The degree of reaction for a Parson's reaction turbine is:

Explanation:

The above-mentioned figure is a schematic diagram of Parson's reaction turbine

• The velocity triangles for inlet & outlet conditions are symmetrical.
• ∴  \(\left\{ {\begin{array}{*{20}{c}} {{\alpha _1} = {\beta _2}}\\ {{\alpha _2} = {\beta _1}} \end{array}} \right\}\;\;\;\;\;\;\;\& \;\;\;\;\left\{ {\begin{array}{*{20}{c}} {{V_{{r_1}}} = {V_2}}\\ {{V_{{r_2}}} = {V_1}} \end{array}} \right\}\)

• This consists of a symmetrical stator and rotor blades.

• And the degree of reaction is defined as the (ratio of enthalpy drop in the rotor to that of the stator) 

• the pressure drop is equally shared by the stator and the rotor for a turbine.

• It is a  very widely used reaction turbine also known as a 50% reaction turbine

Degree of reaction \({\rm{R}} = \frac{{{\rm{Static\;enthalpy\;drop\;or\;rise\;in\;rotor}}}}{{{\rm{Stagnationenthalpy\;drop\;or\;rise\;in\;stage}}}}\)

Or, \(R = \frac{{static\;pressure\;drop\;or\;rise\;in\;rotor}}{{stagnation\;pressure\;drop\;or\;rise\;in\;stage}}\)

In the case of impulse turbine, there is no change in the enthalpy in the rotor, so the degree of reaction is zero. R = 0.

And for Parson’s turbine or 50% reaction turbine R = 0.5

And for Hero’s turbine R = 1.