Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 15 - The Laws of Thermodynamics - Problems - Page 440: 38

Answer

$1310 saved per year About 11 years $11,200 over 20 years

Work Step by Step

The $\$2000$ worth of heat that the electric heater provides is $Q_H$. The amount of energy W required for the heat pump to deliver that amount of heat is $Q_H$ divided by the coefficient of performance. $COP=\frac{Q_H}{W}$ $W=\frac{Q_H}{COP}$ Therefore, the cost of providing $Q_H$ from the electric heater should be divided by the COP to calculate the cost of running the heat pump. The annual savings is $\$2000$ minus that. Savings = $\$2000-\frac{\$2000}{COP}=\$2000-\frac{\$2000}{2.9}=\$1310$ Calculate the payback time in years. $$\frac{\$15000}{\$1310/year}\approx 11 years$$ The savings over 20 years is the savings in electricity minus the investment in the heat pump. $\frac{\$1310}{year}(20 years) - \$15000 = \$11200$
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