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Heat Transfer
Shubber Jawad
Heat Transfer
The original purpose of a building is to provide shelter and to maintain a
comfortable or at least liveable internal temperature. Other purposes include
security, privacy and protection from wind and weather. To feel comfortable in a
thermal sense, a human has to be able to release a well-defined amount of Heat.
If this gets difficult, a person will either feel cold or hot. The human body
operates as a chemical reactor that converts chemical energy of food and
respiratory oxygen into mechanical work and heat. Heat output can vary from
about 100 W for a sedentary person to 1000 W for an exercising person. To
maintain body temperature within a narrow band, the heat produced by an
occupant must be released to the indoor environment. If too much heat is lost,
room temperature should be increased or warmer clothes be worn. The heat
transfer on the human skin, the indoor temperature and the heat transfer through
the building envelope are factors that influence thermal comfort.
Heat Loss of a Building
According to the Second Law of Thermodynamics, heat transfer is only
possible in the direction from a higher temperature to a lower one. It becomes
zero if temperatures are equal. The heat loss through an envelope should
therefore be proportional to the difference (Tinside – Toutside), or to a positive power
of it for small differences. For a simple formula, a linear dependence on
temperature difference is sufficient. Accepting further that heat loss grows
linearly with surface area A, one finds:
q= AU (Tinside – Toutside)
The constant of proportionality, U, is the Overall Heat Transfer Coefficient in
W/(m2K).
The equation above suggests three ways to reduce heat loss:
1) As the heat loss is proportional to the inside-outside temperature difference,
the set-point for the indoor temperature can be reduced during the heating
season;
2) The insulation of the envelope can be improved to reduce the overall heat
transfer coefficient U; and
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3) If possible, the surface area should be reduced without changing the enclosed
volume. A spherical igloo would be optimal, but a cubical shape is still better
than an elongated building with many wings. The opposite is true for the design
of heat exchanger surfaces or fin-tubes, where the effective surface should be
maximized.
Modes of Transport Heat
In building heat transfer, many different types of energy transport are effective.
Often, heat is transported by different modes to or from the same place. Energy
that reaches a point via different paths and modes may be added up for the heat
balance. For instance, the heat loss of a human body is the sum of convection,
radiation, and latent heat released by sweating and so forth.
Primary heat transport modes are:

Conduction: In Conduction, heat transfer takes place due to a temperature
difference in a body or between bodies in thermal contact, without mixing
of mass.

Convection (heat conveyed as internal thermal energy of mass that is
displaced by mean or turbulent motion);

Radiation (heat transfer by electromagnetic waves such as infrared or
visible light).
Heat transfer through a wall can be calculated as
q = U A ΔT
(1)
where
q = heat transfer (J/s, W)
U = overall heat transfer coefficient (W/m2 oC)
A = wall area (m2)
ΔT = (T1 - T2)
= temperature difference over wall (oC).
U and R-values
The overall heat transfer coefficient - the U-value - describes how well a
building element conducts heat or the rate of transfer of heat (in watts or Btu/hr)
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through one unit area (m2 or ft2) of a structure divided by the difference in
temperature across the structure.
U-value (or U-factor) is a measure of the rate of heat loss or gain through a
construction of materials. The lower the U-factor, the greater the material's
resistance to heat flow and the better is the insulating value. U-value is the
inverse of R-value.
The overall U-value of a construction consisting of several layers can be
expressed as
U=1/∑R
(2)
Where:U = heat transfer coefficient (Btu/hr ft2 oF, W/m2 oC)
R = the resistance to heat flow in each layer (hr ft2 /Btu or, m2 oC /W)
The R-value of the single layer can be expressed as:
R=1/C=s/k
(3)
where
C = layer conductance (Btu/hr ft2 or, W/m2 oC)
k = thermal conductivity of layer material (Btu in/hr ft2 or, W/m oC)
s = thickness of layer (inches, m)
Note - in addition to resistance in each construction layer - there is a resistance
from the inner and outer surface to the surroundings.
U = 1 / Ri + R1 + R2 + R3 + ….. + Ro
(4)
where
Ri = thermal resistivity surface inside wall (m2 oC /W).
R1.. = thermal resistivity in the separate wall/construction layers (m2 oC /W).
Ro = thermal resistivity surface outside wall (m2 oC /W).
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Condensation in buildings
Condensation happens in buildings when warm, moist air comes into contact
with cooler surfaces that are at or below the dew point, such as windows,
and water condenses on those surfaces.
When air cools, it is less able to 'hold' moisture, that is, the
saturation water vapour density falls, and so the relative humidity rises. When
the relative humidity reaches 100%, the air will be saturated. This is described as
the dew point. If the air continues to cool, moisture will begin to condense.
Condensation can
occur
on
surfaces,
or
can
be interstitial
condensation occurring between the layers of the building envelope, typically as
a result of air diffusing from the warm interior of the building to the cool
exterior and reaching its dew point within the building fabric. This interstitial
condensation, when trapped within the structure by impervious materials is often
erroneously referred to as rising damp.
Condensation affects the performance of buildings, causing problems such as:

Mould growth which is a cause of respiratory allergies.

Mildew.

Staining.

Slip hazards.

Damage to equipment.

Corrosion and decay of the building fabric.

