Rough Electrical Load calculations for electrical equipments.docx

Electrical Load calculation

1. TOPIC : INTRODUCTION TO BUILDING SERVICES-II
FACULTY : MOHAMMED KAUSAR IQBAL


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Type of Electricity Supply
 Alternating Current – AC - Normal Supply system
 Direct Current – DC - Batteries – Car, Flash Lights, Mobile Charges
(convert AC to DC) – Invertors & UPS (Battery charge is converted from
DC to AC)

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FACULTY : MOHAMMED KAUSAR IQBAL


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How do we use electricity :
We connect the source to the object of use :
Source is :
 DC – Supply on Battery
 AC Supply
Object of use :
 Lights / Lamps
 Fans
 AC’s
 Gadgets – Micro, Oven, Mixer Grinder, Washing Machine
 Computer and its periphery
How do we connect the source and object of use
 Wires
 Sockets, holders for lamps, Ceiling Rose

3. TOPIC : INTRODUCTION TO BUILDING SERVICES-II
FACULTY : MOHAMMED KAUSAR IQBAL


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How do we control :
 Switches

Circuits Series and Parallel :

Resistance is Series : R = R1+R2+R3.....Rn

Parallel is : 1/R = 1/R1 + 1/R2 + 1/R3 ....1/Rn

4. TOPIC : INTRODUCTION TO BUILDING SERVICES-II
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Power Formula :
 P = VI P= Power in Watts. V= Voltage is Volts & I=Current in Amps
 Loss of Energy in Circuit = I
2
R
Say in circuit the power consumption is 2300 Watts at 230 Volts, What is
the Current = P=VI or I=p/v or 2300/230 = 10 Amps
Power loss in circuit = I
2
R = 10x10xR or 100xR

Say in circuit the power consumption is 2300 Watts at 23 Volts, What is the
Current = P=VI or I=p/v or 2300/23 = 100 Amps

Power loss in circuit = I
2
R = 100x100xR or 10000xR

Sources of Electrical Energy :
DCV Source

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 Batteries
Alternating Current – AC – Direct Generation
 Thermal Power Station
 Hydro Power Station
 Nuclear Power
 Small Standalone Sets using Diesel or Bio Fuel – DG Sets etc

Alternating Current – AC – In-Direct Generation
 Solar Power – Here power is generated in DC and converted to AC for
use.

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Supply Voltages for buildings ;






 230 Volts - Single Phase ( 1Ø) - Phase & Neutral (P&N) +Earth (E) -
upto connected load of 5 KW
 Formula for 1Ø = “P=VxI
Where P= Total Power in Watts

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V = Voltage
I= Current in Amps
Total Connected Load = 6 KW
Supply Voltage = 230 Volts, 1Ø
Find Current ? so I= P/V = 6000/230 = 26.0869 Amps

 400 Volts - Three Phase (3 Ø) - R (Red) - Y (Yellow) - B(Blue) +N
(Neutral) +E(Earth) for connected load from 5+ KW to 20 KW
 Formula for 3Ø = “P=√3xVxI
Where P= Total Power in Watts
V = Voltage
I= Current in Amps
√3= 1.732
Total Connected Load = 6 K

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Supply Voltage = 400 Volts, 3Ø
Find Current ? so I= P/=√3xV = 6000/1.732x400 = 8.66 Amps

 *11000 Volts - Three Phase (3 Ø) - R (Red) - Y
(Yellow) - B(Blue) +N (Neutral) +E(Earth) for
connected load from +20 KW
11000 Volts is to be converted with the help of Step
Down Transformer 11/0.4KV) and then supply is
effected.
Requirements :
 Space for Transformer
 Space for Panels

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 Size for transformer depends on the capacity in KW

Phase to neutral Voltage is 230 Volts
Phase to Phase Voltage is 400 Volts.

Neutral is always Black
Earth is always Green

R-N , Y-N, B-N voltage is 230 Volts
R-Y, Y-B, B-R - 400 Volts - Motor Application

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Wiring with respect to connection

Wiring with respect to installation
 Surface
 Cleat Wiring - Temporary wiring
 Wooden Batten - More or less outdated - not used any longer
 PVC Casing Caping
 Pipes
 PVC Pipes - Light, Medium, High - Mechanical Stress
 Steel Pipes

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 Concealed
 Pipes
 PVC Pipes - Light, Medium, High - Mechanical Stress
 Steel Pipes


Wiring Materials
 Differentiate between wiring material and equipments


Various Wiring Materials:
 Meter and Main Switch - Various rating as per load of the building -
 Wires - Various sizes, types etc.
 Conduit Pipes

