Substation Battery Sizing Calculation Made Easy

Batteries are the lifeline to substations, providing backup power. I’m going to go over a typical substation battery sizing calculation.

I’ll guide you through each step and highlight the key factors to keep in mind for all your different substation loads. But before we dive into the nitty-gritty of the calculation, let’s take a moment to appreciate the crucial role these batteries play in keeping power flowing.

The lifeline of substations is batteries

Substations are the unsung heroes of the power grid. They convert voltage from high to low and vice versa, keeping the lights on and the electricity flowing.

In fact, power grids around the world would be helpless without substations. This is why they need batteries to ensure all critical substation loads can operate at any time.

But why, you ask? Well, while the primary source of power comes from the AC power supply, it’s not always reliable. If a transmission line or generation source goes offline, you risk losing power altogether. And this is where batteries come in – providing backup power to keep critical equipment running when you need it most.

So let’s take a look at a schematic diagram to better understand the magic of substation batteries.

You can see how the 120V AC power supply feeds the 21kV switchgear protection loads. But the protection loads are all powered by 48V DC.

For this reason, we use a battery charger. The battery charger converts the 120V AC supply to 48V DC, while also keeping the floating 48V DC battery charged and ready to go.

What’s float charging?

It’s just like standby use. The battery acts as a backup power source if the AC power supply fails, so it must always be fully charged and ready to go.

How does the battery system work?

The battery is always on standby, ready to power your system. When you’re connected to the AC power supply, the following two things happen:

The battery charger trickle charges the battery as needed.
The AC power supply powers the loads through the battery charger.

If you need an extra boost of power, the battery’s got you covered. And if the AC power supply suddenly goes out, the battery will take over without skipping a beat.

This is the beauty of having an independent power source like a battery in your system – no more worries about power outages.

Low voltage power schematic with a DC panel

The following schematic is similar to the previous one. But now, the battery charger feeds a 125V DC panel, which in turn powers all the critical DC loads.

Once again, a floating battery is in place as a backup in case of a loss of AC power.

Looking at these schematics, it’s easy to see how batteries play a vital role as backup to the AC power supply. But it doesn’t stop there – batteries also do the following:

Keep microprocessor relays, control circuits, emergency lighting, and communication equipment powered up
Ensure power is uninterrupted when you need it most
Trigger power circuit breaker trip and close coils
Keep motor-operated air breaks, valves, pumps, and fans in motion

Without reliable DC backup power, you risk damage to your substation and downstream equipment, not to mention safety concerns.

Substation battery sizing calculation

Let’s do some math to figure out the size of a flooded cell, lead-acid battery for a substation. The battery will be rated 125V DC nominal and have an amp-hour capacity rated for an 8-hour rate of discharge.

The 8-hour rate of discharge is the typical discharge rate for substations. This gives operators an 8-hour window to fix any AC power supply issues before things start going haywire.

Important Note: The substation battery sizing calculation we’ll be using is per IEEE Standard 485. This standard is a method for defining DC loads and sizing lead-acid batteries.

But before we get started, the following are the two types of loads in a substation:

Continuous loads typically run for 8-hour periods.
Momentary loads operate for less than 1 minute.

Now, we’ll go over the loads for our substation battery sizing calculation.

Continuous load list

Auxiliary relays & timers: (10) x 5VA = 50VA
Indicating lights (LED): (20) x 5.5VA = 110VA
Multifunction relays: (18) x 5VA = 90VA
Power meter: (16) x 10VA = 160VA
SCADA allowance: 1000VA

Total: 50VA + 110VA + 90VA + 160VA + 1000VA = 1,410VA

$\Rightarrow \dfrac{1410VA}{125VDC} = 11.28A\: at\: 125VDC$

Important Note: The “Multifunction relays” consist of all your regular protection relays. For example, 86L, 86B, 86T, 151T/151N, 87A, 51AT, and so on.

