About EFF

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Hi,


I am confused about the use of EFF, if ACTIVITY LEVEL=CAP*AFA*CAP2ACT,

then where will the EFF be accounted for? How should I derive that value, given the energy consumption and other parameters?

Best,
Xiao

[Antti-L]

EFF is used for the main process transformation, which typically defines the conversion from the inputs to the activity (or from the inputs to the outputs). EFF is one alternative for defining the main process transformation, others are Input/Output, ACT_EFF, ACT_FLO, FLO_FUNC, and FLO_EFF. If the process transformation is left undefined (and your process is output-normalized) your process would thus not consume any inputs.

For example, if you have a LED lamp process representing typical household lighting with each lamp having luminous power of 800 lumen and input electrical power of 10 watts, you could define EFF = 800/10 = 80, assuming that the process is output-normalized and the activity represents the luminous energy (in petalumen•second). Note that although the efficiency looks high, it is just because in terms of energy, lumen-second is a unit much smaller than joule:
1 (lm•s) = 1.464x10-3 joules at 555nm.

[[email protected]]

EFF is used for the main process transformation, which typically defines the conversion from the inputs to the activity (or from the inputs to the outputs). EFF is one alternative for defining the main process transformation, others are Input/Output, ACT_EFF, ACT_FLO, FLO_FUNC, and FLO_EFF. If the process transformation is left undefined (and your process is output-normalized) your process would thus not consume any inputs.

For example, if you have a LED lamp process representing typical household lighting with each lamp having luminous power of 800 lumen and input electrical power of 10 watts, you could define EFF = 800/10 = 80, assuming that the process is output-normalized and the activity represents the luminous energy (in petalumen•second). Note that although the efficiency looks high, it is just because in terms of energy, lumen-second is a unit much smaller than joule:
1 (lm•s) = 1.464x10-3 joules at 555nm.

Hi,

Thank you for the reply.

So if the input and output are all with a unit of PJ, generally I can define the EFF as 1 (if without any energy loss, like heat loss). 

On the other hand, is the EFF same as the technique efficiency? 

For example, the following denoted that the efficiency for residential cooling process are 12%-13% (RSDELC as the input while the R_ES-SD-SpCool as the output, both with a unit of PJ), does that mean that the EFF is 0.12 or 0.13?

New-Unit-Efficiencies

Room (EER)1) 9.4 12

Central (SEER)2) 10.3 13





1) Energy Efficiency Ratio.
2) Seasonal Energy Efficiency Ratio.

also my understanding is INPUT commodity (PJ)=Activity level=STOCK*CAP2ACT*AFA, and the OUTPUT commodity (PJ)=STOCK*CAP2ACT*AFA*EFF, right?

Best,
Xiao

[Antti-L]

> So if the input and output are all with a unit of PJ, generally I can define the EFF as 1 (if without any energy loss, like heat loss). The EFF is same as the technique efficiency ?

Right, if the process has no conversion losses, you could define EFF = 1.

> my understanding is INPUT commodity (PJ)=Activity level=STOCK*CAP2ACT*AFA, and the OUTPUT commodity (PJ)=STOCK*CAP2ACT*AFA*EFF, right?

Wrong. In general, process equations are generated for each timeslice, but assuming that EFF and AFA have indeed been defined, we could write in simplified terms on the ANNUAL level (region and process index omitted, no vintaging, all commodities of same type, default PCG):

  (1) SUM({s in S, c in Inputs} VAR_FLO(t,c,s) × ACT_EFF(t,'ACT',s)) = SUM({s in S} VAR_ACT(t,s))
  (2) SUM({s in S} VAR_ACT(t,s)) = SUM({s in S, c in Outputs} VAR_FLO(t,c,s) / PRC_ACTFLO(t,c))
  (3) SUM({s in S} VAR_ACT(t,s)) ≤ VAR_CAP(t) × PRC_CAPACT × NCAP_AFA(t,'UP')

You can see that the capacity-activity relations (3) are, of course, by default inequalities.
EFF(r,y,p,s) is a VEDA shortcut for ACT_EFF(r,y,p,'ACT',s).

[Antti-L]

Concerning your EER/SEER question, explanation from Wikipedia:
The EER uses mixed units, so it does not have an immediate physical sense and is obtained by multiplying the COP by the conversion factor from BTUs to watt-hours: EER = 3.41214 × COP (see British thermal unit).

In other words, those numbers you presented (12–13) are not percentages, but ratios that do not have any immediate physical sense.

For example, if EER= 12, it means that EER = 3.41214 × COP = 12, i.e. COP = 12 / 3.41214 = 3.517.  COP is the ratio of the useful energy output to the electrical input of the heat pump, and therefore, in essence, the useful cooling energy efficiency would thus be 352%, and not 12% like you seemed to assume. A COP of 3.517 (useful energy efficiency 352%) would indeed be quite reasonable for a cooling heat pump.