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 PVC
 Steel
 Conduit accessories such as bend, elbows, junctions, circular boxes,
bar and saddles, base boxes for switches
 Switches
 available in various Current rating for various equipments such as
lights, fans etc to be controlled by lower rating switches and AC,
Geyser etc to be controlled by higher rating switches.
 Normal rating are 5-6 Amps, 16-20 Amps, 32 Amps
 Function to MAKE or BREAK a circuit in other words ON-OFF or CLOSE-
OPEN - to control a circuit / equipment use - using mechanical force -
switch cannot operate automatically on it own
 Types
 Tumbler - now a days outdated

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 Piano Key Type switches mounted on Bakelite sheets
 Modular Switches - most commonly used now a days
 Sockets outlets to be used for various types of plugs -
 Use - for connection equipments such as kitchen equipments, Iron,
Washing machine, mobile chargers, TVs, AC,s Geysers, Computers etc.
 Normal rating are 5-6 Amps, 16-20 Amps
 serve at connection to various equipments
 Fan Regulators / Light Dimmers
 Holders - Angle & Batten
 Ceiling Rose

DISTRIBUTION BOARD

14. TOPIC : INTRODUCTION TO BUILDING SERVICES-II
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 FUSE
 What is fuse -
 Safety device
 Protecting against OVERLOAD & SHORTCIRCUIT which melts due
to heat thus breaking the circuit
 If fuse blows out then it has to be repalced.
 Distribution Board with MCB's or Miniature Circuit Breaker and protection
Devices - we have to study this in detail

15. TOPIC : INTRODUCTION TO BUILDING SERVICES-II
FACULTY : MOHAMMED KAUSAR IQBAL


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Two Bed room house :
 Drawing - Fan (Ceiling Rose or connection) , TV+ (Sockets outlets) ,
Lights -2 (Angle Holder / Batten holder, Ceiling Rose), Extra Multipurpose 5
Amps Sockets
 Dinning - Lights-2, Fan-1, Fridge, +Extra 5 Amps Sockets
 Kitchen - Lights, Chimney / Exhaust, Ceiling Fan, Kitchen Equipments -
Cluster of 3 to 4 Sockets - Heavy Duty (15 Amps)
 Toilets - Lights, Exhaust, Geyser
 Bed Room - Lights, Fan, Socket, Air Conditioners
 Master Bed Room with Toilet Light Fan Socket
 Balcony - Lights, Fans, Sockets,

16. TOPIC : INTRODUCTION TO BUILDING SERVICES-II
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Bring Main Supply to building
 Meter
 Main Switch
 Distribution Board (Fuse or MCB) - Number of Circuits
 Switch Boxes / Junction Boxes for each area


Fuse is more or less being replaced by MCB - Miniature Circuit Breaker

Switch - purpose to make or break circuit - mechanically
Circuit Breaker - Break automatically in case of faults but make
mechanically

17. TOPIC : INTRODUCTION TO BUILDING SERVICES-II
FACULTY : MOHAMMED KAUSAR IQBAL


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what is fault ?
 Overload - Drawing more than the rated current
 Short Circuit - Negligible resistance between Phase & Neutral in other
words the phase and neutral is joined
 Earth Fault - Shock from Electric Current
 Surge (Sudden abnormal rise of voltage for a very short period) -
Lightning from thunder storm / Operation of heavy equipment in nearby
area

OVERLOAD FAULT : More current flowing in circuit than the
designed load current but less that short circuit current.

18. TOPIC : INTRODUCTION TO BUILDING SERVICES-II
FACULTY : MOHAMMED KAUSAR IQBAL


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Heat in Electrical Circuit is = I
2
R whre I= Current & R=Resistance
In fault condition the circuit shall break.
How? - By melting of Fuse or automatic tripping / opening of
MCB (Miniature Circuit Breaker) thus breaking the Circuit.

MCB's can trip in OVERLOAD condition with a time lag
depending upon the percentage of Overload and
instantly in SHORT CIRCUIT

P=VI - for Single Phase Circuit or (P=VI CosØ)
P=√3VI for Three Phase Circuit (P=√3VI CosØ)
CosØ - Power Factor

19. TOPIC : INTRODUCTION TO BUILDING SERVICES-II
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P= Total Power in watts, I= Current in Amps & V= Voltage
MCB is an equipment to protect against Overload and Short Circuit
FAULTS.

