ut final project

7/31/2019 Ut Final Project

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I. INDRODUCTION

In this project, the students were asked to design a 3 phase low voltage distribution for two

double storey houses. Apart from that designing, members of the group also have learnt how to

balance the needs of consumers and providers of energy in order to ensure safe and reliable

supply.

The low voltage distribution system of this project consists of loads, cables, proactive device

and distribution boxes. The main objective of this project is to draw a single line diagram design

and this design layout of each floor of the two houses was presented in this project. In this task

each group was given the permission to choose the electrical loads of their houses; the estimation

electrical load values in this design were given in the calculation parts of this project, this was to

make sure that the work of each group must be different than the works of the other groups.

For the best results in electrical installation design and LV distribution requires a study of a

proposed electrical installation that requires an adequate understanding of all governing rules and

regulations. The total power demand can be calculated from the data relative to the location and

power of each load, together with the knowledge of the operating modes (steady state demand,

starting conditions, non simultaneous operation, etc.).

From these data, the power required from the supply source and (where appropriate) the

number of sources necessary for an adequate supply to the installation is readily obtained. Local

information regarding tariff structures is also required to allow the best choice of connection

arrangement to the power-supply network, e.g. at medium voltage or low voltage level.

The whole installation distribution network is studied as a complete system. The distribution

equipment (panel boards, switchgears, circuit connections ...) are determined from building plans

and from the location and grouping of loads. The type of premises and allocation can influence

their immunity to external disturbances.


2/35

This dissertation consists of four main parts. First part is an introduction; which describes the

electrical structure and loadings for the LV distribution system. Second part is the calculation

part; this explains all the methods applied in calculating the connected load, maximum demand

and cable sizes. Third part is the discussion and analyses of the results obtained in this project.

Last part is the appendices, this part is used to show all the calculations and schematic diagrams

since the technical report pages are limited.

II. CALCULATIONSHOUSE TYPE C

A) TYPE: HOUSE C GROUND FLOOR

Table1 shows the number of Nos in each load connected to the ground floor

TABLE 1: Ground floor loads

Load Nos Estimated power

(W)

Total Power (W)

Down light 13 100 1300

Ceiling fan 2 60 120

Wall light 2 100 200Wall fan point 1 60 60

Water heater point 1 1000 1000

1X13A S/S/0(300mm.A.F.F.L)

7 250 1750

1X13A weather proofS/S/0

2 250 500

2 X13A S/S/0(300mm.A.F.F.L)

2 250*2 1000

1X13A S/S/0 atkitchen and yard

(1500mm A.F.F.L)

4 250 1000

1X15A S/S/0 at

kitchen (1500mm

A.F.F.L)

3 500 1500

15A S/S/O AIR

COND POINT

2 746 1492

TOTAL CONNECTED LOAD 9922


3/35

PMD=PCL x DF

Assume DF=65%x

PMD= 9922W x 0.65 = 6448W

cos3

,3For

)(

LLMDVIP

MD

AV

WI

MD11

85.04003

6488)(3

AV

WI

MD84.32

85.0231

6448)(1

(PLEASE REFER TO APPENDIX A, FOR THE FORMULAE AND LINE CURRENT

CALCULATIONS)

Line Total load per line Current

(A)

Cable size Circuit

breaker

Current (A)

R1 4 NOS DOWNLIGHT+ I NOS CEILING FAN 2.3 2 x 1.5mm ) PVC

cable6

B1 4 DOWN LIGHT + I NOS CEILING FAN + WALL

LIGHT

2.82 2 x 1.5mm ) PVC

cable6

Y1 5 NOS DOWN LIGHT+I NOS WALL FANPOINT+1NOS WALL LIGHT

3.36 2 x 1.5mm ) PVCcable

6

R8 I NOS 15A S/S/0 AIR COND POINT 3.8 2 x 4.0mm ) PVC 20

B7 I NOS 15A S/S/O AIR COND POINT 3.8 2 x 4.0mm ) PVC 20

Y8 I NOS WATER HEATER 15A S/S/O 5.1 2 x 4.0mm ) PVC 20

B2 2 NOS 1X13A S/S/O (300MM.AF.F.L) 2.55 2 x 2.5mm ) PVC 20A

B3 2 NOS 1X 13A S/S/O AT KITCHEN AND YARD

(1500mm A.F.F.L)

