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SECTION 050

CIRCUIT LOADS AND DEMAND FACTORS

Introduction (See Figure 050-(1)). Section 050 is considered as a

General section and must be read together with other sections that are
supplementary or modifying of the General Sections. As it pertains to the
Section 050, the following should be taken into account:
(a) For the calculation of the feeders where motors exist, they must be taken into
count the Rules 160-110 and 160-204, and the requirements must be added of
other loads, with the demands calculated according to the provisions set forth in Section 050.

(b) To size the conductors and the protection, when they are going to be connected
transformers and capacitors, the requirements of Section 150 have
priority over those in Section 050.

Distribution Network Connection:

Transport the energy from the connection point
with the distribution network up to the main
protection and/or control device.
Take-off box and/or
demotion The number and sizing of the conductors of
connections are determined according to the
Main device of characteristics of the system and the factors of
protection/disconnection applicable demands.

Wh Feeder:
Team of Transport the energy from the load side of the
measurement main protective device until the entry of
the protection devices of the derived circuits.
The number and sizing of the conductors of the
feeder, is determined according to a the
General Board characteristics of the system and the demand factors
applicable.
Devices of
protection against Derived Circuits:
overcurrents They transport the energy from the last device
protection up to the outlets for the devices or
electrical equipment.

Figure 050(1)
Identification of the Circuits

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Scope
Support of Rule 050-000. This section specifies the nominal values for the
electric equipment that powers various types of loads. Specifically, it covers the
conductivity capacity of the conductors for the connections, the feeders and
the derived circuits, just like the nominal values of the conductors of
power supply or that feed electrical equipment.

Purpose of Rule 050-000. If the electrical installations are designed only

according to the nominal values of the equipment nameplates,
it would simply take the sum of the nominal currents and ignore the type of
electrical equipment and the environment where they will be installed. In some cases, this
it would result in a lower installation (for example, a circuit derived from a stove)
electric central) and in others above the necessary (for example, in a connection
commercial with an electric ambient heating system). In some installations
In electrical systems, we cannot use tangible values, as the final load connected does not
will know until the structure is completed and the connection is installed (by
example, shopping centers). Section 050 aims to address a variety of
situations and environments, and assigns demand values to certain loads, taking into
It states that due to its operational diversity, they should not reach maximum loads.
same time as other loads (for example, electric ambient heating and air
conditioned). Some types of loads are covered in individual sections (by
For example, Section 160 for motors and Section 220 for welders.

This section not only applies demand factors lower than those of values
nominal data of the characteristic plates (for example, an electric stove of 12 kW
in a housing unit is evaluated at 6 kW), it also allows for other demands
are applied to existing demands (for example, if this same kitchen
electricity is in an apartment block, the demand of 6 kW for a
small apartment in a series of more than 20 housing units is taken at 10%,
So the 12 kW electric stove according to the plate now has a demand of
0.6 kW.

Rule 050-000(b) warns that Section 050 also refers to the amounts
minimum positions on the boards of the derived circuits for units of
housing. In this section, care must be taken to understand the difference between a
single-family housing and a housing unit. Putting it simply:
(a) each single-family home is a housing unit (e.g., duplex, triplex,
bungalow); but
(b) each housing unit is not a single-family dwelling (for example, a
small apartment in a block of apartments.
However, both are residential properties.

Generalities
Sustenance and Purpose of Rule 050-100. Section 160 considers the various
types of loads in watts or volt-amperes, for various types of uses. This
the rule proposes to establish a standard voltage base with which all calculations
from currents, based on the loads expressed in watts or volt-amperes, must be
made. By increasing or decreasing the loads in the utilization system it can
affect the system's tension. Using the standardized tensions without having in

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It accounts for the provided tension, the Code establishes a standard basis for
ensure a safe installation within acceptable voltage ranges.

A good example is a single-phase heating load of 20 kW, 240 V, where the

The service voltage of the connection is 220 V, the calculated current is more or
less 90 A instead of the 83 A rated current from the nameplate specifications. If
the designer uses the lower value for the derived circuits and the calculations of
feeder, this may reduce the sizing of the conductors, which
they would be overloaded if the nominal voltage of the equipment were applied. This
It also allows the designer/contractor/inspector to make their calculations without having to
obtain the value of the installation voltage to arrive at obtaining the value of the
current.

Support of Rule 050-102. The impedance of a derived circuit creates a drop

of tension in the circuit. This can result in a lower application of tension than what
accepted, unless appropriate preventive measures are taken. In general, a
low voltage decreases the operational efficiency of electrical equipment, such as
motors, heating systems, and lighting systems. The establishment of
criteria for the maximum acceptable voltage drop in a circuit ensures the use
of acceptable tensions to achieve optimal performance of the electrical equipment.

Purpose of Rule 050-102. (See Figure 050-102). Parameters are established.

basics to ensure that the use of voltage for electrical equipment is within
the prescribed values. For the designer, the voltage drop is a calculated value, for
it has been established the requirement that the calculation of voltage drop is
based on the calculated demand load. Subrule (3) covers a branch circuit
where the connected load is unknown and puts the demand at 80% of the nominal value
from the overcurrent protection devices of the circuit.

The allowed percentage of voltage drop is based on two values:

a maximum total voltage drop of 4% for the feeder plus circuit
derived; that is to say from the connection point to the energy meter up to the
last point of use; and
a maximum voltage drop of 2.5%, both for the feeder and for the circuit
derived.

Considering a typical installation with a service connection, feeder, and a circuit

derived (See Figure 050-102), a maximum voltage drop is allowed

- 1% in the connection (a);

- 4% at most between the feeder (b) + the derived circuit (c).
So there is a voltage drop of 11 V (220x0.5) across the entire installation.
with a supply of 220V, from the delivery point to the last point of use.
The maximum voltage drop for any feeder or derived circuit is 2.5%, in
in this case 5.5 V (220 x 0.025). However, Subrules (1) and (2) establish the
parameters for the 4% (8.8 V) voltage drop distribution between the
feeder and the derived circuit. That is, if the feeder (b) has a drop of
2.5% tension, then the derived circuit (c) can have a maximum of a
voltage drop of 1.5% (or vice versa), so that the total voltage drop must not
greater than 4%.

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In this way, the sizing of the conductors becomes a matter

economic. If the section of a feeder produces a voltage drop of 4
V (1.8%), the remaining 4.8 V (8.8 V - 4 V) (2.2%) can be used in the circuit.
derived.

Distribution Network

Delivery Point
1% Maximum
(a) Service conductor (See Standard of
BT connections
take-off box and/or
demediation
Energy Meter

(b) Feeder
2.5% Maximum
4% Maximum

(c) Derived Circuit 2.5% Maximum

Maximum Load or
MaximumDemand

Figure 050-102
Maximum Allowable Voltage Drops in a Circuit

Support of Rule 050-104(1). A common method is necessary to determine the

capacity of the nominal current of a connection, feeder, or derived circuit.
Some circuits may have a conductor with a nominal current in excess of
overcurrent device (voltage drop requirements) and other circuits
they can have an overcurrent device exceeding the nominal currents of the
conductor (motor circuits).

Purpose of Rule 050-104(1). It must be established that, in order to determine the

capacity of the nominal current of a service connection, feeder, or derived circuit.
the lower capacity between the conduction capacity of the
circuit conductor, or the current overload capacity of the device.

Support and Purpose of Rule 050-104(2). The calculated load, as it

It is determined in this Section, it must not exceed the nominal current capacity of the
circuit, to ensure that the circuits are not overloaded (See the
This applies to both existing and new installations.

The correct procedure is to calculate the load according to Section 050 and then determine
the dimensions of the conductor and the overcurrent device according to the sections

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corresponding to the Code (for example, Section 030 for Drivers, Section 080
for overcurrent devices).

Support and Purpose of Rule 050-104(3). Parameters are established for a

continuous load and a non-continuous load. Sometimes a load is considered
continuous is that which is lit for a relatively long period of time and
then it turns off, and a non-continuous load is one that is controlled by the
thermostatic or what goes through cycles. This is a mistake.

The calculated load is considered continuous, unless it meets one of

the criteria of paragraphs (a) or (b) of this subrule, or is specified as not
continues with other rules of the Code.

In paragraph (a), the criterion for continuous loads is that they must be on for a total
for 50% of the time (1 hour) or more in a period of 2 hours, for loads not exceeding
the 225 A.

Similarly, in paragraph (b), the criterion for continuous loads is that they are
lit for a total of 50% of the time (3 hours) or more in any 6-hour period
for loads exceeding 225 A.

A lighting system in a retail store or office would fall under the designation
it continues, just like a commercial water heater, since during normal use
it is on more than 50% of the time, even though it is thermostatically controlled
controlled.

It may be necessary to separate continuous loads from non-continuous loads when

calculate a connection or feeder, to avoid unnecessarily increasing the
size of the electrical equipment. Rules 050-200(3) and 050-202(2) suggest that, for the
application of this rule the electrical installations of the type housing units
residential areas are not considered as continuous.

Support and Purpose of Rule 050-104(4). (See Figure 050-104(4)). The

electric equipment is certified to operate continuously at 100%, when the
the nominal current capacity of its overcurrent devices allows it.
Such equipment must be marked or should indicate on their nameplate
techniques, as suitable for such conditions of use.

The maximum continuous load supplied by the circuit must not exceed 100% of the
nominal currents of overcurrent devices when used the
nominal currents of Table 2 (conductors in conduits), if:

the capacity of the conductors is equal to or greater than the rated current of the
overcurrent devices;

The insulation in the conductors has a nominal temperature of 70 ºC or

90 ºC (Table 2); and

the capacity of the nominal current of the conductors is determined from

from Table 2.