EARTH FAULT - Current flowing to the earth through a body following
a least resistance path
 Connect the metallic body of the equipment (Fans, Kitchen
equipments, motors, etc. to the earth through a earth wire in 3 Pin
socket or directly to the body in case of industrial equipments and
Ceiling fan, light fitting etc.

Device for protection : ELCB (Earth Leakage Circuit Breaker) / RCCB
(Residual Current Circuit Breaker)

20. TOPIC : INTRODUCTION TO BUILDING SERVICES-II
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 Earthing - for protection - all devises are to be connected to earth
 Earth Resistance - shall be ideally ZERO
 Earthing Methods
 Devices for Earth Fault Protection
Type of Earthing : Earthing is governed by IS Code 3043 of 1987
and the types are :
1. Pipe Earthing
2. Plate Earthing
3. Mat / Grid Earthing

The Earth Resistance value shall be ideally ZERO but never more
than 5Ω (5 Ohms)

21. TOPIC : INTRODUCTION TO BUILDING SERVICES-II
FACULTY : MOHAMMED KAUSAR IQBAL


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Next Class we shall read
 Earthing Installation details as per IS
 Symbols for electrical drawing

Purpose of Earthing :
Since potential of Earth is ZERO, it acts as a safe passage for fault
current by connecting the body of the equipment to the earth
electrode thus providing safety from shock in case of faults.

GI Pipe 3 Meters long - 40 to 50mm in diameter or Cast Iron Plate
1200mmX1200mmX12mm or Copper plate of 600mmX600mmX6mm is
buried in the ground at a depth of 3 meters and the surrounding of the

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pipe / plate is filled with a mixture of Salt & Charcoal. It also has pipe
arrangement for watering to keep the ground moist.
Purpose of Salt & Charcoal :
 Salt - to increase conductivity
 Charcoal to retain moisture

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24. TOPIC : INTRODUCTION TO BUILDING SERVICES-II
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25. TOPIC : INTRODUCTION TO BUILDING SERVICES-II
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Protection from Lightning generated from Thunder Storm / Rain
 Lightning Arrestor - is installed on building / structure top and
connected to the earth / earthing pipe - It is not connected in the
electrical circuit - The lightning arrestor / spikes are installed on
top of the building to prevent STRUCTURAL DAMAGE to the
building - it does not provide protection to electrical equipments'
inside the building since both are not interconnected.
 The number of Lightning Arrestor / Spike is calculated as per
IS:2309 and the number depends on the following factors:
 Roof Top Area
 Height
 Number of Thunder Storm Days as per table in IS Code

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 It is governed by IS Code 2309 (1989) - PROTECTION OF BUILDINGS
AND ALLIED STRUCTURES AGAINST LIGHTNING - CODE OF
PRACTICE
Total Number of Major Items in Lightning Arrestor
1. Spike or Lightning Arrestor – TRISHUL
2. Conductor Connecting the LA and Earth
3. Earthing

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 Surge Protector - Is connected in line in the circuit and protects the
equipments - the equipment is SPD or Surge Protection Device. It
is connected between the live terminals and the earth. It is not
connected to the lightning arrestor.

 SPD or Surge Protection Device is connected in the main to protect
the Equipments from Surge spikes as follows :






SPD
Live Line

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One pole and Two Pole Switch :

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One pole switch - only connects / disconnects the phase -
Application - controlling equipments
Two pole switch - connects / disconnects the phase & Neutral ;
Application main incoming supply, Heavy equipments such as Ac,
Geyser etc.
Three pole switch - connects / disconnects all the three phases :
application is in motors / industrial
Four Pole Switch : connects / disconnects all the three phases and
Neutral : application is main incoming supply

Special Notes ;
 Two Pole or Four Pole switch is used to disconnect the neutral

30. TOPIC : INTRODUCTION TO BUILDING SERVICES-II
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 Equipment switch is always placed in phase why? To isolate
the equipment thus preventing any charge to the equipment
and ease of repair / maintenance.
 Equipment such as Air Conditioner / Geyser i.e. heavy
equipment shall be provided with Double pole switch.
 Providing Switch in the Phase is safety measure.

PROTECTION DEVICES :
 MCB - Miniature Circuit Breaker - Protection is Overload and Short
Circuit and with add on modules – for extra protection against
Under voltage & Over voltage
 Versions
 Single Pole(SP), - Controls Phase Only

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 Single Pole & Neutral (SP&N) - Controls Phase &
Neutral
 Double Pole (DP) - Controls Phase & Neutral
 Three Pole (TP) - Controls Three Phases only
 Three Pole & Neutral (TP&N) - Controls Three Phases &
Neutral
 Four Pole (FP) - Controls Three Phases & Neutral

 Ratings - upto 63 Amps in all poles for both Single Phase - 240
Volts and Three Phase - 400 Volts

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 RCCB (Residual Current Circuit Breaker) or ELCB (Earth Leakage
Circuit Breaker) - 2Pole and Four Pole Version available upto
63 Amps and 30mA, 100mA & 300mA versions.
 SPD - Surge Protection Device - 2 Pole and 4 Pole versions.