2.55 2 x 2.5mm ) PVC 20A

B4 1 NOS 1X13A weather proof S/S/0 + 1X13A

S/S/0 (300mm.A.F.F.L)

2.55 2 x 2.5mm ) PVC 20A

B5 1NOS 1X13A S/S/O (300MM.AF.F.L) 1.27 2 x 2.5mm ) PVC 20A

B6 1NOS 1X15A S/S/0 at kitchen (1500mmA.F.F.L)

2.55 2 x 2.5mm ) PVC 20A

R3 I NOS 1X13A S/S/0 (300mm.A.F.F.L) + I NOS

1X13A S/S/0 at kitchen and yard (1500mm

A.F.F.L)

2.55 2 x 2.5mm ) PVC 20A

R4 1 NOS 1X13A S/S/0 (300mm.A.F.F.L) 1.27 2 x 2.5mm ) PVC 20A

R5 1 NOS 2 X13A S/S/0 (300mm.A.F.F.L) 2.55 2 x 2.5mm ) PVC 20A

R7 1NOS 1X15A S/S/0 at kitchen (1500mm 2.55 2 x 2.5mm ) PVC 20A


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A.F.F.L)

Y3 I NOS 1X13A S/S/0 at kitchen and yard(1500mm A.F.F.L) + 1NOS 1X15A S/S/0 at

kitchen (1500mm A.F.F.L)

3.8 2 x 2.5mm ) PVC 20A

Y4 1 NOS 1X13A weather proof S/S/0 + 1NOS

1X13A S/S/O (300MM.AF.F.L)

2.55 2 x 2.5mm ) PVC 20A

Y5 1 NOS 2 X13A S/S/0 (300mm.A.F.F.L) 2.55 2 x 2.5mm ) PVC 20A

Y7 1NOS 1X15A S/S/0 at kitchen (1500mm

A.F.F.L)

2.55 2 x 2.5mm ) PVC 20A

(PLEASE REFER TO APPENDIX E, FOR THE SCHEMATIC DIAGRAM OF THE

GROUND FLOOR)

B) TYPE: HOUSE C FIRST FLOORTable2 shows the number of Nos in each load connected to the ground floor

TABLE 2: LOADS IN FIRST FLOOR

PMD=PCL x DF

Assume DF=65%x

PMD = 9528W x 0.65 = 6193.2W

No Load Item No. NOS Estimate

Power (W)Total Power

(W)

1 Down Light 12 100 1200

2 Ceiling Fan 4 60 240

3 Wall Light 1 100 100

4 1x13A S/S/0 (300mm A.F.F.L) 5 250 1250

5 2x13A S/S/O (300mm

A.F.F.L)

3 250*2 1500

6 15A S/S/O Air Cond Point 3 746 2238

7 15A S/S/O Water Heater point 3 1000 3000

8 Total Connected load 9528


5/35

cos3

,3For

)(

LLMDVIP

MD

A

V

WI

MD52.10

85.04003

6193.2)(3

cos

,1For

)(1 VIP

MDMD

AV

WI

MD54.31

85.0231

6193.2)(1

The table below shows the calculated current for each line (per phase) and the preferred

circuit breaker with the cable size of type C fist floor.

(PLEASE REFER TO APPENDIX B, FOR THE FORMULAE AND LINE CURRENT

CALCULATIONS)

LINE TOTAL NOS CONNECTED CURRENT (A) CABLE SIZE Circuit breaker(A)

R 1 (3 Nos down lighting +1nos wall lighting +1 Nos

ceiling fan)

