Auniversity campus has 200 classrooms and 400 faculty offices. the classrooms are equipped with 12 fluorescent tubes, each consuming 110 w, including the electricity used by the ballasts. the faculty offices, on average, have half as many tubes. the campus is open 240 days a year. the classrooms and faculty offices are not occupied for an average of 4 h a day, but the lights are kept on. if the unit cost of electricity is $0.115/kwh, determine how much the campus will save a year if the lights in the classrooms and faculty offices are turned off during unoccupied periods.

The campus would save

$218.04.

Explanation: Total number of flourescent tubes in classroom is 12 and office is 6. The Electric Power consume by a single tube is 110W.

But,

Energy = Power * Time

For classroom energy consumed in a day that is 24hours would be

Energy = 12*110*24 = 31.7kwh

For office,

Energy = 6*110*24= 15.8kwh

Total Energy consumed at the campus for 24hrs{a full day} will be

Total Energy = 31.7kwh + 15.8kwh

= 47.5kwh.

In a year of activities at the campus given as 240days,

Energy consumed= 47.5kwh * 240

= 11400kwh.

The cost of 1kwh from the question is $0.115.

11400kwh will cost; 11400*$0.115

=$1,311

If the florescent tube are switched off for the 4hrs in a day the campus is not on session. Then we will have just 20hrs of usage in a day. So let's calculate;

Classroom,

Energy = 12*110*20= 26.4kwh

Faculty office,

Energy consumed = 6*110*20

= 13.2kwh

Total energy consumed in a day {20hrs}

= 26.4kwh + 13.2kwh= 39.6kwh.

In a year of activities at the campus given as 240days

Total Energy consumption would be

{39.6*240}kwh= 9504kwh.

The cost of 1kwh of electricity is $0.115.

Therefore, 9504kwh will cost;

$0.115 * 9504 = $1092.96.

This is more cheaper compared the the first one we calculated for 24hrs in a day.

So the campus will save;

$1,311 - $1092.96 = $218.04.

Which is the cost of energy of putting ON the flourescent tubes for 24hrs in a day minus the cost of putting them ON for just 20hrs in a day.

colorectal cancer may also develop from areas of abnormal cells in the lining of the colon or rectum. this area of abnormal cells is called dysplasia and is more commonly seen in people with certain inflammatory diseases of the bowel such as crohn's disease or ulcerative colitis.

continuous conducting path for current, power supply