summarized in Appendix B of this text. This tabulation also includes the modu-lus of elasticity, the shear modumodu-lus, and the coefficient of thermal expansion for each metal. These fundamental properties are referred to throughout this book, and it is convenient to have typical values readily available. It should be men-tioned that some materials do not have totally unique properties. For critical cal-culations the precise physical properties should be obtained from a metallurgical reference source, or specific tests of the metal.
Fig. 3–7 Hollow Circular Cylinder
With Physical Dimensions Length (
L) Outer Diameter (Do) Inner Diam.(Di)
Do = 2 Ro Di = 2 Ri
Aannulus π×Ro2–π×Ri2 π
4---×(Do2–Di2)
= =
Vannulus L×Aannulus π×L
---4 ×(Do2–Di2)
= =
Wannulus ρ×Vannulus π ×L ×ρ
---4 ×(Do2–Di2)
= =
Mass and Support Distribution 71
For some calculations, it is necessary to know the mass of the mechanical part. For the annulus under discussion, the mass is easily determined by divid-ing the weight from equation (3-4) by the acceleration of gravity G as follows:
(3-4) In many instances, machinery shafts are not hollow, and they are fabricated of solid metal. If the inner diameter Di is set equal to zero in equations (3-1) through (3-4), the following expressions for solid machine shafts with an outer diameter of D are easily developed:
(3-5)
(3-6)
(3-7)
(3-8) In order to be perfectly clear on the dimensional aspects of equations (3-1) to (3-8), the defined variables and their respective English engineering units are summarized as follows:
L = Cylinder Length (Inches)
D = Solid Cylinder Outer Diameter (Inches) Do = Hollow Cylinder Outer Diameter (Inches)
Di = Hollow Cylinder Inner Diameter (Inches) Ro = Hollow Cylinder Outer Radius (Inches)
Ri = Hollow Cylinder Inner Radius (Inches) ρ = Cylinder Material Density (Pounds / Inch3) A = Cylinder Cross Sectional Area (Inches2) V = Cylinder Volume (Inches3)
W = Cylinder Weight (Pounds)
M = Cylinder Mass (Pounds-Second2 / Inch)
G = Acceleration of Gravity (= 386.1 Inches / Second2)
These variables and associated units will be used in the next sections on inertia, plus throughout the remainder of this text.
Mannulus Wannulus
---G π ×L ×ρ 4×G
---×(Do2–Di2)
= =
Asolid π×D2 ---4
=
Vsolid π ×L×D2 ---4
=
Wsolid π ×L ×ρ×D2 ---4
=
Msolid π ×L ×ρ×D2 4×G
---=
72 Chapter-3
Case History 3: Two Stage Compressor Rotor Weight Distribution
Consider the compressor rotor depicted in Fig. 3-8. This high speed rotating assembly consists of a shaft with various diameters, plus two midspan impellers, and a thrust collar. Since the impellers are mounted back to back, this particular design does not include a balance piston. As indicated on the diagram, the rotor weights 282 pounds, and there is a moderate overhang on the thrust end of the rotor. Normal operating speed varies between 15,000 and 17,500 RPM. This com-pressor is a drive through unit with a stream turbine coupled to one end, and another compressor coupled to the opposite (thrust) end. Historically, the bearing on the drive end of the rotor seldom exhibited any damage, but the journal bear-ing located at the thrust end was often found to be in a distressed or damaged condition. On other occasions, the bearings would fail during operation, and the rotor would be severely damaged. Depending on the severity of the failure, the shaft was often chrome plated. After severe failures, the entire rotating assembly was totally replaced.
This machine operates in a very dirty and corrosive environment. The rotor was constantly subjected to large unbalance forces due to the accumulation of
Fig. 3–8 Two Stage Cen-trifugal Compressor Rotor With Weight And Moment Diagram
1.11# 4.83# 10.58# 36.41# 19.21# 37.18# 32.40# 37.18# 17.37# 32.67# 10.40# 9.13# 17.51# 13.84# 2.09#
Seal
Mass and Support Distribution 73
material in the impellers, plus random corrosion of the impeller vanes. During a typical run, the compressor would startup smoothly after an overhaul. However, synchronous running speed vibration response would always deteriorate with the passage of time.
An initial step in the analysis of this problem required the determination of static bearing loads. To achieve this goal, the rotor was divided into fifteen sec-tions, and the weight of each section was calculated. Since this was a solid shaft, equation (3-7) was used to compute the weight of each shaft section. The weight of the thrust collar was determined with (3-3), and the two impellers were weighed separately. Next, the thrust collar and wheel weights were combined with their respective shaft section weights, and the weight distribution summa-rized in Fig. 3-8.
This information was then combined with the distance from the center of the drive end bearing to the centroid of each rotor segment. The complete array of weights and distances are shown in Fig. 3-8. From this sketch, it is possible to perform a summation of moments around the center or transverse axis of the drive end journal in the following manner:
(3-9)
Since the total rotor weight is 282 pounds, the next force balance applies:
Thus, the drive end bearing has a 116 pound static load, and the thrust end journal carries 166 pounds. Although the differential force is only 50 pounds, it is an appreciable percentage difference. Ultimately, it was determined that the bearings were only marginally sized to accommodate the rotor weight. However, they were considerably undersized when the additional unbalance forces due to foreign objects were included. It was also determined that the available load capacity for the drive end was barely acceptable, whereas the load carrying capa-bility for the thrust end journal bearing was unacceptable. Based upon these conclusions, both bearings were increased in size and load capacity.
The larger bearings reduced the number of machine failures per year, and overall reliability was substantially improved. Further improvements in machine longevity would require changes in the chemical plant process. Unfortu-nately, the required alterations to the processing scheme could not be economi-cally justified. Hence, the compressor rotor was occasionally sacrificed to meet
Momentsccw
∑
=∑
Momentscw17.54 Inch-Pounds+Wb×40.68 Inches = 6 768.74, Inch-Pounds Wb×40.68 Inches = 6 751.20, Inch-Pounds
Wb = 166 Pounds
Wa+Wb = 282 Pounds Wa+166 Pounds = 282 Pounds
Wa = 116 Pounds
production quotas.
From this example, it is clear that even a simple analysis of rotor weight distribution and bearing static loads may be beneficial. In some cases this may solve a problem, or it may provide insight into prospective solutions. It must also be recognized that the dynamics of the rotating system must be considered. This includes the effects of mass unbalance that serves to deform the mode shape, plus the effects of inertia, stiffness, and damping of machine elements.