Although the theory of centrifugal pumps gives many qualitative results, the most important indicator of a **pump’s performance** lies in **extensive hydraulic testing**.

In industry, the characteristics of all pumps are usually read from their **Q-H curve** or **performance curve **(flow rate – height). As can be seen, the performance charts use a **discharge – Q** (usually in m^{3}/h) and **pump head – H** (usually in m) as basic performance variables.

## System Head

In the chapter on head loss, it was determined that both **major losses** and minor losses in piping systems are **proportional to the square of the flow velocity**. The **system head loss** must be directly proportional to the square of the volumetric flow rate because the volumetric flow rate is directly proportional to the flow velocity.

It must be added that the **open hydraulic systems** contain not only the friction head but also the elevation head, which must be considered. The elevation head (static head) represents the potential energy of a fluid due to its elevation above a reference level.

**In many cases,** the total head of a system is a combination of elevation head and **friction head,** as shown in the figure.

Most of the hydraulic systems are** closed hydraulic loops** in nuclear engineering, and these systems only have** friction head** (no static head).

## Pump Head – Performance Curve

**In fluid dynamics**, the term **pump head** is used to measure the kinetic energy which a pump creates. Head is a measurement of the **height of the incompressible fluid column** the pump could create from the kinetic energy that the pump gives to the liquid. The **head and flow rate** determine the **performance** of a pump, which is graphically shown in the figure as the** performance curve** or **pump characteristic curve**. The main **reason** for using **head instead of pressure** to determine the **performance of a centrifugal pump** is that the **height of the fluid column is not dependent on the specific gravity (weight) of the liquid**. In contrast, the pressure from a pump will change. In terms of pressure, the **pump head** (**ΔP _{pump}**) is the difference between system backpressure and the pump’s inlet pressure.

The **maximum pump head** of a centrifugal pump is mainly determined by the **outside diameter of the pump’s impeller** and the **shaft angular velocity** – speed of the rotating shaft. The head will also change as the volumetric flow rate through the pump is increased.

When a centrifugal pump is operating at a **constant angular velocity**, an increase in the **system head** (back pressure) on the flowing stream causes a **reduction in the volumetric flow rate** that the centrifugal pump can maintain.

The relationship between the **pump head** and the **volumetric flow rate (Q)** that a centrifugal pump can maintain is dependent on various physical characteristics of the pump as:

**the power supplied to the pump****the angular velocity of the shaft****the type and diameter of the impeller**

and the used fluid:

**fluid density****fluid viscosity**

This relationship is very complicated, and its analysis lies in **extensive hydraulic testing** of certain centrifugal pumps, as seen in the picture below.

## Operating Characteristics of a Hydraulic Loop

When we put together the** frictional characteristics** (system head) of a hydraulic loop and the **performance curve,** the result will describe the **characteristics of the entire system** (e.g.,, **one loop of the primary circuit**). The following figure shows a typical performance curve for a centrifugal pump related to the frictional system head.

On the vertical axis, **the pump head **is the **difference** between **system backpressure** and the **pump’s inlet pressure** (**ΔP _{pump}**). On the horizontal axis, v

**olumetric flow rate (Q )**is the fluid flow through the pump. As can be seen, the head is approximately constant at low discharge and then drops to zero at

**Q**. At low discharge, the characteristics can be unstable (with a positive slope of the pump head). These are undesirable characteristics because an unstable pump may start to oscillate between the two possible combinations of flow rate, and the pipeline can vibrate.

_{max}At flow rate **Q1,** the pump gains more head than consumes the frictional losses. Therefore the flow rate through the system will **increase**. The flow rate will **stabilize** itself at the point where the **frictional losses intersect the pump characteristics**.

The following terms are defined to characterize the performance of centrifugal pumps:

- Shut-off Head
- Pump Efficiency
- Best Efficiency Point – BEP
- Brake Horsepower
- Net Positive Suction Head

## 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 the **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**