2.342 x 1.5mm ) PVC

cable

6

Y1 4 Nos down lighting +2

Nos ceiling fan

2.65 2 x 1.5mm ) PVC

cable

6

B1 5 Nos down lighing +1

Nos ceiling fan

2.85 2 x 1.5mm ) PVC

cable

6

R 6 1 Nos air condition with

double poles switch

3.80 2 x 4.0mm ) PVC 20

Y6 1 Nos air condition with

double poles switch

3.80 2 x 4.0mm ) PVC 20

B6 1 Nos air condition with

double poles switch

3.80 2 x 4.0mm ) PVC 20

R7 1 NOS water heater with

Double

5.09 2 x 4.0mm ) PVC 20

Y7 1 Nos water heater with

double poles switch

5.09 2 x 4.0mm ) PVC 20


6/35

(PLEASE REFER TO APPENDIX F, FOR THE SCHEMATIC DIAGRAM OF THE

FIRST FLOOR OF HOUSE TYPE C)

C) TYPE: HOUSE C2 Ground FloorThe table 3below shows the loads and number of Nos connected to the ground floor

Table 3: loads connected to the ground floor in c2

PMD=PCL x DF

B7 1 NOS water heater with

Double

5.09 2 x 4.0mm ) PVC 20

B2 2 NOS 1x13A S/S/O 2.54 2 x 2.5mm ) PVC 20

Y3 1 NOS 1x13A S/S/O 2.54 2 x 2.5mm ) PVC 16


R3 2 NOS 1x13A S/S/O 2.54 2 x 2.5mm ) PVC 20

Y4 1 NOS 2x13A S/S/O 2.54 2 x 2.5mm ) PVC 20


No Load Item No. NOS Estimate Power(W)

Total Power(W)

1 Down Light 14 100 1400



4 1x13A S/S/0 (300mm

A.F.F.L)

7 250 1750

5 2x13A S/S/O (300mm

A.F.F.L)

2 250*2 1000

6 15A S/S/O Air Cond Point 2 746 1492

7 1x13A S/S/O Wether proof 2 250 500

8 1x13A S/S/O at kitchen 4 250 1000

9 1x15A S/S/O at kitchen 3 500 1500

10 15A S/S/O water heater point 1 1000 1000

11 Total connected Load 9962


7/35

Assume DF=60%x

PMD= 9962W x 0.60 = 5977.2W

cos3

,3For

)( LLMD VIP MD

A

V

WI

MD15.10

85.04003

5997.2)(3

cos

,1For

)(1 VIP

MDMD

AV

WIMD

54.3085.0231

597.2)(1

The table below shows the calculated current for each line (per phase) and the preferred circuit

breaker with the cable size of type C2 ground floor.

(PLEASE REFER TO APPENDIX C, FOR THE FORMULAE AND LINE CURRENT

CALCULATIONS)

(PLEASE REFER TO APPENDIX G, FOR THE SCHEMATIC DIAGRAM OF THE

GROUND FLOOR OF HOUSE TYPE C2)

LINE TOTAL NOS

CONNECTED

CURRENT (A) CABLE SIZE

R 1 5 No of down light

and 1 No of fan

2.85 2 x 1.5mm ) PVC

cable

Y1 1 No of wall light,5No of down light and

1 No of fan

3.36 2 x 1.5mm ) PVCcable

B1 1 No of wall

light,54No of down

light

2.54 2 x 1.5mm ) PVC

cable

Y4 water heater point 5.9 2 x 4.0mm ) PVC

B7 Air condition point 3.4 2 x 4.0mm ) PVC

13A S/S/O 2.54 2 x 2.5mm ) PVC

15A S/S/O 7.63 2 x 2.5mm ) PVC


8/35

D) TYPE: HOUSE C2 First FloorThe table4below shows the loads and number of Nos connected to the ground floor

Table 4: loads connected to first floor in c2

PMD=PCL x DF

Assume DF=60%x

PMD= 9678W x 0.60 = 5806.8W

cos3

,3For

)(

LLMDVIP

MD

AV

WI

MD5.9

85.04003

5806.8)(3

No Load Item No. NOS Estimate Power(W)

Total Power

(W)

1 Down Light 11 100 1400



4 Water heater point 3 1000 3000

4 113A s/s/o 6 250 1500

5 213A s/s/o 3 500 1500

6 15A s/s/o 3 746 2238

7 Total connected Load 9678w


9/35

AV

WI

MD57.29

85.0231

5806.8)(1

The table below shows the calculated current for each line (per phase) and the preferred

circuit breaker with the cable size of type C2 fist floor.

