In: Electrical Engineering
ANSWER 16: COP = HEAT REMOVED(Q)/ ELECTRICAL WORK INPUT (W) = Q/W
FOR OLD 20 kW AC RTU WHICH OPERATES FOR 3000 HOURS A YEAR,
ELECTRICAL WORK INPUT = W = POWER RATING * TIME OF OPERATION = 20 kW * 3000 hrs
W = 60000 kWh/YEAR
ELECTRIC COST FOR A YEAR = 60000 * 0.10 = $ 6000
NOW, COP = 2 = Q/W
THUS, Q = COP * W = 2 * 60000 = 120000
NOW WE WANT SAVING OF $1500 ANNUALLY, SO OUR ELECTRIC COST WILL BE 6000 -1500 = $4500
IN $4500 AT $0.10/kWh WE CAN USE 4500/0.10 = 45000 kWh/YEAR.
SO, NEW ELECTRICAL WORK INPUT W = 45000 kWh/YEAR
BUT OUR Q= 120000 IS FIXED.
TO MAKE SURE OUR SAVING IS $1500 ANNUALLY,
NEW COP = Q/W = 120000/45000 =2.667
SO MINIMUM COP FOR THE UNIT TO SAVE ATLEAST $1500 ANNUALLY IS 2.667.
ANSWER: 3.5 (NEAREST OPTION TO 2.667)
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ANSWER 18: TRUE
R134a CONTAINS NO CHLORINE AND HAS NO OZONE DEPLETION POTENTIAL.
ANSWER 19: TRUE