ComAp InteliMains 210 BTB, Глобальное руководство

Страница: 90 / 576

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