In our calculation, we group them together for simplicity. Each relay usually has a load no greater than 5VA, but it’s essential to verify each relay’s load individually.

Momentary load list

Auxiliary relays: (5) x 5VA = 25VA at 1 second
Breaker close coil: (20) x 150VA = 3,000VA at 0.05 seconds
Breaker spring winding motor: (20) x 500VA = 10,000VA at 5 seconds
Lockout relays: (6) x 325VA = 1,950VA at 0.01 seconds
Main breaker trip coil: (20) x 150VA = 3,000VA at 0.03 seconds

Total summed design load

Peak load ( $S_{p}$ ) = Continuous load + Coincident intermittent load
$= 1,410VA + (25 + 3,000 + 10,000 + 1,950) VA$
$= 1,410VA + 14,975VA = 16,385VA$

Important Note: The “main breaker trip coil” and “breaker close coil” shouldn’t both be included in the peak load calculation. You can only activate one at a time. Therefore, we only include one in our calculation to avoid overestimating the load.

Design load $(S_{d}) = S_{p}(1 + K_{g})(1 + K_{c})$ , where:

$K_{g}$ = contingency for future load growth (assume 10% = 0.1)
$K_{c}$ = design contingency margin (assume 10% = 0.1)

$(S_{d}) = S_{p}(1 + K_{g})(1 + K_{c})$
$= 16,385VA (1 + 0.1)(1 + 0.1)$
$= 19,825.85VA$

$\Rightarrow \dfrac{19825.85VA}{125VDC} = 158.61A\: at\: 125VDC$

Design energy demand calculation

$E_{d} = E_{t}(1 + K_{g})(1 + K_{c})$ , where:

$K_{g}$ = contingency for future load growth (assume 10% = 0.1)
$K_{c}$ = design contingency margin (assume 10% = 0.1)

$E_{t} = (1,410VA \times 8 \: hours) + (25VA \times 1 \: min \times \dfrac{1 \: hour}{60 \: min}) + (3,000VA \times 1 \: min \times \dfrac{1 \: hour}{60 \: min}) + (10,000VA \times 1 \: min \times \dfrac{1 \: hour}{60 \: min}) + (1,950VA \times 1 \: min \times \dfrac{1 \: hour}{60 \: min})$

$E_{t} = 11,280VAh + 0.42VAh + 50VAh + 166.67VAh + 32.5VAh$
= 11,529.59VAh

$E_{d} = E_{t}(1 + K_{g})(1 + K_{c})$
= 11,529.59VAh (1+.1)(1+.1)
= 13,950.8VAh

$C_{min}$ = Minimum battery capcaity in Ah

$C_{min} = E_{d}\dfrac{(K_{a} \times K_{t} \times K_{c})}{V_{dc}}\times K_{dod} \times K_{e}$ , where:

$E_{d}$ = design energy demand (VAh)
$V_{dc}$ = nominal battery bank voltage (Vdc)
$K_{a}$ = battery ageing factor (%). This value captures a battery’s decrease in performance due to age.
$K_{t}$ = temperature correction factor (%). This value captures the ambient installation temperature. Use the temperature correction factor table from IEEE-485.
$K_{c}$ = capacity rating factor (%). This value captures the voltage depressions from battery discharge.
$K_{e}$ = system efficiency (%). Allowance for losses in the battery.
$K_{dod}$ = maximum depth of discharge (%).

Below, we’ll assign values to our above-listed variables:

$E_{d}$ = 13,950.8VAh
$V_{dc}$ = 125VDC
$K_{a}$ = 1.25 aging factor
$K_{t}$ = 1.19 (50°F ambient)
$K_{c}$ = 1.1
$K_{e}$ = 0.97
$K_{dod}$ = 0.8 (80%)

$C_{min} = 13,950.8VAh \times 1.25 \times \dfrac{1.19 \times 1.1}{125VDC} \times 0.8 \times 0.97=141.71Ah$

Next, we increase the calculated battery capacity by 125%. This ensures the battery meets our design requirements throughout its entire life.

