Velocity diagrams for different values of Rd appear as shown below:
(i) V1>V2, Vr1>Vr2; Rd<0 (Rd is negative)
(ii) Vr1=Vr2; Rd=0
(iii) V1=Vr2; Vr1=V2; Rd=0.5
(iv) V1=V2 ; Rd=100%
(v) V2>V1; Vr2>Vr1 ; Rd>100%
We know that for utilization factor ε to be maximum, the exit velocity V2 should be minimum.
For a given rotor speed U, the minimum value of V2 is obtained only if V22 is axial and the velocity triangles would look as shown:
Velocity triangle for maximum utilization factor condition:
Form the expression, it is clear that εmaximum will have the highest value if α1 = 0.
But α1 = 0, results in V2 = 0 which is not a practically feasible condition. The zero angle turbines which would have α1 = 0 appears as shown:
Though the zero angle turbines are not practically feasible it represents the ideal condition to be aimed at. In a Pelton wheel, we arrive at a condition wherein the jet is deflected through an angle of 165 to 170 degrees. Though an angle of 180 degrees would be the ideal condition as in case of a zero angle turbine. Impulse turbine designed for maximum utilization.
The ratio is referred to as a blade speed ratio φ which will have limiting value of 0.5 for a zero angle turbine. But in practical situation, α1 is in between 20 to 25 degrees. But φ varies from 0.45 to 0.47. The blade speed ratio is very useful performance parameter and it may be noted that the closer its value is to 0.5, the better it is.
Expression for power output:
We know that,
For a fixed value of α1, as Rd increases εmaximum.
But for Rd=1 (100% reaction turbine), this equation doesn’t holds good.
Let us examine how εmaximum is affected by Rd.
Case (1): Rd=1,
Hence by Euler’s turbine equation
For maximum utilization V2 needs to be axial. If V2 is to be axial, then V1 also should be axial which means that the denominator of the expression becomes equal to infinity which reduces to to zero. This only means that α1 should be as low as possible to get meaning full values of. This represents contradicting condition and hence Rd = 1 is not preferred.
Case (2): Rd>1
As Rd> ε tends to zero.
In this case, V2>V1 and hence V2 can never be axial and hence the condition for εmaximum [An axial orientation for V2 can never be met]
The utilization factor ε is given by
As Rd increases, ε decreases.
This means that the stator has to function to not only diffuse V2 to as low a value as possible but also turn the fluid through a very large angle. This results in the poor flow efficiency and hence Rd greater than 100% is not practically preferred.
Case (3): Rd<0 [negative Rd]
For this condition, it is noticed that rd is negative denominator increases, ε decreases.
Vr2<Vr1, also means that the pressure is increasing as fluid passes through the rotor.
i.e. the rotor is acting like a diffuser. This is not preferred since pressure always has to decrease along the flow path for good flow efficiency. Hence, Rd < 0 is not practically preferred.
Case (4): Rd = 0.5,
We know that for a 50% reaction turbine, velocity triangles are similar and for maximum utilization condition the triangle would appear as shown.
The angles are identical but reversed for the rotor and the stator. From the practical view point, the manufacturing of blades becomes simple. Since the same blade can be used for either the stator or the rotor by merely reversing the direction. It can also be shown that in a multistage turbines 50% reaction gives maximum stage efficiency. Since Vr2>Vr1, pressure reduces along the flow path in the rotor resulting in high flow efficiency. In general, n Rd value between 0 and 1 is preferred due to practical considerations.
From the velocity triangle it can be noted that Vw1=U
Vw2 = 0 (for maximum utilization factor condition)
∴ P = U2
Comparing the energy transfer achieved by 50% reaction turbine with an impulse turbine when both are designed for εmaximum condition and operating with the same rotor velocities. We notice that an impulse turbine transfers twice as much energy as 50% reaction turbine gives the better flow efficiencies.
If multi staging is attempted, then for a given value of energy transfer, a 50% reaction turbine would need twice the number of stages as that of impulse turbines. In actual practice, when multistage is attempted, the initial stages are designed for an impulse turbine when maximum fluid velocity is available. The subsequent stages are 50% reaction stages.
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