MCB - upto 63 Amps
P=√3VI for Three Phase Circuit - V=400Volts, I=63Amps P=?
P= 1.732x400x63 = 43646 Watts = 43.646 KW

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Beyond 40 KW for large loads we use
MCCB - Moulded Case Circuit Breaker or ACB - Air
Circuit Breaker
Versions are Three Pole for Motor Applications and Four Pole of other
application
Ratings are from 40 Amps to 2500 Amps (MCCB)
Ratings are from 400 Amps to 6000 Amps (ACB)

In short all have two words in common i.e. CIRCUIT BREAKER
31-01-2022 End

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 Various Electrical Equipment Appliance

 Various Electrical Installation Materials
 Switch
 Switch Board – Plate & Box
 Wires
 For Installation of Wires
 Wooden Batten (outdated)
 Casing Caping of PVC / Wooden
 Conduit – Metallic / Non Metallic (PVC)
 Junction Boxes with covers
 End Treminals such as
 Angle Holder

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 Batten Holder
 Ceiling Rose
 Plug Socket Outlets
 Distribution Board with MCB or Fuse
 Main Switch
 Meters

XXXXXXXXXX Next Class
Load Calculation
Load Factor
Tariff - Electricity Bill
Load Calculation :

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we are connecting a equipment in a circuit and put the switch
ON - what will happen
P=VI for single phase and
P=√3VI for Three Phase Circuit - V=400Volts, I=63Amps P=?

Power Consumption :
Watts = Volts x Amps – for Single Phase equipments

Watt-Hour
i.e. one 100 Watts lmap if ON for 10 Hours consumes 100Wattx10Hr
= 1000 Watts Hour or 1 KWH
Power x Time = Watt-Hour
1000 Watts Hour = 1 KWHr or 1 Unit.

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Say your fan runs for 10 Hours (Rating of Fan is 70 Watts)
Power Consumed in Units : 70Watt x 10 Hours = 700 Watt- hour
In Kilowatt-Hour = 0.7 KWh = 0.7 Units
Say rate of Energy per unit is Rs. 5.00 Therefore the Amount is
Rs. 5x0.7 Units = Rs. 3.50 - Bill

Say your AC runs for 10 Hours (Rating of AC is 1500 Watts)
Power Consumed in Units : 1500Watt x 10 Hours = 15000 Watt- hour
In Kilowatt-Hour = 15 KWh = 15 Units
Say rate of Energy per unit is Rs. 5.00 Therefore the Amount is
Rs. 5x15 Units = Rs. 75.00 - Bill

38. TOPIC : INTRODUCTION TO BUILDING SERVICES-II
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A 70 Watts Fans – What is the current – Voltage is 230V
P=VI or 70=230xI or I= 70/230 = 0.3 Amps

Formulas :
P = V x I or I = P / V

If V = 230 Volts and P = 2000 Watts
therefore I = P/V = 2000/230 = 8.69 Amperes

Current Consumption by Geyser :

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Rating of Equipment :
Label of any electrical equipment contains the following data
V= 230 Volts - +/- 6% or 180-250 Volts
F= 50Hz of 50cps
P= 2000 Watts
I = xxx Amps
Rating of various common electrical Equipment in WATTS :

1. Lamps - Starts from 3 Watts (LED) to 1000Watts (Halogen)
Standard ratings are LED Lamps/ Bulbs - 3W, 5W, 9W, 12W,
15WW-18W, 24W and more
GLS Lamps (Old Glass lamps) 15W, 25W, 40W, 60W, 100W and
250Watts

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CFL Lamps - Starts from 5 Watts to 70Watts
Tube Lights - 40 Watts normally
2. Fans - Ceiling Fans

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 LED Lamps are available from 3 Watts to 70 watts. Normal is 10-15
Watts.

