按工作周期选定固定式蓄电池
依据工作周期选定固定式蓄电池及充电器。
尝试一个工程 4
每项工作都会预填示例数值。请替换为您自己的数值。
- Size a substation battery by duty cycleThe Ah and charger of a 125 V battery for a trip and three hours of load.
- Size a 48 V radio site batteryThe Ah and charger for four hours of radio load and a transmit burst.
- Size a 24 V control panel batteryThe Ah for two hours of PLC load and a valve close at the end.
- Find the charger for a faster rechargeThe charger current to put a 110 V battery back in 8 h, not 12.
输入
使用默认输入值的示例计算。
- 电池标称电压
- 110 V
- 串联电池数
- 55
- 负荷可接受的最低电压
- 96 V
- 连续周期
- 3 行
- 倍率系数
- 1
- 温度系数
- 1
- 设计裕度
- 1.1
- 老化系数
- 1.25
- 充电器在充电时的负荷
- 10 A
- 充电时间
- 12 h
- 充电系数
- 1.1
按安时计的容量
112.1Ah
- 工作负荷的安时数
- 81.5Ah
- 充电器电流
- 17.5A
- 每节电池终止电压
- 1.75V
- 工作负荷中的最大电流
- 70A
- 额定电压下的能量
- 12.33kWh
此结果仅供设计参考。请对照现行标准进行核验。
计算步骤
- 周期 1:70 A,持续 1 min1.17Ah
- 周期 2:10 A,持续 478 min79.67Ah
- 周期 3:40 A,持续 1 min0.67Ah
- 工作负荷的安时数81.5Ah
- 工作时长480 min
- 额定系数,取自厂商表格×1.00
- 温度系数×1.00
- 设计裕量×1.10
- 老化系数×1.25
- 按安时计的容量结果112.1Ah
- 每节电池终止电压
96 V ÷ 55 个电池 = 1.745 V
- 充电器电流17.5A
- 额定电压下的能量12.33kWh
关于此计算器
How this is calculated
A stationary battery carries a site's d.c. loads when the charger loses its supply: protection relays, breaker trip and close coils, emergency lights, radios, controls. Those loads change through the outage. The duty cycle lists them in order, as a current for a number of minutes.
The tool adds the ampere-hours of each period: the current times the minutes over sixty. That is the charge the duty takes out of the battery.
A battery gives less than its rated capacity when it discharges faster than its rated hours, or to a higher end voltage. So the tool multiplies the duty by a rate factor you read from the maker's table. It then applies a temperature factor for a cold battery room, a design margin for growth and unknowns, and an ageing factor so the battery still does the duty near the end of its life. The product is the capacity by ampere-hours. When a later period draws more than the one before it, that sum is too small, and the tool warns.
The end voltage per cell is the lowest voltage your loads accept, at the battery terminals, over the number of cells in series. Read the maker's capacity at that end voltage.
The charger must carry the standing load and put back what the duty took, with a margin for charge losses, in the recharge time you set. The tool adds the continuous load to the duty's ampere-hours times the recharge factor over the hours.
AmpSize does not hold IEEE 485 or any cell's capacity factors. With the maker's Kt on every period, the tool works the last section, each change in current times its Kt, and that figure governs. Every factor is an assumption you set, and the result says so.
| Step | Table or clause | What it decides |
|---|---|---|
| Duty | Σ I × t ÷ sixty | The ampere-hours the periods take out |
| Capacity | Duty × rate × temperature × margin × ageing | The battery capacity by ampere-hours |
| Sections | Σ ΔI × Kt × temperature × margin × ageing | The capacity with the maker's Kt, which governs |
| End voltage | Lowest voltage ÷ cells | The end voltage per cell to read the maker's table at |
| Charger | Continuous load + duty × recharge factor ÷ hours | The charger current |
Worked examples
- A 110 V substation battery for a trip, eight hours of standing load and a close needs 112.1 Ah. The duty takes 81.5 Ah, with a charger of 17.5 A to recharge in 12 hours.
- A 125 V battery for a trip and three hours of lights and controls needs 159.2 Ah. The duty takes 91.5 Ah, with a charger of 25.1 A.
- A 48 V radio site with a transmit burst needs 49.7 Ah. The duty takes 36.17 Ah, with a charger of 12 A. The burst draws more than the radio load, so the tool warns that the sum understates the battery.
- A 24 V control panel with a valve close at the end needs 11.7 Ah. The duty takes 8.5 Ah, with a charger of 5.2 A. The valve close draws more than the load before it, so the tool warns that the sum understates the battery.
Limits
Without Kt values, the tool adds ampere-hours. A battery's capacity falls at high rates, so a heavy load of a minute or two can need far more than its ampere-hours show. The rate factor carries that, and only the maker's table gives it.
It does not pick a cell, a chemistry or a float voltage, and it does not check ventilation or the battery room. Follow the maker's data for those.
See also: Battery, Fire alarm battery and DC cable size.
问题
结果仅供设计参考。请依据现行标准核对。
如何根据蓄电池工作周期选择蓄电池?
逐时段列出蓄电池须承载的负荷:电流和分钟数。累加安时数。再乘以制造商表中的倍率系数、温度系数、设计余量和老化系数。若后一时段的负荷大于前一时段,则上述求和偏小:请使用制造商的 Kt 和分段法。
为何短时大负荷所需的容量不止安时数?
蓄电池在高倍率放电时提供的安时数少于其额定倍率下的数值。1 分钟的脱扣负荷所需容量可能远超其在工作周期中的份额。在终止电压下,从制造商表中取得的倍率系数可对此进行修正。
每个电池单体的终止电压应取多少?
将负荷可接受的最低电压(在蓄电池端)除以电池节数。查取制造商在该终止电压下的容量。终止电压越高,同一电池可提供的容量越小。
充电器应选多大容量?
充电器承担持续负荷,并在充电时间内补回被消耗的安时数,并略有富余。工具将持续负荷与工作制安时数乘以充电系数,再除以小时数,两者相加。
本工具是否遵循 IEEE 485?
AmpSize 未收录 IEEE 485,也未收录任何电池的 Kt 值。请在每个时段输入制造商的 Kt,本工具即可计算该方法的最后一部分。若未输入,则本工具累加安时数,应用您的系数,并在后续时段耗电增大时发出警告。
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