Question

1. House of 40 ’x 40’, with a voltage of 120/240 V. Determine the load by:

a) Classic Method

b) Optional Method

In addition to the basic loads, the house has:

• 3 units of 12 kBTU air conditioners (240 V, 10.41 A)

• 1⁄2 Hp tank pump (120 V, 5 A)

2. What is the difference between the methods? What is the correct method?

Answer 1

1) for a house of 40*40 with voltage 120/240

for such a hose there will be 2 bedrooms 1 kitchen 2 bathrooms and a sitout also PF be 0.6 assuming this

  a) Classic method

list of all appliace used in the house is given below approximately

power used by 12kbtu air conditioner = 240*10.41*0.6=1500W (where 0.6 is PF)

power used by 1/2 hp tank pumb = 120*5*0.6=360W

SL no	ELECTRICAL APPLIANCE	QUATITY	WATTAGE	HOUR	KILO WATT HOUR
1	20 W TUBE LIGHT	3	20	8	480
2	10 W BULB	5	10	8	400
3	CEILING FAN	4	60	8	1920
4	EXHAUST FAN	1	100	2	200
5	12 KBTU AIR CONDITIONER	3	1500	2	9000
6	1/2 HP tank pump	1	360	0.5	180
			TOTAL	=	12180W

Now as we know the electrical load of this building will be the total electricity consumed by the above appliances, so the first thing for is to know the electricity consumed by these individual appliances

So the total energy consumed by all the appliances in a day is = 12180W

but for practical consideration only 60% of this load is taken into consideration

hence = 12180 * 0.6

= 7308 W =7.308KW

So in the above example the total energy consumed by the house is 7.308 KWH or 7.308units.

current = 7380/(240*0.6) = 50.5A

b) Use the optional calculation to find the load on the house

40*40 house so it is approximately

1600 sq. ft

240V, 10.41A air conditioner

360W 1/2 hp pump

Step 1: Multiply the sq. ft. area by 3 VA per Sq. ft.

1600 sq. ft. X 3 VA = 4800 VA (VA = volt amperes)

Step 2:

Add in 1500 VA for each 2-wire, 20-amp small appliance branch circuit and the laundry circuit

1,500 VA X 3 = 4,500 VA

Step 3:

Add in the appliances loads at nameplate value.

240*10.41=2498VA(AIR CONDITIONER)

120*5 =600VA(PUMP)

Step 4: Add all appliance loads together. Total = 12398VA approx from the above table

Step 5: Take the first 10 kW at 100%. 10,000 VA Take the remainder (2398VA) at 40%.

2398 VA X .40 = 959.2VA

Step 6: Add the two values from step 5 together to find the general load. 10,000 VA + 959.2 VA = 10959.2VA

Step 7:

Compare the heating load to the AC load and take the larger of the two loads.

AC load at 100%.

10.41 amps X 240 volts = 2498.4 VA

Step 8: Add the general load to the largest of the AC .

General load = 10952.2 VA

AC load = 2498 VA

Total = 13450.2VA

Step 9:

Divide the load in VA by the voltage.

13450.2 VA ÷ 240 = 56.04amps.

2)optional method is more practical but classic method only gives rough idea of the load so the more accurate or correct method is optional method