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FACULTY : MOHAMMED KAUSAR IQBAL


SUBJECT CODE SEMESTER BATCH DATE

KiiT SCHOOL OF ARCHITECTURE & PLANNING
BUILDING SERVICES-II AR-423 4TH 2019-24 14-12-2020

Connected Load of a House 2 Bed Room House:

Area Light
Point
Fan /
Exhaust
point
5 Amps
Plug
Heavy
Duty Plug
AC Geyser W/M Fridge TV
Drawing 2 1 2 1
Dinning 2 1 2 1
Bed Room-1 2 1 2 1 1
Bed Room-2 2 1 2 1 1
Kitchen 2 1 1 1
Balcony 1 1 1
Toilet Attached 1 1 1
Toilet Common 1 1 1
Total : 13 7 10 1 2 2 1 1 3

44. TOPIC : INTRODUCTION TO BUILDING SERVICES-II
FACULTY : MOHAMMED KAUSAR IQBAL


SUBJECT CODE SEMESTER BATCH DATE

KiiT SCHOOL OF ARCHITECTURE & PLANNING
BUILDING SERVICES-II AR-423 4TH 2019-24 14-12-2020


Connected Load :
Light Points - 13 Nos @ 30 Watts = 390 Watts
Fan Point - 7 Nos @70 Watts = 490 Watts
5 Amps Plugs - 10 Nos @ 100 Watts = 1000 Watts
Heavy Duty Plug - 1 Nos @ 1000 Watts = 1000 Watts
AC - 2 Nos @ 2000 Watts = 4000 Watts
Geyser - 2 Nos @ 2000 Watts = 4000 Watts
Washing Machine - 1 Nos @ 500 Watts = 500 Watts
Fridge - 1 Nos @ 125 Watts = 125 Watts
TV - 3 Nos @ 80 Watts = 240 Watts
Grand Total Load = 11,745 Watts
Or Say 11.745 KW or 12 KW

45. TOPIC : INTRODUCTION TO BUILDING SERVICES-II
FACULTY : MOHAMMED KAUSAR IQBAL


SUBJECT CODE SEMESTER BATCH DATE

KiiT SCHOOL OF ARCHITECTURE & PLANNING
BUILDING SERVICES-II AR-423 4TH 2019-24 14-12-2020


LOAD FACTOR
Ratio of
1. Load in Use / Total Connected Load
2. Instantaneous Load / Total Connected Load
3. Energy Consumed / Energy which could have been consumed.

Maximum Energy which can be consumed in a month
=Total connected Load x Number of Hours in Month
30 Days Month = 24x30 = 720 Hours
31 Days Month = 24x31 = 744 Hours

46. TOPIC : INTRODUCTION TO BUILDING SERVICES-II
FACULTY : MOHAMMED KAUSAR IQBAL


SUBJECT CODE SEMESTER BATCH DATE

KiiT SCHOOL OF ARCHITECTURE & PLANNING
BUILDING SERVICES-II AR-423 4TH 2019-24 14-12-2020

For 12 KW Load = 12x720 = 8640 KWH @ 5/- = Rs. 43,200.00






 All the equipments in house office etc work at 230 Volts Single
Phase except heavy equipments like large size water pump
motor, AC more than 3 Ton Capacity, Lifts etc. (They work on
400 Volts, 3 Phase)
 The voltage is constant but current varies according to the
capacity of the equipment.

47. TOPIC : INTRODUCTION TO BUILDING SERVICES-II
FACULTY : MOHAMMED KAUSAR IQBAL


SUBJECT CODE SEMESTER BATCH DATE

KiiT SCHOOL OF ARCHITECTURE & PLANNING
BUILDING SERVICES-II AR-423 4TH 2019-24 14-12-2020


Following is the Electrical data of a house in respect of a house with connected load
of each point in watts, number of points, hours of use of each point. The supply to
the house is done through Single Phase, 230 Volts, 50Hz, AC.
Appliance Rating
in
Watts
Total
Numbers
Duration
used in
Hours
PER DAY
Connected
Load
Energy
Consumption
per day in
WH
Current of
Equipment is
I=P/V
(2) / 230
(1) (2) (3) (4) (5)=(2x3) (6)= (4x5)
Lights 20 15 8 300 2400 0.09
Ceiling Fan 70 4 15 280 4200 0.30
Electric Iron 750 1 1 750 750 3.2
Geyser 2000 1 2 2000 4000 8.69

48. TOPIC : INTRODUCTION TO BUILDING SERVICES-II
FACULTY : MOHAMMED KAUSAR IQBAL


SUBJECT CODE SEMESTER BATCH DATE

KiiT SCHOOL OF ARCHITECTURE & PLANNING
BUILDING SERVICES-II AR-423 4TH 2019-24 14-12-2020

Exhaust Fan 80 2 2 160 320 0.34
Air
Conditioner
(1.5 Ton)
1550 1 8 1550 12400 6.74
Personal
Computer
250 1 8 250 2000 1.01
Washing
Machine
750 1 2 750 1500 3.26
Refrigerator 250 1 12 250 3000 1.01
Television 125 2 8 250 2000 0.54
Micro Oven 2000 1 1 2000 2000 8.69
Kitchen
Chimney
70 1 6 70 420 0.3

