I am modeling a school in which we are using several water-to-water geothermal heat pumps for our domestic water heating. I am assuming I should be able to get significant energy savings in comparison to a standard electric water heater. How do I model this between my baseline and proposed? Can't I use the ASHRAE minimum required efficiencies for an electric water heater based on the BTUH's needed and compare the energy usage to my W2W Heat pumps? My LEED reviewer did not like this approach the first time around. Thanks!
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Marcus Sheffer
LEED Fellow7group / Energy Opportunities
LEEDuser Expert
5912 thumbs up
October 17, 2013 - 5:54 pm
Table G3.1-11 Baseline (b) indicates your baseline should be electric resistance. See Table 7.8 to determine the efficiency of the baseline system. Not sure what your reviewer saw?
David Mirabile
LEED AP, BD+C68 thumbs up
October 18, 2013 - 9:30 am
The reviewer comment was essentially "Provide additional information to justify all water heater savings" and then described a method for low-flow credit of which I was not attempting.
With the volume of a water to water heat pump being minimal, what can I compare my efficiencies to between the baseline and proposed? So if I have roughly 20 gal of Volume between all my heat pumps, I am looking at about a 90% EF requirement? And my savings would be any efficiency above that 90% that my W2W heat pumps are actually producing? Maybe I am just not understanding the process and comparing the incorrect systems, can you enlighten me?
Marcus Sheffer
LEED Fellow7group / Energy Opportunities
LEEDuser Expert
5912 thumbs up
October 18, 2013 - 10:06 am
Sounds like you are comparing the right systems.
So your service water heating savings are basically all related to the efficiency difference of the systems correct? If so the percent energy savings you are claiming should be roughly proportional to the percent difference in efficiency. If these values are in alignment then just explain that to the reviewer. If not then check your modeling inputs for any potential issues.
David Mirabile
LEED AP, BD+C68 thumbs up
October 18, 2013 - 10:38 am
That is correct, or at least that was my intent, to base my savings on the efficiency differences. But the efficiency differences have to both be based on a heat pump correct? I cannot use an electric 20 gal water heater versus my heat pump? I know the focus of the energy savings is on the HVAC side, but it seems like LEED is not giving the savings on the SHW side its due. Outside of figuring in the low flow fixtures, is there any other approach to increase savings on the SWH side? Thanks for all you help and insight.
Marcus Sheffer
LEED Fellow7group / Energy Opportunities
LEEDuser Expert
5912 thumbs up
October 18, 2013 - 11:14 am
See my first comment. Your baseline is clearly electric resistance IMO. Even the "heat pump" listed in table 7.8 uses the same formula as the electric resistance water heater so there should be no difference.
Demand reduction is a very common way to show additional hot water savings. Other strategies (extra tank and pipe insulation, heat traps, etc.) tend to produce very minor additional savings.
David Mirabile
LEED AP, BD+C68 thumbs up
October 18, 2013 - 11:19 am
But that doesn't allow me to model a Volume difference in the baseline, based on a typical storage tank achieving the same capacities as my proposed, does it? (Thus lowering the efficiency requirement of my baseline and greatly increasing my savings).
Marcus Sheffer
LEED Fellow7group / Energy Opportunities
LEEDuser Expert
5912 thumbs up
October 18, 2013 - 12:25 pm
I am not understanding why you would have a significant volume difference. I would think that the w-to-w heat pumps would have the need for storage since their recovery would be slower? It would not be a fair comparison to compare your system with a 20 gallon electric water heater if that single heater could not meet the hot water demand. Is there storage associated with the heat pumps? If so the volume of the storage would be used for comparison in the baseline.
You size the baseline system based on 7.4.1 which indicates you size it based on the hot water demand as calculated for the proposed design. If this results in different storage volumes in both cases then you could attempt to justify a difference based on the storage needs in both cases. I think you would need to demonstrate why there is a difference.
David Mirabile
LEED AP, BD+C68 thumbs up
October 23, 2013 - 12:09 pm
OK, I had to investigate this by going to the engineer of design. So for my proposed, I have a 147.7 mbh output capacity on my W2W heat pump with an input of 7.148 kW and COP of 6.1. This uses a 318 gallon storage tank. Therefore, using 0.93-0.00132V for my req'd minimum efficiency, EF= 0.51. Then I can model against a baseline water heater of COP=0.51?
So,
COP=out/input
0.51=147,700 btuh / 3413 * kW,
or
147,700 btuh / 0.51 * 3413 = kW
84.9=kW
I therefore am supplying the same amount of HW capacity using (84.9kW-7.148kW) 77.8kW less energy?
Marcus Sheffer
LEED Fellow7group / Energy Opportunities
LEEDuser Expert
5912 thumbs up
October 24, 2013 - 9:12 am
The size category in Table 7.8 is based on the baseline system capacity as calculated to meet the proposed hot water demand. It is not based on the proposed input heat pump capacity. In your case it appears highly likely the the baseline size category will be over 12 kW. So you use the stand-by loss formula and assume the electric coil in the tank has a COP of 1.