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The maximum continuous load supplied by the circuit must not exceed 85% of the
nominal current of overcurrent devices when the currents are used
nominal values from Table 1 (outdoor conductors), if any:

the capacity of the conductors is equal to or greater than the nominal current of the
overcurrent devices;

the insulation on the conductors has a nominal temperature of 70 ºC or

90 ºC (Table 1); and

(3) the capacity of the nominal current of the conductors shall be determined from
from Table 1.

Note: In Annex B it indicates that the electrical equipment that is not marked or does not have as
plate data, whose nominal current is continuous, should be considered suitable for operation
continuously at 80% of the nominal current capacity of their devices
overcurrent.

For application only with 50/60 Hertz

Maximum Current (A)

Symmetrical current in rms.
Volts Interruption Short Circuit

600 50,000 25,000

480 100,000 25,000
240 100,000 25,000

Suitable for continuous operation at 100% of the nominal value.

Figure 050-104(4)
Typical Continuous Operation Plate

Examples:
(a) A switch with a 1,000 A fuse is rated for continuous operation at
100% of the rated current capacity of its devices
overcurrent. If the conductors of this switch are going to be laid
In a conduit, what size of continuous load can it supply?

response
This switch with a fuse can power a continuous load of 1,000 A yes
is that:
The conductors have an insulation temperature of the current.
nominal of 70 ºC or 90 ºC; and
the nominal current capacity of the conductors is 1,000 A
determined from Table 2 (in channels).

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A switch with a 1,000 A fuse is rated for continuous operation at

100% of the nominal current capacity of your devices
overcurrent. If the conductors of this switch are to be laid
outdoors, what magnitude of continuous load can it power?

Response:
This switch with fuse can supply a continuous load of 1000x85%
= 850 A, yes it is that:

Conductors have an insulation temperature of the current

nominal of 70 ºC or 90 ºC; and

The nominal current capacity of the conductors is 1,000 A.

determined from Table 1 (outdoor).

Note: The reduction of the nominal current capacity in Subrule 050-104(4)(b) is

applies to the entire circuit. This means that the conductors supplying this load
continue from 850 A, they must be evaluated at 1,000 A according to Table 1.

Sustenance and Purpose of Rule 050-104(5). When electrical equipment is

prepared to operate continuously at only 80% of the current capacity
nominal of its overcurrent devices. Said equipment must have a plate or
proper registration. If there is no mark, as noted in Appendix B, the
The equipment should be considered suitable for continuous operation at 80%.

The maximum continuous load supplied by the circuit must not exceed 80% of the
nominal currents of overcurrent devices when using the
nominal currents of Table 2 (conductors in conduit), if:

the capacity of the conductors is equal to or greater than the nominal current of the
overcurrent devices;

(2) the insulation has a nominal temperature of 70 ºC or 90 ºC (Table

2); y

the capacity of the nominal current of the conductors is determined at

starting from Table 2.

The maximum continuous load supplied by the circuit cannot exceed 70% of the
nominal current of overcurrent devices when using the currents
nominal of Table 1 (outdoor conductors), if:

the capacity of the conductors is equal to or greater than the nominal current of the
overcurrent devices;

(2) the insulation in the conductors has a nominal temperature of 70 ºC or

90 ºC (Table 1); and

(3) the capacity of the nominal current of the conductors is determined at

starting from Table 1.

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On the contrary, if a load is not classified as continuous, the Subrules

(4) and (5) do not apply, and the electrical equipment can be charged up to its current
nominal regardless of the wiring method used.

When a service entrance box, a switch, a circuit breaker, or a panel

from panels, will supply continuous and non-continuous loads, the total load that must be
Setting overcurrent devices must be determined using the
continuous loads with the demand factors from Subrule (4) or (5), plus the loads
do not continue.

Examples:
A switch with a 1,000 A fuse is rated for continuous operation.
to 80% of the nominal current capacity of your devices
overcurrent. If the conductors are going to be laid in a
conduit, What magnitude of continuous load can this feed?
switch?
respuesta
This fuse switch can supply a continuous load of 1,000 x
80% = 800 A, if it is that:

1) conductors have an insulation temperature to current

nominal of 70 ºC or 90 ºC; and
2) the capacity of the nominal current of the conductors is 1,000 A,
determined from Table 2 (in channels).

A 1,000 A fuse switch is rated for continuous operation at

80% of the nominal current capacity of their devices
Overcurrent. If the conductors are going to be laid outdoors, what
What size of continuous load can this switch handle?
Response:
This fuse switch can power a continuous load of:
1000 x 70% = 700 So it is that:
Conductors have an insulation temperature to the current
nominal of 70 ºC or 90 ºC; and
The nominal current capacity of the conductors is 1000 A.
how it is determined based on Table 1 (outdoors).

(c) A switch with a 1,000 A fuse is rated for continuous operation.

at 80% of the nominal current capacity of its devices
overcurrent. What magnitude of non-continuous load can feed this
switch?
Response:
This fuse switch can supply a non-continuous load of 1,000 A,
yes it is that:

1) the conductors have a nominal insulation temperature of 70 ºC or

90 ºC; and

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The current rating capacity of the conductors is 1,000 A.

how it is determined from Tables 1 and 2; and
The load is determined according to Rule 050-104(3).

A switch with a 1,200 A fuse is rated for continuous operation at

80% of the nominal current capacity of its devices
overcurrent. The wiring method will be an outdoor conductor and the
conductors will have a nominal insulation temperature of 70 º C.
Can this fuse switch supply a continuous load of 500 A and
a non-continuous load of 480 A?

Response:
Continuous load calculated (500 / 70%) = 714.28 A
Continuous load + 480.00 A
Total combined load 1,194.28 A

A switch with a 1,200 A fuse can supply a combined load.

from 1 194 A, if the conductors have an insulation of 70 ºC and have
a nominal current capacity of 1,200 A, determined from the
Table 1.

Support and Purpose of Rule 050-104(6). The reductions in capacity of the

nominal current of the conductors is applied in various sections of the Code. Only
the largest reductions in the rated current capacity must be applied
to any driver.

Example:
Section 030 applies the reductions of the nominal current capacity between the
80% up to 50%. If 80% of Rule 050-104 is applied and a reduction already exists.
from the capacity of the nominal current of 70% applied to the conductor according to the Rule
030-004(1), 80% of Rule 050-104 should be ignored and only 70% should be applied.
from Section 030.

Note: The reductions of the nominal current capacity of Rule 030-004(8) and (9)
are additional to those of Rule 030-004(1) or (2).

Support and Purpose of Rule 050-104(7). Drivers with capacity

nominal assigned in excess of the allowed nominal currents for loads
continuing under Rule 050-104(4)(b) and (5)(b) may introduce unwanted effects
of heating in the envelopes, covers under Section 050.

It is important to ensure that the nominal currents applied to the conductors,

not specifically covered by Rules 050-104(b) and (5)(b), are not in excess of
the nominal currents of the conductors.

When an underground cable or cables feed into a junction box, switch,

automatic switch or panel, the maximum continuous load must not exceed:

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85% of the nominal current of overcurrent devices when the

the equipment has a continuous operation at nominal current at 100%; or
70% of the rated current of the overcurrent devices when the
equipment has a continuous operation of nominal current at 80%, or if the equipment
not marked.

Service Considerations for Rule 050-104. When they are going to be determined
the maximum permitted currents must take into account the considerations of
Services listed in Annex B under Notes for Rule 050-104(7).

Example:
What section of unipolar AC 90 cable (armored cable - 90 °C) can be used for
a continuous load of 450 A, using continuous operation equipment
nominal at 100%, when the cable is to be buried directly as in Detail 1
from Diagram B4-1?

Steps:
Determine the total of the continuous and non-continuous load.
Continuous load — 450 A
Non-continuous load — Not applicable

Determine the continuous operation of the equipment's nominal current.

100%

(3) Determine the dimensions of the conductors using Table D8B- Detail 1
100%, and the Diagram B4-1
Table D8B - 185 mm2AC 90 °C

(4) Determine the capacity of the conductors, using Rule 050-104(4)(b), the
wiring method, the nominal continuous operating current of the equipment, the
loads of the Step (1), and the nominal insulation temperature of the conductor.

Continuous Load—450 A
Divided by the % of 050-104(4)(b) — 85%
529.4 A
But the load does not continue — Not applicable
Driver Capacity 529.4 A

Determine the size of the conductor using Table 1

Table 1 — 185 mm2-90 °C

(6) Select the conductor section from Steps 3 or 5 according to Rule 050-
104(7)
Step 3 or 5 — 185 mm2AC-90 °C

Support and Purpose of Rule 050-106(1). The switches and conductors are
classified into standard dimensions and in some cases it may be advisable
the use of switches and conductors with rated currents lower than the load
calculated.

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With the calculated load, according to this Section, the dimensions must be determined.
minimums of conductors and switches. Rarely, the calculated load is equal to the
standard nominal currents of the switches and conductors. They
establish the conditions under which an electric device of size can be accepted
standard, which may be a little lower than the calculated load. For example, if you
It has a calculated load of 210 A; 5% of 210 = 10.5 A and 210 - 10.5 = 199.5 A.

Basis and Purpose of Rule 050-106(2). The Code is based on

minimum requirements and does not attempt to limit the design. It is important that the
service entrances and feeders are sized appropriately for the loads
what they must carry. The connections where, except for those calculated according to the
Rules 050-200 and 502-202, the electrical installation is based on requirements.
higher than those of Section 050, the capacities of the connections and of the
Feeders must be increased to meet the requirements of the
real load, despite the requirements of Section 050.