49. TOPIC : INTRODUCTION TO BUILDING SERVICES-II
FACULTY : MOHAMMED KAUSAR IQBAL


SUBJECT CODE SEMESTER BATCH DATE

KiiT SCHOOL OF ARCHITECTURE & PLANNING
BUILDING SERVICES-II AR-423 4TH 2019-24 14-12-2020

8610 34810 37.44

1. Calculate the Connected load in KW = 8610 Watts or 8.610 KW
2. Calculate monthly bill (30 DAYS) for electric consumption using the above data,
the rate of Energy being Rs. 5.00 per UNIT.
3. Calculate the load Factor
4. Calculate the maximum current in Amps drawn for total connected
Load = P=VI or I=P/V = 8610/230 = 37.434 Amps.
 Per Day Energy Consumption is 34810 Watts Hour or 34.810 KWH
 Therefore Energy Consumption for 30 Days = 34.810x30 = 1044.3 KWH
 Bill @ Rs. 5.00 / KWH = Rs. 5.00 x 1044.3 = 5,221.50
 Load Factor
 Connected Load = 8610 Watts
 So maximum energy Consumption for 30 Days = 8610x24x30 = 6199200
Watts Hour or 6199.20 KWHR

50. TOPIC : INTRODUCTION TO BUILDING SERVICES-II
FACULTY : MOHAMMED KAUSAR IQBAL


SUBJECT CODE SEMESTER BATCH DATE

KiiT SCHOOL OF ARCHITECTURE & PLANNING
BUILDING SERVICES-II AR-423 4TH 2019-24 14-12-2020

 Therefore Load Factor =1044.3/6199.20 = 0.16845 or 16.485%


Load of ONE Standard Room
Light 3 Nos x 20 Watts = 60 Watts = 60
Fan 1 Nos x 70 Watts = 70 Watts = 70
Exhaust Fan = 1 Nos x 70 Watts = 70 Watts = 70
Plug Misc 1 Nos x 20 Watts = 20 Watts = 20
Plug 1 Nos x 750 Watts = 750 Watts = 750
Total Watts = = 970 I = 4.2

AC 1 Nos x 1600 Watts = 1600 Watts =1600 6.95
Geyser 1 Nos x 2000 Watts = 2000 Watts = 2000 8.69


Standard Rating of Transformers :

51. TOPIC : INTRODUCTION TO BUILDING SERVICES-II
FACULTY : MOHAMMED KAUSAR IQBAL


SUBJECT CODE SEMESTER BATCH DATE

KiiT SCHOOL OF ARCHITECTURE & PLANNING
BUILDING SERVICES-II AR-423 4TH 2019-24 14-12-2020

Wire Size & Current Capacity :

8. (a) An office has connected load of 60 KW. The office working hours is
from 9.00 AM to 6 PM from Monday to Friday and is closed on
Saturday and Sunday. The load factor is as follows:
 Working Hours : 90%
 Non Working Hours : 5%

(a) Calculate the Energy Bill for a 30 Day Month which has 4 Saturdays
& Sundays. Assume the rate of Energy as Rs. 8.00/KWH
[4]
(b) Calculate the Average Load Factor for the month [2]
(c) Calculate the current at Connected load at 400 Volts, 3 Phase. [2]

9 AM to 6 PM = 9 Hours – Working from Monday to Friday i.e. 5 Days

52. TOPIC : INTRODUCTION TO BUILDING SERVICES-II
FACULTY : MOHAMMED KAUSAR IQBAL


SUBJECT CODE SEMESTER BATCH DATE

KiiT SCHOOL OF ARCHITECTURE & PLANNING
BUILDING SERVICES-II AR-423 4TH 2019-24 14-12-2020

6 PM to 9 AM = 15 Hours – Closed from Monday to Friday i.e. 5 Days
Saturday & Sunday – Full Closed i.e. 24 Hours for 2 Days per week
Connected Load 60 KW
At 90% Load Factor what is the load?
During working Hours - Load is 54 KW (90% of 60 KW)
During Idle / Closed Hours - Load is 3.0 KW (5% of 60 KW)

