Chapter 20, Problem 20.56P

DATA For a refrigerator or air conditioner, the coefficient of performance K (often denoted as COP) is, as in Eq. (20.9), the ratio of cooling output |Q_C| to the required electrical energy input |W|, both in joules. The coefficient of performance is also expressed as a ratio of powers,

$K=\frac{\left|Q_{\text{C}}\right|/t}{\left|W\right|/t}$

where |Q_C|/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 |Q_C| 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 95°F outside temperature and an 80°F return air temperature. Calculate EER for a Carnot device that operates between 95°F and 80°F. (c) You have an air conditioner with an EER of 10.9. Your home on average requires a total cooling output of |Q_C| = 1.9 × 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 unit’s 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?