Poor performance of insulation.
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Example 1: Compute (U) value and temperature at inner at outer face for a brick
wall with a thickness of (105) mm if inside air temperature is 22 ºC and outside
air temperature is 2 ºC , Kbrick = 1.2 W/m ºC, Rsi = 0.123 m2 ºC /W, Rso= 0.053
m2 ºC /W.
Ans.
R= s / k
Rbrick =
= 0.0875 m2 oC /W
U=1/∑R =
3.795W/m2 oC
=
Rate of heat loss (q) = U A ΔT
A= 1m2
ΔT= (Tout-Tin) = (2-22) = -20 oC
q = 3.795(1) (-20) = -75.9 W/m2
Note:- (-) sign refers to heat losing
Heat at inner face of brick = -75.9 =
(ΔT1)
ΔT1= - 9.336 oC = (T2- T1)
T2 = (-9.336+ 22) = 12.664 oC
Heat at outer face of brick = -75.9 =
(1) (ΔT2)
ΔT2= - 6.64 oC = (T3- T 2)
T3 = (-6.64+ 12.664) = 6.024 oC
For checking
q = -75.9 =
(1) (ΔT3)
ΔT3= - 4.023 oC = (T4- T 3)
T4 = (- 4.023+6.024) ≈ 2 oC…………….. o.k
or
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Heat Transfer
q = -75.9 =
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(1) (ΔT3)
ΔT3 = - 4.023 oC = (T4- T 3)
T4 = (- 4.023 - 2) = 6.023 oC …………….o.k
Example 2: A brick wall with a thickness of 105 mm covered from inside by
wooden sheet with a thickness of 25 mm, Kbrick = 1.2 W/m ºC, Kwood= 0.09 W/m
ºC, Rsi = 0.123 m2 ºC /W, Rso= 0.053 m2 ºC /W. inside temperature is 22 ºC and
outside temperature is 2ºC. If dew point of inside air is 5 ºC, at which position of
the wall the condensation may happen?
Ans.
R= s / k
Rbrick =
= 0.0875 m2 oC /W
Rwood =
= 0.277 m2 oC /W
U=1/∑R=
1.847 W/m2 oC
=
q= U A ΔT = 1.847(1) (2-22) = - 36.94 W/m2
Heat at inner face of wood = -36.94 =
(ΔT1)
ΔT1= -4.543 oC
T2= 22 - 4.543 = 17.457 oC>5 oC no condensation will happen
Heat at outer face of wood = -36.94 =
(ΔT2)
ΔT2 = -10.232 oC = (T3- T2)
T 3 = (-10.232+17.457) = 7.22 oC >5 oC no condensation will happen
Heat at outer face of brick = -36.94 =
(ΔT3)
ΔT3 = - 3.232 oC = (T4- T 3)
T4 = (-3.232+ 7.22) = 3.987 oC < 5 oC condensation will happen
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Example 3: A building with inner air temperature of 20 oC and outer air
temperature of 5 oC has a wall with area of 20 m2 consist of the following layer:
1- Outer layer of brick with a thickness of 120 mm and K= 0.84 W/m ºC.
2- Insulating layer with a thickness of 50 mm and K= 0.026 W/m ºC.
3- Inner layer of brick with a thickness of 120 mm and K= 0.84 W/m ºC.
4- Inner finishing layer of plaster with a thickness of 10 mm and K=0.5 W/m ºC.
If Rsi = 0.123 m2 ºC /W and Rso= 0.055 m2 ºC /W, compute the following:A- The value of heat transferred for each 1 m2 of wall, gain or loss in heat will
happen for the building?
B- The temperature of inner face for inner brick layer.
C- Insulating layer thickness required to reduce transferred heat to a quarter of
its value.
Ans.
R= s / k
R brick =
= 0.142 m2 oC /W
= 1.923 m2 oC/W
R insulating =
R plaster =
= 0.02 m2 oC/W
U=1/∑R=
0.415 W/m2 oC
=
1- q= U A ΔT = 0.415(1)(5-20) = - 6.225 W/m2
(-) sign refers to heat losing
2- q =
-6.225 =
(1)(T2-T1)
(T2-20)
Heat at inner face of plaster (T2) = 19.23 oC
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Heat Transfer
q=
(1) (T3-T2)
-6.225 =
(1) (T3- 19.23)
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The temperature of inner face for inner brick layer (T3) = 19.1 oC
3-
= 1.556 Watt/m2 (heat transferred to a quarter)
q = U (A) (Tout-Tin ) = 1.556 = U (1) (5-20)
U = 0.1037 =
R insulating= 9.155 W/m2 oC
9.155 =
t = 0.238 m = 238 mm
Thermal Comfort in Buildings
BS EN ISO 7730 defines thermal comfort as '…that condition of mind which
expresses satisfaction with the thermal environment', i.e. the condition when
someone is not feeling either too hot or too cold. An environment can be said to
achieve 'reasonable comfort' when at least 80% of its occupants are thermally
comfortable. This means that thermal comfort can be assessed by
surveying occupants to find out whether they are dissatisfied with their thermal
environment.
The human thermal environment is not straight forward and cannot be
expressed in degrees. Nor can it be satisfactorily defined by
acceptable temperature ranges. It is a personal experience dependent on a great
number of criteria and can be different from one person to another within the
same space.
The thermal comfort variables used here are defined as follows:
Dry Bulb Temperature of Air (DBT)
The temperature of the air measured by the ordinary thermometer is called as the
dry bulb temperature of air, commonly referred as DBT. When ordinary
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thermometer is exposed to the atmosphere, it indicates the dry bulb temperature,
which is nothing but the atmospheric temperature.
Wet Bulb Temperature of Air (WBT)
The wet bulb temperature of air is also measured by the ordinary thermometer,
but the only difference is that the bulb of the thermometer is covered by the wet
cloth. Temperature of the ordinary air measured by the thermometer when it is
covered by wet cloth or wick is called as the wet bulb temperature, commonly
referred to as WBT. When the air comes in contact with the wet cloth it absorbs
some moisture and gives up some heat, sue to which the temperature of the air
reduces. This reduced temperature measured by the thermometer is called as the
wet bulb temperature. Figure 1- below shows sling thermometer.
Figure 1- Sling psychrometer and its usage. Air motion encourages evaporation from the
moist cloth, lowering the wet-bulb temperature below the surrounding air temperature,
whereas the dry-bulb temperature stays constant at the surrounding air temperature.
(At 100% RH, WB and DB temperatures will be equal.)
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The Dew point Temperature (DP)
The Dew point temperature is the temperature at which the air can no
longer "hold" all of the water vapor which is mixed with it, and some of
the water vapor must condense into liquid water. The dew point is
always
lower
than
(or
equal
to)
the
air
temperature.
If the air temperature cools to the dew point, or if the dew point rises to
equal the air temperature, then dew, fog or clouds begin to form. At this
point where the dew point temperature equals the air temperature, the
relative humidity is 100%.
Relative Humidity (RH)
Relative humidity is the amount of water vapor (vapor pressure) that is in the
air. It is a percentage of how much moisture the air could possibly hold. The
amount of vapor that can be contained in the air increases with temperature. The
higher the percentage of relative humidity, the more humid (moist) the air feels,
while a lower percentage usually feels drier. Saturation occurs when air is
holding the maximum amount of water vapor possible at the existing pressure
and temperature. Saturation is equal to 100% relative humidity, resulting in
precipitation
Specific Humidity or Humidity ratio (HR) or (MC)
Is the ratio of the mass of water vapour to the mass of dry air in a given volume
of mixture, measured by Kg water vapour/Kg dry air
Enthalpy
The total heat in the mixture measured above zero degree Fahrenheit and
including the latent heat of water vapour.
British Thermal Unit (BTU)
Is the amount of heat required to raise the temperature of one pound of water to
one degree Fahrenheit.
Sensible Heat
Is the energy required to change the temperature of a substance with no change
in substance phase.
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Latent heat