125% of 141.71 Ah = 177.14Ah

Important Note: Select a battery amp-hour capacity, which is greater than the minimum capacity calculated. Additionally, specify the battery discharge rate, also known as the C rating. For example, if you need a battery, which can discharge for 24 hours, use C24 in your battery call-out.

Our battery will have a minimum rating of 200Ah at an 8-hour rate, or C8.

Sizing the substation battery charger

Required charger rating: $A = \dfrac{kC}{H} + L_{c}$ , where:

A = required charger output rating
k = efficiency factor (1.1 for lead acid batteries)
C = calculated Ah discharge from the battery based on the duty cycle
H = recharge time (8 hours)
$L_{c}$ = continuous load in amps

$A = \dfrac{1.1\times141.71Ah}{8} + 11.3 = 30.79$ Amps

125% x 30.79 amps = 38.49 Amps. So, the battery charger will have a minimum output rating of 40 amps.

Important Note: All batteries have internal resistance, which causes the cell voltage to decrease as it depletes. The 125% multiplier accounts for this voltage drop to ensure the battery is not undersized.

Important battery notes

1) Heavy discharge: Lead-acid batteries prefer intermittent loads over continuous loads. Intermittent loads give batteries a rest period to recompose their chemical reaction.

2) Battery room ventilation: Lead-acid batteries release hydrogen gas when recharging. Without proper ventilation, hydrogen gas builds up and increases explosion risks.

3) Battery room temperature: Optimum temperature is normally between 68-77° Fahrenheit. Battery life will shorten if the temperature goes too much higher.

4) Slow charging: Lead-acid batteries charge slowly. The typical charging time is 14 to 16 hours.

5) Full state of charge: Lead-acid batteries need to always remain fully charged. A low charge causes sulfation, leading to attenuated performance.

6) Charging voltage: To maintain optimal battery performance, it’s important to use the correct charging voltage limits. A low voltage limit can cause poor battery performance and sulfation buildup on the negative battery plate. On the other hand, a high voltage limit can improve performance, but it can also cause grid corrosion to form on the positive battery plate.

It’s crucial the battery charger applies the appropriate voltage to a fully charged battery to prevent overcharging and avoid any potential damage.

Substation battery sizing calculation wrap up

Calculating the size of your substation’s batteries is a breeze. All you need is your trusty load information. Although, I’ll admit, gathering all the load info can be a bit of a drag.

When it comes to a powerhouse like a substation, redundancy is key. Think about it, every critical component in a machine needs a backup plan, right? Well, the same goes for a power substation, which is essentially a massive machine.

So, to ensure smooth sailing, it’s essential to include properly sized batteries and chargers in your substation. By doing so, you’ll never have to worry about dreaded downtimes, which could potentially cost you a fortune. Keep your substation running like a well-oiled machine, and you’ll be in the clear.

Do you have any substation battery sizing tips? In what other systems have you seen batteries provide backup to the main power source?

SUBSCRIBE TO ENGINEER CALCS NEWSLETTER

Get daily articles and news delivered to your email inbox

Engineer Calcs

Koosha started Engineer Calcs in 2020 to help people better understand the engineering and construction industry, and to discuss various science and engineering-related topics to make people think. He has been working in the engineering and tech industry in California for over 15 years now and is a licensed professional electrical engineer, and also has various entrepreneurial pursuits.

Koosha has an extensive background in the design and specification of electrical systems with areas of expertise including power generation, transmission, distribution, instrumentation and controls, and water distribution and pumping as well as alternative energy (wind, solar, geothermal, and storage).

Koosha is most interested in engineering innovations, the cosmos, our history and future, sports, and fitness.

