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Water Heating Equipment Performance Requirements per ASHRAE 90.1-2007

asked 2017-03-31 09:16:55 -0600

markos gravatar image

updated 2017-05-08 16:10:23 -0600

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

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answered 2017-03-31 14:52:30 -0600

I think the issue is that you are confusing energy factor with thermal efficiency; they are not the same.

From the DOE Water Heater Website

The energy factor (EF) indicates a water heater's overall energy efficiency based on the amount of hot water produced per unit of fuel consumed over a typical day. This includes the following:

  • Recovery efficiency โ€“ how efficiently the heat from the energy source is transferred to the water
  • Standby losses โ€“ the percentage of heat loss per hour from the stored water compared to the heat content of the water (water heaters with storage tanks)
  • Cycling losses โ€“ the loss of heat as the water circulates through a water heater tank, and/or inlet and outlet pipes.

So thermal efficiency doesn't take into account standby losses, but energy factor does.

Below, I've laid out the steps to calculate water heater properties for an electric water heater with a capacity of less than 12kW per Enhancements to ASHRAE Standard 90.1 Prototype Building Models - Appendix A: Service Water Heating. This example assumes you are using EnergyPlus as the simulation engine, but these inputs should be found in the water heater model for other simulation engines as well.

# Water heater properties
volume_gal = 528.4
capacity_w = 9000

# Fixed electric water heater efficiency of 100% per document
water_heater_eff = 1 # 100%

# Calculate the minimum Energy Factor (EF)
base_ef = 0.93
vol_drt = 0.00132

ef = base_ef - (vol_drt * volume_gal)
ef = 0.93 - (0.00132 * 528.4) = 0.2325 # Which matches what you had

# Calculate the skin loss coefficient (UA)
ua_btu_per_hr_per_f = (41094 * (1 / ef - 1)) / (24 * 67.5)
ua_btu_per_hr_per_f = (41094 * (1 / 0.2325 - 1)) / (24 * 67.5) = 83.74

# Convert UA to SI units (for EnergyPlus)
ua_w_per_k = 44.17

# Set the water heater input properties

# Efficiency
setHeaterThermalEfficiency(water_heater_eff)

# Skin loss
setOffCycleLossCoefficienttoAmbientTemperature(ua_w_per_k)
setOnCycleLossCoefficienttoAmbientTemperature(ua_w_per_k)
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Thank you for your reply! So you are saying that EF is just for determining the skin loss and the water heater eff. is still 100% for electric heaters right? (which makes sense of course). I thought that EF according Table 7.8 is heater efficiency adjusted by the tank volume.

So once again: it doesn't matter what simulation engine is used, however the water heater eff. in baseline case is 100% (in case of electric water heater) and skin loss is determined by EF. Please confirm if it is true. I need to know it once and for all.

Thank you very much.

markos gravatar image markos  ( 2017-04-03 06:40:58 -0600 )edit

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Asked: 2017-03-31 09:16:55 -0600

Seen: 1,685 times

Last updated: Mar 31 '17