# How to Size a Load Center, Panelboards and Distribution Board?

**Sizing Load Center, Panelboards, Distribution Board & Consumer Unit According to NEC and IEC?**

Planning is the first and key priority for all electrical wiring projects and a proper estimation and analysis based on accurate calculation is must while designing and installation of a distribution system in residential or commercial applications. It is because the properly determined size of load center and panel boards will carry and handle the current as well as the future load (if any) safely and smoothly.

This way, load centers and distribution boards should be properly sized according to NEC and IEC or other applicable regional codes (similarly sizing a proper wire and cable, selecting right size MCBs or calculating the rating of plugs and outlets etc.). In today’s step by step guide, we will show how to select a right size panelboard (load center, distribution board or circuit breaker panel) according to NEC and IEC with solved examples.

**Panelboard, Load Center & Distribution Board or Consumer Unit**

Different terms are used for the same i.e. load center, panel board, distribution board or consumer unit. In short, a panel board or distribution board is a combination of multiple protective devices such as circuit breakers which is used to safely control and distribute the eclectic power to the load points such as lighting points and final circuits.

**Terms used in the US**: Panelboard, Load center, Breaker box, service panel or main electric panel.**Terms used in UK & EU**: Distribution board (commercial), Consumer unit (residential), circuit breaker box, main panel etc.

We have already discussed that in our previous post in detail, let’s move forward to the step by step tutorial. The following example can also be used to **determine the load capacity of the main circuit breaker** as well as it can be used to **calculate the the overall electrical load in a home**.

Related Posts:

*How to Determine the Number of Circuit Breakers in a Panelboard?**How to Determine the Right Size Capacity of a Subpanel?**How to Find the Proper Size of Circuit Breaker? Breaker Calculator & Examples*

Table of Contents

### How to Size a Panel Board & Load Center? 120/240V – NEC?

The common voltage levels in the USA for residential application are 120V and 240V single phase. The three wires (identified as Hot1 as Black color, Hot 2 as Red color and Neutral as White color) from the secondary side of the split transformer enters into the meter box and main service panel (Main switch breaker).

This way, the available voltage in a single phase distribution system is as follow:

- Voltage between any Hot (Hot 1 or Hot 2) and Neutral = 120V
- Voltage between Hot 1 and Hot 2 = 240V

The following figure shows the overall overview of a panel box, level of voltage between different conductors and number of circuit breakers.

Click image to enlarge

**Note 1:** The following example is based on VA (Volt x Amperes) which is known as apparent power. You may use the active or real power (in Watts) which is equal to Apparent power x power factor or VA x PF as power factor in residential buildings are almost unity (1). In this case, the apparent power in VA is equal to real power in W “Watts”.

Now lets see the following example to determine the suitable size of panelboard (load center or distribution board).

**Note 2:** This example is based on NEC which is applicable in North America, especially in US and Canada which follows NEC and CEC. Check the other examples for IEC and UK/EU right after this example.

Following is a general overview of a main service panel and its different parts including the space for future load points and 120V & 240V circuits.

Click image to enlarge

**Example:**

Calculate the right size of load center or distribution board for an 1500 ft^{2} (square foot) or 139.35 m^{2} (square meters) home floor plan having the following load points:

- Air conditioner: 240V x 25A = 6000 VA = 6 kVA
- Electric range: 240V x 35A = 9600 VA = 8.4 kVA
- Electric heater: 240V x 30A = 7200 VA = 7.2 kVA
- Clothes dryer: 240V x 15A = 3600 VA = 3.6 kVA
- Dishwasher: 120V x 10A = 1200 VA = 1.2 kVA
- Garbage disposal = 120V x 8A = 960VA = 0.96 kVA
- Two Small appliances circuits in the kitchen for refrigerator, blinder, etc.
- General lighting, fans, bathroom appliances and future load etc.

**Solution:**

Let us find and calculate the power rating for different home appliances based on floor plan without basement and garage using NEC codes and related tables.

**General Lighting Load:**

The minimum general lighting load including non-appliance receptacles for home appliances e.g. TV, table light etc) for a dwelling is 3 VA per ft^{2} (NEC Table 220.12).