Latent heat is the energy absorbed by or released from a substance during a
phase change from a gas to a liquid or a solid or vice versa without changing in
its temperature.
Environmental factors Effecting on Thermal Comfort
1- Air temperature
The temperature of the air that a person is in contact with, measured by the dry
bulb temperature (DBT).
2- Air velocity
The velocity of the air that a person is in contact with (measured in m/s). The
faster the air is moving, the greater the exchange of heat between the person and
the air (for example, draughts generally make us feel colder).
3- Radiant temperature
The temperature of a person surroundings (including surfaces, heat generating
equipment, the sun and the sky).
4- Relative humidity (RH)
The higher the relative humidity, the more difficult it is to lose heat through
the evaporation of sweat.
Criteria for thermal comfort
In winter 22.2 ºC (72ºF), and relative humidity (30%-60%)
In summer 25.5 ºC (78ºF), and relative humidity (30%-60%)
The design temperature for room air conditioning depends on the types of
human activiy
Living room (20-21) ºC
bedroom
(13-16) ºC
kitchen 16 ºC
offices 20 ºC
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shops 19 ºC
classroom ºC 17.
Air Conditioning
Heating, Ventilating and Air Conditioning (HVAC), generally means the
provision of an acceptable thermal environment within buildings. It includes
heating, cooling, humidifying, dehumidifying, filtering, and distribution of air at
suitable conditions for the maintenance of human comfort or for the undertaking
of a particular process.
In order to design an air conditioning system the appropriate heating, cooling,
and other environmental loads must first be calculated and a variety of other
factors such as initial and running costs, plant room and distribution space,
control requirements etc. must be assessed. In practice, there are a wide variety
of air conditioning systems available and selection is often dictated by factors
other than the air conditioning loads.
Air Conditioning Psychrometrics
Comfort conditions are maintained within a building by supplying conditioned
air at an appropriate state to overcome the incident heating and cooling loads.
The required state will have specified conditions of temperature and moisture
content and the processes required to achieve that supply state can be evaluated
by air conditioning psychrometrics.
The changes in psychrometric properties of a sample of air as it passes through
an air conditioning system can best be visualized by use of a psychrometric chart
which, for a given barometric pressure, presents the psychrometric properties of
moist air over a range of conditions. The standard psychrometric chart
as Figure 2. Any point on the chart is defined by two psychrometric properties
and from the chart all other properties of the state point can be determined. A
schematic representation of the various air conditioning processes is given
in Figure 3.
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Figure 2- The standard psychrometric
Figure 3- Schematic representation of air conditioning processes.
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Example 1: Design the capacity of AC for a house with 6 persons according to
the data given below:1- Outdoor air DBT 95 ºF, WBT 75 ºF, indoor air DBT 75 ºF, WBT 58 ºF,
mixing air DBT 76 ºF.
2- Ventilating air 20 cfm/person, daily change temperature (low)
3- Windows, regular single glass inside shading, 130 ft2 west in direction, 145 ft2
north in direction, 200 ft2 east in direction.
4- Wood door 88 ft2 , walls, masonry brick plastered 190 ft2 , ceiling 4 inch
concrete insulation light 2848 ft2, metal door 44 ft2 (U = 0.42).
5- Lighting and kitchen 2500 Watt, use temperature swing factor of 0.75.
Ans.
Calculating of heat gain:1- RSH=300 × 6= 1800 Btuh
2- Windows use table (10-4):
Area
HTM
RSH (Btuh)
E
200
89
17800
W
130
89
11570
N
145
31
4495
3-walls use table (10-3) = 190× 9.7= 1843 Btuh.
4- Ceiling use table (10-3) = 2848× 2.4= 6835 Btuh.
5- Metal door = 44× 0.42 × (95-75) = 369.6 Btuh.
6- Wood door = 88× 14 = 1232 Btuh.
7- Ventilating = 120 cfm × 22 = 2640 Btuh.
8- Light and kitchen = 2500 × 3.41= 8525 Btuh.
∑RSH = 1800+ (17800+11570+4495) + 1843+6835+369.6+1232+2640+8525
= 57109.8 Btuh × 0.75 = 42832.35 Btuh.
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RLH= 90 × 6= 540 Btuh
SHR =
SHR=
= 0.98
Exterior air or ventilation air (Q1) =20 × 6 = 120 cfm.
Mixing air (Q3) = exterior air (Q1) + returned air (Q2)
DBT mixing air=
[
(
)
DBT mixing air = 76 ºF =
((
(
)] [
(
)
) (
(
(
)
(
(
)]
)
))
Q3 = 2400 cfm
Q2 = 2400-120 = 2280cfm.
WBT mixing air=
[
WBT mixing air =
(
(
)
) (
)
(
)] [
(
)
(
(
)
(
)]
)
= 58.85 ºF.
From psychrometric chart:For mixing air DBT= 76 ºF and WBT= 58.8 ºF
ΔT=
E = 25.7 Btu/lb
= 17.84
=
DBT of A.C = 75-17.84 = 57.16
From psychrometric chart:DBT of A.C = 57.16
, SHR= 0.98
WBT of A.C= 46.4 ºF, E =18.4 Btu/lb
Btuh (A.C) = 60 × 0.074 × Q3 × (E before A.C-E after A.C)
= 60 0.074
(25.7-18.4) = 77788.8 Btuh
1 ton = 12000 Btu
A.C =
= 6.5 ton
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Example 2: If total room sensible heat gain (RSH) = 90000Btuh, latent heat gain
(RLH) = 300 Btuh/person, sensible heat ratio 0.75, ventilation rate 1500 cfm,
outdoor air DBT= 95 ºF, WBT =75 ºF, indoor air DBT= 75 ºF, WBT =62.5 ºF,
A.C air DBT= 60 ºF and E =22.5 Btu/lb, find the following;1- No. of persons in the room.
2- Ventilating rate / person.
3- Quantity of air returned.
4- Capacity of A.C required.
Ans.
SHR =
, RLH= 30000 Btuh.
0.75 =
RLH = latent heat/ person
No. of persons
No. of persons = 30000/ 300 = 100 person.
Ventilation rate/ person = 1500/ 100 = 15 cfm/ person.
Total air quantity required (Q3) =
=
(
= 6000 cfm.
)
Quantity of air returned (Q2) = 6000-1500 = 4500 cfm.
DBTmixing air=
[
(
DBTmixing air =
WBTmixing air=
(
[
WBTmixing air =
)
) (
(
(
)] [
(
)
)
)
) (
(
(
)
(
(
)]
)
= 80 ºF
(
)] [
(
)
)
(
(
)
(
)
= 65.62 ºF.
From psychrometric chart E = 30.5 Btu/lb.
Btuh (A.C) = 60× 0.074× Q3× (E before A.C-E after A.C)
Heat to be removed = 60× 0.074 6000 (30.5 – 22.5) = 211392 Btuh.
Capacity of A.C = (211392/ 12000) = 17.62 ton.
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Mechanism of cooling and heating
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Electric Services
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Electric services
Ohm's Law deals with the relationship between voltage and current in an
ideal conductor. This relationship states that:
The potential difference (voltage) across an ideal conductor is
proportional to the current through it.
The constant of proportionality is called the "resistance", R.
Ohm's Law is given by:
V=IR
where V (volt) is the potential difference between two points which include a
resistance R(Ohm's ). I (ampere) is the current flowing through the resistance.
Question
A source of 6.0V is connected to a purely resistive lamp and a current of 2.0
amperes flows. What is the resistance of the lamp?
V=IR
R=V/I
R = 3.0
Power Factor
In AC circuits, the power factor is the ratio of the real power that is used to do
work and the apparent power that is supplied to the circuit.
Electrical Power formulas in Single Phase AC Circuits
Real Power(watt) = V x I x (PF)
where
PF = cos Φ = power factor (0.7 - 0.95)
For pure resistive load: PF = cos Φ = 1
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
resistive loads converts current into other forms of energy, such as heat

