UPS
Wednesday, 22 May 2013
Black Start
Start-up of the UPS (inverter part only) to provide power to the load, even when the Main is not present on the input of the UPS.
This option is made of a key switch to force the battery circuit breaker Q5 to close. In this way, the battery circuit is powered, thus allowing the inverter to start thanks to the battery current.
Low Voltage disconnect contactor LDV
Protect the battery from deep discharges and thus enhance battery lifetime.
The LDV option includes an output contactor controlled by voltage relay in order to disconnect the load at the end of
battery autonomy period. Reconnection of the load is automatic at the charger restoration and upon the resumption of normal conditions.
MAKE BEFORE BREAK CHANGEOVER
Under normal operation switches A and C
will be closed and switch B will be open. The inverter will be feeding
the load through its static switch at a phase and frequency which is
synchronized with the bypass supply.
To initiate a manual transfer the
inverter is first switched off after
verification that the inverter is synchronized with the bypass.
This will force the static switch to affect an immediate no-break transfer to
the bypass supply (can expand on this if required). Now the manual bypass
switch B can be safely closed consequently paralleling the bypass through 2 circuits:
the manual bypass switch and the bypass static switch. All that then
remains is to isolate the UPS for maintenance by opening switches A and C.
Retransfer to inverter can by carried
out by effectively reversing the above procedure.
2 x 50% BATTERY
Assume we have a110VDC installation of 2 x 50% batteries, previously sized at 82 x LBE600P (600Ah) per 50% battery. Both batteries are connected by a normally open DC bus-tie A. Should the need to arise to close the bus-tie, what would be the magnitude of the circulating current between a potentially fully charged 50% battery and a fully discharged 50% battery:
On the basis that the battery complies
to DEP standards, i.e. 110VDC +10%/-15% without dropping diodes, the end cell
voltage after a full discharge will be 1.14V per cell. On closing the
bus-tie A there will be a potential difference between the 2 battery halves of
(82 x 1.2V) – (82 – 1.14V) = 4.92V.
Each LBE600P cell has an internal
resistance of 0.3mOhm per cell when fully charged. On discharging this
will increase significantly so a fully discharged battery will have a
significantly higher internal resistance. For the purposes of this
example we’ll assume the worst case which is the lowest resistance. So
the complete DC circuit comprises 82 cells x 2 batteries x 0.3 mOhm = 0.0492
Ohm
For a short period immediately after the
bus-tie is closed a current will circulate from the fully charged half battery
to the fully discharged half battery of 4.92V / 0.0492 Ohm = 100A. This
will very quickly diminish as the battery voltages become more equal.
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