• # question_answer The isentropic enthalpy drop in moving blade is two - thirds of the isentropic enthalpy drop in fixed blades of a turbine. The degree of reaction will be:     A) 0.4                               B) 0.6              C) 0.66                             D) 1.66

${{R}_{d}}=\frac{{{(\Delta h)}_{m}}}{{{(\Delta h)}_{m}}{{(\Delta h)}_{f}}}=\frac{\frac{2}{3}{{(\Delta h)}_{f}}}{\frac{2}{3}{{(\Delta h)}_{f}}+{{(\Delta h)}_{f}}}$ $=\frac{\frac{2}{3}}{\frac{2}{3}+1}=\frac{2}{5}=0.4$