inductive loads use magnetic fields like motors.
apparent power= KVA = V x I / 1000
Electrical Power formulas in Three Phase AC Circuits
P(watt) =√3 x V x I x PF ,
√3=1.732
KVA = V x I x 1.73/1000
*Where
I = Current in Amperes (A)
V = Voltage in Volts (V)
P = Power in Watts (W)
R = Resistance in Ohm (Ω
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Example:
An operating three phase motor has voltages measured with a voltmeter on
each phase of 400 volts, amperage measured on each phase with an ammeter
are 14.1, 13.9, and 13.8 amps, power factor was measured as 0.82 compute
power and KVA.
Voltage = (400+400+400) / 3 = 400V
Ampacity =(14.1 + 13.9 + 13.8) / 3 = 13.9A
Power = (400V x 13.9A x 0.82 x 1.732) / 1000 = 7.89 Kwatts
KVA=(400V x 13.9A x 1.732) / 1000= 9.63 KVA
Typical load connection on 3-phase 240/400 Volts system
Single phase connection : phase to neutral 240V (2-wires)
Single phase connection : phase to phase 400V (2-wires)
Single phase connection : (240/400)V (3-wires)
Three phase connection : phase to phase 400V ( 3wires)
Three phase connection: (240/400)V (4wires)
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Show the connection of the following appliances
1 - single phase 240V motor
2- 3- phase 400V motor
3- heater single phase 400V
4- air conditioner (240/400)V 3- phase
5- air conditioner (240/400)V single phase
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Branch Circuit
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Branch Circuits
The portion of the wiring system extending past the final overcurrent device. These
circuits usually originate at a panel and transfer power to load devices.
Any circuit that extends beyond the final overcurrent protective device is called a
branch circuit. This includes circuits servicing single motors (individual) and circuits
serving many lights and receptacles (multi wire). Branch circuits are usually low
current (30 amps or less), but can also supply high currents. The purposes of this
division are:a)
To prevent danger in the event of a fault by insuring that the fault current needed
to operate the protective device is reasonably low.
b)
To enable parts of the installation to be switched off for testing and maintenance
without effecting others.
c)
To prevent a fault on one part from shutting down the complete installation.
A basic branch circuit as shown below is made up of conductors extending from the
final overcurrent protective device to the load. Some branch circuits originate at safety
switches (disconnects), but most originate at a panel board. The following are several
branch circuit classifications:
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• Individual branch circuit—A branch circuit that supplies a single load.
• Multi outlet branch circuit—A branch circuit with multiple loads.
• General purpose branch circuit—A multi outlet branch circuit that supplies
multiple outlets for appliances and lighting.
Circuits of Socket Outlets
Plugs and socket outlets conforming with BS1363 & BS546 are of the following
sizes:- 2; 5; 13; 15, 30 Amp ( The 13 A size is always fused )
The above sizes are very widely used for domestic and commercial installations.
In addition, other sizes ( 16, 32, 63, 125 amp.) conforming to BS4343 are
available for industrial applications. The ring and radial circuits are both used for
providing power supply to socket outlets. The following guidelines must be taken into
account: Floor area served should not exceed the values indicated on diagrams for the type of
circuit used.
Circuit breaker (or fuse) and wire sizes depend on the circuit used and shall be as
given on diagrams.
 A separate circuit (ring or radial) should be provided for the kitchen.
 If two or more ring circuits are installed in the same location, the socket outlets should
be reasonably shared among them.
 Various appliances can be individually protected by a fuse of a suitable rating (2, 3, 5,
or 10 amp.) installed in the plug. Such a fuse will protect the plug, the flexible cord
and the appliance connected to it.
 Permanently connected appliances with a relatively high power demand (such as
water heaters, window-type air conditioners, split-unit type air conditioners, electric
cookers, etc, should not be connected to a ring or radial circuits, they must be
supplied by their own separate circuits.
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Various authorities have given recommendations concerning the appropriate number
of socket outlets to be installed in various locations:-
Location
Number of outlets
Kitchen
4-5
Living Room
3-4
Dining Room
1-2
Master Bedroom
3-4
Single Bedroom
2-3
Landing/Stairs/Hall
1-2
Garage
1
Store/Workroom
1
Lighting Circuits
 Circuits intended to provide power supply to lighting fixtures are called
Lighting Branch Circuits.
 Circuits supplying lighting fixtures with some other small appliances such as
electric clocks, time stamps, bell transformers, shaver sockets, ceiling & exhaust
fans, etc…, are ALSO called Branch Circuits.
 NEC permits loading of a circuit to its ampere rating, However, it is a better
design practice to limit the circuit loading to 80% of its rating.
 When a lighting branch circuit is protected by 7.5 or 10 amp. CB or Fuse, the
wire size that should be used in 1.5 mm2 PVC insulated copper conductor.
 When a lighting branch circuit is protected by 15 or 20 amp. CB or Fuse, the
wire size that should be used is 2.5 mm2 PVC insulated copper conductor.
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 In all cases the lighting branch circuit must be assumed to require the connected
load, but with a minimum of 100W for each light-point outlet (0.46 amp. Per
outlet).
Branch-Circuit Rating
A branch circuit is sized for the load it will supply. Sizing the circuit for additional
future loads is good practice. The rating of a branch circuit depends on the rating of
the overcurrent device protecting the circuit. Branch circuits serving only one device
can have any rating, while a circuit supplying more than one load is limited to ratings
