Pump Head Calculation – Pump Performance Calculation

 
Pump Head - Pump Performance
In fluids 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, while the pressure from a pump will change. In terms of pressure the pump head (ΔPpump) is difference between system back pressure and the inlet pressure of the pump.

pump head - performace curve - chartThe 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 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 pump. As can be seen from the picture below.

performace curve - brake horsepower

Pump Head Calculation – Pump Performance Calculation

Pump Head CalculationIn this example, we will see, how to predict

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

Shaft torque:                         Tshaft    =                                     ρQ(r2Vt2 – r1Vt1)

Water horsepower:             Pw         =     ω . Tshaft        =      ρQ(u2Vt2 – u1Vt1)

Pump head:                            H           =     Pw / ρgQ      =      (u2Vt2 – u1Vt1)/g

Given are the following data for a centrifugal water pump:

  • diameters of the impeller at the inlet and outlet
    • r1 = 10 cm
    • r2 = 20 cm
  • Speed = 1500 rpm (revolutions per minute)
  • the blade angle at inlet β1 = 30°
  • the blade angle at outlet β2 = 20°
  • assume that the blade widths at inlet and outlet are: b1 = b2 = 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):

Vr1 = u1 tan 30° =  ω r1 tan 30° = 2π x (1500/60) x 0.1 x tan 30° = 9.1 m/s

Radial component of flow velocity determines how much the volume flow rate is entering the impeller. So when we know Vr1 at inlet, we can determine the discharge of this pump according to following equation. Here b1 means the blade width of the impeller at inlet.

Q = 2π.r1.b1.Vr1 = 2π x 0.1 x 0.04 x 9.1 = 0.229 m3/s

In order to calculate the water horsepower (Pw) required, we have to determine the outlet tangential flow velocity Vt2, because it has been assumed that the inlet tangential velocity Vt1 is equal to zero.

The outlet radial flow velocity follows from conservation of Q:

Q = 2π.r2.b2.Vr2  ⇒ Vr2 = Q / 2π.r2.b2 = 0.229 / (2π x 0.2 x 0.04) =  4.56 m/s

From the figure (velocity triangle) outlet blade angle, β2,  can be easily represented as follows.

cot β2 = (u2 – Vt2) / Vr2

and therefore the outlet tangential flow velocity Vt2 is:

Vt2 = u2 – Vr2 . cot 20° = ω r2 – Vr2 . 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:

Pw  = ρ Q u2 Vt2 = 1000 [kg/m3] x 0.229 [m3/s] x 31.4 [m/s] x 18.9 [m/s] = 135900 W = 135.6 kW

and the pump head is:

H ≈ Pw / (ρ g Q) = 135900 / (1000 x 9.81 x 0.229) = 60.5 m

 
References:
Reactor Physics and Thermal Hydraulics:
  1. J. R. Lamarsh, Introduction to Nuclear Reactor Theory, 2nd ed., Addison-Wesley, Reading, MA (1983).
  2. J. R. Lamarsh, A. J. Baratta, Introduction to Nuclear Engineering, 3d ed., Prentice-Hall, 2001, ISBN: 0-201-82498-1.
  3. W. M. Stacey, Nuclear Reactor Physics, John Wiley & Sons, 2001, ISBN: 0- 471-39127-1.
  4. Glasstone, Sesonske. Nuclear Reactor Engineering: Reactor Systems Engineering, Springer; 4th edition, 1994, ISBN: 978-0412985317
  5. Todreas Neil E., Kazimi Mujid S. Nuclear Systems Volume I: Thermal Hydraulic Fundamentals, Second Edition. CRC Press; 2 edition, 2012, ISBN: 978-0415802871
  6. Zohuri B., McDaniel P. Thermodynamics in Nuclear Power Plant Systems. Springer; 2015, ISBN: 978-3-319-13419-2
  7. Moran Michal J., Shapiro Howard N. Fundamentals of Engineering Thermodynamics, Fifth Edition, John Wiley & Sons, 2006, ISBN: 978-0-470-03037-0
  8. Kleinstreuer C. Modern Fluid Dynamics. Springer, 2010, ISBN 978-1-4020-8670-0.
  9. U.S. Department of Energy, THERMODYNAMICS, HEAT TRANSFER, AND FLUID FLOW. DOE Fundamentals Handbook, Volume 1, 2 and 3. June 1992.
  10. White Frank M., Fluid Mechanics, McGraw-Hill Education, 7th edition, February, 2010, ISBN: 978-0077422417

See above:

Centrifugal Pumps