Derivation of Euler's turbine equation

Euler's equation is the basic equation of all turbomachines. It follows from the angular momentum principle: the torque on the rotor equals the rate of change of angular momentum of the fluid passing through it.

Velocity triangles

At inlet (1) and outlet (2): $V$ = absolute velocity, $u$ = blade (peripheral) velocity, $V_r$ = relative velocity, $V_w$ = whirl component (along $u$), $V_f$ = flow component (radial for a radial machine), $\alpha$ = angle of $V$ with $u$, $\beta$ = angle of $V_r$ with $u$ (blade angle at inlet is $\theta$, at outlet $\phi$).

  Triangle (same form at inlet 1 and outlet 2)

              apex
             /|\
        V   / | \  Vr
           /  |  \
          /   |Vf \
         / a  |  b \
        *-----+-----*------> blade motion
        |<-Vw->|
        |<----- u ----->|

  V  = u + Vr (vector sum),  Vw = V cos a,  Vf = V sin a

Derivation

Let a mass flow $m=\rho Q$ pass through the runner per second. Inlet radius $r_1$, outlet radius $r_2$.

Angular momentum per second at inlet: $m V_{w1} r_1$. At outlet: $m V_{w2} r_2$.

Torque on the runner:

$$T = \rho Q\,(V_{w1} r_1 - V_{w2} r_2)$$

Angular velocity $\omega = u_1/r_1 = u_2/r_2$, so the power given to the runner is

$$P = T\omega = \rho Q\,(V_{w1} u_1 - V_{w2} u_2)$$

Work done per second per unit weight of water (weight flow $= \rho g Q$) is the Euler head:

$$\boxed{H_e = \frac{1}{g}\,(V_{w1} u_1 - V_{w2} u_2)}$$

Notes

• If the water leaves without whirl ($V_{w2}=0$, radial discharge), $H_e = V_{w1}u_1/g$.
• For an axial machine $u_1=u_2=u$, so $H_e = u(V_{w1}-V_{w2})/g$.
• $V_{w2}$ is taken negative if it is opposite to the blade motion.
• For a pump the sign is reversed: $H_e = (V_{w2}u_2 - V_{w1}u_1)/g$.
• Hydraulic efficiency of a turbine: $\eta_h = H_e/H$, where $H$ is the net head.