of 15, 20, 30, 40, or 50 amps.
Fractions of an Ampere. Except where the calculations result in a major fraction of
an ampere (0.5 or larger), such fractions are permitted to be dropped (NEC 2017) .
Conductor Size and Ampacity
The amperage rating of branch-circuit conductors must be greater than the maximum
load the circuit will provide. Where a branch circuit supplies continuous loads or any
combination of continuous and noncontinuous loads, the minimum branch-circuit
conductor size shall have an allowable ampacity not less than the noncontinuous load
plus 125 % of the continuous load, or the conductor operate at 80% of their rating at
continuous loading (NEC 2017).
Continous Load: A load where the maximum current is expected to continue for 3
hours or more like electric space heating equipments, air conditioners, refrigerators,
motors (NEC 2017).
Overcurrent Protective Device (OCPD)
Where a branch circuit supplies continuous loads or any combination of continuous
and noncontinuous loads for non-motor loads, the rating of the overcurrent device
(circuit breaker or fuse) shall not be less than the non-continuous load plus 125
percent of the continuous load.
OCPD for motors
Motors except Hermetically Sealed 250% of motor full load current (NEC430-52)
Motors Hermetically Sealed, air conditioners and heat pumps not exceeding 175%225% of motor full load current NEC 440.22-2017.
Fig 3.5 contains circuit breaker standard size.
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Example 1:
Load Furnace 5 KW works for 4 hours continuously, I phase, 220V, find current,
KVA, PVC wire size, circuit breaker size, P.F. = 1.0
I = KW/ KV = 5KW/.23KV = 22.7 Amp
Wire Size
Wire Size = Load/ 0.8 = ampacity = 22.7 Amp /0.8 = 28.375 Amp
Or 22.7 × %521 =28.375 Amp
from (table C.52.1 of IEC 60364-5-52), use (2× 6 mm2), 2pvc wire in conduit
Circuit Breaker Sizing
CB =22.7× %521 =28.375 Amp
Next higher CB size is 32 Amp. see Fig 3.5
Example 2
Load: Unit Heater 10 KW, 400 V, 3 phase, find I, PVC wire size, circuit breaker size
I= 10×1000/(400√3)= 14.43 Amp
Wire Size: 14.43 A× 1.25 = 18 Amp
(table C.52.1 of IEC 60364-5-52), Use 3×4 mm2 , 3pvc wire in conduit
Circuit Breaker = 14.43 A× 1.25 = 18 Amp, so use 20 Amp. 3 pole
Example 3
Branch circuit includes a 9 Amp. microwave with a 4 Amp. Refrigerator, find PVC
conductor size and circuit breaker size.
Circuit ampacity= ( 9 × 100% + 4 × 125%)= 14 Amp.
From (table C.52.1 of IEC 60364-5-52), use 2× 2.5 mm2, 2pvc wire in conduit
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Circuit breaker rating = ( 9 × 100% + 4 × 175%)= 16 Amp.,
From Fig 3.5 use 16 Amp. circuit breaker.
Example 4:
Load: Motor 3 HP( I = 17A), 1 phase, 230V find PVC wire size, circuit breaker size
Wire Size = 17A× 1.25= 21.5A
From (table C.52.1 of IEC 60364-5-52), use (2× 4 mm2), 2pvc wire in conduit
CB size = (I)( 2.5) NEC 430-52
CB Size = (17 A) (2.5) = 42.5A USE next size up 50A circuit breaker.
Example5: For an air-conditioning unit with a three phase, 230-volt, (15 amps), what
is the required size of the branch-circuit wire and overcurrent protective device?
Solution: This load must be increased by 125% for the conductor ampacity:
Conductor ampacity = 15 A × 1.25= 18.75 A
From (table A.52.4 of IEC 60364-5-52), use (3×4 mm2), 3pvc wire in conduit
Circuit Breaker = 15 A × 1.75 = 26 Amp so use next size up 32 Amp. 3 pole
Example 6: a branch circuit consist of 3KW electric space heating and 20 Amp. air
conditioner find PVC wire size and circuit breaker required.
Ampacity of electric space heating = 3000/220 = 13.6
Air conditioner ampacity is larger than electric space heating ampacity, so omit
electric space heating
The ampaciy of wire= (1.25×20) =25 Amp.
From (table C.52.1 of IEC 60364-5-52)), use (2× 6 mm2), 2pvc wire in conduit.
CB size = (1.75×20) =35 Amp., from fig 3.5 use 40 Amp. CB
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Voltage Drop of Cables
The impedance of circuit conductors is low but not negligible: when carrying load
current there is a voltage drop between the origin of the circuit and the load terminals.
The correct operation of a load (a motor, lighting circuit, etc.) depends on the voltage
at its terminals being maintained at a value close to its rated value. It is necessary
therefore to determine the circuit conductors such that at full-load current, the load
terminal voltage is maintained within the limits required for correct performance.
Maximum voltage drop limit
Typical values of maximum allowable voltage-drop for LV installations are given
below in table G25.
Table. G25: Maximum voltage-drop between the service-connection point and the
point of utilization (IEC60364-5-52 table G.52.1)
These voltage-drop limits refer to normal steady-state operating conditions and do
not apply at times of motor starting.
The value of 8 %, while permitted, can lead to problems for motor loads; for example:
a- In general, satisfactory motor performance requires a voltage within ±5 % of its
rated nominal value in steady-state operation,
b- Starting current of a motor can be 5 to 7 times its full-load value (or even higher). If
an 8 % voltage drop occurs at full-load current, then a drop of 40 % or more will occur
during start-up. In such conditions the motor will either:32
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1- Stall (i.e. remain stationary due to insufficient torque to overcome the load torque)
with consequent over-heating and eventual trip-out
2- Or accelerate very slowly, so that the heavy current loading (with possibly
undesirable low-voltage effects on other equipment) will continue beyond the normal
start-up period.
Finally an 8 % voltage drop represents a continuous power loss, which for
continuous loads will be a significant waste of (metered) energy. For these reasons it is
recommended that the maximum value of 8 % in steady operating conditions should
not be reached on circuits which are sensitive to under-voltage problems (see Fig.
G26).
Calculation of Voltage Drop