This way, the general lighting load for 1800 ft^{2} (given in example):

3 VA x 1500 ft^{2} = 4500 VA = **4.5 kVA**

**Small Appliances Load**

There should be at least two 120V, 20A small appliance circuits i.e. in the kitchen for small appliances such as coffee maker & toasters etc. (NEC Article 210.11(C)(1). These circuits should be rated at 1.5 kVA (NEC Article 220.52(A). This way, the small appliances loads in the given example house:

2 x 1500 VA = 3000 VA = **3 kVA**

**Laundry Circuit**

There should be at least one 120V, 20A circuit for the laundry area (NEC Article 210.11(C)(2). The minimum VA rating of the laundry circuit should be 1.5 kVA (NEC Article 220.52(B). This way, the load rating in the laundry area:

1500 VA = **1.5 kVA**

This way, the total general lighting and small appliances including laundry circuit rating:

- General lighting = 4.5 kVA
- Small appliance load = 3 kVA
- Laundry circuit = 1.5 kVA

Total of general lighting and small appliances = 4.5 kVA + 3 kVA + 1.5 kVA = **9 kVA**

**Demand Factor**

As we know that all electrical appliances are not operational at once i.e. (only one can be used as either electric heater or refrigerator depending on the temperature). Similarly, all equipment are not always ON continuously such as electric iron, water heater, lighting, fans etc. For this reason, the first 3 kVA is rated at 100% while the remaining load can be rated at a demand factor of 35% (NEC Table 220.42). This way

- The first 3 kVA at 100% = 3 kVA
- Remaining 6 kVA (9 kVA – 3 kVA) at 35% = 2.1 kVA

Net total of general lighting and small appliances = 3 kVA + 2.1 kVA = **5.1 kVA**

**Large Appliance Loads**

High power rating e.g. large appliances with continuous and non-continues operation should be handled differently. We have the following high power rated appliances in the above example:

- Air conditioner: 240V x 25A = 6 kVA
- Electric range: 240V x 35A = 8.4 kVA
- Electric heater: 240V x 30A = 7.2 kVA
- Clothes dryer: 240V x 15A = 3.6 kVA

As we have already mentioned above, an air conditioner or electric heater can be used at the same time i.e only one appliance is needed to operate based on the temperature. In this case, appliances with larger ratings should be taken into account (NEC® Article 220.82(C)). In our example, the rating of electric heater (7.2 kVA) is greater than the air-conditioner (6 kVA), so we will consider the heater then, i.e. 7.2 kVA

The rest of the appliances should be rated at 100% except electric range as it is used for a short time i.e. it is very non-continuous as compared to other appliances. The allowable demand factor for 7.2 kW electric heater is 5.76 kW (NEC Table 220.55). We assumed the power factor is unity i.e. “1” where apparent power = real power e.g. kVA = kW. This way;

- Electric range: = 8.4 kVA
- Electric heater = 5.76 kVA
- Clothes dryer: = 3.6 kVA

Net total of large rated appliances = 8.4 kVA + 5.76 kVA + 3.6 kVA = **17.76 kVA**

**Miscellaneous Loads**

The miscellaneous loads given in the example are:

- Dishwasher = 1.2 kVA
- Garbage disposal = 0.96 kVA

Net miscellaneous load rating = 1.2kVA + 0.96 kVA = **2.16 kVA**

**Total Load**

Load Points |
kW or kVA Rating |

General lighting, laundry & small load | 5.1 kVA |

Net large appliance load | 17.76 kVA |

Miscellaneous appliance load | 2.16 kVA |

Total Load |
25.02 kVA |

**Needed Service**

The common supply voltage levels in the US supplied to residential homes are 120V/240V. Thus, we may use the highest level of voltage to determine the required service (for amps) by using the following formula.

I = P / V

Where:

Putting the values;

I = 25.02 kVA / 240V

**I = 104.25A**

It means, the required service is 105A. But we have to add the future expansion and safety factor as well.

