Site power: the numbers that are quietly optimistic

Rated capacity is quoted at a discharge rate you will never use, and a voltage drop measured against nominal hides the only comparison that matters.

The label is not the answer

A battery rated at 100 Ah is one that delivers 5 A for twenty hours. That is the convention: capacity is quoted at a slow discharge, almost always the twenty-hour rate. It is a perfectly reasonable way to label a cell, and it is not the number you want when a site loses mains and empties its string in two hours.

Ask that same 100 Ah cell for 50 A and it does not give you two hours. The chemistry cannot keep up with the demand; the usable charge falls. Peukert's equation describes how far. With an exponent around 1.25, typical for VRLA, a two-hour discharge yields about 56% of the rated amp-hours — so the naive amp-hours-over-amps calculation promises 120 minutes and the string delivers about 67.

The important part is that this is not a fixed fudge factor. At the twenty-hour rate the correction vanishes completely, because that is the rate the label describes. The error is a function of how hard the string is being pushed, which is why it is invisible in a slow bench test and obvious in a real outage.

The derates compound

The rate correction comes first, because it applies to the rated capacity at the rated rate. Everything else comes off after it, and they multiply rather than add:

• Depth of discharge. Taking a lead-acid string below about 20% remaining shortens its life sharply, so only 80% is usable in practice.
• Temperature. Cold cells hold less — roughly 0.8% per degree below 25 °C for lead-acid. An unheated cabinet at 0 °C has lost a fifth of its capacity before anything else is counted.
• Age. A string near end of life at 80% of rating is still considered serviceable, and that 80% multiplies everything above.

A four-year-old VRLA string in a cold cabinet, discharged fast, delivers roughly a third of what the label suggests. Each factor on its own looks like a detail. Together they are the difference between riding out an outage and not.

Low voltage means large current

Power is voltage times current. At −48 V a 3 kW load draws 63 A; the same load on 230 V AC draws 13 A. Everything downstream scales with the current, not the power: conductor cross-section, lug size, fuse rating, busbar, and the heat dissipated in all of them.

Two arithmetic points follow, and both are commonly got wrong in the optimistic direction.

First, both conductors carry the current, so the resistive length of a run is twice the physical distance. Using the one-way length halves the calculated drop.

Second, the current to size on is the current at the low-voltage disconnect, not at nominal or float. A constant-power load draws more current as the voltage falls, so the largest current a conductor ever sees is at the moment the battery is nearly flat — the same moment the voltage headroom is smallest.

Why a 5% drop can be fatal

Voltage drop is conventionally quoted as a percentage of nominal, and against 48 V a 2.7 V drop is under 6%. That sounds like a comfortable margin. It is not the relevant comparison.

The equipment does not care about nominal. It cares about the voltage at its own terminals at the worst moment, which is the end of a battery discharge. At that point the supply is already down at the low-voltage disconnect — 42 V on a −48 V system — and typical equipment stops at around 40 V. That leaves two volts of headroom, and a 2.7 V drop uses all of them and more.

This failure is systematically missed because everything looks fine in normal operation. On float the supply is at 54 V and the same drop is irrelevant. The cable is only inadequate during an outage, which is exactly when nobody wants to discover it.

Rectifiers carry the load and the recharge

A rectifier plant sized on the site load is undersized. After an outage the battery has to be put back, and that recharge current runs alongside the load for hours. Charging is not perfectly efficient either — lead-acid wants roughly 1.1 times the charge it gave out.

A plant sized on load alone survives one outage. The battery is then flat, or nearly so, for the second one an hour later. Adding the recharge term is what turns a plant that works once into one that works repeatedly.

Oversizing a generator is not free

Diesel engines want to be loaded. Below about 30% of rating they wet-stack: fuel does not fully burn, unburnt diesel passes into the exhaust, cylinders glaze and the turbo fouls. A generator chosen with a generous margin for a site that never grew into it can fail from being too lightly loaded.

Fuel consumption is not proportional to load either. An engine burns fuel simply to turn over, so a rough model is a fixed quarter of the full-load rate plus a load-proportional part. Runtime at low load is therefore worse than proportional reasoning suggests.

Everything becomes heat

Essentially all the electrical power drawn by equipment in a room leaves it as heat. The fraction that departs as light down a fibre or as radio waves off an antenna is negligible. So the cooling figure is the power figure, converted: one watt is 3.41 BTU per hour, and twelve thousand BTU per hour is a ton of refrigeration.

Rectifier and UPS losses sit on top of the DC load rather than inside it. A plant at 95% efficiency delivering 3 kW draws about 3.16 kW and puts the extra 160 W into the room along with everything else.

What this is not

Conductor sizing here considers voltage drop and nothing else. Current-carrying capacity, grouping, installation method, ambient derating, fault-current withstand and protection coordination can each demand a larger conductor than the drop alone, and the applicable electrical code governs. Nothing here certifies an installation or replaces a qualified electrical design.

Battery figures are model output. Vendor discharge curves are the authority, they differ between products, and the condition of a real string is measured rather than calculated.

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Frequently asked questions

Why does a 100 Ah battery not give two hours at 50 A?

Because rated capacity is quoted at a slow discharge, almost always the twenty-hour rate. Peukert's equation describes what happens at a faster one: with an exponent around 1.25 a two-hour discharge yields about 56% of the rated amp-hours, so 120 promised minutes become about 67. At the twenty-hour rate the correction disappears entirely.

Does the same correction apply to lithium?

Much less. Lithium iron phosphate has a Peukert exponent near 1.05, so it delivers most of its rating even at a fast discharge. Its problems are elsewhere — charging below freezing, and a battery management system that can disconnect the string entirely.

Why is the run length doubled for voltage drop?

Both conductors carry the current, so the resistive length is twice the physical distance. Using the one-way length halves the answer, and always in the optimistic direction.

A 5% drop sounds fine. Why is it flagged?

Because the percentage is against nominal voltage and the equipment is not. What matters is the voltage at the equipment terminals at the end of a battery discharge, when the supply is already at the low-voltage disconnect. Between a 42 V disconnect and a 40 V equipment minimum there are two volts of headroom, and a 2.7 V drop — under 6% of nominal — uses all of them.

Which current should I size conductors on?

The current at the low-voltage disconnect. A constant-power load draws more current as voltage falls, so the largest current a conductor sees is when the battery is nearly flat — the same moment the voltage headroom is smallest.

Why size rectifiers above the site load?

Because the plant carries the load and recharges the battery simultaneously, and charging is not perfectly efficient. A plant sized on load alone survives one outage; the battery is then flat for the second.

Can a generator be too big?

Yes. Below about 30% load diesel engines wet-stack: unburnt fuel glazes the cylinders and fouls the exhaust. Oversizing is not a free safety margin, and a generator that never sees real load can fail because of it.

Why is the cooling load the same as the electrical load?

Because virtually all the electrical power drawn by equipment becomes heat in the room. Rectifier and UPS losses are on top of the DC load, not included in it.

Is this an electrical design tool?

No. It works out voltage drop, load, autonomy and heat. It does not consider current-carrying capacity, grouping, installation method, derating, fault withstand or protection coordination, and it certifies nothing. The applicable electrical code and a qualified designer govern.