Voltage drop can be calculated by using Figure G28, which gives, with an adequate
approximation, the phase-to-phase voltage drop per km of cable per ampere, in terms
of:
1- Kinds of circuit use: motor circuits with cos ϕ close to 0.8, or lighting with a cos ϕ
close to 1.
2- Type of circuit; single-phase or 3-phase
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Voltage drop in a cable is then given by:
ΔU=K x IB x L
ΔU is Voltage drop in volts
K is given by the table, V/A/Km
IB is the full-load current in amps,
L is the length of cable in km.
The column motor power “cos ϕ = 0.35” of Figure G28 may be used to compute
the voltage drop occurring during the start-up period of a motor.
The % of voltage drop can be calculated as follow:Phase/phase:- ΔU% =
˟100%
Phase/neutral:-ΔU% =
˟ 100%
Balanced 3-phase: (3 phases (with or without neutral)): ΔU% =
˟100%
Where:
Un: phase-to-phase voltage
Vn: phase-to-neutral voltage
Example1: A single-phase lighting circuits 240 V, of 1.5 mm2 c.s.a. copper 25 m
long, and passing 15 A. Examine the adequacy of wire for this circuit.
Ans.
From Figure G28, K= 32 V/A/Km
ΔU= K x IB x L= 32 x 15 x (25/1000) = 12 V phase to neutral
ΔU% =
˟ 100% =
˟ 100% = 5% < 3% not ok, use next larger wire
For 2.5mm2, K= 19 V/A/Km
ΔU= 19 x 15 x (25/1000) = 7.125 V phase to neutral
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ΔU% =
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˟ 100% = 2.9% < 3% ok,
use 2.5 mm2 wire
Example 2
A 3-phase 4-wire copper line of 70
mm2 c.s.a. and a length of 50 m passes
a current of 150 A. The line supplies 3
single-phase lighting circuits, each of
2.5 mm2 c.s.a. copper 20 m long, and
each passing 20 A. as shown in figure.
It is assumed that the currents in the 70
mm2 line are balanced and that the three lighting circuits are all connected to it at the
same point. What is the voltage drop at the end of the lighting circuits?
Ans.
1-Voltage drop in the 4-wire line:
Balanced 3-phase: ΔU% =
˟100%
Figure G28 shows K=0.59 V/A/km
ΔU line = 0.59 x 150 x 0.05 = 4.4 V phase-to-phase
which gives:
= 2.54 V phase to neutral.
2- Voltage drop in any one of the lighting single-phase circuits:
Figure G28 shows K=19 V/A/km
ΔU for a single-phase circuit = 19 x 20 x 0.02 = 7.6 V
The total voltage drop is therefore
7.6 + 2.54 = 10.1 V
ΔU% =
˟100% =
˟100% = 4.2% < 6%
This value is satisfactory, being less than the maximum permitted voltage drop of 6 %.
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Example 3
A three-phase 25 mm2 copper cable 50 metres long connected from a LV public power
distribution network supplies a 400 V motor taking:
a- 100 A at a cos ϕ = 0.8 on Normal service load
b- 500 A at a cos ϕ = 0.35 during start-up
What is the voltage drop of motor cable at:
1- Normal service load?
2- During start-up?
Ans.
1- Voltage drop in Normal service load:
Figure G28 shows K= 1.4 V/A/km so that:
ΔU for the cable = 1.4 x 100 x 50/1000= 7 V
ΔU% =
˟100% =
˟100% = 1.75 % < 5%, ok, it is satisfactory
2- During start-up
Figure G28 shows 0.7 V/A/km so that:
ΔU for the cable = 0.7 x 500 x 50/1000= 17.5 V
ΔU% =
˟100% =
˟100% = 4.37% < 5%, it is satisfactory during motor starting
Example 4:
Somebody wants to use local generator to supply 6A, 220V single phase 1 ton air
conditioner in his house by using of 2.5mm2 copper wire for a distance of 100 meter.
Examine the adequacy of wire for this circuit. Assume start-up current 5 x IB.
1- Voltage drop in Normal service load:
For 2.5mm2, K= 15.3 V/A/Km
ΔU= 15.3 x 6 x (100/1000) = 9.18 V phase to neutral
ΔU% =
˟ 100% = 4.17% < 8 % ok,
2-Voltage drop during start-up
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For 2.5mm2, K= 6.8 V/A/Km
ΔU= 6.8 x ( 6 x 5) x (100/1000) = 20.4 V phase to neutral
ΔU% =
˟ 100% = 9.27% < 8 %, not ok, use next larger wire (4 mm2 )
1- Voltage drop in Normal service load:
For 4 mm2, K= 9.6 V/A/Km
ΔU= 9.6 x 6 x (100/1000) = 5.76 V phase to neutral
ΔU% =
˟ 100% = 2.61% < 8 % ok,
2-Voltage drop during start-up
For 4 mm2, K= 4.3 V/A/Km
ΔU= 4.3x ( 6 x 5) x (100/1000) = 12.9 V phase to neutral
ΔU% =
˟ 100% = 5.8% < 8%, ok,
Using of (4 mm2 ) wire it is satisfactory during motor starting.
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Maximum Demand: It is the greatest of all demands which have occurred
during a specified period of time. It is usually expressed in kW, kVA, Amp.,
etc.
Demand Factor :- The ratio of the maximum coincident demand of a system,
or part of a system, to the total connected load of the system.
Demand factor (in IEC, Max. Utilization factor (Ku))
Demand Factor is express as a percentage (%) or in a ratio (less than 1).
Demand factor is always < =1
The lower the demand factor, the less system capacity required to serve the
connected load.
Example1: an oversized motor 20 Kw drives a constant 15 Kw load whenever
it is ON. The motor demand factor is then 15/20 =0.75= 75 %.
Example2: if a residence having 6000W equipment connected has a maximum
demand of 3300W, then demand factor = 3300W / 6000W = 55%.
(1) A Residence Consumer has 10 No’s Lamp of 400 W but at the same time It
is possible that only 9 No’s of Bulbs are used at the same time. Here Total
Connected load is 10×40=400 W. Consumer maximum demand is 9×40=360
W. Demand Facto of this Load = 360/400 =0.9 or 90%.
(2) One Consumer have 10 lights at 60 Kw each in Kitchen, the load is 60 Kw x
10 = 600 KW. This will be true only if All lights are Turns ON the same time
(Demand factor=100% or 1)
For this Consumer it is observed that only half of the lights being turned ON
at a time so we can say that the demand factor is 0.5 (50%). The estimated load
= 600 Kw X 0.5 = 300 Kw.
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Use of demand factors:
Feeder conductors should have sufficient Ampere Capacity to carry the load.
The Ampere Capacity does not always be equal to the total of all loads on
connected branch-circuits. The demand factor permits a feeder-circuit ampere to
be less than 100% of the sum of all branch-circuit loads connected to the feeder.

