Question 1) For the noon time period, calculate the rate of total internal gains

Question

Question 1) For the noon time period, calculate the rate of total internal gains for the café (in Btu/h). (Round to the nearest integer.)

Question 2) For the evening time period, calculate the rate of total internal gains for the café (in Btu/h). (Round to the nearest integer.)

Because the café is very popular, it usually has enough people inside to keep the indoor temperature above 72°F. When internal heat gains are insufficient, the heater will turn on, but you can ignore that situation in this problem. But when internal heat gains are too high, and it gets too warm, the café has a variable-speed fan rated at 3000 cfm (cubic feet per minute) to bring in outside air for ventilation and cooling. To simplify the calculations, assume the envelope is perfectly insulated, and the only heat transfer is from air exchange. Evening 12pm 1pm 9pm-10pm Noon Average number of customers in cafe (350 Btu/h each) Number of employees (500 Btu/h each) Lights (fluorescent bulbs, 25 Watts each) Espresso machine (Btu/h) Number of times front door is opened during the hour (300 ft3 of air exchange each time) Outside Air Temperature (F) 80 45 25 8,000 65 12 8,000 95 70 51

Explanation / Answer

a)

for noon time period.add all the loads and customers interms of Btu/h

in the problem watts is given so convert into Btu/hr

1watt=3.214Btu/hr

total internal gain for the cafe

=350*80+500*5+25*3.214*12+8000

=28000+2500+964.2+8000

=39464.2Btu/h

b)

for evening time period

total internal gains for the cafe

=350*45+500*3+25*3.214*25+8000

=15750+1500+2008.75+8000

=27258.75Btu/h

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