Regarding your calculations above - the EF includes stand-by losses so it is not the same thing as COP. As a result the rest of your calculations are not valid.
David Mirabile
LEED AP, BD+C68 thumbs up
October 28, 2013 - 3:50 pm
Im sorry, I am still confused. With a proposed hot water demand of 147.7mbh being provided with a proposed 7.148kW w2w heat pump at COP=6.1 with a proposed 318 gal storage tank, not sure what that gets me when using the SL formula. A SL=644 btuh? What does that tell me in comparing baseline and proposed savings.
Marcus Sheffer
LEED Fellow7group / Energy Opportunities
LEEDuser Expert
5912 thumbs up
October 28, 2013 - 5:29 pm
Use the proposed demand to size an electric storage tank hot water system for the baseline. What size electric element is needed to meet the proposed demand? The storage tank size is identical to the proposed. That value you calculated is the stand-by loss of the storage for the baseline system. The COP is 1. These are the inputs you need to model the baseline system to determine the savings.
Francesco Passerini
engineer90 thumbs up
July 6, 2015 - 1:06 pm
Hello Marcus, I'm modeling the same situation that David Mirabile modeled (in the designed system the service hot water is heated by a geothermal heat pump).
You wrote: "The storage tank size is identical to the proposed."
I don't understand why.
In the proposed model the storage tank size shall be equal to the designed size, while in the baseline model it shall be autosized, correct?
In EnergyPlus we can autosize the storage tank size through the input parameter "Time Storage Can Meet Peak Draw": "This field provides the time, in hours, that the tank’s volume can sustain a peak draw. It is used to size the tank’s volume which is the simple product of peak draw volume flow rate and the draw time."
Then EnergyPlus can autosize the heating capacity. We can use the parameter "Time for Tank Recovery": "This field provides the time, in hours, that tank’s heater needs to recover the volume of the tank. The temperatures used to define recovery are a starting temperature of 14.4ºC (58ºF) and a final temperature of 57.2ºC (135ºF)."
Do you think it is an acceptable method according to paragraph 7.4.1?
As default, EnergyPlus uses 0,6 hours (36 minutes) for both parameters.
Best Regards
Marcus Sheffer
LEED Fellow7group / Energy Opportunities
LEEDuser Expert
5912 thumbs up
July 6, 2015 - 1:49 pm
90.1 does not regulate the size of the tank. Table G3.1-11 Baseline (a) is very clear that the system capacities must be identical to the Proposed.
Why - not exactly sure but many parameters are held neutral to eliminate game playing.
7.4.1 applies to the proposed design and the 90.1 citation above applies to the baseline.
Francesco Passerini
engineer90 thumbs up
July 7, 2015 - 3:40 am
I think that G3.1-11 Baseline (a) is valid for existing systems, i.e. for systems that existed before the new design.
In my case, the system is completely new and I think that I shall apply G3.1-11 Baseline (b):
"the system shall be sized according to the provisions of Section 7.4.1 and the equipment shall match the minimum efficiency requirements in Section 7.4.2".
Marcus Sheffer
LEED Fellow7group / Energy Opportunities
LEEDuser Expert
5912 thumbs up
July 7, 2015 - 11:23 am
I stand corrected, (a) applies to existing systems. Let me try again.
Are you claiming any savings for reduced hot water demand from low flow fixtures?
I would think that the capacity would be greater for a heat pump water heater compared to an electric resistance water heater.
Francesco Passerini
engineer90 thumbs up
July 7, 2015 - 11:37 am
No, I am claiming no savings for reduced hot water demand from low flow fixtures.
Marcus Sheffer
LEED Fellow7group / Energy Opportunities
LEEDuser Expert
5912 thumbs up
July 7, 2015 - 11:56 am
Then according to Table G3.1-11 Baseline (i) the service water loads and usage must be identical. So the peak draw must be the same in both models. The time for tank recovery with a heat pump water heater is usually slower so the tank must be larger than the Baseline electric water heater.
Are you actually producing any additional savings when you auto-size the baseline storage capacity?
If you model the storage the same it will not raise any flags to the reviewer. If you model it different then I would suggest you model it as an exceptional calculation.
Francesco Passerini
engineer90 thumbs up
July 10, 2015 - 4:59 am
1.In my baseline model EnergyPlus is autosizing the storage tank volume according to the input parameter "Time Storage Can Meet Peak Draw": "This field provides the time, in hours, that the tank’s volume can sustain a peak draw. It is used to size the tank’s volume which is the simple product of peak draw volume flow rate and the draw time.".
Is it an acceptable method? Or could I impose as tank size the same value of the proposed model?
The volume of the proposed model was designed by the mechanical designers.
2.In my baseline model EnergyPlus is autosizing the heating capacity as:
Heating Power = TankVol * WaterDensity * WaterSpecificHeat * 42.8°C / TimeforTankRecovery
Is it an acceptable method?
Now EnergyPlus is autosizing 17.8 kW, i.e. according to Table 7.8 the standby loss shall be equal to
20 + 35 V^0.5 [W]
Because of this equation, if I decrease the volume of the baseline model, then I decrease the standby loss of the baseline model.
Best Regards