(PLEASE REFER TO APPENDIX D, FOR THE FORMULAE AND LINE CURRENT

CALCULATIONS)

(PLEASE REFER TO APPENDIX H, FOR THE SCHEMATIC DIAGRAM OF THE

FIRST FLOOR OF HOUSE TYPE C2)

III. DISCUSION AND ANALYSIS

A) DISCUSION

In this project the students were asked to draw the schematic diagrams of two types of

houses; which are house type C and house type C2. Each house consists of two parts which are

LINE TOTAL NOS

CONNECTED

CURRENT (A) CABLE SIZE Circuit beaker

R 1 (3 Nos down

lighting +1 noswall lighting +1

Nos ceiling fan)

2.32 x 1.5mm )

PVC cable

6A

Y1 4 Nos down

lighting +2 Nos

ceiling fan

2.65 2 x 1.5mm )

PVC cable

6A

B1 4 Nos down

lighing +1 Nosceiling fan

2.34 2 x 1.5mm )

PVC cable

6A

R6 1 Nos water heaterwith double poles

switch

5.1 2 x 4.0mm )

PVC

20A

R7 1 Nos air

condition with

double polesswitch

3.8 2 x 4.0mm )

PVC

20A


10/35

ground floor and first floor. There were seven members in the group; four of them were given to

design house type c while the other three were given to design house type c2.

To achieve the objectives of the project; members counted all the numbers of Nos in each

floor according to the task division of the group. An estimated power was chosen for each load asstated in section II (calculation part). In house C demand factor of 65% was chosen while house

C2 60% of demand factor was chosen. The total connected load of each floor was calculated and

then the demand factor was obtained by using relevant formula as shown in the calculation part.

The number of nos in each line were counted in order to obtain the current in each line of

the four floors. As result of line current, then the current of each circuit breaker was obtained

according to the standard and safety regulations. In addition to that, the cable size was

determined.

There are two types of cables that available in the market, which are copper wire and

alloy aluminium cable. In this project, a copper wire was chosen because copper cable is

recognized as high quality product and it is good conductor compare to aluminium , copper has

greater tensile strength compare to aluminium, even though aluminium is electricity conductive.

The group members have discussed the type of miniature circuit breaker to use, they have

agreed upon to choose a suitable one by considering the load and overload current. MCB is

divided into two; which are domestic and commercial types. In this project, domestic one was

chosen because of its general use of residential purposes.

Heavy load such as air conditional and water heater, was assigned to own a separate

MCB, due to their high load and high overload current, there for, they require high load power

supply, which may spoil other equipment if combine with other appliance that has lower load

current or lower power supply. Similar to socket, the socket which provide to high load appliance

will also has their own MCB, similar load appliance will has similar load current and power

supply, when over current or short circuit occur, the MCB able to protect the cable and

equipment if the MCB use is in suitable current rating.


11/35

Two different demand factors were estimated due to make comparison between the

maximum demands of each floor. House C was given 65% demand while 60% demand factor

was given house type C2. This estimation has shown that when the demand factor is high themaximum demand becomes also high. Hence ground floor of house type C has higher maximum

demand than the ground floor of house type C2. However, there is also total connected load

which is another factor that affects the maximum demand of the floor; higher connected gives

higher maximum demand and vice versa.

B) ANALYSIS

1) House type C

The total connected load, maximum demand, and line currents were obtained by using

relevant formulae. The total power consumption in the ground floor is 9922w while the first floor

is 9528w. it can be seen clearly, that the ground floor consumed huge power compare to first

floor, this is because the kitchen in the ground floor has additional sockets that require bulk of

power.

At the side of voltage drop, line of R7, Y8, and B6 in the ground floor can show the

highest voltage drop, because these lines has sockets such as 1x15A S/S/O at the Kichen, 15A

S/S/O water heater point, and generally these loads are called heavy loads since they consume

huge power.