Support for Rules 050-106 (3), (4) and (5). Lower load demand is
produced
(a) when the loads are controlled to prevent the total load from being
operation at the same time;
(b) when environmental systems such as environmental heating are installed
electricity and air conditioning, which are not used simultaneously; or
(c) when the loads can be in operation at the same time, but are of
cyclical nature and do not produce a demand equal to the total connected load.

Purpose for Rules 050-106 (3), (4) and (5). In Subrule (3), it is indicated that
they must use interlocking controls between the loads, so that two or more
loads cannot operate at the same time. In this case, the load that produces the highest
demand is used to determine the total load.

In Subrule (4), it is indicated that in the case of installing heating loads

Electric and air conditioning environmental controls do not need to be blocked.
the switches, since it is assumed by common sense that they do not heat up and cool down
simultaneously. It is indicated that the greatest of the ambient heating loads
electric or air conditioning, should be considered in the demand calculation.

Subrule (5) is a bit more difficult and subjective. Here we have a number of loads.
they are of a cyclical nature. This means that the loads can be
physically connected and that it is possible to operate them at the same time, but the operation
The system is such that in normal operation they should not operate simultaneously. You are
you can call this, the diversity of the system's operation. A good example can
to be a warehouse with vending machines, or a made-to-order manufacturing plant, where,
Although there are many machines, only a few can be in operation.
simultaneous. Since not all machines can be in operation simultaneously
depending on the business cycle, any decrease in the requirements of the
The burden must be carefully discussed between the parties involved. In such
in cases, it is better to err in having too large a team; otherwise, if it is that
the installed electrical equipment is too small for the load, a modification
very expensive it may be necessary.

Support and Purpose of Rule 050-106 (6). The rules of the sections
individuals are supplemental or amend the General Sections and by

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the following take precedence over such General Sections of the Code, (See the Preface and
Rule 020-002). An example of this would be that, despite the Rule 050-104(5)
it says that the load should not exceed 80% of the nominal current
overcurrent devices, other sections may allow higher values
(example, Section 150 of capacitors).

Support of Rule 050-106 (7). It may be that the application of the factors of
demand on some loads, results in the conductors of a feeder or a
the derived circuit exceeds the requirement of the service conductors.

Purpose of Rule 050-106(7). When several demand factors are applied

to the connections, feeders, and derived circuits, the required capacity of the
feeders and the derived circuits may be greater than the service connections or
feeders from which they are supplied (for example, when the feeder supplies to
motor loads just like other types of loads, or when Table 14 applies
with other loads). In such cases, the capacity of the feeder or the branch circuit,
it does not require exceeding the capacity of the supply or the feeder that supplies them.

Support of Rule 050-106(8). When loads are added to a connection or

existing feeder, care must be taken that the electrical equipment is not
overloaded. Generally, the exact information about is not available
installed loads in the last 12 months, which makes it difficult to perform calculations.
Suppliers generally measure the demands of larger customers and
they may have that information available.

Purpose of Rule 050-106(8). With the maximum demand of the facilities

existing, resulting from measurements in the last 12 months, plus the loads
additional, with the demand factors as allowed in the Code, it can be
determine the new calculated load that should be considered for a connection or
existing feeder. These existing load or maximum demand figures,
generally they can be obtained from suppliers, or if the increase in
proposed load is considered in the long-term planning; the installation of
The new measuring or service equipment must be carried out in accordance with the standard.
of Low Voltage Connections. The new calculated load that is being added
concerning the existing supply line or feeder, it must be subject to the requirements
from Rules 050-104(4) and (5), regarding the nominal current of the equipment
electric

Support of Sub-rule 050-106 (9). The smaller a unit is,

housing and less the diversity of its burdens, the probability that all the burden is
connected at the same time is higher, for this reason the demand factor for the units
The housing provisions outlined in this Subrule is 1.0. Furthermore, the probability of a
the growth of housing is very fast and also very high for the most housing
small (90 m2and 3 kW); this possibility is covered by a minimum section of
the service conductors inside the dwelling; this minimum section, for
Another part must conform to other rules of the Code such as Rule 040-302.
(Reference: Subrule 050-110 (2)).

Purpose of Subrule 050-106(9). It is required that the service entrance drivers

of the homes have a conduction capacity no less than the current of the
load they supply; and that the minimum section of these conductors be 4 mm2
for mechanical reasons and to suit the possibility of load growth

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of the smallest houses. In the case of the feeders, the minimum section
it should be 2.5 mm2.

I support Rule 050-108(1). The insufficiency of derived circuits may result

in an overload of the conductors, which can cause overheating and failure
subsequent of the electrical equipment.

Purpose of Rule 050-108(1). For single-family homes, the

requirements to be provided on the board, a minimum number of spaces for the
devices against overcurrent in branch circuits, based on capacity
of the service or feeder conductors.

The following table outlines the minimum number of spaces for devices against
overcurrent of the branch circuits, for the common capacities of the
feed or supply line in single-family homes. It must be taken into account the
different requirements if a centralized electric stove is used.

Capacity of the Number of Spaces for the

Heating
Connection or Overcurrent Devices
Central
Feeder Derived Circuits of 220 V
0 – 60 A No 16
60 – 100 A No 24
100 – 125 A Yes 24
100 – 125 A No 30
125 – 200 A Yes 30
125 – 200 A No 40

Support of Rule 050-108(2). The Code presents a minimum standard and when
There is the possibility of installing additional electrical equipment; space must be planned.
additional resources to be able to feed them.

Purpose of Rule 050-108(2). It must be ensured that space is reserved for

two bipolar overcurrent devices of 15 A, regardless of the
requirements of branch circuits in single-family homes (for example, for
an electric stove and a clothes dryer). There must be space for all the others.
overcurrent devices, and at least two additional spaces for future
extensions. The Code makes no distinction between a house with a basement
finished and one without a basement. If the basement is finished, more should be used
derived circuit positions.

It is considered as an example a standard bungalow heated without electricity, with a

board for 24 circuits, 100 A. If more than 22 circuits are required for
this house, leaving less than two available spaces, should use a board of 30
circuits.

Support and Purpose of Rule 050-108(3). The limitation of the equipment is recognized.
electricity in apartments and similar buildings and therefore, there is less
minimum space requirements for bipolar overcurrent devices
of 220 V and 35 A. However, when using three-phase equipment or cooking, it must
reserve space for three-phase overcurrent device.

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In apartments and similar buildings, no additional spaces are required, already

that no increases are expected. The words 'at least' are important here,
even more so than for a single-family home, which has all the options covered in
Subrule (2). For a small department with electric baseboard heating,
individual heating for the washing machine, the dryer, and water, the requirements of
derived circuits can easily exceed twelve.

Support of Subrule 050-108(4). It is necessary to ensure that the units of

housing, even the smallest ones, have adequate places and spaces for the
connection equipment and the internal distribution board according to its load
(Reference: Rule 050-110(2).

Purpose of Subrule 050-108(4). For housing up to 90 m 23 kW of load

and single-phase supply, a place should be provided inside for the panel of
distribution with space for at least three 15 A bipolar switches.

For homes up to 150 m2, with a load greater than 3 kW and up to 5 kW, and supply
single-phase, space must be provided for five bipolar switches, of which
one must be 20 A.

For homes up to 200 m2with a load between 5 kW and 8 kW, with supply
single-phase, space must be planned for seven switches, of which two must
be of 30 A; or two 15 A tripolar if it were the case of three-phase supply.

Support of Subrule 050-110. It is necessary to ensure a consistent method to

calculate the habitable area for the buildings covered by Rules 050-200 and
050-202.

Purpose of Subrule 050-110. The requirements for calculating are established.

consistently the habitable areas in single-family homes and buildings of
departments.

This rule is very straightforward. The total habitable area of a single-family home is
determined
(1) Calculating the area of the first floor based on the internal dimensions;
(2) Calculating the area of the upper floors based on the internal dimensions;
y
If the building has a basement, calculating 75% of the basement area
based on the internal dimensions.

Instead of the area being calculated with the internal dimensions, it also
you can use the covered area, since it is a data available in the blueprints
architectural.

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Example:
A two-story single-family home measures 10 m x 12 m with a basement that has
200 mm thick walls. What is the total habitable area of the house?
(1) Area first floor [10 - (2 x 0.200)] x [12 - (2 x 0.200)] 111.36 m2
(2) Area of the upper floors [10 - (2 x 0.200)] - [12 - (2 x 0.200)] = 111.36 2 m
% m2
(3) Basement area [10 (2 x 0.200)] – [12 - (2 x 0.200)] x 7583.52
_______________________
Total 306.24 m2

Support of Subrule 050-110(2). It is necessary to consider the cases of housing.

with loads and conditions that correspond to the reality of the country. The criterion
the selected one has been to relate the housing area with the installed load,
widely used in the rating of urban developments and electrification of groups
existing housing. From the review of available documented information and of the
Through the analysis of experiences from other countries, the figures provided in this report have been reached.
subrule.
Note:
Reference 1: RBT, Low Voltage Electrotechnical Regulation of the Ministry of Industry and
Energy Spanish; Instruction MIE-BT-10 Section 2: Degree of Electrification of Homes.
Reference 2: CENERGIA: "Study of the Characterization of the Electrical System Load of the City"
of Trujillo.

Purpose of Subrule 050-110(2). For single-family housing units,

when there is no information available about the loads connected or to be installed, one
a minimum load of 3 kW must be assigned to homes of up to 90 m2, 5 kW to
housing between 90 m2y 150 m2and 8 kW to homes of 150 m2and 200 m2.

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Services and Feeders

Support and Purpose of Rule 050-200(1). In a single-family home, there
wait for all installed electrical equipment to operate simultaneously. To
regarding, a basic criterion is established with which the capacity can be determined
of the conductors of a service connection or from the feeder.