13 thoughts on “Substation Battery Sizing Calculation Made Easy”

Anonymous

February 4, 2021 at 11:36 am

clear explanation with example
- Engineer Calcs
  
  February 4, 2021 at 3:24 pm
  
  Glad you found it helpful.
Daniel Pecsoy

April 15, 2021 at 6:09 am

Hello Sir, I find your article very useful and relevant to my work as a maintenance personnel in a distribution Utilities. I have a query sir. 1) In our 20 MVA Substation, we are going to replace our 48 VDC Batteries (4-12 VDC batteries, 100AH into 2Volts, 200AH batteries (24pcs.). This is to have a longer years for the batteries to operate. My question is: a) Is there a need to upgrade the charger since we will upgrade the battery’s AH? the charger is rated at 25A, 48V output, and 240V input. The continuous load is at 3.8 A. thank you.
- Engineer Calcs
  
  April 16, 2021 at 2:53 am
  
  Glad you found it helpful!
  
  As an off-the-cuff reply, you probably would need to replace the charger, but there are other variables you need to consider too. Thus, you need your engineer on record to review the existing charger’s compatibility.
A. S. D. Shuaibu

July 9, 2021 at 9:19 am

I found it very helpful
- Engineer Calcs
  
  July 9, 2021 at 2:22 pm
  
  Glad you found it to be helpful.
Michael

September 9, 2021 at 12:20 pm

Batteries with step loads as you have described will inherently decrease in voltage as they progress further into their 8 hours cycle. While you might float the batteries at 130VDC you typically have an end voltage of somewhere around 105VDC. Your calculations utilize the nominal voltage of 125VDC but battery load profiles will show you that the battery voltage will continuously taper off during the load draw. IEEE recommends you use the lowest voltage of your battery to calculate all your loads and during sizing. Additionally if you use any large battery manufacturer sizing programs, you will find that they as well utilize the end voltage to convert VA(W) to Amps. While I agree using the end voltage is ultra conservative, the method of using the nominal voltage is incorrect and results in undersized batteries for true use. You have masked that by adding all of your % increases but ultimately it will not be sized sufficiently for the full life cycle.
- Engineer Calcs
  
  September 14, 2021 at 5:14 pm
  
  Michael, thanks for your great comment.
  
  Yes, all batteries have an internal resistance, and the cell voltage decreases as it depletes.
  
  The 125% you’re referencing, accounts for the slack with the voltage drop. This percent actually outputs a more conservative value than using 105V. While noting, 105V is another assumption value similar to the 125%. Using the nominal value with a thought-out multiplier is more practical than thinking you have precision with the 105V value.
  
  To also point out, the VAs are not constant, as each device is based on a range of voltages. While noting, as the voltage drops on a DC system, the current drops as well (assuming a constant resistance). This needs to be considered as well.
  
  Your comment highlights how the post needs some added clarity with the percentage though. I added a note.
Josh

August 23, 2022 at 2:45 pm

Hello,

Thank you for this very informative article. I was wondering if the 8 hour rate of discharge is discussed in IEEE standards as a typical rate to use in substations or this is just an industry practice?
- Josh
  
  August 23, 2022 at 2:47 pm
  
  Also, are there any regulatory requirements (i.e. NERC, FERC, etc.) that mentions how long the rate of discharge should be?
  - Engineer Calcs
    
    September 6, 2022 at 3:00 am
    
    Generally speaking, with substations you want a short duration high discharge battery. Also, you want to ensure the battery has the capacity to support the load and trip per IEEE 450 (e.g. maintaining a minimum battery voltage).
- Engineer Calcs
  
  September 6, 2022 at 12:39 am
  
  Hi Josh,
  
  It’s an industry standard, but you’ll come across various requirements from customer to customer. There are recommended battery sizing practices in IEEE 485 as well.
  
  In short, you want ample backup power allowance for operators to restore AC power or make any necessary control decisions.
Mina

January 1, 2023 at 10:18 am

Short and useful