In this house, demand factor of 65% was estimated in order to compare with the other

house of type c which its demand factor was estimated as 60%. Since the demand factor is 65%

for house, the maximum demand was obtained by using formula as shown in section II. Hence,

the maximum demand of ground and first floors are 6448W and 6193.2W respectively.

Maximum demand of the ground floor is higher than the first floor although they have the same

demand factor, this is due to the total connected load of the ground floor is higher than that of the

first floor.


12/35

2) House type C2

Same as house type C, the ground floor of house C2 has higher power

consumption than the first floor. This is due to that the ground floor was equipped with

more kitchen sockets and appliances which consumed more power. This can be seen from

the total connected load of the ground floor which is 9962, while the total connected of

the first floor is 9678.

Although the floors of this house have higher total connected loads compare to

their respective floors in house type C, but the maximum demand of the ground floor of

house C2 is lower than that of the house type C ground floor, and the same goes to the

first floor. This is due to the demand factor, because the demand factor of house type C is

higher than that of house type C2 as shown in the calculation part.

3) As generally

To chose the correct rating of circuit breaker we have to determine the current requirements for

each line, which also enables us to chose the correct cable size in order to not to affect the cost

as well. This current is known as Design current, either specified by the manufacturer or can be

calculated by the formulae.

cos

,1currentlineFor

1 VIP

For current calculation in each line refer appendix A85.0231

1

V

PI


13/35

For all floors of both type of houses, the lights, fans and bell are used ( 2 x 1.5mm ) PVC

cable. For the installations of air conditioners, water heater and spare are used ( 2 x 4.0mm )

PVC cable while the socket outlet 13A is use ( 2 x 2.5mm ) and ( 2 x 4.0mm ) PVC cable ring

type. Using ring type is intended to save the use of protective fuse. Each cable is installed using a

conduit in the wall constructions. All the wiring has installations of fuse to give protections and

this fuse is accordance to the standards. For the light, fan and bell, the used fuse is 6A and for the

water heater, air conditioner and spare, the used fuse is 20A except for wiring socket outlet ( 2 x

2.5mm ) PVC cable. But for the socket outlet ( 2 x 4.0mm ) PVC cable, the used fuse is 32A.

Starting from TNB supply, cut-off fuse is installed before the meter KWH. A protection begins

with the installations of main switch 40A/60A and then divided into two parts. The first part is

earth leakage circuit breaker (ELCB) 40A. This section is used to protect the general power

circuit such as lights, fans, air conditions, water heater and spare. On the water heater placed

another ELCB 20A to avoid short circuit incidents. The second part is dividing ELCB 40A to

power socket outlet circuit.


14/35

Appendix A (current calculation)

Type c ground floor

Loads of R1 (4 NOS DOWNLIGHT + 1 NOS CEILING FAN)

Given data:

Load Power Total power

4 NOS DOWNLIGHT 100 4100=400

1 NOS CEILING FAN 60 60Total 460

The current following in R1 is IL_N

IL_N=

IL-N =

IL_N =2.3 A

The loads of B1 (4 NOS DOWNLIGHT +1NOS CEILING FAN+ 1NOS WALL LIGHT)

Given data:

Load Power (w) Total power


1 NOS CEILING FAN 60 60

1 NOS WALL LIGHT 100 100

TOTAL 560

The current following in B1 is

IL_N=

IL-N =

Thus, IL_N =2.82A


15/35

The loads in Y1 (5 NOS DOWNLIGHT +1 NOS WALL FAN POINT + 1 NOS WALL LIGHT)

Given data:

Load Power (w) Total power (w)


1 NOS WALL FAN POINT 60 60

I NOS WALL LIGHT 100 100

TOTAL 660

The current following in Y1 is

IL_N=

IL-N =

IL_N =3.36A

Loads of R8 (1 NOS 15A S/S/O AIR COND POINT)

Given data:

Load Power(w) Total power (w)

1 NOS 15A S/S/O AIR

COND POINT

746 746

Total 746


IL_N=

IL-N =

IL_N =3.80 A

Loads of Y8 (1 NOS 15A S/S/O WATER HEATER POINT)

Given data:


