Converted the 110W (j/s) into kWh

[TyErd]

Homework Statement

I have attached the question.

Homework Equations

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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.

 

[wukunlin]

looks right to me

 

[TyErd]

are the numbers correct?

 

[legendary_]

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