Electrical Design Estimation and Costing
Term work: 5
Design and draw wiring diagram for residential installation by
using Inverter as backup supply
Dear Students,
In the term-work 4 i. e. Residential Wiring you have prepared estimation and
costing of a bungalow (your dream house). In the present task 5 we have to select
suitable UPS/Inverter for your bungalow, estimate the quantity of material
required for providing inverter/UPS connection to various points/appliances,
and calculate the cost involved.
Guidelines for using inverter as a back up:
Inverter systems are a common feature in our homes and workplace where
they play a prominent role in the ensuring uninterruptible power to sensitive
loads and devices. For home applications, there is the need to adequately size
your inverter to be able to meet the expected load demand.
Inverters convert DC voltage to AC voltage. They have a battery system
which provides adequate backup time to provide continuous power in the home.
The inverter system then converts the battery voltage to AC voltage through
electronic circuitry. The inverter system also has some charging system that
charges the battery during utility power. During utility power, the battery of the
inverter is charged and at the same time power is supplied to the loads in the
house. When utility power fails, the battery system begins to supply power via
the inverter to the loads in the home as shown below:
How to Size and Calculate the Inverter Power Requirement
Inverter power is rated in VA or KVA.
Power in VA = AC Voltage x AC Current in Amps
Power in KVA = AC Voltage x AC Current in Amps/1000
Power in Watts = AC Voltage x AC Current in Amps x PF
Where PF = power factor
Power in KW = AC Voltage x AC Current in Amps x PF/1000
Also, Power in W = Power in VA x PF
Power in KW = Power in KVA x PF
Sample Calculations:
Calculate Size of Inverter for following Electrical Load .Calculate Size of Battery
Bank and decide Connection of Battery.
Electrical Load detail:
 2 No of 60W, 230V, 0.8 P.F Fan.
 1 No of 200W, 230V, 0.8 P.F Computer
 2 No of 30W, 230V, 0.8 P.F Tube Light.
Inverter / Battery Detail:
 Additional Further Load Expansion (Af)=20%
 Efficiency of Inverter (Ie) = 80%
 Required Battery Backup (Bb) = 2 Hours.
 Battery Bank Voltage = 24V DC
 Loose Connection/Wire Loss Factor (LF) = 20%
 Battery Efficiency (n) = 90%
 Battery Aging Factor (Ag) =20%
 Depth of Discharge (DOD) =50%
 Battery Operating Temp =46ºC
Temp. °C
80
70
60
50
40
30
20
Factor
1.00
1.04
1.11
1.19
1.30
1.40
1.59
Calculations:
Step 1: Calculate Total Load:
 Fan Load= No x Watt =2×60=120 Watt
 Fan Load=(No x Watt)/P.F=(2×60)/0.8= 150VA
 Computer Load= No x Watt =1×200=200 Watt
 Computer Load=(No x Watt)/P.F =(1×200)/0.8= 250VA
 Tube Light Load= No x Watt =2×30=60 Watt
 Tube Light Load=(No x Watt)/P.F =(2×30)/0.8= 75VA
 Total Electrical Load=120+200+60 =380 Watt
 Total Electrical Load=150+250+75= 475VA
Step 2: Size of Inverter:
 Size of Inverter=Total Load+(1+Af) / Ie VA
 Size of Inverter= 475+(1+20%) / 80%
 Size of Inverter= 712 VA
Step 3: Size of Battery:
 Total Load of Battery Bank= (Total Load x Backup Capacity) / Battery
Bank Volt
 Total Load of Battery Bank=(380 x 2) / 24 Amp Hr
 Total Load of Battery Bank= 32.66 Amp Hr
 Temperature Correction Factor for 46ºC (Tp)=1
 Size of Battery Bank=[ (Load) x (1+LF) x (1+Ag) x Tp] / [n x DOD]
Amp/Hr
 Size of Battery Bank= (32.66 x (1+20%) x (1+20%) x 1) / (90% x 50%)
 Size of Battery Bank= 101.3 Amp/Hr
Step 4: Connection of Battery:
If We Select 120 Amp Hr , 12V DC Battery for Battery Bank:
Series Connection:
 Series configurations will add the voltage of the two batteries but keep the
amperage rating (Amp Hours) same.
 Condition-I :
 Selection of Battery for Voltage = Volt of Each Battery <= Volt of Battery
Bank
 Selection of Battery for Voltage =12< 24
 Condition-I is O.K
 No of Battery for Voltage = Volt of Battery Bank / Volt of Each Battery
 No of Battery for Voltage =24/12 = 2 No’s
 Condition-II :




