TURBOMACHINES. VIJAYAVITHAL BONGALE Associate Professor and Head Department of Mechanical Engineering Malnad College of Engineering, Hassan

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1 TURBOMACHINES VIJAYAVITHAL BONGALE Associate Professor and Head Department of Mechanical Engineering Malnad College of Engineering, Hassan Mobile No: vvb@mcehassan.ac.in 1

2 Dimensional Analysis: It is a logical formal procedure which identifies the variables involved and groups them into nondimensional numbers much lesser in number than the variables themselves. In a design problem or performance test these nondimensional quantities (dimensionless numbers ) are varied instead of the large number of variables forming these groups. Some dimensionless groups give an idea of the type of machine and its range of operation. Dimensional analysis helps us in determining a convenient arrangement of variables in a physical relationship. 2

3 BUCKING ham s п theorem: This theorem states that in a given problem if the number of variables is n, the greatest number of non-dimensional groups or dimensionless numbers (also known as п terms) is given by, п=n k (1) Where k m and m is the number of primary dimensions and in majority of situations k will be taken equal to m. 3

4 Let, there be 3 dependent variables ( y1, y2, and y3) and five independent variables (x1, x2, x3, x4 and x5) Then, y1, y2, y3 = ƒ(x1, x2, x3, x4, x5) (2) If three primary dimensions M, L and T are involved then the п theorem gives п = n m = ( 3+5) 3 = (3) Thus the number of dimensionless groups or п terms is five. This can be expressed by the following relation. П1, П2 = ƒ (П3, П4, П5) (4) By using these п -terms more dimensionless groups can be formed and the selection of п terms depends on the behaviour of the given machine. 4

5 Application of п theorem: For applying the theorem m repeating variables are selected from among the n variables. In problems in which all the three fundamental dimensions are involved, it is usually expedient to select repeating variables using the following guidelines. 1. Select the first repeating variable from those describing the geometry of flow. ( Size and shape of the fluid passage or of the moving body such as diameter, length etc) 2. Select the second repeating variable from those representing the fluid properties. ( Fluid properties such as density, viscosity, surface tension, elasticity and vapour density) 3. Select the third repeating variable from those characterising the fluid motion. ( The kinetic and dynamic characteristics of flow such as velocity, acceleration, discharge, pressure, force, power ) It is usually conventional to select a length dimension, representing category 1, 5 density from category 2, and velocity form category 3 as repeating variables.

6 Example 1. The resisting force F of a plane during flight can be considered as dependent upon the length of air craft L, velocity V, air viscosity μ, air density ρ and bulk modulus of air k. Express the functional relationship between these variables using dimensional analysis. Explain the physical meaning of the dimensionless group. Solution: The functional relationship between the dependent variable F and the independent variables can be written as, F = ƒ (L, V, μ, ρ, k ) (1) 6

7 Express these variables in terms of their dimensions in the tabular form below. Sl. No. Name of the variable Symbol Dimensions 1 F M L T -2 2 Resistance force Length of aircraft L L 3 Velocity V LT Air viscosity Air density Bulk modulus of air μ ρ ML-1T -1 ML-3 k ML-1T

8 From the table, it is seen that, the total number of variables, n = 6, the number of fundamental dimensions, m = 3. Therefore the total number of п terms to be obtained, п = n m = 6 3 = (2) Selecting ρ, V and L as repeating variables and using Bucking hamп theorem, we can write, П1 = F ρa V b Lc (3) П2 = μ ρe V f Lg (4) П3 = k ρx V y Lz (5) 8

9 Expressing П1 = F ρa V b Lc dimensionally, i.e. M 0L 0T 0 = (MLT -2) (ML-3) a (LT -1) b (L) c (6) Equating the powers of M, L and T, we get For M: 0 = 1+ a For L: 0 = 1-3a + b + c For T: 0=-2-b On solving we get a = -1, b = -2 and c = -2. Substituting the values of a, b and c in equation 3 we get П1 = F ρ V L = F 2 ρl V (7) 9

10 Similarly, expressing П2 = μ ρe V f Lg dimensionally, i.e. M 0L 0T 0 = (ML-1T -1) (ML-3) e (LT -1) f (L) g (8) Equating the powers of M, L and T, we get For M: 0=1+e For L: 0 = -1-3e + f + g For T: 0 = -1 - f On solving we get e = -1, f = -1 and g = -1. Substituting the values of e, f and g in equation 4 we get μ П2 = μ ρ-1 V -1 L-1= ρ L V (9) 10

