In an impulse reaction turbine stage, the heat drops in fixed and moving blades are 28.32 kJ/kg and 18.61 kJ/kg, respectively. The degree of reaction for the stage will be: 

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JSSC JE Mechanical Re-Exam 23 Oct 2022 Official Paper-II
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  1. 0.396 
  2. 0.603 
  3. 0.657  
  4. 1.522 

Answer (Detailed Solution Below)

Option 1 : 0.396 
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Detailed Solution

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Concept:

Degree of Reaction (R):

  • The degree of reaction is the ratio of the enthalpy drop in the rotor to the enthalpy drop in a stage.
  • In the case of an impulse turbine, there is no change in the enthalpy in the rotor, so the degree of reaction is zero (R = 0).
  • For Parson's turbine or 50% reaction turbine R = 0.5 and for Hero's turbine R = 1.

 

\(R = \frac{\Delta h _{mb}}{\Delta h_{stage}} = \frac{\Delta h _{mb}}{\Delta h_{mb} + \Delta h_{fb} }\)

where, ∆hmb = Enthalphy Drop across Moving Blade, ∆hfb = Enthalphy Drop across Fixed Blade

Calculation:

Given:

 ∆hmb = 18.61 kJ/kg, ∆hfb = 28.32 kJ/kg

\(R = \frac{18.61}{18.61 + 28.32}\)

R = 0.396

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