An air conditioner operates exactly as a refrigerator; it removes heat from a co

Question

An air conditioner operates exactly as a refrigerator; it removes heat from a cold place (a bedroom, for example) and conveys it to a warmer place (usually the outside). Quantities of practical importance in understanding the efficiency of air conditioners are the rate of heat removal and the electric power input to the unit (i.e., the electric energy consumed per unit time). Find an expression for the coefficient of performance of an air conditioner that has a rate of heat removal and a power consumption .


Express your answer in terms of and .
=



For most commercial air conditioner systems, the performance of the refrigerator unit is rated according to a quantity called the energy efficiency rating or EER. The EER of an air conditioner is defined as the ratio of the rate of heat removal to the power input required for operation. Therefore, it is essentially equivalent to the coefficient of performance , although, commercially, is expressed in British thermal units per hour (Btu/hour) and in watts.
Part B
Given the definition of EER, find the EER of an 8000 air conditioner that requires a power input of 1500 .
Express your answer numerically in British thermal units per hour per watt.
EER =




Part C
What is , the rate of heat absorbtion by an air conditioner with an EER of 8.3 that operates at 1800 ? Recall that .
Express your answer in joules per minute.
=




Part D
Obviously, from the point of view of efficiency, the higher the EER the better, but a higher EER is tyically accompanied by a higher price. Consider two 12,000-air conditioners, one with an EER of 7.8 and the other with an EER of 10. The difference in price is about 70 . Assuming that the air conditioner will be operated for 6 hours daily during a period of 4 months every year, how long will it take to break even financially if you purchase the more expensive unit? Take the cost of electricity to be 0.10 .
less than a year
about two years
about three years
about four years
more than five years

Explanation / Answer

A) K= H/P

H = Q/T and P= W/ T

Q=HT and W=PT

By substitution,

K=Q/W --> K= HT/PT --> K=H/P

B) 5.33

8000 (Btu/hour) / 1500 W = 5.33

C) 2.63*10^5

8.3 (Btu/W*hour)*1800 W *(1/60 hour/minute) * (1055 Btu/ J) =

2.63E5 J

D) About three years

EER1=7.8; EER2=10

EER1=7.8=12000Btu/(W?h)

EER2=10.0=12000 Btu/(W?h)

Convert W*h to dollar/kWh

For EER_1 1540W?h? (1 kWh/1000 W?h)?0.10 dollar/kWh=0.154 dollar/hour

For EER_2 (Use same conversion) --> 0.12 dollar/hour

Convert to how much money is spent in the four months of operation per year

For EER_1 $0.15/hr * 6 hours/day *30 days/month

(REMEMBER- rough estimates are allowed for this problem)

So with EER_1, you would spend $108.00 per year

Thus, for EER_2, you would spend $86.40 per year

Since you save $21.60 per year with EER_2, it takes approximately 3 years ($70.00/$21.60) to break even financially Thus, about three years.

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