11 Expressing П3 = k ρx V y Lz dimensionally, i.e. M 0L 0T 0 = (ML-1T -2) (ML-3) x (LT -1) y (L) z (10) Equating the powers of M, L and T, we get For M: 0=1+x For L: 0 = -1-3x + y + z For T: 0 = -2 - y On solving we get x = -1, y = -2 and z = 0. Substituting the values of x, y and z in equation 5 we get k П3 = k ρ-1 V -2 L0 = (11) ρv 11

12 Where, П1 is the ratio of resistance force F and dynamic force П2 is the reciprocal of Reynolds number П3 is the reciprocal of square of the Mach number. Example 2. Assuming that rate of discharge Q of a centrifugal pump is dependent upon the mass density ρ of fluid, pump speed N( rpm), the diameter of impeller D, the pressure p and the viscosity of fluid μ. Show using the Bucking ham s п theorem that it can be represented by, gh ν 3 Q (ND )φ 2 2, 2 N D ND where H is head and υ is kinematic viscosity of fluid. 12

13 Solution: From the given data, we have Q = ƒ (ρ, N, D, p, μ) (1) Express these variables in terms of their dimensions in the tabular form below. Sl.No. Name of the variable Symbol Dimensions 1 Discharge Q L3 T-1 2 Fluid density ρ ML-3 3 Pump speed N T -1 4 Impeller diameter D L 5 Pressure P M L-1 T-2 6 Fluid viscosity μ ML-1 T -1 13

14 Example 3. Solution: Express these variables in terms of their dimensions in the tabular form below. 14

15 From the table, it is seen that, the total number of variables, n = 7, the number of fundamental dimensions, m = 3. Therefore the total number of п terms to be obtained, п = n m = 7 3 = (2) 15

16 16

17 Example 4 : Solution: 17

18 18

19 he Performance of a turbomachine for incompressible flow and hydraulic machines can be expressed as a function of: i) density of the fluid ρ ii) Speed of the rotor N iii) Characteristic diameter D iv) Discharge Q v) Gravity head (gh) vi) Power developed P and (vii) Viscosity µ. 19

20 Each group of variables in equation (2) is truly dimensionless and all are used in hydraulic turbomachinery practice. 20

21 In case of hydraulic machine it is found that the Reynolds number is usually very high and therefore the viscous action of the fluid has very little effect on the power output of the machine and the power coefficient of the machine remains only a function flow coefficient and head coefficient. For model studies for similar turbomachines, one can use Definitions: It signifies the volume flow rate of fluid through a turbomachine of unit diameter runner, operating at unit speed. It is constant for similar rotors. Hence for a fan or pump of a certain diameter running at various speeds, the discharge is proportional to the speed and this is called Ist fan law. 21

22 It represents the ratio of the kinetic energy of the fluid spouting under the head H to the kinetic energy of the fluid running at the rotor tangential speed. It is constant for similar impellers. For a machine of specified impeller diameter, the head varies directly as the square of the speed. This is the second law of fan. 22

23 Specific Speed: Ns Parameter of great importance in selecting the type of machine required for a given duty. Pump or Hydraulic turbine: Design parameters :? Pump Head H, Volume flow rate Q and Rotational speed N Turbine Power P, Head H, and Rotational speed N For a set of geometrically similar turbines or pumps, the specific 23 speed will have the same value

24 We have, Ns is the specific speed and is justified to the extent that N s is 24 directly proportional to the speed N.

25 Specific speed as defined above is at the point of maximum efficiency of a turbomachine. Where, i) Q is in m3 / s ii) H in m iii) N in rad/s iv) P in Watts v) ρ in kg/ m3 and vi) g in m /s2 25

26 Significance of specific speed 26

27 27

28 SIMILARITY CONDITIONS FOR MODEL AND PROTOTYPE: Hydraulic Turbines: 28

29 5. The hydraulic and volumetric efficiencies of the model and prototype must be equal Therefore power developed at the shaft must be equal 29

30 Pumps: 30

31 31

32 Problems: 1. A quarter scale turbine model is tested under a head of 10 m. The full scale turbine is required to work under a head of 28.5 m and runs at 415 rpm. At what speed must the model be run if it develops 94 kw and uses 0.96 m3/s at this speed. What power will be obtained from the full scale turbine and name the type of turbine. Solution: Given: 32

33 We have, The speed ratio of two turbines must be equal Specific speed of two similar turbines must be equal 33

34 Pp can also be obtained by using the following relation 34

35 Type of turbine: 35

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