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试卷: LEED AP ID+C 模拟试卷C[查看] The interior lighting power allowance of a tenant space is 23,280 watts. What minimum lighting power reduction must be achieved for the project to earn 6 points?
A: 10,476 watts B: 8,148 watts C: 6,984 watts D: 9,312 watts 参考答案: D
本题解释: Reference: Energy and Atmosphere Credit, Optimize Energy Performance-Lighting Power, Calculations
To earn 6 points the project needs to achieve a 40% reduction. 5 points would be earned for Energy and Atmosphere Credit, Optimize Energy Performance-Lighting Power and 1 point for exemplary performance under ID Credit 1.
To calculate the lighting power reduction use each answer choice to determine what will yield the 40% reduction.
Percent reduction = Lighting power reduction (watts) / Interior lighting power allowance (watts)
9,312 watts / 23280 watts = 40% YES
10,476 watts / 23280 watts = 45% YES, but the question asks for the minimum reduction
8,148 watts / 23280 watts = 35% NO, earns 5 points
6,984 watts / 23280 watts = 30% NO, earns 4 points
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