For a refrigerator or air conditioner, the coefficient of performance K (often denoted as COP) is, as
Question:
where |QC|/t is the cooling power and |W|/t is the electrical power input to the device, both in watts. The energy efficiency ratio (EER) is the same quantity expressed in units of Btu for |QC| and W h for |W|.
(a) Derive a general relationship that expresses EER in terms of K.
(b) For a home air conditioner, EER is generally determined for a 95F outside temperature and an 80F return air temperature. Calculate EER for a Carnot device that operates between 95F and 80F.
(c) You have an air conditioner with an EER of 10.9. Your home on average requires a total cooling output of |QC| = 1.9 Ã 1010 J per year. If electricity costs you 15.3 cents per kW h, how much do you spend per year, on average, to operate your air conditioner? (Assume that the units EER accurately represents the operation of your air conditioner. A seasonal energy efficiency ratio (SEER) is often used. The SEER is calculated over a range of outside temperatures to get a more accurate seasonal average.)
(d) You are considering replacing your air conditioner with a more efficient one with an EER of 14.6. Based on the EER, how much would that save you on electricity costs in an average year?
Step by Step Answer:
University Physics with Modern Physics
ISBN: 978-0133977981
14th edition
Authors: Hugh D. Young, Roger A. Freedman