Explanation Determine the lighting power saved in refrigerator. lighiting power saved = l light bulb − l energy-efficient bulb (I) Here, the energy consume by the light bulb is l light bulb and the energy consume by the energy-efficient bulb is l energy-efficient bulb . Determine the lighting energy saved a year by the energy efficient bulb. lighting energy saved = ( lighting power saved ) × ( operating hours ) (II) Determine the amount of energy saved in refrigeration. Refrigeration energy saved= lighting energy saved COP (III) Here, the coefficient of performance of the refrigeration is COP. Determine the total energy saved by the energy efficient light bulb. Total energy saved = ( lighting energy saved + refrigeration energy saved ) (IV) Determine the amount of money saved by the energy efficient light bulb. Money saved = ( total energy saved ) × ( unit cost of energy ) (V) Conclusion: Substitute 40 W for l light bulb and 18 W for l energy-efficient bulb in Equation (I). lighiting power saved = ( 45 W ) − ( 18 W ) = 22 W Substitute 22 W for lighting power saved and 60 h / year for operating hours in Equation (II). lighting energy saved = ( 22 W ) × ( 60 h / year ) = 1320 Wh / year = 1320 Wh / year × ( 1 kWh / year 1000 Wh / year ) = 1

9th Edition

ISBN: 9781259822674

Author: Yunus A. Cengel Dr., Michael A. Boles

Publisher: McGraw-Hill Education

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Chapter 6.11, Problem 122P

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The energy savings of the proposed light bulb justify its cost.

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Chapter 6.11, Problem 121P

Chapter 6.11, Problem 123P

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Chapter 6 Solutions

Thermodynamics: An Engineering Approach

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The energy savings of the proposed light bulb justify its cost.

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