For the time period when the fan is not adequate to provide cooling, if the fan

Question

For the time period when the fan is not adequate to provide cooling, if the fan is running at its maximum 3,000 cfm, what will the resulting indoor temperature be (°F)? (Round to the nearest tenth of a degree).

12-1pm
OAfan + OAdoor = 3,475 cfm
?T = Tin - 70°F
Qint (Btu/h ) = 1.08 * cfm * ?T

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

BTU due to men = 350*80 =28000

BTU due to espresso = 8000

BTU due to employee = 500*5 =2500

BTU due to Lamp =12*25*3.412 =1023.6

So -

28000+2500+1023.6+8000+95*0.4299*1.005*1.2*0.064*300*(Tin-70) = 945.669(Tin-70) + 39523.6 =1.08*3475*(Tin-70)

Tin = 70+14.07 =84.07F

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