A rotating blade row exchanges torque with a flowing fluid by changing the fluid's angular momentum about the shaft axis. For steady flow with one effective inlet and one effective outlet, the axial control-volume angular-momentum balance gives

$$\tau_{z,,\text{rotor on fluid}} =\dot m\left(r_2V_{\theta2}-r_1V_{\theta1}\right),$$

where

• $r$ is radius from the shaft axis;
• $V_\theta$ is the tangential component of the absolute fluid velocity;
• subscripts 1 and 2 denote inlet and outlet of the rotor.

This is the torque form of the Euler turbomachinery equation.

From torque to shaft work

Let the rotor spin with angular velocity $\Omega$. The blade speed at radius $r$ is

$$\boxed{U=\Omega r}.$$

Mechanical power transferred from the rotor to the fluid is

$$\dot W_{\rm rotor\to fluid}=\Omega\tau_{z,,\text{rotor on fluid}}.$$

Substituting the angular-momentum relation gives

$$\dot W_{\rm rotor\to fluid} =\dot m\Omega\left(r_2V_{\theta2}-r_1V_{\theta1}\right).$$

Since $U_i=\Omega r_i$,

$$\boxed{\frac{\dot W_{\rm rotor\to fluid}}{\dot m} =U_2V_{\theta2}-U_1V_{\theta1}}.$$

The quantity on the left is shaft work transferred to the fluid per unit mass. Thus

$$\boxed{w_{\rm rotor\to fluid}=U_2V_{\theta2}-U_1V_{\theta1}}.$$

This is the work form commonly called Euler's turbomachinery equation.

Pumps and compressors

In a pump or compressor, the rotor normally supplies net mechanical energy to the fluid. Under the sign convention above,

$$w_{\rm rotor\to fluid}>0.$$

The fluid leaves the rotor with a larger value of $rV_\theta$ in the sense of rotor rotation than it had on entry, so the rotor exerts positive torque on the fluid.

Turbines

In a turbine, the fluid supplies net mechanical energy to the rotor. The fluid's angular momentum in the direction of rotation decreases through the rotor, so

$$r_2V_{\theta2}-r_1V_{\theta1}<0.$$

and therefore

$$w_{\rm rotor\to fluid}<0.$$

The positive specific shaft work extracted from the fluid is then

$$\boxed{w_{\rm out}=U_1V_{\theta1}-U_2V_{\theta2}}.$$

The same angular-momentum law therefore covers both energy-adding and energy-extracting turbomachines; only the direction of work transfer changes.

Worked example: idealized pump rotor

A pump handles

$$\dot m=20,\mathrm{kg/s}$$

and rotates at

$$\Omega=100,\mathrm{rad/s}.$$

At the rotor inlet,

$$r_1=0.10,\mathrm m,\qquad V_{\theta1}=0,$$

while at the outlet,

$$r_2=0.25,\mathrm m,\qquad V_{\theta2}=18,\mathrm{m/s}.$$

The blade speeds are

$$U_1=\Omega r_1=10,\mathrm{m/s},$$

$$U_2=\Omega r_2=25,\mathrm{m/s}.$$

The ideal specific work transferred from rotor to fluid is

$$w_{\rm rotor\to fluid} =U_2V_{\theta2}-U_1V_{\theta1}$$

$$=(25)(18)-(10)(0) =\boxed{450,\mathrm{J/kg}}.$$

The corresponding power is

$$\dot W =\dot m w =(20)(450) =\boxed{9.0,\mathrm{kW}}.$$

The torque exerted by the rotor on the fluid is

$$\tau=\frac{\dot W}{\Omega} =\frac{9000}{100} =\boxed{90,\mathrm{N,m}}.$$

The same result follows directly from

$$\tau =\dot m(r_2V_{\theta2}-r_1V_{\theta1}) =(20)(0.25)(18)=90,\mathrm{N,m}.$$

Mechanical versus thermodynamic viewpoints

The Euler turbomachinery equation determines ideal rotor work from the change in angular momentum of the flow. A thermodynamic steady-flow energy balance instead relates shaft work to changes in enthalpy, kinetic energy, potential energy, and heat transfer across a complete device.

For an ideal adiabatic rotor, the two viewpoints are compatible and can be combined. In real machines, losses and downstream components determine how rotor work appears in pressure, enthalpy, velocity, and entropy changes.

Euler's equation is therefore not a replacement for the energy balance. It is the mechanical law that explains how rotating blades exchange torque and shaft work with a flowing fluid.