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Hi all,

can someone help me please to determine the correct EF (energy factor) for baseline electric water heater according to ASHRAE 90.1-2007 per Table 7.8? The design case water heaters parameters are as follows: V = 2 x 1000 L = 2000 L (528,4 gal US) Electric Input = 2 x 4,5 kW = 9 kW

According to these parameters the equation in my opinion should be "0,93 - (minus) 0,00132 * V", because the electric input is less than 12 kW and overall storage volume is more than 20 gallons: 0,93 - 0,00132 * 528,4 = 0,232 EF = 23,2%

Please let me someone know if it is correct calculation. Obviously it is very bad efficiency and LEED reviewer insist that it should be corrected and rewised in energy model. But I don't know how it should be corrected, because it's just simple math. Maybe I'm wrong in the equation, or should be used another equation?

Thanks in advance for any advices and suggestions.

Martin

Hi all,

can someone help me please to determine the correct EF (energy factor) for baseline electric water heater according to ASHRAE 90.1-2007 per Table 7.8? The design case water heaters parameters are as follows: V = 2 x 1000 L = 2000 L (528,4 gal US) Electric Input = 2 x 4,5 kW = 9 kW

According to these parameters the equation in my opinion should be "0,93 - (minus) 0,00132 * V", because the electric input is less than 12 kW and overall storage volume is more than 20 gallons: 0,93 - 0,00132 * 528,4 = 0,232 EF = 23,2%

Please let me someone know if it is correct calculation. Obviously it is very bad efficiency and LEED reviewer insist that it should be corrected and rewised in energy model. But I don't know how it should be corrected, because it's just simple math. Maybe I'm wrong in the equation, or should be used another equation?

Thanks in advance for any advices and suggestions.

Martin

Hi all,

can someone help me please to determine the correct EF (energy factor) for baseline electric water heater according to ASHRAE 90.1-2007 per Table 7.8? The design case water heaters parameters are as follows: V = 2 x 1000 L = 2000 L (528,4 gal US) Electric Input = 2 x 4,5 kW = 9 kW

According to these parameters the equation in my opinion should be "0,93 - (minus) 0,00132 * V", because the electric input is less than 12 kW and overall storage volume is more than 20 gallons: 0,93 - 0,00132 * 528,4 = 0,232 EF = 23,2%

Please let me someone know if it is correct calculation. Obviously it is very bad efficiency and LEED reviewer insist that it should be corrected and rewised in energy model. But I don't know how it should be corrected, because it's just simple math. Maybe I'm wrong in the equation, or should be used another equation?

Thanks in advance for any advices and suggestions.

Martin

Hi all,

can someone help me please to determine the correct EF (energy factor) for baseline electric water heater according to ASHRAE 90.1-2007 per Table 7.8? The design case water heaters parameters are as follows: V = 2 x 1000 L = 2000 L (528,4 gal US) Electric Input = 2 x 4,5 kW = 9 kW

According to these parameters the equation in my opinion should be "0,93 - (minus) 0,00132 * V", because the electric input is less than 12 kW and overall storage volume is more than 20 gallons: 0,93 - 0,00132 * 528,4 = 0,232 EF = 23,2%

Please let me someone know if it is correct calculation. Obviously it is very bad efficiency and LEED reviewer insist that it should be corrected and rewised in energy model. But I don't know how it should be corrected, because it's just simple math. Maybe I'm wrong in the equation, or should be used another equation?

Thanks in advance for any advices and suggestions.

Martin