# Converted the 110W (j/s) into kWh

1. Sep 10, 2011

### TyErd

1. The problem statement, all variables and given/known data
I have attached the question.

2. Relevant equations
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3. The attempt at a solution
Well i first converted the 110W (j/s) into kWh which is 0.11kWh. so each tube used 0.11kWh an hour. now because there is 12 in each classroom and 6in each faculty room and 200 class rooms and 400 faculty rooms. That means there are 2400 tubes classroom tubes and 2400 faculty room tubes. that means for the 240 days it is a total power usage of: (0.11 x 2400 x 24 x 240) x 2 = 3041280 kWh. the cost then is 3041280 x 0.082 = $249384.96 When the lights are turned off during the unoccupied 4hours the total power usage then becomes: (0.11 x 2400 x 20 x 240) x 2 = 253400kWh, so the cost is 253400 x 0.082 = 207820.80. which means the campus will save$41564.16

Is this correct??? the only thing that worries me is the numbers. $249384.96 is really high but then again it is an awful lot of fluorescent tubes. #### Attached Files: • ###### aaaaaaaaa.png File size: 64.8 KB Views: 81 2. Sep 10, 2011 ### wukunlin Re: power looks right to me 3. Sep 10, 2011 ### TyErd Re: power are the numbers correct? 4. Sep 10, 2011 ### legendary_ Re: power 1. Given: 1 classroom : 12 fluorescent 1 fluorescent : 110 Watts 1 office : 6 fluorescent 1 year : 240 days$0.082 : 1 kWh
1 unoccupied period : 4h per day
$? [cost] : 1 year 2. Let's use the quantity of units for analysis. The objective here is to match the units on the right side to the required unit on the left side. We start with the unit$. In the given,
we have a ratio between $and KWh so we wrote them down on the first term of the right side of the equation. We do the process until the uncancelled term is$/yr. We attach the values for each unit afterwards.

$/yr =$/kwh * W/flourescent * flourescent*(classroom+offices)
* days/yr * h/day

C = 0.082/1000 * 110 * (12*(200+400)) * 240 * 4 = answer