Everyone, is there a relationship written down somewhere that removes fan energy from a heat pump COP value to develop a heating EIR? Something like the DX cooling EER equation that removes fan energy from EIR cooling values? It looks like the eQuest wizard is doing something like that when a heat pump heating COP is entered, but I am not sure what.
Hi Kathryn,
This is what we use:
EER adjusted = EER/(0.9585-0.01217*EER)
The conversion for EER to COP is: COP = EER/3.412.
Kristy Walson, PE, LEED AP
Mechanical Engineer / Sustainable Design
You can use this link to access discussion on this topic.
http://lists.onebuilding.org/pipermail/equest-users-onebuilding.org/2010-May/004407.html
Heating EIR = (1/(COP*3.413)-0.012167)/((1/3.413)-0.012167)
and
Heating EIR = (1/((HSPF*0.28+1.13)*3.413)-0.012167)/((1/3.413)-0.012167)
Hope this helps,
Ana
Date: Mon, 3 Jan 2011 08:12:26 -0800
From: kathryn.kerns at bceengineers.com
To: equest-users at lists.onebuilding.org
Subject: [Equest-users] Converting Heat Pump COP to EIR
Everyone, is there a relationship written down somewhere that removes fan energy from a heat pump COP value to develop a heating EIR? Something like the DX cooling EER equation that removes fan energy from EIR cooling values? It looks like the eQuest wizard is doing something like that when a heat pump heating COP is entered, but I am not sure what.
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Here's a good reference for use too. See attached,
Pasha
Hi Kathryn,
The document that Pasha included in her email will work if your heat pumps are terminal heat pumps (System 2) and are rated using the ARI 240/260 rating method. However, if you are modeling packaged rooftop heat pumps (System 4), these heat pumps follow a different rating method, ARI 340/360. I haven?t reviewed this in detail but my guess is that the fan power is a little different than the 240/260 method.
If you know the gross and net cooling capacities from the equipment cutsheet, you can calculate the fan power per CFM and an adjusted COP (or EIR) for the compressor/condenser power only. The difference between net and gross is the cooling required to overcome the fan heat and can be converted to Watts and used as the fan power. See the snapshot below from the 90.1-2007 Users Manual. It is an example on how to separate the two values.
Alternatively, you could review the 340/360 standard to see how fan power is accounted and use those numbers. It could be as simple as the method described in Pasha?s email or by assuming a value for total static pressure, fan efficiency and motor efficiency and the calculating the fan power directly.
John
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John T. Forester, P.E., LEED AP