1 NOS 15A S/S/O WATERHEATER POINT

1000 1000


16/35

Total 1000


IL_N=

IL-N =

IL_N =5.1 A

Loads of Y8 (1 NOS 15A S/S/O WATER HEATER POINT)

Given data:

Load Power(w) Total power (w)1 NOS 15A S/S/O WATER

HEATER POINT

1000 1000

Total 1000


IL_N=

IL-N =

IL_N =5.1 A

Loads of B7 (1 NOS 15A S/S/O AIR COND POINT)

Given data:


1 NOS 15A S/S/O AIR

COND POINT

746 746

Total 746


IL_N=

IL-N =

IL_N =3.80 A


17/35

Appendix B (current calculation)

Type C first floor

Determine the current requirements of the circuit. This current is known as Design current, either

specified by the manufacturer or can be calculated by the formulae.

1. THE LOADS CONNEDCTED TO R1 (3 NOS DOWNLIGHT +1 NOS WALL LIGHT

+1 NOS CEILING FAN)

Given data:

Formula used:

cos

,1For

1 VIP

AV

WI 34.2

85.0231

4601

Load Power Total power3 NOS DOWNLIGHT 100 w 3100w=300w

1 NOS WALL LIGHT 100w 100w

1 NOS CEILING FAN 60w 60w

Total 460


18/35

2. THE LOADS CONNEDCTED TO Y1 (4 NOS DOWNLIGHT +2 NOS CEILING FAN)

Given data:


4 NOS DOWNLIGHT 100 w 4100w=400W

2 NOS CEILING FAN 60w 120W

Total 520W

Formula used:

cos

,1For

1 VIP

AV

WI 65.2

85.0231

2051

3.

THE LOADS CONNEDCTED TO B1 (5 NOS DOWNLIGHT +1 NOS CEILING FAN)

Given data:


4 NOS DOWNLIGHT 100 w 5100w=500W

1 NOS CEILING FAN 60w 60W

Total 560W

Formula used:

cos

,1For

1 VIP

AV

WI 85.2

85.0231

5601


19/35

4. THE LOADS CONNEDCTED TO R6 ( 1 NOS AIR CONDITION WITH DOUBLE

POLES SWITCH )

Given data:


1 NOS water heater with

Double

746w 746w

Formula used

cos

,1For

1 VIP

A

V

WI 80.3

85.0231

7461

5. THE LOADS CONNEDCTED TO Y6 ( 1 NOS AIR CONDITION WITH DOUBLE

POLES SWITCH )

Given data:



Double

746w 746w

Formula used


20/35

cos

,1For

1 VIP

AV

WI 80.385.0231

7461

6. THE LOADS CONNEDCTED TO B6 ( 1 NOS AIR CONDITION WITH DOUBLE

POLES SWITCH )

Given data:

Load Power Total power1 NOS water heater withDouble

746w 746w

Formula used

cos

,1For

1 VIP

AV

WI 80.3

85.0231

7461

7. THE LOADS CONNEDCTED TO R7(1 NOS WATER HEATER WITH DOUBLE

POLES SWITCH)

Formula used:



Double

1000w 1000w=1000w


21/35

cos

,1For

1 VIP

AV

WI 09.5

85.0231

00011

8. THE LOADS CONNEDCTED TO Y7(1 NOS WATER HEATER WITH DOUBLE

POLES SWITCH)

Formula used:

cos

,1For

1 VIP

AV

WI 09.5

85.0231

00011

9. THE LOADS CONNEDCTED TO B7(1 NOS WATER HEATER WITH DOUBLE

POLES SWITCH)

Formula used:


1 NOS water heater withDouble

1000w 1000w=1000w



Double

1000w 1000w=1000w


22/35

cos

,1For

1 VIP

AV

WI 09.5

85.0231

00011

For the power side we used either 13A S/S/O or 15A S/S/O, so that we can take one in each

S/S/O which can simply be determined the cable size and assume the others as well.