Subrule (1)(b) indicates that the minimum nominal current of the conductors should be
A board can be installed with a minimum number of space according to the
Rule 050-108 (4).

For the purposes of Rule 050-104, Subrule (3) of Rule 050-200 proposes
that the charges in Subrules (1) and (2) are not considered continuous; for
there is no need to apply any reduction of the nominal current.

Steps to follow to calculate the minimum capacity of the conductors of the

connection or of the feeder that supplies electrical power to a unit of
single-family housing

(1) from 050-110 - Calculate the total covered area in m²2Covered area that
is indicated in the Location Plan of
Architecture.

Recommendation: If the excess area fraction is less than or equal to 1 m2,

it is recommended not to consider this fraction.

(2) of 050-200(1)(a)(i) - 2,500 W for the first 90 m2of covered area;

more
(3) of 050-200(1)(a)(ii) - 1,000 W for every 90 m2or excess fraction of the
first 90 m2; more

Note: The basic load 050-200(1)(a)(i) and the additional load 050-200(1)(a)(ii)
they consider the loads corresponding to the lighting and outlets of the
housing unit.
It should be noted that the loads for outlets are only for the
that do not exceed 1,500 W or those loads that are characterized by being
installations that have independent branch circuits. For these cases
see the notes from step (7).

(4) of 050-200(1)(a)(iii) - A. Calculate the amount of heating load

electric environmental using demand factors
from Rule 270-116(2)(a) and (b). The first 10 kW to
100% and the rest at 75%;
B. Calculate the amount of charge of the air
conditioned with a demand factor of 100%;
Note: Use the greater of A or B according to Rule 050-106(4); more

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Note: The heating loads, among other similar ones, correspond to:
Heating of surfaces, floors, baseboards.
Pipe heating.
Heating for ice melting.
Sauna (Rule 279-500).

(5) from 050-200(1)(a)(iv) Calculate

- the required load for an electric kitchen.
Use 6,000 W for the kitchen's rated current
electric, but 40% of the nominal current of the kitchen
electric that exceeds 12 kW; more
(6) of 050-200(1)(a)(v) - Calculate the required load for water heaters
for pool and individual or common bathrooms with a
demand factor at 100%; more
Note: The loads of water heaters, among other similar ones, correspond to:
Water heater for bathroom (therma).
Electric shower.
Hot tub heater.
Pool heater.

(7) of 050-200(1)(a)(vi) - Calculate for the remaining loads greater than 1,500 W
with a demand factor of 25%, if it has been planned
an electric stove.
If electric cooking has not been planned, it should be considered.
the sum of the loads that exceed 1,500 W up to
a total of 6,000 W at 100%, and the excess of the
6,000 W at 25%.
Note 1: Additional loads over 1,500 W, among others similar
correspond to:
Clothes dryer.
Special outdoor lighting.
Garden lighting.
Facade lighting.
Pool lighting.
Elevators (apply Section 160).

Note 2: Additional loads less than or equal to 1,500 W must be

considered with a demand factor of 100%. These loads among others
similar ones correspond to:
Dishwasher.
Water pump.
Drain pump.
Pool electric pump.
Lift door.
Hydromassage.
Waste disposer.
They are characterized by being facilities that have derived circuits.
independents).

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(8) from 050-200(1) - To determine the capacity of the conductors of the

feeder or power supply, use the largest of
loads calculated in 050-200(1)(a) or the minimum load
stipulated in paragraph (b).
Note: The loads are non-continuous according to Rule 040-200(3), and it is not necessary to apply
the reduction factor of the nominal current capacity.
Select a conductor from Tables 1 or 2, and take into account the other tables
related.
(10) Determine the capacity of overcurrent protection devices.

Example 1:

A two-story single-family home, with non-electric heating, with the dimensions

external dimensions of 12 m x 10 m and a basement with external walls 200 mm thick.
Installed electrical equipment: 4,000 W clothes dryer
4,000 VA air conditioning unit

050-110 Housing unit area 306 m2

A. Heating with automatic control:

First 10 kW with a power factor of 100%, and the balance
with F.D. 75%
A.C. F.D. 100%
Note: Use the greater of A or B according to the Rule.
050-106(4).

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Example 2:

3,000 VA air conditioning unit (Rule 050-106 (4))

4000 W hot tub heater

4,000 W clothes dryer

Waterheaterforbathroom

The load of the housing unit is 39,300 W, which is equivalent to 103.26 A with supply.
three-phase of 220 V and considering power factor 1.

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Example 3.1:
Three-story housing unit - With Electric Kitchen
A three-story house plus a basement (4 levels), with a covered area of 470 m².
100% of the area of the 1st, 2nd, and 3rd floor 390 m²
75% of the basement area 80 m² 60 m²
Total covered area 450 m²
Steps:

Waterheaterforbathroom

The load of the housing unit is 37,040 W, which is equivalent to 97.32 A, with three-phase supply of
220 V and considering a power factor of 1.

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Example 3.2:
Three-story housing unit - Without Electric Kitchen
A three-story house with a basement (4 levels), which has a covered area of 470 m².
100% of the area of the 1st, 2nd, and 3rd floor 390 m²
75% of the basement area 80 m² 60 m²
Total covered area 450 m²
Steps:

The load of the housing unit is 37,740 W, which is equivalent to 91.28 A, with three-phase supply of
220 V and considering a power factor of 1.

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Example 4.1:
Two-story housing unit - With Electric Kitchen
A two-story house without a basement, which has a covered area of 180 m².
100% of the area of 1erfloor 85 m²
100% of the area of 2ofloor 95 m²
Total covered area 180 m²
Steps:

And considering a power factor of 1.

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Example 4.2:
Two-story housing unit - Without Electric Stove
A two-story house without a basement, which has a covered area of 180 m².
100% of the area of 1erfloor 85 m²
o
100% of the area of 2 floor 95 m²
Total covered area 180 m²
Steps:

220 V and considering a power factor of 1.

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Example 5.1:
Housing unit of an apartment - With Electric Kitchen
A one-story house without a basement, which has a covered area of 90 m².
100% of the area of 1erfloor 90 m²

Steps:

220 V and considering a power factor of 1.

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Example 5.2:
Housing unit of an apartment - Without Electric Kitchen
A one-story house without a basement, which has a covered area of 90 m².
100% of the area of 1erfloor 90 m²

Steps:

220 V and considering a power factor of 1.

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Support of Rule 050-200(2) and (3). When the same service line or feeder
supplies energy to two or more housing units of a condominium, the factors
demand can be applied to the service connection or to the feeder according to the diversity
operational of the housing units.

Purpose of Rule 050-200(2) and (3). In the case of a condominium, it must be applied
the same principles that apply to apartments in buildings, in what
regarding Rules 050-202(3)(a) (i) to (v).

Rule 050-104 and Subrule (3) of Rule 050-200 suggest that the loads of
Subrules (1) and (2) are not considered continuous; therefore, it is not necessary to apply
no reduction of the nominal current capacity.

The steps to calculate the minimum capacity of the service conductors or

of the feeder for two or more housing units of a condominium are the
following:

(1) from 050-200(1) - Calculate the total load for each housing unit
of the condominium in watts;

(2) from 050-200(2) and -Exclude any environmental heating load

050-202(3)(a) and air conditioning in the housing units,
calculate the total load of all units of
condominium housing, based on the following
demand factors:
a) 100% of the load of the housing unit plus
big; more
b) 65% of the loads of the next 2 units
of larger housing; more
c) 40% of the loads of the next 2 units
larger housing; more
d) 30% of the loads of the next 15 units
larger housing; more
e) 25% of the loads of housing units
remaining;

(3) of 050-202(3)(b) - Calculate the total ambient heating load

electricity for all housing units using
the demand factors of 270-116(2)(a) and (b) - the
first 10 kW at 100%, the rest at 75%;

(4) of 050-202(3)(c) - Calculate the total load of the air conditioning for all
the housing units, using the factor of
demand at 100%

Note: According to Rule 050-106(4), only the one that results in greater of the
application of Steps (3) or (4) can be used.

(5) of 050-202(3)(d) - Calculate any additional loads not located in the

housing units, considering a factor of
demand of 75%.

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Calculate the total load by summing Steps (2), (3), (4), and (5);
Note: The loads are not continuous according to Rule 050-202(2); and it is not necessary to apply
a reduction factor.

Size the equipment, overcurrent devices, and conductors.

Note: If there are two or more alternating or cyclical operation teams, it must be applied the
Rule 050-106 (5).

Example:
Eight Housing Units in a Condominium - Without Electric Heating
The condominium has 3 kW in lighting for pathways and gardens.
Each housing unit has: 4,000 W for clothes dryer, 4,000 W for air conditioning
air conditioning, 11 kW for electric stove and 2.5 kW for water heater.
Water pump 747 W and lift door 373 W.
Covered area of each house 330 m²
100% of the area of 1hefloor 120 m²
100% of the area of 2ofloor 130 m²
2
75% of the basement (of 80 m) ) 60
Total covered area 310 m²

Calculation of the load of each housing unit

220 V and considering a power factor of 1.

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Calculation of the total load of the eight housing units:

Steps:
050-200(1) The load of each housing unit is 19,620 W

Note.- All loads are non-continuous, according to Rule 050-200(3).

Support for Rule 050-200(4). A criterion is necessary to determine the

minimum capacity of the conductors, in amperes, corresponding to the part of the
service entrance or feeders installed inside single homes with loads
relatively low, when specific information about them is not available.
(Reference: Subrule 050-110(2))

Purpose of Subrule 050-200(4). For single-family homes, the conductors of

the service connection and the feeder up to the main panel must have a capacity
minimum driving as indicated in Subrule 050-106(9).

