UNIT 5
Principle of similarity in fluid machinery
In order to evaluate the performance characteristics of a turbomachine under varying conditions of flow, speed, pressure ratio, power etc. it involves the conduct of large number of experiments which is time consuming and costly. In order to reduce the time and cost, these variables are grouped into few dimensionless quantities than the actual number of variables themselves. Then the experiments are performed by varying these dimensionless quantities to predict the performance of a turbomachine.
The design and development of an actual (prototype) machine is very costly. Usually the geometrically similar models are designed and tested by varying the non-dimensional quantities which helps in predicting the performance of prototype machines. This is called as model testing.
Therefore, the use of dimensionless parameters has following advantages:
Dimensional Analysis
The procedure involved to get dimensionless quantities is called the dimensional analysis.
The fundamental quantities considered are mass (M), length (L) and time (T) since these have no direct relation between themselves.
The quantities derived from fundamental quantities are called secondary quantities e.g. surface area is (length)2. Thus, the dimension of area is L2.
Dimensions of Some Physical Quantities
Some examples for determining the dimensions of certain physical quantities are given below:
Dimensional Homogeneity
Dimensional analysis is based on the principle that the variables in a physical phenomenon is arranged properly to give an equation which is dimensionally homogeneous.
The equation in which dimensions of left-hand side are equal to the dimensions of right-hand side is called dimensional homogeneity.
Example: Check the dimensional homogeneity of the Darcy's equation
h = fLV2 / 2gd
Dimension of L.H.S. = h = L
Dimension of R.H.S. = fLV2 / 2gd = {1 x L x (L/T)2} / {2 x (L/T2) x L}
= {L3/T2} / {L2/T2} = L
(Neglect 2 in denominator and consider constant f as 1.)
Dimension of L.H.S = Dimension R.H.S.
h = fLV2 / 2gd is dimensionally homogeneous equation, so it can be used in any system of units.
Applications of Dimensional Homogeneity
In case of number of variables involved in a physical phenomenon, then the relations among the various variables can be determined by the following methods:
1. Rayleigh method.
2. Buckingham's π-theorem.
Above methods are discussed below.
This method is quite convenient to use in case the expression for a variable depends upon 3 or 4 variables only. When the number of independent variables exceed more than four, this method cannot be used conveniently to find the expression for the dependent variable.
Method
Consider a variable X which depends on the three variables X1, X2 and X3 variables. Therefore, according to Rayleigh method the variable X is the function of variables X1, X2 and X3.
Mathematically,
X = f(X1, X2, X3)
Above expression can also be written as:
X = constant (k) X1a X2b X3c
Where a, b, and c are arbitrary powers of the variables X1, X2 and X3
By using the concept of dimensional homogeneity i.e. by equating the powers of the fundamental dimensions on both sides, an expression for dependent variable (X) can be obtained in terms of variables (X1, X2, X3).
2. Buckingham Pi (π) Theorem
Buckingham Pi-theorem states that, "If there are n variables (dependent and independent variables) in a dimensionally homogeneous equation and if these variables contain m fundamental dimensions (M, L, T), then the variables are arranged into (n - m) dimensionless terms. These dimensionless terms are called 'π terms'.
Step 1:
Mathematically, if any variable X1 depends on independent variables X2, X3, X4,…....... Xn
Then, the functional relationship between the dependent and independent variable is expressed as
X1 = f (X2, X3, X4,…....... Xn) ... (1)
Equation (1) can be written as,
f (X1, X2, X3, X4,…....... Xn) = 0 … (2)
It is dimensionally homogeneous equation and has n variables.
Step 2:
Equation (2) can be written in terms of number of dimensionless terms which is equal to (n - m).
f (π1, π 2, π 3, π 4,…....... π n-m) = 0 … (3)
Step 3:
Each π term contains m + 1 variables, where m is number of fundamental dimensions and it is also called as repeating variable.
Step 4:
If X2, X3, X4 are repeating variables, then each π-term can be written as
π1 = X2a1 X3b1 X4c1 X1
π2 = X2a2 X3b2 X4c2 X5 … (4)
πn-m = X2an-m X3bn-m X4cn-m Xn-m
Step 5:
Where each equation in Equation (4) is solved by principle of dimensional homogeneity and the values of a, b, c are obtained.
Step 6:
Substitute the value of a, b, c in their corresponding π1, π 2, π 3, π 4 etc. in Equation (4).
Step 7:
The value of π1, π 2, π 3, π 4 etc. are substituted in Equation (3)
Step 8:
The required expression can be obtained by expressing any one of the π- terms as a function of others.
π1 = F [ π2, π3 π4 …......πn-m ]
Selection of Repeating Variables
The following points should be considered for selecting the repeating variables.
(i) Geometric properties i.e. length, diameter, height etc.
