I've received a comment from the LEED reviewer after they denied my approach in calculating the LPD for the proposed building (dwelling units). They recommended to follow the "The Energy Star Multifamily High Rise Program Simulation Guidelines, section 3.6.2.3 says: - "Hardwired fixtures in rooms, such as bedrooms and living rooms, that may be supplemented by lighting that is connected to receptacles must be estimated to provide illumination at a rate of no more than 3 ft2 per Watt." But in some spaces (all bedrooms) the designed lighting provide illumination at a higher rate (over 7 ft2 per watt)? Shall I only use 3 ft2 per my provided watts and account for the rest of the area using the 1.1 Baseline LPD? Also the In-unit lighting worksheet of the Performance Path Calculator, calculates the Proposed LPD. Should this number be applied for all the proposed spaces identically (ex; all spaces would get 0.8 LPD)? although it is a combination of the actual design layout for the hardwired fixtures (0.5 LPD for some spaces like bedrooms) and the plug-ins lighting (1.1 LPD for the spaces with no fixtures on the layout like living rooms). I hope someone here who went through this before can assist me to get over it too. Thank you in advance.

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Marcus Sheffer

LEED Fellow

7group / Energy Opportunities

LEEDuser Expert

4454 thumbs up

Tue, 03/22/2016 - 18:00

The proposed is modeled as designed. So if it is greater than the baseline allowance you model it that way. Residential in-unit lighting is not regulated by 90.1. So you must model it identically in both models. If you wish to claim savings you follow the guidelines above. You can only claim savings for hardwired fixtures. The spaces must meet the illumination requirements in the guidelines and you need to demonstrate with room photometrics that they do meet those illumination levels. The proposed is modeled as designed. Any spaces or supplemental lighting in hardwired spaces must be modeled identically in both models at 1.1 W/sf. The baseline is 1.1W/sf. The lighting schedule is also restricted to 2.34 h/d. The issue of 3 sf/W as a restriction limits the illumination from hardwired fixtures. If you have 7sf/W that means you have a lower rate of illumination, not a higher one. If that is the case i would be concerned that you will not be able to meet the illumination requirements at all. If you can't meet the illumination requirements you cannot use this methodology. By far the hardest part of this methodology is demonstrating that you meet the minimum illumination requirements since photometrics are not often generated for this type of project. I would start there and if you meet the illumination requirements then try and figure out the wattage per area to include in the units.

Khaled Fouda

Design Engineer, LEED AP (BD+C)

Energy Concepts

LEEDuser Basic Member

5 thumbs up

Wed, 03/30/2016 - 14:47

Thank you, Marcus for your response. I have some points to discuss here; - I got a bit confused when you said "The proposed is modeled as designed. Any spaces or supplemental lighting in hardwired spaces must be modeled identically in both models at 1.1 W/sf. The baseline is 1.1W/sf." Well the hardwired spaces are the spaces with designed fixtures, do I model these ones as designed (0.5 W/sf) or identically to the baseline as (1.1 W/sf). I do understand that I should design the spaces with NO specified hardwired lighting in the proposed model as (1.1 W/sf). - The way I apply the 3sf/W restriction limit is that I only apply it to a specific area in the space to match this rule and I apply 1.1 W/sf for the rest of the space (ex: if the fixture is 5 watts and the area is 20 ft2, I apply the 3ft/W for only 15 ft2 and the rest of the area would be modeled as 1.1 W/sf, so in average the space LPD will be 0.525 W/sf). The MFHR Performance Path Calculator does it in this way to find the average LPD of the space.

Marcus Sheffer

LEED Fellow

7group / Energy Opportunities

LEEDuser Expert

4454 thumbs up

Wed, 03/30/2016 - 15:21

- In those spaces with hardwired fixtures you model the Proposed as designed. - I do not think that is how it works. Let's say you have a hard wired fixture in a 10x10 bedroom. Your fixture would be limited to 33 watts.The bedroom must meet the illumination requirements without any supplemental lighting. If it doesn't you can't claim any savings in that room. Perhaps if your room was L shaped and one leg had a hardwired fixture and the other did not you would separate the hardwired area from the non-hardwired area and the non-hardwired area would be 1.1 W/sf. That is my guess about what the calculator is trying to do. It would make sense when you had a space with a mix of hardwired and non-hardwired lighting. It would not make sense in a space with only hardwired lighting.

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