Support of Rule 050-202(1), (2) and (3). Despite the fact that the individual units
housing can be treated as single-family homes, the operational diversity
from a group of housing units feeding from a single feeder or
connection allows for the application of demand factors.

Purpose of Subrule 050-202(1), (2) and (3). The applicable factors are indicated.
allowed demands when calculating the minimum capacity of the conductors of the
incoming or feeder, which supplies housing units in a building of
departments. The basic principles for a single-family home are applied
according to Subrule (1)(a), with the sole exception of the step added when calculating the load
basic for recognizing a small one-bedroom apartment.

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Subrule (3) specifies the applicable demand factors allowed for the
calculate the minimum capacity of the service or feeder conductors,
from the main connection in an apartment and similar buildings. In order to
use this Subrule to calculate the minimum capacity of the conductors
The feeder or supply must have a minimum of two loads of units.
housing connected to the service or feeder conductors. When it
use this Subrule to calculate the minimum capacity of a feeder that has a
minimum load of two housing units from the main connection, others
rules can be applied (for example, 160-110 for motor loads). It must be
use Subrule (4) when the service connection or the feeder conductors
they only supply the loads outside of the housing units (for example, loads of
passages and external light, loading laundry rooms, elevator loads.

The steps to follow to calculate the minimum capacity of the conductors of the
the connection or the main feeder of an apartment building are the
next:

Calculate the load of each housing unit or

department of different size or type;
(a) from 050-110(3) - Determine the usable area of each different type of
housing unit or apartment in m2
based on the internal dimensions of each
housing unit or apartment;
(b) of 050-202(1)(a)(i) - 1,500 W for the first 45 m2of living area
the housing unit or apartment; more
(c) of 050-202(1)(a)(ii) - 1,000 W for the next 45 m2of living area
from the housing unit or apartment; more

(d) from 050-202(1)(a)(iii) 1,000

- W per 90 m2or portion of the remaining area;
more

Note: The basic load (i) and the additional loads (ii) and (iii) consider the loads
corresponding to the lighting and outlets of the apartment.

(e) from 050-202(1)(a)(iv) To

- calculate the load of ambient heating
electricity using demand factors according to the
Rule 270-116(2)(a) and (b) – the first 10 kW at
100%, and the rest at 75%.
B. calculate the air conditioning load with factor
demand at 100%;
Note: Use the greater of A or B according to Rule 050-106(4); more

Note: The heating charges, among similar ones, correspond to:

Surface heating, floors, baseboards.
Pipe heating.
Heating for ice melting.
Sauna.

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Note: The loads of water heaters, among other similar ones, correspond to:
Water heater for bathroom.
Electric shower.
Hot tub heater.
Pool heater.

(f) of 050-202(1)(a)(v) - Calculate the required load for an electric kitchen.

Use the nominal power of 6000 W for the kitchen
electric, plus 40% of the rated power of the kitchen
electric over 12 kW; more
(g) from 050-202(1)(a)(vi) Calculate
- for the remaining loads greater than 1,500 W
with a demand factor of 25%, if it has been anticipated
an electric stove.
If electric cooking has not been planned, it should be considered.
the sum of the loads that exceed 1,500 W up to
a total of 6,000 W at 100%, and the excess of the
6,000 W at 25%.

Note 1: Additional loads exceeding 1,500 W, among others similar

correspond to:
Clothes dryer.
Special outdoor lighting.
Garden lighting.
Facade lighting.
Pool lighting.
Water heater for bath (therma) > 1,500 W.
Electric shower.
Hot tub heater.
Pool heater.

Note 2: Additional loads less than or equal to 1,500 W must be

considered with a demand factor of 100%. These loads among others
similar ones correspond to:
Water heater for bathroom (therma)≤ 1,500 W.
Water electric pump.
Drain pump.
Swimming pool pump.
Upward door.
Hydromassage.
Waste disposer.
(They are characterized by being facilities that have exclusive derivative circuits).

(h) of 050-202(1)(b) - Use the highest of the calculated loads in 050-202 (1)
(a) or (b) 25 A.
Note: The loads are not continuous according to Rule 050-202(2); and it is not necessary to apply.
a reduction factor.

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(2) of 050-202(3)(a) - Exclude any ambient heating load

electric and air conditioning in the units of
housing or apartments; calculate the total load
of all housing units or apartments
based on the following demand factors:
100% of the load of the housing unit plus
big; more
(b) 65% of the charges of the following two
larger housing units; more
(c) 40% of the charges of the following two
larger housing units; more
30% of the loads of the next 15 units
larger housing; more
(e) 25% of the loads of the remaining units of
housing.
(3) of 050-202(3)(b)- Calculate the total load of the ambient heating
electricity of all housing units using
the demand factors of Rule 270-116(2)(a) and
the first 10 kW at 100%, and the rest at 75%;
(4) of 050-202(3)(c) - Calculate the total load of the air conditioning for all
the housing units using the factor of
demand of 100%;
Note 1: According to Rule 050-106(4), only the greater of Steps (3) and (4) should be used.
Note 2: The loads in Steps (2), (3), and (4) are discontinuous according to Rule 050-
202(2); it is not necessary to apply the current capacity reduction factor.

(5) of 050-202(3)(d) - Calculate any additional load not installed inside

of the housing units or apartments with a
demand factor of 75%

(6) Determine, from the loads in Step (5), which loads are continuous and which are not.
do not continue according to 050-104(3);

(7) Apply the current capacity reduction factor according to Rule 050-
104(4) or (5) depending on the nominal current of the equipment used and of the
Tables used to determine the capacity of conductors:

Calculate the total load by adding the loads calculated from Steps (2), (3), (4) and
(7).

Dimension the equipment, the protective devices against overload and

the drivers.

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Example 1.1:
Calculation of loads for a typical 45 m apartment2With Electric Cooking
Typical apartment with 45 m² of covered area.

Calculation of the Department load

Steps:

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Example 1.2:
Calculation of loads for a typical 45 m apartment2With Electric Kitchen
Typical apartment with 45 m² of covered area.

Calculation of the Department load

Steps:

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Example 2.1:
Load calculation for a typical 90 m Department2With Electric Kitchen
Typical apartment with 90 m² of covered area.

Calculation of the Department load

Steps:

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Example 2.2:
Calculation of loads for a Typical Department of 90 m2Without Electric Stove
Typical apartment with 90 m² of covered area.

Calculation of the Department's workload

Steps:

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Example 3.1:
Calculation of loads for a building with 8 typical apartments of 180 m2With Electric Kitchen
Typical apartment with 180 m² of covered area.
The building has a basement of 650 m² for parking and has the following loads:
Exterior lighting 10 points of 100 W each.
02 Elevators of 5,500 W each.
2 water electric pumps of 1190 W each (with alternating operation).
2 drainage electric pumps of 750 W each (with alternating operation).
CO extractor of 1,500 W.
2 lift doors of 375 W each.
Central intercom, 1 point of 500 W.
Alarm Central, 1 point of 800 W.

Calculation of the load of each housing unit

Steps:

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Calculation of the total load of the Apartment Building:

Using the steps of Rule 050-200(2) and (3), the minimum capacity of the conductors is calculated
power supply that provides energy to the condominium of the eight (8) housing units.
Steps:
050-202(1)(b) The load of each department is: 17 497 W

All loads outside the departments are continuous.

Considering a protection device with continuous operation at 100%

and taking a conductor from Table 2 (100%)

they must support a current of 298.5 A, with a three-phase power supply of 220 V and considering a
power factor 1.

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Example 3.2:
Calculation of loads for a building with 8 typical apartments of 180 m2Without Electric Stove
Typical apartment with 180 m² of covered area.
The building has a basement of 650 m² for parking and has the following loads:
Exterior lighting 10 points of 100 W each.
2 elevators of 5,500 W each.
02 water pumps of 1,190 W each (with alternating operation).
02 drainage electric pumps of 750 W each (with alternating operation).
1,500 W CO extractor.
02 sliding doors of 375 W each.
Central intercom, 1 point of 500 W.
Alarm control center, 1 point of 800 W.

Calculation of the load of each housing unit

Steps:

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Calculation of the total load of the Apartment Building:

Using the steps of Rule 050-200(2) and (3), the minimum capacity of the conductors is calculated
supply line that provides energy to the condominium of eight (8) housing units.
Steps:
050-202(1)(b) The load of each department is: 15 247 W

they must withstand a current of 274.8 A, with a three-phase power supply of 220 V and considering a
power factor 1.

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Example 3.3:

The conductors of the service feeder or supply line must withstand

a current of 70.26 A, with a three-phase electric supply of 220 V and considering a
power factor 1.

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Example 4.1:
Using the previous steps:

DEPARTMENTS
Type A
4
80
12
3
4
4

The building requires a panel to supply the following loads:

- 15 kW of 220 V, single-phase for each electric heater with a thermostat

own in the common area;
- 6 kW at 220 V for general lighting (incandescent);
- Three of 6 kW, 220 V, single-phase for dryers;
- 4.5 kW for washing machines;
- 8.5 kW for fluorescent lighting in the garages;
- Four 5 kW, 220 V, single-phase for hot water tanks;
- 20 free power outlets for car heaters;
- An elevator motor of 20 A, 380/220 V, three-phase.

