**Euler’s turbomachine equation**, or sometimes called

**Euler’s pump equation**, plays a central role in

**turbomachinery**as it connects the

**specific work Y**and the geometry and velocities in the impeller. The equation is based on the concepts of

**conservation of angular momentum**and

**conservation of energy**.

**Euler’s turbomachine equations** are:

**Shaft torque: T _{shaft} = ρQ(r_{2}V_{t2} – r_{1}V_{t1})**

**Water horsepower: P _{w} = ω . T_{shaft } = ρQ(u_{2}V_{t2} – u_{1}V_{t1})**

**Pump head: H = P _{w} / ρgQ = (u_{2}V_{t2} – u_{1}V_{t1})/g**

where

**r**and_{1}**r**are the_{2}**diameters of the impeller**at the inlet and outlet, respectively.**u**and_{1}**u**are the_{2}**absolute velocities of the impeller**(u_{1}= r_{1}. ω) at the inlet and outlet, respectively.**V**and_{t1}**V**are the_{t2}**tangential velocities of the flow**at the inlet and outlet, respectively.

**Euler’s turbomachine equations** can predict the impact of changing the impeller geometry on the head. It does not matter when we deal with a pump or with a turbine. If torque and angular velocity are **like a sign**, work is being done on the fluid (a pump or compressor). If torque and angular velocity are **of opposite sign,** work is being extracted from the fluid (a turbine). Thus for the design aspect of turbines and pumps, the Euler equations are extremely useful.

## Example: Pump Performance Calculation

In this example, we will see how to predict

**the design discharge****water horsepower****the pump head**

of a centrifugal pump. This performance data will be derived from **Euler’s turbomachine equation:**

**Shaft torque: T _{shaft} = ρQ(r_{2}V_{t2} – r_{1}V_{t1})**

**Water horsepower: P _{w} = ω . T_{shaft } = ρQ(u_{2}V_{t2} – u_{1}V_{t1})**

**Pump head: H = P _{w} / ρgQ = (u_{2}V_{t2} – u_{1}V_{t1})/g**

Given are the following data for a centrifugal water pump:

**diameters of the impeller**at the inlet and outlet**r**_{1}= 10 cm**r**_{2}= 20 cm

**Speed = 1500 rpm**(revolutions per minute)- the blade angle at inlet
**β**_{1}= 30° - the blade angle at outlet
**β**_{2}= 20° - assume that the blade widths at inlet and outlet are:
**b**._{1}= b_{2}= 4 cm

Solution:

First, we have to calculate the **radial velocity of the flow** at the outlet. From the velocity diagram, the radial velocity is equal to (we assume that the flow enters exactly normal to the impeller, so tangential component of velocity is zero):

**V _{r1}** = u

_{1}tan 30° = ω r

_{1}tan 30° = 2π x (1500/60) x 0.1 x tan 30° =

**9.1 m/s**

The radial component of flow velocity determines how much the **volume flow rate is entering the impeller**. So when we know **V _{r1}** at the inlet, we can determine

**the discharge**of this pump according to the following equation. Here b

_{1}means the blade width of the impeller at the inlet.

**Q** = **2π.r _{1}.b_{1}.V_{r1} **= 2π x 0.1 x 0.04 x 9.1 =

**0.229 m**

^{3}/sIn order to calculate the **water horsepower (P _{w})** required, we have to determine the

**outlet tangential flow velocity V**, because it has been assumed that the inlet tangential velocity V

_{t2}_{t1}is equal to zero.

The outlet radial flow velocity follows from **conservation of Q**:

**Q = 2π.r _{2}.b_{2}.V_{r2}** ⇒

**V**

**= Q / 2π.r**

_{r2}_{2}.b

_{2}= 0.229 / (2π x 0.2 x 0.04) =

**4.56 m/s**

From the figure (**velocity triangle**) outlet blade angle, β_{2}, can be easily represented as follows.

**cot β _{2} = (u_{2} – V_{t2}) / V_{r2}**

and therefore the outlet tangential flow velocity V_{t2} is:

**V _{t2}** =

**u**= ω r

_{2}– V_{r2}. cot 20°_{2}– V

_{r2}. cot 20° = 2π x 1500/60 x 0.2 – 4.56 x 2.75 = 31.4 – 12.5 =

**18.9 m/s.**

The water horsepower required is then:

**P _{w} = ρ Q u_{2 }V_{t2} **= 1000 [kg/m

^{3}] x 0.229 [m

^{3}/s] x 31.4 [m/s] x 18.9 [m/s] = 135900 W =

**135.6 kW**

and the pump head is:

** H ≈ P _{w} / (ρ g Q)** = 135900 / (1000 x 9.81 x 0.229) =

**60.5 m**