Let W= work done per kg of steam;

$C_i \text{= Absolute velocity of steam at inlet to moving blades;} \\ C_0\text{=Absolute velocity of steam at exit to moving blades;} \\ C_{wi}\text{=Tangential component of Ci called as velocity of whirl at the inlet to moving blades, m/s} \\ C_{wo}\text{=Velocity of whirl at the exit of moving blades, m/s} \\ C_{ri}\text{=Relative velocity of steam to moving blades at inlet, m/s} \\ C_{r0}\text{=Relative velocity of steam to moving blades at outlet, m/s} \\ C_{fi}\text{= Axial component of Ci called as velocity of flow at inlet, m/s} \\ C_{fo}\text{= Velocity of flow at exit, m/s} \\ α \text{= Exit angle of nozzles or the angles to the tangent of wheel with which the steam at Ci enters the moving blades, degrees} \\ β \text{= Inlet angle of fixed blades or the angle to the tangent of wheel at which the steam at Co leaves the moving blades} \\ θ\text{= Inlet of moving blades, degrees;} \\ Ф\text{=Exit angle of moving blades, degrees;} \\ ṁ\text{= Mass flow rate of steam, kg/s;} \\ K\text{= Blade velocity coefficient or friction factor=} \dfrac{C_{ro}}{C_{ri}} \\ \text{=1, if friction is neglected} \\ \text{d= Drum diameter,m;} \\ \text{N= Speed of turbine, r.p.m;} \\ \text{Blade velocity,} C_b=\dfrac{π.d.N}{60} \\ \text{Work done per kg of steam is given by,} W=(C_{wi}+C_{wi} ) C_b \\ C_{wi}=C_i \cosα \\ And C_{wo}=C_{ro} \cosФ-C_b \\ =C_i \cosα-C_b \\ ∴W=C_b (2C_i \cosα-C_b ) \\ Or \\ W=C_i^2 \bigg[\dfrac{2C_b.C_i \cosα}{C_i^2} -\dfrac{0C_b^2}{C_i^2}\bigg ] \\ \text{Let the ratio of blade velocity to steam velocity called, blade velocity ratio, be represented by ‘s’ i.e.} \\ \text"{Blade velocity ratio, s}=\dfrac{\text{Blade velocity,}(C_b)}{\text{Steam Velocity,}(C_i)} \\ \text{Using above equation, the equation for work becomes} \\ W=C_i^2 (2s \cosα-s^2 ) \\ \text{Energy supplied per kg of steam}=\dfrac{C_i^2}{2}+\dfrac{C_{ro}^2-C_{ri}^2}{2} \\ =\dfrac{C_i^2-C_{ri}^2}{2} \\ \text{Since} C_{ri}^2=C_i^2+C_b^2-2C_b.C_i \cosα \\ \text{Energy supplied per kg of steam=}C_i^2-\text{C_i^2+C_b^2-2C_b.C_i \cosα}{2} \\ =\dfrac{C_i^2-C_b^2+2C_b.C_i \cosα}{2} \\ =\dfrac{C_i^2}{2} [1+2s \cosα-s^2 ] \\ \text{The blade efficiency for the stage, } \\ η_b=\dfrac{\text{Work done}}{\text{Energy supplied}}=\dfrac{C_i^2 (2s \cosα-s^2)}{\dfrac{C_i^2}{2} (1+2s cosα-s^2 )} \\ =\dfrac{2(2s \cosα-s^2)}{(1+2s \cosα-s^2)}= \dfrac{2(1+2s \cosα-s^2)-2}{(1+2s cosα-s^2)} \\ =2-\dfrac{2}{(1+2s \cosα-s^2)} \\ \text{Therefore, ηb becomes maximum when factor (1+2s cosα-s^2) becomes maximum The required condition is,} \\ \dfrac{d(1+2s \cosα-s^2)}{ds}=0 \\ Or \\ 2 cosα-2s=0 \\ Or \\ s=\cosα \\ \text{Condition for maximum efficiency,} \\ s=\dfrac{C_b}{C_1} =\cosα \\ \text{Substituting the value of s in the equation of efficiency,} \\ (η_{b\max} )=2-\dfrac{2}{1+2s \cosα \cosα-\cos^2 α} \\ (η_{b\max} )=\dfrac{2\cos^2 α}{1+\cos^2 α}$