1. The line current of(2 No 13A S/S/O )

AV

WI 54.2

85.0231

5001


23/35

Appendix C(current calculation)

Type c2 Ground floor

1. The line current of R1 (connected to 5 No of down light and 1 No of fan)

Pt = (5x100) + (1x60)=560W

AV

WI 85.2

85.0231

5601

2. The line current of Y1 (connected to 1 No of wall light,5 No of down light and 1 No of fan)

Pt =(5x100)+(1x100)+(1x60)=660W

AV

WI 36.3

85.0231

6601

3. The line current of B1 (connected to 1 No of wall light,54No of down light )

Pt = (4x100) + (1x100)=500W

A

V

WI 54.2

85.0231

500

1

4. The line current of Y4 (water heater point)

Pt = (1x1000) =1000W

AV

WI 9.5

85.0231

10001

5. The line current of B7 (15A S/S/O 1 No of Air condition point)

Pt = (1x1000) =1000W

For the power side we used either 13A S/S/O or 15A S/S/O, so that we can take one in each S/S/O which

can simply be determined the cable size and assume the others as well.

2. The line current of(2 No 13A S/S/O )

AV

WI 4.3

85.0231

7461


24/35

AV

WI 54.2

85.0231

5001

3. The current for 15A S/S/O

AV

WI 63.7

85.023115001

Appendix D (current calculation)

Type c2 first floor

CALCULATION FOR FIRST FLOOR DOUBLE STORY BUILDING

LOAD CALCULATION:

10.THE LOADS CONNEDCTED TO R1 (3 NOS DOWNLIGHT +1 NOS WALL LIGHT

+1 NOS CEILING FAN)

Given data:


3 NOS DOWNLIGHT 100 w 3100w=300w

1 NOS WALL LIGHT 100w 100w


Formula used:

pf)power/(v=IL_N

0.8560w)/231v+100w+(300w=N-IL

Thus, IL_N =2.3 A

11.

THE LOADS CONNEDCTED TO Y1 (4 NOS DOWNLIGHT +2 NOS CEILING FAN)

Given data:





25/35

Formula used:

IL_N=

IL-N =

Thus, IL_N =2.65A

12.THE LOADS CONNEDCTED TO B1 (4 NOS DOWNLIGHT +1 NOS CEILING FAN)

Given data:




Formula used:

IL_N=

IL-N =

Thus, IL_N =2.34A

13.THE LOADS CONNEDCTED TO R6 (1 NOS WATER HEATER WITH DOUBLE

POLES SWITCH)

Given data:



Double

1000w 1000w=1000w


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Formula used:

IL_N=

IL-N =

Thus, IL_N =5.1A

14.THE LOADS CONNEDCTED TO Y6 (1 NOS WATER HEATER WITH DOUBLE

POLES SWITCH)

Given data:



1000w 1000w=1000w

Formula used:

IL_N=

IL-N =

Thus, IL_N =5.1A

15.THE LOADS CONNEDCTED TO B6 (1 NOS WATER HEATER WITH DOUBLE

POLES SWITCH)

Given data:



1000w 1000w=1000w

Formula used:

IL_N=

IL-N =

Thus, IL_N =5.1A


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16.THE LOADS CONNEDCTED TO R7 ( 1 NOS AIR CONDITION WITH DOUBLE

POLES SWITCH )

Given data:



746w 746w

Formula used

IL_N=

IL-N =

Thus, IL_N =3.8A

17.THE LOADS CONNEDCTED TO Y7 (1 NOS AIR CONDITION WITH DOUBLE

POLES SWITCH)Given data:



746w 746w

Formula used

IL_N=

IL-N =

Thus, IL_N =3.8A

18.THE LOADS CONNEDCTED TO B7 (1 NOS AIR CONDITION WITH DOUBLE

POLES SWITCH)

Given data:



746w 746w

Formula used

IL_N= IL-N =

Thus, IL_N =3.8A


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SCHEMATIC DIAGRAMS

Appendix E

GROUND FLOOR HOUSE TYPE C


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Appendix F

FIRST FLOOR HOUSE TYPE C


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Appendix G

GROUND FLOOR HOUSE TYPE C2


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Appendix H

FIRST FLOOR HOUSE TYPE C2


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ut final project

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