Steps:
(1) Calculation of the load for each type of apartment:
Type A
Not applicable (N/A) --
(b) (1)(a)(i) First 45 m2of area 1,500
Next 45 m2 of area 1,000
(d) (1)(a)(iii) - Next 95 m2 or fraction --
e) (1)(a)(iv) - Electric heating space loads
according to Rule 207-116(2) (a) and (b)
First 10 kW at 100% 4,000
balance at 75% --
(f) Load of kitchens, first 12 kW 6,000
balance at 40% (C: 40%*2 kW)
(g) Other loads over 1,500 W at 25% of the demand factor
Clothes dryers 4 kW x 25% 1,000
Water heaters 3 kW x 25% 750
Total in W: 14,250
(h) Sizing of feeders
220 V current, three-phase 37.4

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Total load minus ambient heating in W:

the housing unit with the highest burden 12 050x100%

(b) The next two 2x12 050x65%
(c) The two following (12,050 + 11,050) x 40%
the last seven (3x11 050+ 4x10 250)x30%

Type A4 x 4000 = 16,000

TypeB4 x 5000 = 20,000
Type C 4 x 6 000 = 24,000
Total 60,000
Using the demand factors of Rule 270-116(2)(a) and (b)
First 10 kW at 100% 10,000
The remainder (60,000 - 10,000) at 75% 37,500
Total 47,500

Electric heating of common areas 15,000 x 75% 11,250 W

General lighting 6,000 x 75% 4,500 W
Dryers (3 x 6,000) = 18,000 x 75% 13,500 W
Washing machines 4 500 x 75% 3 375 W
Garage lighting 8 500 x 75% 6,375 W
Hot water tanks (4x5,000) x 75% 15,000 W
Power outlets / cars 20x1 200x 75% 18,000 W
Elevator motor 18.91 x 220 x 1.73 x 75% 5,398 W
Total 77 398 W

Note - All external loads on the departments are assumed to be continuous loads.

Total continuous loads of step (5); 77 398 W

From Table 2, Rule 050-104(5) is required:

96.747 96,748 W

Total load Step (2) 59 200 W

Step (3) 47,500 W
Step (7) 96 748 W
Total 203 448 W

The minimum rated current of the equipment

with three-phase supply of 220 V and with factor of
power 1, will be:

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Support of Rule 050-202(4). The demand factors applicable to the loads,

of the connections that supply housing units and loads located outside of the
housing units are not applicable for feeders that only supply
loads located within the housing units. These loads do not have the same
diversity.

Purpose of Rule 050-202(4). The feeders that supply power to

loads not located in the housing units, after applying any factor
the demand permitted by the Code must not be less than the nominal current
of the installed equipment. In the previous Example 4.1 according to Rule 050-202(1), (2) and
(3), the total load outside the housing units for the sizing of the
the building's service connection is 77,398W, while in the following Example 4.2,
using the same example, the total load outside of the housing units only for the
The feeder of the building loads is 101,246 W.

The requirements of Rule 050-104(4) and (5) must be applied to the loads.
outside of the housing units, which are continuous according to Rule 050-202(2).

Example 4.2:
Using the loads from Example 4.1 of Rule 050-202(1), (2), and (3), the calculation is made.
minimum capacity of the feeder conductors required to supply the
loads outside the housing units.

Building feeder (loads outside the apartments)

The sizing of the building feeder using Rule 050-202(4):
Electric ambient heating
use demand factors of 270-116(3)(b)
11,250 = 11 250 W
general lighting 6,000 W
dryers (3 x 6 000) = 18,000 W
washing machines 4,500 W
garage lighting 8 500 W
hot water tanks (4 x 5,000) 20,000 W
power outlet for cars 20x1 200 24,000 W
elevator motor 10.95 x 380 x 1.73 x 1.25 = 8 996 W
Total 101 246 W
If the equipment is not marked or does not indicate on the data plate that
can operate continuously at 100% of its nominal capacity according to 050-104(4) and is
use a conductor with PVC insulation at 70 °C from Table 2, and all the loads are
continuing, the minimum load in watts is 101 246 / 80% = 126 558 W;

The minimum nominal current of the equipment is 126 558/(380 x 1.73) = 192 A.

If the equipment is set for continuous operation at 100% of its current

nominal according to Rule 050-104(4) and a conductor with PVC insulation is used
70 °C from Table 2, and all loads are continuous, the minimum load in watts is
101 246 W;

The minimum nominal current of the equipment is 101 246/(380 x 1.73) = 154 A.

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Support of Subrule 050-202(5). A criterion is necessary to determine the

capacity of the conductors for the housing connections in buildings of
departments, with relatively low loads, when information is not available
specific about them.
(Reference: Subrule 050-110 (2))

Purpose of Subrule 050-202(5). For apartments in residential buildings

the conductors of the service connections and feeders to the main panel
each department must have a minimum driving capacity like the
indicated in Subrule 050-106(9).

Support of Rule 050-204. A basic criterion is necessary to calculate the

minimum acceptable capacity, required by the feeder drivers or of the
feed line supplying a school.

Purpose of Rule 050-204. A method is established by which it can be

install the drivers of the connection or the feeder that supplies power to
a School or College, with the adequate capacity and that can handle the loads
building electrical systems safely.

The Subrule (1) indicates the method for determining the minimum capacity of the
conductors of the service or feeder for a school. Subrule (2) indicates
the demand factors that need to be applied to the minimum capacity of the connection
or of the drivers calculated in Subrule (1).

The steps to follow to calculate the minimum capacity of the conductors of the
feed or the main feeder of a school or college are the
next:

(1) from 050-204(1) - determine the total area of the school in m2, based on
the external dimensions;

(2) from 050-204(1) - determine the total area of the school's classrooms in m2;

(3) of 050-204(1)(a) - calculate the basic load in watts for the area of classrooms,
multiplying the total area of halls in m2for 50 W/m2;

(4) of 050-204(1)(b) - calculate the load of the remaining area of the school,
multiplying the remaining area in m2by 10 W/m2.

(5) from 050-204(1)(c) calculate

- the total load in watts by summing, the heating
electrical environment, the air conditioning, and the loads of
power, based on the nominal power of the equipment
installed;

(6) calculate the total load of the building, summing the steps (3),
(4) and (5);

(7) subtract any electric environmental heating load

from the total load of the building calculated in step (6);

(8) from 050-204(2)(a)- apply the following demand factors for a

school with an area up to and including 900 m2

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based on the external dimensions, apply the

demand factors of Subrule (2)(b) – see step (9);

(a) calculate the total loads of ambient heating

electric applying the demand factors of the
Section 270;

(b) calculate the remaining load of the building taking into account the
75% of the remaining total load of the building, minus
any load of electric ambient heating (see
step (7);

(c) calculate the minimum capacity of the connection or of the

feeder conductor, adding the steps (8)(a) and
(8)(b);

(9) of 050-204(2)(b) apply the following demand factors for a

school with an area of more than 900 m2, based on
the external dimensions;

(a) calculate the total loads of ambient heating

electric applying the demand factors of the
Section 270;

(b) calculate the nominal load per square meter by dividing

the rest of the total load of the building minus the loads of
the electric ambient heating (see Step 7), by the area
total of the school (see Step 1);
i) calculate the load for the first 900 m2of the area
total of the school in watts, taking 75% of the load
per square meter multiplied by 900;
50% of the load per square meter multiplied by
the total area of the school exceeds 900 m2;
(c) calculate the minimum capacity of the connection of the
feeder conductor, adding the steps (9)(a),
(9)(b)(i) and (9)(b)(ii);

(10) determine the wiring method that will be used, and how much of the load it is
continue according to 050-104(3), and from there apply the requirements of 050-104(4) or
(5);

dimension the conductors of the feeder or the service connection and the devices
about overcurrent.

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Example 1:
Determine the minimum capacity of the service conductor for a school with a total area
built of 900 m2of which 750 m2it is the total area of classrooms. The other loads in the school
They total 65,000 W, with an electric ambient heating load of 90,000 W. The voltage
The supply of the connection is 380/220 V, three-phase with 4 wires.
Steps:
Pot. Dem.
Inst. Max.
(W) (W)

050-
204(1)(c)
Other school burdens 65,000

Total load of the building minus any load of

heating

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Example 2:

Steps:

050-
204(1)(c)

Adding steps (3), (4) and

(5)
Total load of the building minus any load from
heating

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Support of Rule 050-206. A basic standard is required when calculating the

minimum acceptable capacity of the service or feeder conductors, which
they are required to supply energy to a hospital.

Purpose of Rule 050-206. A method is established by which it is determined the

adequate capacity of the conductors of the supply or feeder that
supplies energy to a hospital, to carefully manage the electrical loads
in the building.

Subrule (1) provides the method for determining the minimum capacity of the conductors.
of the feeders or the supply for a hospital. Subrule (2)
give the demand factors that should be applied to the minimum capacities of the
incoming or feeder calculated in Subrule (1).

