EasyManua.ls Logo

ComAp InteliMains 210 BTB

ComAp InteliMains 210 BTB
576 pages
To Next Page IconTo Next Page
To Next Page IconTo Next Page
To Previous Page IconTo Previous Page
To Previous Page IconTo Previous Page
Loading...
InteliMains 210 Global Guide
90
Image 5.24 Run hours equalization - case #2
step 0 1 2 3 4 5 6 7 8 9 10 11 12 13
RHE1 0 11 11 11 11 33 33 33 33 55 55 55 55 77
RHE2 0 0 11 11 22 22 33 33 44 44 55 55 66 66
RHE3 0 0 0 22 22 22 22 44 44 44 44 66 66 66
Run G1 (ΔRHE1) 0 11 0 0 0 22 0 0 0 22 0 0 0 22
Run G2 (ΔRHE2) 0 0 11 0 11 0 11 0 11 0 11 0 11 0
Run G3 (ΔRHE3) 0 0 0 22 0 0 0 22 0 0 0 22 0 0
Case #3:
Gen-set 1 running hours = 250 -> running hours considered in RHE = 100 (150-RunHoursBase)
Gen-set 2 running hours = 450 -> running hours considered in RHE = 200 (250-RunHoursBase)
Gen-set 3 running hours = 750 -> running hours considered in RHE = 250 (500-RunHoursBase)
The gen-set 1 has the lowest RHE1 = 100 h. By applying the SwapTime formula, we get the run time of gen-set
2 before next swapping:
SwapTimeG1 = 200 100 + 10 + 1 = 111
Till the step 5, the evolution of the gen-set swapping is the same as in the case #1, just gen-set 1 and gen-set 2
involve. In the step 6 the gen-set 2 can run only 17 hours (previously 22 hours) because the gen-set 3 involves.
The evolution of RHE1, RHE2 and RHE3 is shown on the figure below.
Image 5.25 Run hours equalization - case #3

Table of Contents

Related product manuals