This factor must be applied to each individual load, with particular
attention to electric motors, which are very rarely operated at full load.

Demand factor can be applied to calculate the size of the sub-main which
is feeding a Sub panel or a fixed load like a motor etc. If the panel have
total load of 250 kVA, considering a Demand factor of 0.8, we can size
the feeder cable for 250 x 0.8= 200 kVA.

Demand factors for buildings typically range between 50 and 80 % of the
connected load.

In an industrial installation this factor may be estimated on an average at
0.75 for motors.

For incandescent-lighting loads, the factor always equals 1.
(2) Diversity factor / (in IEC, simultaneity factor (Ks))
Diversity Factor is ratio of the sum of the individual maximum demands of
the various sub circuit of a system to the maximum demand of the whole
system.
Diversity Factor = Sum of Individual Maximum Demands / Maximum Demand
of the System.
Diversity Factor = Installed load / Running load.
The diversity factor is always >= 1.
Diversity Factor is always >1 because sum of individual max. demands >Max.
demand.
In other terms, Diversity Factor (0 to 100%) is a fraction of Total Load that is
particular item contributed to peak demand. 70% diversity means that the
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device operates at its nominal or maximum load level 70% of the time that it is
connected and turned ON.
It is expressed as a percentage (%) or a ratio more than 1.
If we use diversity value in % than it should be multiply with Load and if we
use in numerical value (>1) than it should be divided with Load.
 Diversity occurs in an operating system because all loads connected to the
system are not operating simultaneously or are not simultaneously operating
at their maximum rating.
 Diversity factor is an extended version of demand factor. It deals with
maximum demand of different units at a time/Maximum demand of the entire
system.
Greater the diversity factor, lesser is the cost of generation of power.
Example I: One Main Feeder have two Sub feeder (Sub Feeder A and Sub
Feeder B), Sub Feeder-A have demand at a time is 35 KW and Sub Feeder-B
have demands at a time is 42 KW, but the maximum demand of Main Feeder is
70 KW.
Total individual Maximum Demand =35+42=77 KW.
Maximum Demand of whole System=70 KW
So Diversity factor of The System= 77/70 =1.1
Example-II: A sub-station has three outgoing feeders:
1- Feeder 1 has maximum demand 10 MW at 10:00 am,
2- Feeder 2 has maximum demand 12 MW at 7:00 pm and
3- Feeder 3 has maximum demand 15 MW at 9:00 pm,
4- While the maximum demand of all three feeders is 33 MW at 8:00 pm.
Ans.
the sum of the maximum demand of the individual sub-systems (feeders) is 10
+ 12+ 15 = 37 MW, while the system maximum demand is 33 MW. The
diversity factor = 37/33 = 1.12.
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Example III :-Two factories A & B with load curves indicated are supplied by
the same feeder. Calculate the load diversity factor between the two factories
S( kVA)
S (kVA)
Factory B
380
Factory A
300
240
200
120
100
0
6
12
18
24
Time (hr)
0
6
12
18
24
Time (hr)
Diversity factor =
Calculate Size of Electrical cable and CB by Demand & Diversity Factor:
 The estimated electrical demand for all feeders served directly from the
service entrance is calculated by multiplying the total connected loads by their
demand factors and then adding all of these together. This sum is divided by
the diversity factor to calculate the service entrance demand which is used
to determine ampacity requirements for the service entrance conductors.
 When used Diversity and Demand Factor in an electrical design it should be
applied as follows, the sum of the connected loads supplied by a feeder-circuit
can be multiplied by the demand factor to determine the load used to size the
components of the system.
 The sum of the maximum demand loads for two or more feeders is divided
by the diversity factor for the feeders to derive the maximum demand load.
Example-1: Suppose We have four individual feeder-circuits with connected
loads of 250 kVA, 200 kVA, 150 kVA and 400 kVA and demand factors of
90%, 80%, 75% and 85% respectively. Use a diversity factor of 1.5.
Calculating demand for feeder-circuits
250 kVA x 90% = 225 kVA
200 kVA x 80% = 160 kVA
150 kVA x 75% = 112.5 kVA
400 kVA x 85% = 340 kVA
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The sum of the individual demands is equal to 837.5 kVA.
However, using the diversity factor of 1.5, the kVA = 837.5 kVA ÷ 1.5 = 558
kVA for the main feeder.
Example-2: For network shown below Calculate the following;1- Size of main circuit breaker and main feeder PVC wire.
2- Size of sub main circuit breaker and sub feeder PVC wire.
Use power factor of 0.8 for all loads.
Maximum Demand of Consumer 1 =32 KW x 0.65 =20.8 KW
As Diversity of Consumer 1 is 1.5 so,
Maximum Demand on sub feeder 1 =20.8 KW/1.5 = 13.86 KW.
Sub feeder 1 current =
√
Select a 25 A triple-pole circuit breaker for Sub- MDB 1
Select from the table a cable size of (4x6) mm2 for sub feeder 1
Sum of Maximum Demand of Consumer 2 =40 KW x 0.75 =30 KW
As Diversity of Consumer Connected on sub feeder 2 is 1.1 so,
Maximum Demand on sub feeder 2 =30 KW/1.1 = 27.27 KW
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Electrical loads
Sub feeder 2 current =
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√
Select a 50 A triple-pole circuit breaker for Sub- MDB 2
Select from the table a cable size of (4x16) mm2 for sub feeder 2
Sum of Maximum Demand of Customer 3 =50 KW x 0.65 =32.5 KW
As Diversity of Consumer Connected on sub feeder 3 is 1.5 so,
Maximum Demand on sub feeder 3 = 32.5 KW/1.5 = 21.66 KW
Sub feeder 3 current =
√
Select a 40 A triple-pole circuit breaker for Sub- MDB 3
Select from the table a cable size of (4x10) mm2 for sub feeder 3
Maximum Demand on Main Feeder =13.86 +27.27 +21.66 / 1.3 =48.3 KW
Main feeder current =
√
Select a 100 A triple-pole circuit breaker for MDB
Select from the table a cable size of (4x25) mm2 for main feeder
Example-3
A three story residential building to be supplied from a unit substation installed
at the electrical room at the ground floor level. Each floor level is composed of
three apartments with maximum demand of 15 kW each. The power factor of all
building is 0.9.
Assume a diversity factor of 1.3 between apartments at each floor level.
Assume a diversity factor of 1.4 between loads of various floors.
Assume individual floor supply with the sub-main distribution boards
1-Calculate pvc wire and circuit breaker size for each apartment
2-Calculate pvc wire and circuit breaker size for each floor
3-Calculate size of service entrance PVC wire and circuit breaker size for main
distribution boards
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Electrical loads
Apartment current =
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√
Select a 25 A triple-pole circuit breaker for each apartment
Select from the table a cable size of (4x6) mm2 for each apartment
Total floor kW =
Floor current =
√
Select a 63 A triple-pole circuit breaker for each floor
Select from the table a cable size of (4x25) mm2 for each floor
Total building kW =
Total building current =
√
Select a 125 A triple-pole circuit breaker for main distribution boards
Select from the table a cable size of (4x35) mm2 for service entrance pvc wire.
Example-4
Calculate size of main circuit breaker and a service entrance PVC Aluminium
wire size for a house supplied with 230 V, single phase had a panel board
containing the following branch circuits:
1- Lighting circuit (10×40 watt) florescent fixture with power factor of 0.8 and
demand factor of 1.
2-Socket outlets (15×13 Ampere) with power factor of 0.8 and demand factor of
0.1
3- Water pump 0.5 hp with power factor of 0.8 and demand factor of 0.75