The steps to follow are:

(1) from 050-206(1) - determine the total area of the hospital in m2, based on
in the external dimensions;

(2) from 050-206(1) - determine the total area of high intensity areas,
like the operating room, of the hospital in m2;

(3) of 050-206(1)(a)- calculate the basic load of the hospital in watts

multiplying the total area of the hospital in m2 by 20
W/m2;

(4) of 050-206(1)(b)- calculate the load of high intensity areas in the

hospital, multiplying the total of the high areas
density in m2 per 100 W/m2
(5) of 050-206(1)(c)- calculate the total load in watts, adding the loads of
the electric ambient heating, of the air
air-conditioned and the power loads based on the
rated power of the installed electrical equipment;

(6) calculate the total load of the building by adding the steps
(3), (4) and (5);

(7) subtract any ambient heating load

electric load of the total building calculated at the
step (6);

(8) of 050-206(2)(a) - apply the following demand factors for a

hospital with an area of up to and including 900 m2,
based on the external dimensions (if the hospital
it has an area larger than 900 m2based on their
external dimensions, apply the demand factors
from Subrule (2)(b) - see step (9));
(a) calculate the total loads of the ambient heating
electric, applying the demand factors of the
Section 270;

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(b) calculate the remaining load of the building taking 80%

of the rest of the total load of the building minus any
electric ambient heating load (see step (7);
(c) calculate the minimum capacity of the connection or of the
feeder conductor adding the steps (8)(a) and
(8)(b);

(9) of 050-206(2)(b) - apply the following demand factors for a

hospital with an area of around 900 m2based on
the external dimensions;
(a) calculate the total loads of ambient heating
electric, applying the demand factors of the
Section 270;
(b) calculate the nominal load per square meter,
dividing the remainder of the total load of the building minus
the charges of the electric ambient heating (see
Step 7), for the total area of the hospital (see Step 1);
(i) calculate the load for the first 900 m2of area
in watts, taking 80% of the load per meter
square multiplied by 900;
65% of the load per square meter multiplied
for the total area of the hospital in excess of 900 m2;

(c) calculate the minimum capacity of the conductors of the

coming from the feeder, adding the step (9)(a),
(9)(b)(i) and (9)(b)(ii);

(10) determine the wiring method that will be used, and how much of the load is
continue according to 050-104(3), and from there apply the requirements of 050-104(4) or
(5);
(11) size the connection or the feeder conductor and the devices of
overcurrent.

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Example 1:
Determine the minimum capacity of the service conductor for a hospital with
total built area of 12,000 m2, of which 300 m2high-intensity areas. The
The supply is 380/220 V, three-phase, 4 wires.
The hospital has the following loads:
Electric ambient heating 1,200,000 W
Air conditioning 100,000 W
Equipment and other charges 165,000 W
Steps:
The total area of the hospital is 12,000 m.2;
The total area of the high-intensity areas is 300 m2;
(3) The basic load of the hospital 20 x = 240,000
12,000 m2 W;
The load of the high intensity areas 100 x 300
= 30,000 W;
The total of the other charges in the hospital
Electric ambient heating 1,200,000 W
Air conditioning is not considered (see 050-106(4)) ------
Equipment and other loads 165,000 W
Total 1,365,000 W;
The total load of the building
Step (3) 240,000 W
Step (4) 30,000 W
Step (5) 1,365,000 W
Total 1,635,000 W;
The total load of the building minus any ambient heating load
electric
1,635,000 - 1,200,000 = 435,000 W;

Since the total area of the hospital is more than 900m2use step (9) to calculate
the demand factors;

(9) (a) The total load of the electric ambient heating with the factors of
demand of Section 270.
1,200,000 x 75% = 900,000 W;
calculate the load per square meter
435,000 / 12,000 = 36.2 W/m2;
2
(b)(i) calculate the load for the first 900 m of the area
80% x 36.2 x 900 = 26 064 W;
(b)(ii) calculate the load for the remaining area of the hospital
65% x 36.2 x (12,000 - 900) = 261 183 W;
(c) calculate the minimum capacity of the supply connection
1,806 A

(10) Determine the wiring method that will be used, the amount of load
continues; and the demand factors according to Rule 050-104(4) or (5).

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Example 2:
Determine the minimum capacity of the conductor for the service connection for a hospital with
total built area of 900 m2, of which 150 m2high intensity areas. The
Other hospital loads have a total of 65,000W, with ambient heating of 90.
The supply connection voltage is 220/380, three-phase with four wires.

Steps:
The total area of the hospital is 900 m2;
The total area of high-intensity areas is 150 m2;
(3) The basic load of the hospital 20 x 900m2 =18,000 W;
(4) The load of the high intensity areas 100 x 15,000
150 W;
The total area of the other loads in the hospital
155000 155,000 W;
The total load of the building
Step (3) 18,000 W
Step (4) 15,000 W
Step (5) 155,000 W
Total 188,000 W;

The total load of the building minus any ambient heating load
electric 98,000
= 98 000 W;

Since the total area of the hospital is more than 900m2, use step (9) to calculate.
the demand factors;

(9) (a) The total load of the electric ambient heating with the factors of
demand of Section 270.
67,500 = 67 500 W;
(b) the load of the rest of the building is
80% of 98,000 78,400 W;
(c) calculate the minimum capacity of the supply connection
(67 500 + 78 4000) / (380 x 1,73) = 222 A;

(10) Determine the wiring method to be used; the amount of load

continues; and the demand factors according to Rule 050-104(4) or (5).

Support for Rule 050-208. A basic standard is required to calculate the

minimum capacity of the feeder conductors or the service connection that
They supply energy to hotels, motels, and similar establishments.

Purpose of Rule 050-208. A method is established by which it can be

determine the appropriate capacity of the feeder or service conductors,
that will be fed to hotels, motels, dormitories and similar establishments. The
Subrule (2) gives the demand factors that should be applied to the capacities
minimums of the conductors of the feeder or the supply line, calculated in the
Subrule (1).

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The steps to follow are:

(1) from 050-208(1) - determine the total area of the hotel, motel, dormitory, or
similar establishments in m2based on the
external dimensions;

(2) of 050-208(1)(a) - calculate the basic load in watts by multiplying the area
total of the building in m2per 20 W/m2;

(3) of 050-208(1)(b) - calculate in watts the total lighting load for the
special areas, such as ballrooms,
based on the rated power of the electrical equipment
installed;

(4) of 050-208(1)(c) - calculate the total load in watts, including the heating
electric environmental, the air conditioning and the loads of
force, based on the rated power of the equipment
installed electric
Note: When calculating the load for motors installed in the feeder, it should be
comply with Rule 160-110(1).

Calculate the total load of the building by adding the steps (2), (3) and (4);
Note: The following demand factors can be applied:

(6) subtract any electric ambient heating load from the total load of the
building calculated in step (5);

(7) of 050-208(2)(a) - apply the following demand factors for a hotel,

motel, bedroom, etc., with an area up to, and
including 900 m2based on the dimensions
externals (if the hotel, motel, dormitory, etc. has a
area greater than 900 m2based on its dimensions
external, apply the demand factors of the Subrule
(2)(b)- see step (8));
(a) calculate the total loads of electric ambient heating, applying
the demand factors of Section 270;
(b) calculate the remaining load of the building by taking 80% of the remaining load
total of the building minus any electric ambient heating load
(see step (7));
(c) calculate the minimum capacity of the service conductor or the
feeder adding the steps (7)(a) and (7)(b);

(8) of 050-206(2)(b) - apply the following demand factors for a hotel,

motel, bedroom, etc., with an area of about 900 m2
based on the external dimensions;
(a) calculate the total loads of the electric ambient heating, applying
the demand factors of Section 270;

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(b) calculate the nominal load per square meter, dividing the remainder of the
total load of the building minus the loads of environmental heating
electrical (see Step 7) by the total area of the building (see Step 1);
(i) calculate the load for the first 900 m2of the total area of the
building in watts, taking 80% of the load per square meter
multiplied by 900;
65% of the load per square meter multiplied by the total area
from the building in excess of 900 m2;
(c) calculate the minimum capacity of the conductor of the service connection or of the
feeder, summing steps 8(a), (8)(b)(i), and (8)(b)(ii);
(9) determine the wiring method that will be used, and how much of the load is
continue according to 050-104(3), and then apply the requirements of 050-104(4)
o (5);

(10) determine the dimensions of the service conductor or feeder and of

overcurrent devices.

Example 1:
Determine the minimum capacity of the service conductors for a hotel with
total built area of 4,800 m2 The supply is 380/220 V, three-phase, 4-wire.

The hotel has the following charges:

Specialized lighting 22,000 W
Electric ambient heating 500,000 W
Air conditioning 300,000 W
Power Loads 340,000 W
Steps:
The total area of the hotel is 4,800 m2;
(2) The basic load of the hotel 20 x 4 800 = 96 000 W;
The loads of specialized lighting are 22,000 W;
The total charges at the hotel are:
Electric ambient heating 500,000 W
Air conditioning is not considered (see 050-106(4)) -----------
Power loads 340,000 W
Total 840,000 W
The total load of the building
Step (2) 96,000 W
Step (3) 22,000 W
Step (4) 840,000 W
Total 958,000 W;

The total load of the building minus any ambient heating load.
electric
958 000 – 500 000 = 458,000 W;

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Since the total area of the hotel is more than 900m2use step (8) to calculate the
demand factors;
(a) The total load of electric ambient heating with the factors of
demand of Section 270.
125000 375,000 W;
(b) calculate the load per square meter
458 000 / 4 800 95.4 W/m2
2
(b)(i) calculate the load for the first 900 m of area
80% x 95,4 x 900 = 68 688 W;
(b)(ii) calculate the load for the remaining area of the hotel
65% x 95,4 x (4 800 – 900) = 241 839 W;
(c) calculate the minimum capacity of the supply connection
(375,000 + 68,688 + 241,839) / (380 x 1.73) = 1,043 A

The following table details the steps to follow to apply step (9). The only step
what has not been considered was how much of the load was continuous and how much was not continuous.
In the example, the calculated load is considered as continuous. This consideration is
it must take into account consulting all parties involved with the electrical installation.

Method of Nominal Current of

Load Fencing Connection Team
Calculated Regime Regime Driver Capacity
Table
Table 2 I continue I continue
1
100% 80%
1,043 A If --- If --- 1,227 A
2. 1 043 A Yes --- --- Yes 1 043 / 70% = 1 490 A
3. 1 043 A --- If Yes --- 1 043 A
4. 1 043 A --- Yes --- Yes 1,043 / 80% = 1,304 A

Support and Purpose of Rule 050-210. This rule provides a basis on which
the minimum capacities of the establishments can be determined, apart from the
covered by Rules 050-200 to 050-208. A method is established by which
can properly determine the capacity of the service conductors or
of the feeder that supplies various types of establishments. The concept,
previously presented, in watts per square meter and the application of various
percentages of demands according to the establishment have been extended through the use
from Table 14.

