Question

The lighting requirements of an industrial facility are being met by 1000 40-W standard fluorescent bulbs. The lamps are close to completing their service life and the facility is examining replacing them with 34-W high efficiency bulbs. A standard 40-W bulb costs $1.77 each and a high efficiency 34-W bulb costs $2.26 each. The facility operates 8:00 AM to 8:00 PM every day of the year, and the lamps are kept on during all operating hours. If the cost of electricity is given by the APS rate structure utilized in class notes (Note Set 1-Introduction ), determine (a) how much energy will be saved per year as a result, (b) how much money will be saved per year from electrical costs, and (c) the simple payback period.

Note: Assume that holidays are split evenly between summer and winter rates. Note: Simple Payback = Implementation Costs (Added Cost required by 34-W bulbs) / Annual Savings

On-Peak Demand Charge: On-Peak Energy Charge: Off-Peak Energy Charge: Summer $8.40 $0.13160 $0.07798 Winter $8.40 0.11017 per kWh per kW $0.07798 per kWh TIME PERIODS The On-Peak time period for residential rate schedules is 3p.m. to8 p.m. Monday through Friday year round. All other hours are Off-Peak hours.

Answer 1

(a)

A standard fluorescent bulb consumes 40 Watt i.e. 40 Joule in one second.

So,, 1000 standard fluorescent bulb will consume 40 x 1000 = 40000 Joule in 1 second.

So, in one year these bulbs will consume 365 x 12 x 60 x 60 x 40000 = 630720 x 10⁶ Joule of energy

But the new high efficiency bulbs consumes only 34 watts i.e. 34 Joules of energy in one second.

So, 1000 new bulbs will consume 34 x 1000 = 34000 Joules of energy in one second

So, in one year these new bulbs will consume 365 x 12 x 60 x 60 x 34000 = 536112 x 10⁶ Joules of energy .

Therefore, the energy saved in one year is

= (630720 - 536112 ) x 10⁶ = 94608 x 10⁶ Joule = 9.46 x 10¹⁰ Joules of energy .

NOTE :

Total number of seconds in one year = 365 x 24 x 60 x 60

But i have used half of it because the on time for the bulb is only 12 hour not 24 hour.

(b)

Electricity consumed by one 40 Watt bulb in one year is

= 40 x 12 x 365 = 175200 Wh = 175.2 kWh

And

the electricity consumed by one 34 Watt bulb in one year is

= 34 x 12 x 365 = 148920 Wh = 148.92 kWh

Assuming that the total number of month in winter is exactly equals to the total number of month in summer .

So, total electricity saved by one bulb in one year = 175.2 - 148.92 = 26.28 kWh

Electricity saved by one bulb in winter = 26.28/2 = 13.14 kW/h

and the electricity saved by one bulb in summer = 26.28 /2 = 13.14 kWh

So, total money saved by one bulb in one year = 13.14 x 0.13160 + 13.14 x 0.11017 = 3.1768578 dollars

Therefore, total money saved by 1000 bulbs in one year = 1000 x 3.1768578 = 3176.8578 dollars

(c) Implementation cost of new bulb = 1000 x cost of one bulb = 1000 x 2.26 = 2260 dollars

So,

  

For any doubt please comment.