Calculate the Total Number of Working hours & Idle Hours in Month
Total Hours in a month of 30 Days = 24x30 = 720 Hours
Total Number of Idle / Closed Days = 4 Saturday + 4 Sundays = 8 Days
Total Number of Working Days = 30 – 8 = 22 Days

53. TOPIC : INTRODUCTION TO BUILDING SERVICES-II
FACULTY : MOHAMMED KAUSAR IQBAL


SUBJECT CODE SEMESTER BATCH DATE

KiiT SCHOOL OF ARCHITECTURE & PLANNING
BUILDING SERVICES-II AR-423 4TH 2019-24 14-12-2020



Working Hours in a month :
22 Days x 9 Hours = 198 Working Hours

Idle / Closed Hours =
22 Days x 15 Hours = 330 Hours
+
8 Days x 24 Days = 192 Hours

Total Number of Idle Hours in the Month = 330+190 = 522 Hours
 Energy Bill
Energy Consumption for Working Hours = 198 Hours x 54 KW = 10692 KWH
Energy Consumption for Idle Hours = 522 x 3 Kw = 1566 KWH

54. TOPIC : INTRODUCTION TO BUILDING SERVICES-II
FACULTY : MOHAMMED KAUSAR IQBAL


SUBJECT CODE SEMESTER BATCH DATE

KiiT SCHOOL OF ARCHITECTURE & PLANNING
BUILDING SERVICES-II AR-423 4TH 2019-24 14-12-2020

Total Energy Consumption for the Month = 12258 KWH
Energy Bill @ Rs. 8.00 for 12258 KWH or Units = Rs. 98,064.00
 Average Load Factor for the Month :
Total Energy Consumption = 12258 KWH
Maximum Energy which could be consumed for a connected load of 60 KW is
60 KW x 30 x 24 = 43200 KWH
Therefore Load Factor = (12258 / 43200) x100 =0.28375 or 28.375%

 Current at 400 Volts, 3 Phase
P=√3VI for Three Phase Circuit
Here P = 60 KW or 60,000 Watts & Voltage is 400 Volts
Therefore I = P/=√3V = 60000 /1.732x400 = 86.60 Amps

55. TOPIC : INTRODUCTION TO BUILDING SERVICES-II
FACULTY : MOHAMMED KAUSAR IQBAL


SUBJECT CODE SEMESTER BATCH DATE

KiiT SCHOOL OF ARCHITECTURE & PLANNING
BUILDING SERVICES-II AR-423 4TH 2019-24 14-12-2020





8. A shopkeeper has connected load of 40 KW. His shop is open on all
days from 10.00 AM to 10.00 PM except on SUNDAY. His load
factor is as follows:
 From 10.00AM to 10.00PM on working days : 90%
 From 10.00 PM to 10.00 AM and Sunday : 5%

(a) Calculate the Energy Bill for a 30 Day Month which has 4 Sundays.
Assume the rate of Energy as Rs. 8.00/KWH
[7]
(b) Calculate the Average Load Factor for the month [3]

56. TOPIC : INTRODUCTION TO BUILDING SERVICES-II
FACULTY : MOHAMMED KAUSAR IQBAL


SUBJECT CODE SEMESTER BATCH DATE

KiiT SCHOOL OF ARCHITECTURE & PLANNING
BUILDING SERVICES-II AR-423 4TH 2019-24 14-12-2020


Calculate Energy Bill @ Rs. 5.00 Per KWH for a 5 KW Connected load at 230
Volts, Single Phase at 30% Load factor for the month for March 2021. Also
Calculate the Current

Energy Consumption at 30% Load Factor for 5 KW = 5x0.3x24x31 = 1116 KWH
Bill = Rs. 5.00 x 1116 KWH = Rs. 5580.00

Current at Single Phase = P=VI or I=p/v = 5000/230 = 21.74 Amps

57. TOPIC : INTRODUCTION TO BUILDING SERVICES-II
FACULTY : MOHAMMED KAUSAR IQBAL


SUBJECT CODE SEMESTER BATCH DATE

KiiT SCHOOL OF ARCHITECTURE & PLANNING
BUILDING SERVICES-II AR-423 4TH 2019-24 14-12-2020



Standard Rating of Transformers as per BIS :

Comes in KVA Rating
KVAx P.F.(Power Factor) = KW or KW/P.F. = KVA
Normally Assume PF = 0.9

16 KVA, 25KVA, 40KVA, 50KVA, 63KVA, 100KVA, 150KVA, 200KVA, 250KVA,
315KVA, 400KVA, 500KVA, 800KVA, 1000KVA, 1250 KVA, 1500KVA