The loads such as electric ambient heating, air conditioning, power and
lighting, they are treated separately from their nominal powers, with factors of
demands permitted by other Sections of the Code, (for example, ambient heating
electric and motor load.
The steps to follow are:
(1) from 050-210 - determine the total area of the establishments in m2
based on the external dimensions;

(2) from 050-210(a) - calculate the basic load in watts, for the type of
establishment using watts per square meter
from Table 14, for the particular type of

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establishment, multiplied by the area of the

establishment of step (1);
(3) of 050-210(a)- apply the demand factor from Table 14 to the load
basic calculated in step (2) for the drivers
from the connection or feeders;

(4) of 050-210(b)- calculate the total special load in watts, such as the
electric ambient heating, air conditioning, the
power loads, lighting of showcases and
stage lighting, based on power
nominal of the installed electrical equipment with any
demand factor allowed by the Code;
Note: When calculating the load for motors installed on the feeder, it should be
comply with Rule 160-110(7).

(5) calculate the minimum capacity of the service conductors or the

feeder, adding the steps (3) and (4);

(6) determine the wiring method used; how much of the load is continuous according to
050-104 (3); and then apply the requirements of 050-104(4) or (5);

(7) determine the dimensions of the service or feeder conductor and of the
overcurrent devices.

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Example 1:
Calculate the minimum conductor capacity for the following establishments.
a shopping center. The voltage is 380/220, three-phase with 4 wires.

Office:
Area 200 m2
Loads 10 kVA air conditioning
electric ambient heating 25,000 W
computer load 27.3 A at 220 V
water heater 1 500 W
Steps:
Area 200 m2;
2
(2) Table 14 - 50 W/m
basic load 50 x 200 = 10,000 W;
(3) Table 14 - 90%, basic load 10,000 x 90% 9,000 W;
(4) Special loads:
air conditioning, is not considered (see Rule 050-106(4)) --------
electric ambient heating 25,000 x 0.75 18,750 W
computer load 27.3 x 220 = 6,000 W
water heater 1,500 W
Total 26 250 W
Minimum capacity of the service connection conductor
(9,000 + 26,250) / (380 x 1.73) = 53.6 A

Mini-market:
- Area - 300 m2
Loads refrigerators 36.4 A at 380 V
showcase lighting 4 000 W
electric ambient heating 30,000 W
30 kVA air conditioning
water heater 4,000 W
Steps:
(1) Area 300 m2;
Table 14 30 W/m2basic load 30 x 300 9,000 W;
Table 14 100%, basic load 9,000 W;
Special loads:
refrigerators 36.5 x 380 x 1.73 24,000 W;
showcase lighting = 4,000 W;
air conditioning 30,000 VA
water heater 4,000 W
Total 62,000 W

Minimum capacity of the service conductor

(9,000 + 62,000) / (380 x 1.73) = 108 A.

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Hobby Store:
Area 100 m2
- Loads:
window display lighting 2,000 W
electric ambient heating 12,000 W
water heater 1 500 W
Steps:
(1) Area- 100 m2;
2
(2) Table 14 - 30 W/m
, basic load 30 x 100 = 3,000 W;
(3) Table 14 - 100%, basic load = 3,000 W;
Special loads:
showcase lighting = 2,000 W;
electric ambient heating 12,000 x 75% = 9 000 W;
water heater 1,500 W;
Total 12,500 W;
Minimum capacity of the connection conductor
(3,000 + 12,500) / (380 x 1.73) = 23.6 A

Bank:

Area - 120 m2
Cargas special lighting 1 500 W
electric ambient heating 15,000 W
3000 W water heater
Steps:
(1) Area- 120 m2;
(2) Table 14 50 W/m2basic load 50 x 120 = 6,000 W;
(3) Table 14 100%, basic load = 6,000 W;
(4) Special loads:
special lighting 1,500 W;
electric ambient heating 15,000 x 75% 11,250 W;
water heater = 3 000 W;
Total 15,750 W;

Minimum capacity of the service driver

(6,000 + 15,750) / (380 x 1.73) = 33.1 A

Pharmacy:
Area 150 m2
Cargas
special lighting 3 000 W
electric ambient heating 15,000 W
water heater 3000 W
10 kVA air conditioning
11 A refrigerators at 380 V

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Steps:
(1) Area- 150 m2;
(2) Table 14 30 W/m2Basic load 30 x 150 = 4,500 W;
(3) Table 14 100%, basic load = 4,500 W;
(4) Special loads:
special lighting = 3,000 W;
electric environmental heating 15,000 x 75% 11,250 W;
water heater = 3 000 W;
air conditioning - is not considered (see Rule 050-106(4)) ----- ---
refrigerators = 7 200 W
Total = 24 450 W;

Minimum capacity of the service connection conductor

(4,500 + 24,450) / (380 x 1.73) = 44 A.

Beauty Salon
150 m2
loads
special lighting 4,000 W
electric ambient heating 10,000 W
water heater 6,000 W
air conditioning 12 kVA
hair dryers 12,000 W
Steps:
2
(1) area - 150 m
;
(2) Table 14 30 W/m2basic load 30 x 150 = 4,500 W;
(3) Table 14 90%, basic load 4 050 W;
(4) special loads:
special lighting = 4 000 W;
ambient heating - not considered (see Rule 050-106(4)) -------
water heater 6000 W;
air conditioning = 12,000 W
hair dryers 12,000 W
Total 34,000 W

minimum conductor capacity of the connection

(4,050 + 34,000) / (380 x 1.73) = 58 A.

Support for Rule 050-212. The area method for calculating the load is not
always practical, especially when the lighting loads are not confined
to a single geographical location within the building. In this case, another method
Apart from the watts per square meter, it should be used.

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Purpose of Rule 050-212. An alternative method is established for calculating the

loads powered from a single panel, but not located in one place
geographical area.

Consider ten circuits of 15 A at 220 V loaded with 4.4 A each and feeding
lighting loads throughout the building (the exits of the signs and of the
lighting can be on the same circuit.

It is calculated that the load is: 10 x 4.4 = 44 A at 220 V, in a 380 V feeder.

three-phase, the current would be 14.6 A.

Derived Circuits
Support of Rule 050-300. Kitchen loads vary in different
establishments and certain criteria are required to establish the appropriate capacity
in the branch circuits that supply the kitchen electrical equipment.

Purpose of Rule 050-300. Subrule (1) establishes the principles for calculation.
the capacities of the derived circuits, for electric stoves in units of
housing, which are similar to those previously found in Rules 050-
200 and 050-202 for incoming and feeder demands, except that in this
The claim is only for the drivers of derived circuits.

In Rule 050-200, the demand for a 12 kW electric kitchen was 6 kW. Here.
it is a little bigger, reaching 8 kW. The calculations for the rated currents.
that exceed 12 kW is the same: 40% of the amount of the rated current over
12 kW.

Example:
A single-family housing unit has a 14 kW electric kitchen, 220 V.

Contribution for supply (6 kW + 40% x 2 kW) 6.8 kW

Demand of the branch circuit (8 kW + 40% x 2 kW) 8.8 kW

The same electric stove in another application that is not a housing unit
unfamiliar, it will have a contribution from the supply based on the demand of
establishment. The capacity of the derived circuit must be the nominal current
from the electric kitchen and in our example, it is 14 kW.

In a housing unit, a 6 mm conductor can be used.2At 90 ºC (Table 2),

to cover 6.8 kW or 8.8 kW.

For any other case that is not a housing unit, it must be used by
less one conductor for 14 kW, which would be:
A 16 mm conductor2At 90 ºC, for a single-phase branch circuit
A 6 mm conductor2At 90 ºC, for a three-phase derived circuit.

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Subrule (2) refers to kitchen units, separate from fixed installation, for example,
a stove or an oven. It should be taken into account that the requirements for the
derived circuits for these units are also covered in Rule 150-744. The
Rule 150-746 covers the supply connections for electric kitchens.
Rule 050-300(1).

If you have a 6 kW burner and a 5 kW oven.

The total would be:
5 + 6 = 11 kW
The capacity of the branch circuit is 8 kW.

In Subrule (3), the parameters are established to reflect the characteristics

operational electric kitchen equipment, installed in establishments
commercial, industrial, and institutional. The demand to calculate the capacities
minimums of the derived circuits in these establishments must be no less than
the nominal current of the equipment.

Subrule (4) recognizes that stoves and similar appliances, which are not fixed installations, that
they connect with cords built into the device, they must not exceed 1,500 W and
They can be operated from a standard kitchen countertop outlet.

Support of Regulation 050-302. Even in well-organized regulations, there are found

some topics that do not have a defined place. These fit well under this rule,
the general title of "connected loads".

Purpose of Rule 050-302. In Subrule (1), parameters are established for

calculate the lighting installation demands for display cases.

The length of the window is measured along the base and 650 W per meter is applied.
to file a lawsuit. The fractions that may remain are not considered, they
you can take advantage of this and say that the figure applies to each full meter. A
A 6-meter window will have a demand of 650 x 6 = 3,900 W.

In Subrule (2), some loads with cyclical or intermittent nature are also
You continue. A water heater is a good example of this. When the tank is
cold, the unit can operate for more than two hours until the
desired water temperature. Depending on the use of water, the unit may be
It is in a frequent cycle. This load can be considered continuous.

In Subrule (3), it is specified that the data processing electric equipment

It must be considered as a continuous load for the application of Rule 050-104.

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