(ii) Flow properties i.e. velocity, acceleration, rotational speed etc.
(iii) Fluid properties i.e. mass density, specific weight, viscosity etc.
Application of Dimensional Analysis as Applied to Turbomachines
The performance characteristics of turbo machines are studied under varying condition of volume flow rate or mass flow rate, speed, power, head or pressure, fluid density, fluid viscosity and various dimensions of rotor etc.
Scale Effect
Specific speed of two similar hydraulic machines will only be equal if their overall efficiencies of model and prototype are equal. However, the efficiency of model is less than the efficiency of prototype for the following reasons:
The difference in the efficiency of model and prototype due to roughness of surfaces, disproportionate leakage and mechanical losses is called Scale Effect.
Discharge, speed, power etc. of a hydraulic machine are all functions of head. Therefore, to compare and evaluate the performance of various hydraulic machines based on experimental results, each parameter is reduced to unit head. These reduced quantities are called as unit quantities.
These unit quantities are
Thus, the unit quantity can be defined as the value of parameter which will be obtained when a hydraulic machine operates under a head of 1 m. These unit quantities are useful in comparing the performance of machines under different operating conditions if characteristic curves are plotted using unit quantities. The meaning and significance of these quantities are as follows:
Unit speed of a turbine is defined as its speed while operating under unit head of 1 m.
The relationship between speed and unit speed can be obtained as follows:
Since, Fluid velocity, V = √(2gH)
thus, V α √ H
Runner velocity, u α √ H (i)
Also, u = πDN/60
For given diameter of turbine, u α N (ii)
On combining (i) and (ii),
N α √ H or N / √ H = constant k1 (iii)
By definition, when H = 1m, N = Nu
Therefore, N / √ H = Nu / 1
Or Nu = N / √ H
Therefore, for similarly designed turbines, working under an unit head of 1m,
Nu = N1 / √ H1 = N2 / √ H2
2. Unit discharge (Qu)
Unit discharge of a turbine is defined as its discharge while operating under unit head of 1 m.
Q = AV , so, Q α V
But V α √ H
So, Q α √ H or Q / √ H = constant k2
By definition, when H = 1m, Q = Qu
Q / √ H = Qu / 1
Qu = Q / √ H
Qu = Q1 / √ H1 = Q2 / √ H2
3. Unit Power (Pu)
Unit power of a turbine is defined as power developed by turbine while operating under unit head of 1 m.
P α QH
And, Q α √ H
On combining the above two expressions, we get,
P α H √ H or P α H3/2
P / H3/2 = constant k3
By definition, P = Pu when H = 1m
P / H3/2 = Pu
For similarly designed turbines,
Pu = P1 / H13/2 = P2 / H23/2
Specific speed represents the speed of a turbine which is identical in shape, geometrical dimensions, blade angles and gate openings etc. to actual turbine which will develop unit power of 1 kW when working under unit head of 1 m.
Significance of specific speed
Specific speed is a tool to compare different types of turbine since each turbine has a different specific speed and it helps in selecting the type of turbine. It also helps in predicting the performance of a turbine if its specific speed is known.
Derivation of specific speed
Overall efficiency, ηo of any turbine is the ratio of shaft power to hydraulic or water power i.e.
ηo = Shaft power / Water power
= P / ρ g QH
Therefore, P = ηo. ρ g QH
Since ηo, ρ and g are constants, we can write,
P α QH (i)
Absolute velocity, V α blade velocity, u
V = √(2gH)
V α √ H
i.e. V α u α √ H (ii)
But, u = πDN / 60
u α D- N (iii)
From Equations (ii) and (iii),
√H α DN
D = √H / N (iv)
Discharge Q from the turbine is given as
Q = Area x Velocity = (B x D) x √(2gH) (But B α D)
∴ Q α D2 √H
From equation (iv),
Q α {√H / N}2 x √H α H3/2 / N2 (v)
Putting in (i),
P α (H3/2 / N2) x H α H5/2 / N2
P = K x H5/2 / N2
When P = 1 kW, H = 1m and N = Ns (specific speed), we get,
1 = K x 1 / Ns2 or Ns2 = K
∴ P = Ns2 (H5/2 / N2)
Ns = (N x √ P) / H5/4
References: -
Text Books:
1. G. T. Mase, R. E. Smelser and G. E. Mase, Continuum Mechanics for Engineers, Third Edition, CRC Press,2004.
2. Y. C. Fung, Foundations of Solid Mechanics, Prentice Hall International, 1965.
3. Lawrence. E. Malvern, Introduction to Mechanics of a Continuous Medium, Prentice Hall international, 1969.
4. Hydrantic Machine by Jagdish Lal
5. Hydraulics & Hydraulic Machines by Vasandari
6. Hydrantic Machine by RD Purohit