4- 2 air cooler 500 watt each one with power factor of 0.8 and demand factor of
0.75
5- Oven of 1300 watt with power factor of 1 and demand factor of 1.
6- Water heater 2000 watt with power factor of 1 and demand factor of 1.
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7- 2 Air conditioner 3200 watt with power factor of 0.8 and demand factor of
0.75
Assume a diversity factor of 1.4 between loads of various branch circuits.
Ans.
1- Max. demand of lighting = [(10 × 40)/0.8] × 1 =320 VA
2- Max. demand of socket outlets=(15 × 13 Ampere)×230×0.1= 4485 VA
3- Max. demand of water pump = [(0.5 × 746 Watt)/0.8] × 0.75= 350 VA
4- Max. demand of air coolers = (2 × 500/0.8) × 0.75= 937.5 VA
5- Max. demand of oven load= (1300/1) × 1= 1300 VA
6- Max. demand of water heater= (2000/1) × 1= 2000VA
7- Max. demand of AC =[ (2 × 3200)/0.8] × 0.75= 6000VA
Total Max. demand = 320+4485+350+937.5+1300+2000+6000=15392.5
As diversity factor between branch circuits is 1.4, so
Max. demand =15392.5/1.4 =10994.6VA
House current= 10994.6/230 = 48 Amp.
From fig.3.5 choose 50 Amp. single pole circuit breaker.
From table.G21a choose (2×10) mm2, PVC Aluminium wire for service
entrance cable.
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Zonal Cavity Method for Lighting Design
The Zonal Cavity Method (sometimes called the Lumen Method) is the
currently accepted method for calculating average illuminance levels for indoor
areas, unless the light distribution is radically asymmetric. It is an accurate hand
method for indoor applications because it takes into consideration the effect that
inter-reflectance has on the level of illuminance. Although it takes into account
several variables, the basic premise that footcandles are equal to luminous flux
over an area is not violated. The basis of the Zonal Cavity Method is that a room
is made up of three spaces or cavities. The space between the ceiling and the
fixtures, if they are suspended, is defined as the “ceiling cavity”; the space
between the work plane and the floor, the “floor cavity”; and the space between
the fixtures and the work plane, the “room cavity.” Once the concept of these
cavities is understood, it is possible to calculate numerical relationships called
“cavity ratios,” which can be used to determine the effective reflectance of the
ceiling and floor cavities and then to find the coefficient of utilization.
Foot-candle:- is a unit of lighting which measures the intensity of light in relation
to its source. The unit foot-candle is defined as the amount of illumination the
inside surface of a one-foot-radius sphere would be receiving if there were a
uniform point source of one candela in the exact center of the sphere.
Alternatively, it can be defined as the illuminance on a one-square foot surface of
which there is a uniformly distributed flux of one lumen.
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Illuminance:- is a measure of how much luminous flux is spread over a given
area. (measured in lumens)
Basic steps in any calculation of illuminance level:
Step 1: Cavity ratios for a rectangular space may be calculated by using the
following formulas:
Where:
hcc = distance in feet from luminaire to ceiling
hrc = distance in feet from luminaire to work plane
hfc = distance in feet from work plane to floor
L = length of room, in feet
W = width of room, in feet
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Step 2: Effective cavity reflectances must be determined for the ceiling cavity and
for the floor cavity. These are located in Table (28.7) under the applicable
combination of cavity ratio and actual reflectance of ceiling, walls and floor. The
effective reflectance values found will then be ρcc (effective ceiling cavity
reflectance) and ρfc (effective floor cavity reflectance). Note that if the luminaire
is recessed or surface mounted, then CCR = 0 and ρcc = selected ceiling surface
reflectance. If the floor is the working plane, then FCR = 0 and ρfc = selected
floor surface reflectance.
Step 3: With these values of ρcc, ρfc, and ρw (wall reflectance), and knowing the
room cavity ratio (RCR) previously calculated, find the coefficient of utilization in
the luminaire coefficient of utilization (CU) table (28.9). Note that since the table
is linear, linear interpolations can be made for exact cavity ratios and reflectance
combinations. The coefficient of utilization found will be for a 20% effective floor
cavity reflectance. Thus, it will be necessary to correct for the previously
determined ρfc. As the following:
1-For %01> ρfc >30% effective floor cavity reflectance, divided standard
CU value by the appropriate factor from table (28.8).
2-For ρfc <30% effective floor cavity reflectance, multiply standard CU
value by the appropriate factor from the table (28.8).
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Step 4: Computation of the illuminance level is performed using the standard
Lumen Method formula.
No. of luminaires =
For large areas, a much more useful figure is the area illuminated per luminaire:
area per luminaire =
Where:fc = Foot candle
CU= Coefficient of utilization.
MF= The total light loss factor.
MF = LLD× LDD
LLD = lamp lumen depreciation table (28.10)
LDD = luminaire dirt depreciation figure (28.26)
Step 5: Check S:MH ratio.
MH = mounting height (height from work surface to light source).
S = Max. space between luminaire centres.
SR (spacing ratio) =
The solution considered correct if calculated
is less than given
selected type of luminaire, if not select another type of luminaire.
Example 1: a typical classroom with a dimension of 25×20×12 ft, a required
footcandles = 70, final reflectance: ceiling 80%, wall 50%, floor 10%. Use type
49
for
Lighting Design
Shubbar Jawad
32, 30 in stem mounting, fluorescent type 40 watt, T12, 4 ft with an initial lumen
output= 3250 per lamp. Working plane at 30 in AFF.
Ans.
hcc= 2.5 ft (30 in stem mounting)
hfc= 2.5 ft (30 in AFF)
hrc= 12-2.5-2.5=7 ft
CCR= 5hcc
= 5(2.5)
FCR= 5hfc
= 5(2.5)
RCR= 5hrc
= 5(7)
=1.1
=1.1
= 3.2
From table 28.7
ρcc=65%, ρw= 50% (given), ρfc = 11%
from table 28.9, type 32,
using ρw= 50% and ρfc=20%, we must double interpolate for RCR and ρcc
ρcc
RCR
RCR
70
0.5
0.45
Interpolating first for RCR of 3.2
CU (70) = 0.5 - 0.05(0.2) = 0.49
CU (50) = 0.42- 0.04(0.2) = 0.41
For RCR = 3.2, ρw = 50
ρcc
70
50
CU
0.49
0.41
50
50
0.42
0.38
Lighting Design
Shubbar Jawad
Interpolating for ρcc = 0.65, we have
CU = 0.49 -
(0.08) = 0.49 – 0.02 = 0.47
Final CU = 0.47 for ρfc = 20%
For ρfc = 11%
since 01 %> ρfc >30%, divided standard CU value by the appropriate factor
from table (28.8) which is 1.044
So, corrected CU =
= 0.45
From table 28.10 LLD = 0.86
From fig. 28.26 LDD = 0.86 (12 months, dirty, category II)
So, MF = 0.86
0.86 = 0.74
Area / luminaire =
= 31ft2
=
No. of luminaires =
SR (spacing ratio) =
= 16 luminaires
==
= 0.89 < 1.5, then ok.
51
Lighting Design
Shubbar Jawad
Example 2: A typical classroom with a dimension of 25×20×12 ft, a required
footcandles = 70, final reflectance: ceiling 80%, wall 50%, floor 10%. Use type
33, the luminaire is recessed, fluorescent type 40 watt, T12, 4 ft with an initial
lumen output= 3250 per lamp. Working plane at 30 in AFF.
Ans.
hcc = 0 ft (recessed or surface mounted)
hfc = 2.5 ft (30 in AFF)
hrc= 12-2.5 = 9.5 ft
CCR= 0
FCR= 5hfc
= 5(2.5)
=1.1
RCR= 5hrc
= 5(9.5)
= 4.3
Since CCR= 0, then ρcc= 80%
From table 28.7
ρw= 50% (given), ρfc = 11%
From table (28.9) for type 33, by interpolating CU = 0.39, for ρfc= 20%
For ρfc = 11%
52
Lighting Design
Shubbar Jawad
since 10%< ρfc >30%, divided standard CU value by the appropriate factor from
table (28.8) which is 1.05
So, corrected CU =
= 0.37
From table 28.10, LLD = 0.86
From fig. 28.26 LDD = 0.73, (12 months, dirty, category IV)
So, MF = 0.86
0.73 = 0.63
= 22 ft2
Area / luminaire =
No. of luminaires =
SR (spacing ratio) =
= 23 luminaires
==
= 0.53 < 1.2, then ok.
53