Forum discussion

Performance requirements for Water Heating Equipment

3

You rely on LEEDuser. Can we rely on you?

LEEDuser is supported by our premium members, not by advertisers.

Go premium for $15.95  »

Mon, 06/21/2010 - 13:39

Not sure I completely understand the question. For your base case, all those numbers should be the ones in Table 7.8. For the design case, the manufacturer should provide the numbers.

Tue, 06/22/2010 - 05:37

Thanks Chris, Ok as you say the base case should use the numbers in the table. Say for Electric water heaters >12kW, table says Performance required is "20+35 √V SL, Btu/h". V is the volume and SL is “stand by loss”. How can I get the SL figure? Secondly is the “Performance Required” equal to COP? Assuming there is no SL then the overalls result would be 20? 20 can not be a water heater COP? Normally we are talking about COP of 85% to 95% for electric water heaters. Would you please enlighten me? Cheers,

Fri, 06/25/2010 - 19:19

I think you've misinterpreted the formula. 20+35*√V=SL in BTU/H. So for a 25 gallon tank (let's say), the maximum Standby Loss (SL) is: 20+35 *√25=20+35*5=195 btu/hr

Sun, 06/27/2010 - 05:25

Thanks Chris, this was very useful.

Thu, 12/01/2022 - 17:21

Hi Chris, Could you do an example for a 50 gallon gas water heater with 76,000 Btu/hr input?  Is the Et term for baseline and proposed supposed to be prescriptive and design values, respectively?   EDIT: I found an example in the 90.1-2013 User's Manual.  :-)

Add new comment

To post a comment, you need to register for a LEEDuser Basic membership (free) or login to your existing profile.