Preliminary calculation:
The maximum number of cells: 490V/2.27V=216 cells;
The minimum voltage per cell at the end of discharge: 378/216=1.75V.
Case1:
discharge with a surge at the end of discharge
The surge power is to be 50kW for 10 minutes, followed by 10kW for 2 hours.
Discharge current:
During the surge:50000W/378V=132 amps, and then:10000W/378V=26 amps
Determining the cell required for the current required:
Current flow during surge: (132Ax10min)/60min=22Ah
Current flow for 4 hour:26Ax2h-52Ah
Total capacity drawn:22Ah+52Ah=74Ah
Equivalent discharge time at 26 amps to supply 74Ah:74/26=2.8hours
From the table of performance characteristics, expressed in terms of the discharge current in
amps for 1.75V end voltage, the cell to give a current of 26 amps for 2.8 hours is the SP12-100
Conclusion:
In this example, it is the total number of Ah required which determine the battery to be used, i.e.
216 cells/36 batteries of type SP12-100.
Case2:
discharge with a surge at the end of discharge (here again, it is the surge which
dictates the battery to be used).
The continuous power is to be 10kW for one hour, followed by a surge of 50kW for 20 minutes.
Discharge current:
During the surge: 50000W/378V=132 amps
Before the surge: 10000W/378V=26 amps
Capacity drawn in 1 hour: 26Ax1h=26Ah
Capacity drawn during surge (20 min)(132 ampsx20min)/60min=43.5Ah
Total capacity drawn: 69.5Ah
Equivalent discharge time at 360 amps to supply 26Ah (69.5/132)x60min-32min
From the table of performance characteristics, expressed in terms of the discharge current in
amps for 1.75V end voltage, the cell to give a current of 132 amps for 32 minutes is the
SP12-150. The battery to be used will consist of 216 cells/36 batteries of type SP12-150.
O
pe
ra
tio
n
an
d
m
ai
nt
en
an
ce
17
Summary of Contents for sp series
Page 1: ......