Selection of Battery for Amp Hr = Amp Hr of Battery Bank <= Amp Hr
of Each Battery
Selection of Battery for Amp Hr =3<=120
Condition-II is O.K
We can use Series Connection for Battery & No of Battery required 2 No’s
Parallel Configuration
 In Parallel connection, the current rating will increase but the voltage will
be the same.
 More the number of batteries more will be the amp/hour. Two batteries
will produce twice the amp/hour of a single battery.
 Condition-I :
 Selection of Battery for Amp Hr = Amp Hr of Battery Bank / Amp Hr of
Each Battery <=1
 Selection of Battery for Amp Hr =101/120 = 0.84=1 No’s
 Condition-I is O.K
 Condition-II :
 Selection of Battery for Voltage = Volt of Battery Bank = Volt of Each
Battery
 Condition-II :Selection of Battery for Voltage for Amp Hr = 24<=12
 Condition-II is Not Full Fill
 We cannot use Parallel Connection for Battery as per our requirement But
If We do Practically It is Possible and it will give more Hours of back
Series-Parallel Connection:
 Connecting the batteries up in series will increase both the voltage and the
run time.
 Condition-I :
 Selection of Battery for Amp Hr = Amp Hr of Each Battery <= Amp Hr of
Battery Bank
 Selection of Battery for Amp Hr =120<=101
 Condition-I is Not Full Fill
 Condition-II :
 Selection of Battery for Voltage = Volt of Each Battery <= Volt of Battery
Bank
 Selection of Battery for Voltage = 12<=24
 Condition-II is OK
 We cannot use Parallel Connection for Battery
If We Select 60 Amp Hr , 12V DC Battery for Battery Bank:
Series Connection:
 Selection of Battery for Voltage = Volt of Each Battery <= Volt of Battery
Bank
 Selection of Battery for Voltage =12< 24
 Condition-I is O.K
 No of Battery for Voltage = Volt of Battery Bank / Volt of Each Battery
 No of Battery for Voltage =24/12 = 2 No’s
 Condition-II :
 Selection of Battery for Amp Hr = Amp Hr of Battery Bank <= Amp Hr
of Each Battery



Selection of Battery for Amp Hr =3<=60
Condition-II is Not Full Fill
We can use Series Connection for Battery
Parallel Configuration
 Condition-I :
 Selection of Battery for Amp Hr = Amp Hr of Battery Bank / Amp Hr of
Each Battery <=1
 Selection of Battery for Amp Hr =101/60 = 1.63=1 No’s
 Condition-I is O.K
 Condition-II :
 Selection of Battery for Voltage = Volt of Battery Bank = Volt of Each
Battery
 Condition-II :Selection of Battery for Voltage for Amp Hr = 24=12
 Condition-II is Not Full Fill
 We cannot use Parallel Connection for Battery as per our requirement.
Series-Parallel Connection:
 Condition-I :
 Selection of Battery for Amp Hr = Amp Hr of Each Battery <= Amp Hr of
Battery Bank
 Selection of Battery for Amp Hr =120<=60
 Condition-I is OK
 No of Battery for Amp Hr = Amp Hr of Battery Bank / Amp Hr of Each
Battery
 No of Battery for Amp Hr = 120/60 =1.68 =2 No’s
 Condition-II :
 Selection of Battery for Voltage = Volt of Each Battery <= Volt of Battery
Bank
 Selection of Battery for Voltage = 12<=24
 Condition-II is OK
 No of Battery for Voltage = Volt of Battery Bank / Volt of Each Battery
 No of Battery for Voltage = 24 / 12 =2 No’s
 No of Battery Required = No of Battery Amp Hr x No of Battery for
Voltage
 No of Battery Required = 2 x 2= 4 No’s
 We can use Series-Parallel Connection for Battery

Decision:
 Total Electrical Load=380 Watt
 Total Electrical Load=475VA
 Size of Inverter= 712 VA
 Size of Battery Bank= 101.3 Amp/Hr
 For 120 Amp/Hr , 12V DC Battery : Series Connection & 2 No’s of Battery
or
 For 60 Amp/Hr , 12V DC Battery : Series-Parallel Connection & 4 No’s of
Battery
Based on the above sample calculations, you have to select suitable inverter for
the bungalow of task 4.
Sample Connections of the inverter are shown below:
You have to select the suitable size of the wire and calculate the length of the
wire.
Last Date of the Submission of Task 5 is
May 25, 2020
Complete your task using Journal Pages if
available otherwise use notebook