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Stage efficiency

A stage of an impulse turbine consists of a nozzle set and a moving wheel. The stage efficiency defines a relationship between enthalpy drop in the nozzle and work done in the stage. η s t a g e = W o r k d o n e o n b l a d e E n e r g y s u p p l i e d p e r s t a g e = U Δ V w Δ h {\displaystyle {\eta _{\mathrm {stage} }}={\frac {\mathrm {Work~done~on~blade} }{\mathrm {Energy~supplied~per~stage} }}={\frac {U\Delta V_{w}}{\Delta h}}} Where Δ h = h 2 − h 1 {\displaystyle \Delta h=h_{2}-h_{1}} is the specific enthalpy drop of steam in the nozzle.

By the first law of thermodynamics: h 1 + 1 2 V 1 2 = h 2 + 1 2 V 2 2 {\displaystyle h_{1}+{\frac {1}{2}}{V_{1}}^{2}=h_{2}+{\frac {1}{2}}{V_{2}}^{2}} Assuming that V 1 {\displaystyle V_{1}} is appreciably less than V 2 {\displaystyle V_{2}}, we get Δ h ≈ 1 2 V 2 2 {\displaystyle {\Delta h}\approx {\frac {1}{2}}{V_{2}}^{2}}. Furthermore, stage efficiency is the product of blade efficiency and nozzle efficiency, or η stage = η b η N {\displaystyle \eta _{\text{stage}}=\eta _{b}\eta _{N}}.

Nozzle efficiency is given by η N = V 2 2 2 (h 1 − h 2) {\displaystyle \eta _{N}={\frac {{V_{2}}^{2}}{2\left(h_{1}-h_{2}\right)}}}, where the enthalpy (in J/Kg) of steam at the entrance of the nozzle is h 1 {\displaystyle h_{1}} and the enthalpy of steam at the exit of the nozzle is h 2 {\displaystyle h_{2}}. Δ V w = V w 1 − (− V w 2) = V w 1 + V w 2 = V r 1 cos β 1 + V r 2 cos β 2 = V r 1 cos β 1 (1 + V r 2 cos β 2 V r 1 cos β 1) {\displaystyle {\begin{aligned}\Delta V_{w}&=V_{w1}-\left(-V_{w2}\right)\\&=V_{w1}+V_{w2}\\&=V_{r1}\cos \beta _{1}+V_{r2}\cos \beta _{2}\\&=V_{r1}\cos \beta _{1}\left(1+{\frac {V_{r2}\cos \beta _{2}}{V_{r1}\cos \beta _{1}}}\right)\end{aligned}}} The ratio of the cosines of the blade angles at the outlet and inlet can be taken and denoted c = cos β 2 cos β 1 {\displaystyle c={\frac {\cos \beta _{2}}{\cos \beta _{1}}}}. The ratio of steam velocities relative to the rotor speed at the outlet to the inlet of the blade is defined by the friction coefficient k = V r 2 V r 1 {\displaystyle k={\frac {V_{r2}}{V_{r1}}}}.

k < 1 {\displaystyle k<1} and depicts the loss in the relative velocity due to friction as the steam flows around the blades (k = 1 {\displaystyle k=1} for smooth blades). η b = 2 U Δ V w V 1 2 = 2 U V 1 (cos α 1 − U V 1) (1 + k c) {\displaystyle \eta _{b}={\frac {2U\Delta V_{w}}{{V_{1}}^{2}}}={\frac {2U}{V_{1}}}\left(\cos \alpha _{1}-{\frac {U}{V_{1}}}\right)(1+kc)}

The ratio of the blade speed to the absolute steam velocity at the inlet is termed the blade speed ratio ρ = U V 1 {\displaystyle \rho ={\frac {U}{V_{1}}}}.

η b {\displaystyle \eta _{b}} is maximum when d η b d ρ = 0 {\displaystyle {\frac {d\eta _{b}}{d\rho }}=0} or, d d ρ (2 cos α 1 − ρ 2 (1 + k c)) = 0 {\displaystyle {\frac {d}{d\rho }}\left(2{\cos \alpha _{1}-\rho ^{2}}(1+kc)\right)=0}. That implies ρ = 1 2 cos α 1 {\displaystyle \rho ={\frac {1}{2}}\cos \alpha _{1}} and therefore U V 1 = 1 2 cos α 1 {\displaystyle {\frac {U}{V_{1}}}={\frac {1}{2}}\cos \alpha _{1}}. Now ρ o p t = U V 1 = 1 2 cos α 1 {\displaystyle \rho _{opt}={\frac {U}{V_{1}}}={\frac {1}{2}}\cos \alpha _{1}} (for a single stage impulse turbine).

Therefore, the maximum value of stage efficiency is obtained by putting the value of U V 1 = 1 2 cos α 1 {\displaystyle {\frac {U}{V_{1}}}={\frac {1}{2}}\cos \alpha _{1}} in the expression of η b {\displaystyle \eta _{b}}.

We get: η b max = 2 (ρ cos α 1 − ρ 2) (1 + k c) = 1 2 cos 2 α 1 (1 + k c) {\displaystyle {\eta _{b}}_{\text{max}}=2\left(\rho \cos \alpha _{1}-\rho ^{2}\right)(1+kc)={\frac {1}{2}}\cos ^{2}\alpha _{1}(1+kc)}.

For equiangular blades, β 1 = β 2 {\displaystyle \beta _{1}=\beta _{2}}, therefore c = 1 {\displaystyle c=1}, and we get η b max = 1 2 cos 2 α 1 (1 + k) {\displaystyle {\eta _{b}}_{\text{max}}={\frac {1}{2}}\cos ^{2}\alpha _{1}(1+k)}. If the friction due to the blade surface is neglected then η b max = cos 2 α 1 {\displaystyle {\eta _{b}}_{\text{max}}=\cos ^{2}\alpha _{1}}.