25 ,63, 100, 250, 500, 800, 1000, 1250, 1500

58. TOPIC : INTRODUCTION TO BUILDING SERVICES-II
FACULTY : MOHAMMED KAUSAR IQBAL


SUBJECT CODE SEMESTER BATCH DATE

KiiT SCHOOL OF ARCHITECTURE & PLANNING
BUILDING SERVICES-II AR-423 4TH 2019-24 14-12-2020

Say a Building has 100 Flats + 4 Lifts + Water Pumps + Common Services

Connected Load is 100x10KW + 4x10KW + 15KW + 10KW = 1065 KW
With PF of 0.9 load in KVA is 1065/0.9 = 1183 KVA
Normally we load any electrical machine / machine upto 80% of its rsated
capacity
Here total requirement is 1183KVA /0.8 = 1479 KVA
So we install transformer of 1500 KVA or 1350 KW

We assume a load factor for connected load say 60%
So connected load becomes at 60% LF = 1065KWx0.6 = 639 KW
639KW at 0.9 PF = 710 KVA
At 80% Loading = 887.5 KVA
Nearest rating is 1000KVA or 900 KW

59. TOPIC : INTRODUCTION TO BUILDING SERVICES-II
FACULTY : MOHAMMED KAUSAR IQBAL


SUBJECT CODE SEMESTER BATCH DATE

KiiT SCHOOL OF ARCHITECTURE & PLANNING
BUILDING SERVICES-II AR-423 4TH 2019-24 14-12-2020



Wires and it sizes :

Types of Wires :
 Electrical Wires for building wiring – internal – IS:694
 Single Core
 Two Core
 Three Cores
 Four Cores

 Electrical Cables – Underground – IS: 1554 & IS:7098

 Overhead Conductors Bare

60. TOPIC : INTRODUCTION TO BUILDING SERVICES-II
FACULTY : MOHAMMED KAUSAR IQBAL


SUBJECT CODE SEMESTER BATCH DATE

KiiT SCHOOL OF ARCHITECTURE & PLANNING
BUILDING SERVICES-II AR-423 4TH 2019-24 14-12-2020


 Overhead Conductor Insulated

 Computer Cables or Data Cables – CAT Cable
 Telephone Cables - In pairs
 TV Cables – Co-Axial Cables for video Signals – TV, CCTV


All wires come is various size is denoted by Sq-mm i.e the cross section area
of the conductor. The cross section area or the size determines the CURRENT
CARRYING CAPACITY of the wire or cable for Electrical


Various Sizes and Construction.

61. TOPIC : INTRODUCTION TO BUILDING SERVICES-II
FACULTY : MOHAMMED KAUSAR IQBAL


SUBJECT CODE SEMESTER BATCH DATE

KiiT SCHOOL OF ARCHITECTURE & PLANNING
BUILDING SERVICES-II AR-423 4TH 2019-24 14-12-2020

1
st
lets understand the sizes :

Size in
Sq-mm
Current
Carrying
Capacity in
Amps
Capacity
with factor
of safety 2
or 50%
Load in
watts at
230 Volts
Use
0.5 4 2 460 Flexible Cord
0.75 8 4 920 Flexible Cords
1.0 12 6 1380 Point wiring
1.5 14 7 1610 Circuit wiring
2.5 20 10 2300

62. TOPIC : INTRODUCTION TO BUILDING SERVICES-II
FACULTY : MOHAMMED KAUSAR IQBAL


SUBJECT CODE SEMESTER BATCH DATE

KiiT SCHOOL OF ARCHITECTURE & PLANNING
BUILDING SERVICES-II AR-423 4TH 2019-24 14-12-2020

4.0 26 13 2990
6.0 34 17 3910





And signal strength for DATA and TCV Cables – Communication cables

63. TOPIC : INTRODUCTION TO BUILDING SERVICES-II
FACULTY : MOHAMMED KAUSAR IQBAL


SUBJECT CODE SEMESTER BATCH DATE

KiiT SCHOOL OF ARCHITECTURE & PLANNING
BUILDING SERVICES-II AR-423 4TH 2019-24 14-12-2020








Wires
Types :
 For use in electrical wiring
 Communication
 Telephone
 TV
 Computers

64. TOPIC : INTRODUCTION TO BUILDING SERVICES-II
FACULTY : MOHAMMED KAUSAR IQBAL


SUBJECT CODE SEMESTER BATCH DATE

KiiT SCHOOL OF ARCHITECTURE & PLANNING
BUILDING SERVICES-II AR-423 4TH 2019-24 14-12-2020

 Automation
 Automobile
Size
 Will be shared in the catalouge