**Future Load:**

It is important to add a space of at least two branch circuits for the future expansion. The minimum of 2 breakers each of min 10A space should be added i.e. 2 x 10A = **20A**

The total Amps = 20A + 104.25A = **124.25A**

**Safety Factor**

It is recommended to add a safety factor of 20% to the total amperage as circuit breakers and their operations in the load center are affected by the rise in the temperature. This way, the total current in the amperes:

Net total Amps = 20% + 124.25A = **149A**

The suitable size of load center or distribution panel = **150 Amperes**

Based on the above calculations, **the right size of load center or ****panelboard**** is 150A** which is nearest available to the calculated value.

Click image to enlarge

*Related Posts:*

*How to Wire 120V & 240V Main Panel? Breaker Box Installation**How to Wire a Subpanel? Main Lug Installation for 120V/240V**How to Wire 277V & 480V, 1-Phase & 3-Phase, Commercial Main Service Panel?*

### How to Size a Distribution Board? 3-Phase, 400V – IEC

In the following example, we will show you how to calculate the right size of three phase 400V distribution board which is mostly applicable in countries following the IEC rules e.g. UK, EU and former British colonies.

#### Example:

What is the right size of a three phase distribution board if the estimated total load in a home is 50 kVA. The load is a combination of single phase and three phase systems (400V & 230V AC) including air-conditioning, refrigerator, electric range, water pumps, washing machines and general lighting points etc. Consider the power factor of 0.9.

**Solution:**

First of all, we will find the required amperes by using the three phase current formula.

P = √3 x V_{ }x I x Cos Ф

I = P / √3_{ }x V x Cos Ф

Putting the values:

I = 50kW x / √3_{ }x 400V x 0.9

I = **80.18A**

**Demand or Diversity Factor:**

The general diversity factor is 80% of the f connected load (see IEC 60439 for more details). In this case,

80% x 80.18A = **64.15 A**.

**Future Expansion:**

The general rule of thumb for safety factor is 20%. So you may add it as well if needed.

20% x 64.15A = **76.98 A**

**Safety Factor**

The minimum safe range of the safety factor is 20-25%. So we will it add it to the calculated value of load current as follows:

25% x 76.98 A = **96.22 A**

Now, **the nearest standard available rating of MCCB shall be 100A** for the three phase 400V distribution board which is the suitable size to handle a 50 kW load.

*Related Posts:*

*How to Wire a Three Phase, 400V Distribution Board? IEC & UK**How to Wire Combo of 3-Φ & 1-Φ, 400V/230V Distribution Board?**How to Wire 3-Phase & 1-Phase Split Load Distribution Board?*

### How to Size a Consumer Unit? Single-Phase, 230V – IEC

The following example will show you how to find the right size of single phase 230V AC consumer unit or garage unit and associated MCB/MCCB to handle the residential load.

**Example: **

Find the proper size of a single phase consumer unit if the estimated total load in a home is 12 kVA.

**Solution:**

Generally, the power factor of residential homes having normal loads is considered as unity “1”. This way, the total load in kVA = kW i.e. the apparent power is equal to the real power in watts due to the absence of power factor.

Now, we will have to first calculate the load current using the general formula of current in amperes for single phase circuits.

First of all, we will find the required amperes by using the three phase current formula.

P = V_{ }x I x Cos Ф

I = P / V x Cos Ф

Putting the values:

I = 12 kW x / 230V x 1

I = **52.17A**

**Demand or Diversity Factor:**

The general diversity factor is 80% of the connected load (You may select the proper % according to the load type in the IEC 60439). In this case,

80% x 52.17A = **41.74 A**

**Future Expansion:**

The general rule of thumb for safety factor is 20%. So you may add it as well if needed.

20% x 41.74A = **50 A**

**Safety Factor**

The minimum safe range of the safety factor is 20-25%. So we will it add it to the calculated value of load current as follows:

25% x 50 A = **62.5 A**

This way, we may select the nearest standard available rating of MCCB which is 63A for the single phase 230V consumer unit. Based on the calculation, this **63A MCB or MCCB** is the right size to handle a 12kW load in residential homes.

*Related Posts:*

*How to Wire Single-Phase, 230V Consumer Unit with RCD? IEC, UK & EU**How to Wire a Garage Consumer Unit?**How to Wire 1-Phase Split Load Consumer Unit? – RCD+RCBO**How to Wire 230V Dual Split Load Consumer Unit? – RCD+MCB*

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