the use for same turbine)
Q QQD H H
D
pi
pi= =
Unit Power pi
=43/2
2
1 3/2
Similarly, this Formula derived by
which finally reduces to
( ignored due to the use for same turbine)
PPH
D
pi
pi=
=
11/2 1/2 1/22
Unit Speed
1Formula derived by reduces to
( and ignored )
ND NNgH H
g Dpi
= = =
-
Dimensionless specific speed for a pump, The performance or operating conditions for a pump handling a
particular fluid are usually expressed by the values of N , Q and H
A parameter involving N , Q and H but not D is obtained by dividing by as below.
-
Dimensionless specific Speed for a turbine
The performance or operating conditions for a turbine handling a particular fluid are usually expressed by the values of N , P and H.
It is important to know the range of these operating parameters covered by a machine of a particular shape (homologous series) at high efficiency. Such information enables us to select the type of machine best suited to a particular application, and thus serves as a machine best suited to a particular application, and thus serves as a starting point in its design.
Therefore a parameter independent of the size of the machine D is required which will be the characteristic of all the machines of a homologous series.
A parameter involving N , P and H but not D is obtained by dividing by .
-
Importance of dimensionless specific speed for model simlitude
Since the dimensionless parameters are found as a combination of basic terms, they must remain same for complete similarity of flow in machines of a homologous series.
Therefore, a particular value of dimensionless specific speed relates all the combinations of N , P and H or N , Q and H for which the flow conditions are similar in the machines of that homologous series. conditions are similar in the machines of that homologous series.
For turbines, the values of N , P and H , and for pumps and compressors, the values of N , Q and H are usually quoted for which the machines run at maximum efficiency. The machines of particular homologous series, that is, of a particular shape, correspond to a particular value of specific speed for their maximum efficient operation.
-
Popular expressions for specific speed for turbines and pumps
Considering the fluids used by the machines to be incompressible, (for hydraulic turbines and pumps), and since the acceleration due to gravity does not vary under this situation, the terms g and rho are taken out from the expressions of dimensionless specific speed.
1/2
Specific speed for a turbine, N NP=ST 5/4Specific speed for a turbine, N NPH
=
1/2
SP 3/4Specific speed for a pump, N NQH
=
For a turbine, specific speed is the speed of a member of the same homologous series as the actual turbine, so reduced in size as to generate unit power under a unit head of the fluid. Similarly, for a pump, it is speed of a hypothetical pump with reduced size but representing a homologous series so that it delivers unit flow rate at a unit head. The specific speed is not a dimensionless quantity
-
Popular expressions for specific speed for turbines and pumps
Units of Ns are not RPM Magnitude of turbine specific speed (Ns ) can differ based on
different units of power In SI system, P is in kW In metric units, P is in metric horsepower(1MHP=0.736kW) Ns(SI)=0.8576Ns(Metric) Also, Ns(SI)=165.8
Magnitude of pump specific speed (ns ) is same in SI and metric units
-
Type of Turbine Ns in SIImpulse (Pelton)
Slow 4 32Normal 16 50Fast 23 65
Radial and Mixed Slow
Slow 65 110Normal 110 200Mixed Slow
flow (Francis and Deriaz)
Normal 110 200Fast 200 400
Axial flow (Kaplan)
Slow 300 450 Normal 450 700Fast 700 1200
-
PRACTICE PROBLEMS/EXAMPLES1.6, 1.7,1.8 AND 1.101.6, 1.7,1.8 AND 1.10(APPROACH DISCUSSED IN EXAMPLE 1.9 IN BOOK IS INCORRECT, PLEASE FOLLOW THE APPROACH DISCUSSED IN SLIDES)
-
3 5
223 5 3 5
2 2 2 2
For power calculationsP
constant
162 1000(125)3600 1000
. . 5.9251500 0.3 3000 0.2
For pressure rise0.75 5.925( gH )Q ( gH ) 166
96 / 3600
N D
Pi e P kW
P kPa
=
= =
= = =
-
Ref. Problem 34/page39A model operates under a head of 5 m at 1200rpm. The power in the laboratory is limited to 8 kW. Predict the power and the limited to 8 kW. Predict the power and the diameter-ratio of a prototype turbine which operates under a head of 40 m at 240 rpm. What type of turbine is the prototype?
-
5 ; 1200 ; 840 ; 240
?; ?; type of turbine=?
m m m
m
H m N rpm P kWH m N rpm
DP D
= = =
= =
= =
5/4 5/4model model
5/4prototype
1200 8 454(So Kaplan turbine)5
454
s
N PNH
N PH
= = =
=
prototype
36210 36.21P kW MW
= =
3 5 3 5model prototype
1/53
336210 1200 14.14
8 240m
P PN D N D
DD
=
= =
-
( )5m3 5 3 5Assume N=250rpm
PP= Prototype Power, 6.5 95724mP P kWN D N D
= =( )3 5 3 5= Prototype Power, 6.5 95724mm m
P P kWN D N D
= =
s 5/4
s 5/4
5/4
Now, specific speed of model will tell the type of turbine250 8.25N = 14.26 (So, Pelton Turbine)
23For calculating Head for prototype,
N =
250 9572414.26 971.5
N PH
H mH
=
= =
-
( )
0.25 0.1
0.25 0.1
Efficiency of prototype turbine
1-1-
1 231 1-0.796.5 971.5
0.91
m m
m
D HD H
=
=
=
Efficiency corrected Power output
Predicted Power
0.91= 95724 110264 110.264
0.79
m
kW MW
=
= =