Examples of calculation of residual unbalance according to ISO 1940/1 Standards for rigid rotors

Example Number 1: Fun impeller 

 

Maximum service speed = 1500 RPM

Mass M = 200 kg

Left, right side correction radius Rs = Rd = 800 mm

Balancing quality G = 6,3

From previous diagram we obtain: 

Total acceptable residual eccentricity et = 40 m

Total acceptable residual unbalance Ut = M·e = 200 kg x 40 m = 8000 gr x mm

 

 

 

Note: The acceptable unbalance per plane has been calculated by simply dividing by two the total acceptable unbalance; this operation is correct because the two balancing planes have almost the same distance from the centre of mass position, which is at the same time almost in the centre of the rotor.

 

 

Example Number 2:  Turbine

 

 

Maximum service speed = 3000 RPM

Rotor mass M = 500 kg

Left side balancing radius Rs = 500 mm

Right side balancing radius Rd = 400 mm

Balance quality G = 2,5

 

From previous diagram we obtain:

Total acceptable residual eccentricity et = 8u

 

 

 

Values within brackets are valid for the quality  G = 1 (quality  g 1 is nowadays commonly required for turbines )

 

Example Number Three:  Impeller of a centrifugal pump 

 

Maximum service speed = 6000 RPM

Mass M = 10 kg

Balancing radius R = 100 mm

Required balancing quality G = 6.3

From previous diagram we obtain:

 

 

 

Note: Since the impeller is thin (reduced axial dimensions ) it is balanced in one plane only ( Static balancing)

 

Example Number 4:  Tool holder dynamically balanced 

The tool holder has a useful length L bigger than 2D (where D is the cone diameter ).

Considering its length it is advisable to balance it on two planes.

Maximum service speed = 24'000 RPM

Tool holder mass M = 5 kg

Correction radius on balancing plane 1 R1 40 mm

Correction radius on balancing plane 2 R2 20 mm

Required balancing quality G = 2.5

(ISO standards specify quality G=2.5 for machine tools spindles and driving systems)

Total acceptable residual eccentricity E = 1 m

Total acceptable residual unbalance Ut = M·E = 5 kg x 1 m = 5 gr x mm

 

 

 

  

 

Note: The total acceptable unbalance has been divided by two because we assumed that tool holder mass is more or less symmetrical with regard to the centre of mass position ,and that the two correction planes contain the centre of mass almost in the middle position.

Example Number 5:  Tool holder balanced in one plane only

 

Let us consider a tool holder which is to be balanced in one plane (static balancing).

Normally the tool holder is balanced in one plane only , if its length L is lower than 2D (D is cone diameter)

Maximum service speed = 12'000 RPM

 

Tool holder mass M = 1 kg

Balancing radius = 20 mm

Balancing quality G = 1

(ISO standards specify quality G 1 for grinding machine spindles)

Total acceptable eccentricity E = 2 m

Total acceptable residual unbalance Ut = M·E = 1 kg x 2 m = 2 gr x mm 

 

 

 

 

 

 

 

 

 

 

 

Cemb Hofmann UK is the leading UK Balancing expert and offers not only high precision balancing machine sales but also a highly acclaimed sub contract dynamic balancing service. 

For more information about any area of balancing, please visit the Cemb Hofmann UK Website or call us today on 0161 872 3123.