This appendix is posted
for review. The author   
welcomes comments.  

Appendix to:

Electricity in The Netherlandsa.

Windmills increase fossil fuel consumption & CO2 emission.
Details about the calculations.


back to article

Old Model.

The thermal efficiency of an electricity generator, η, is defined as:

η = E / Q (1)

in which E = the output of electric energy, [kWh]
Q = the input of thermal energy, same units,
η = the thermal efficiency.

The thermal energy needed to produce 1 kWh is:

1 / η kWh = 1000 / η Wh = 3,6 x 106 / η Ws (J) (2)

Tabel
The power dependency.

η is power dependent
power η
coal fired
η
CCGT
η
OCGT
η
nuclear
100%
80%
60%
40%
0,455
0,45
0,43
n.v.t.
0,59
0,57
0,55
0,49
0,32
~
~
~
0,377
0,37
n.v.t.
n.v.t.

Note: These η-values have been measured while the generator is running stationary.
They do not apply when it is ramping up or down to a different power level!

The figures allow for a calculation of the energy or power needed to produce electric energy or to provide a certain electric power. I.e. multiply with 1/η. Please note: same units. Throughout the article we use MWh & kWh or MW & kW. If necessary we add the suffix "therm", when fuel energy is meant.

Table 1 (article) lists the latent heat of earth gas (32 x 106 J/m3). Therefore 1 kWh electric energy requires 6,10 / 32 m3 = 0,19 m3 gas.
The CO2 emission when burning gas is also indicated in the table (2,5 kg CO2/m3).
This results in 0,475 kg CO2 for 1 kWh electricity.
In the old model this is the saving provided by a wind turbine if it produces 1 kWh.

back to article

Current Model.

The thermal efficiency of a generator depends on the power level. It is maximal at design power and it becomes less at lower power levels. The table above alows us to calculate the relative savings at different power levels as listed in table 3 of the article.

{(power at 100%)/η100% - (power at X%)/ηX%} / {(power at 100%)/η100%} (3)

Relative setbacks have been calculated in a similar way:

(saving old model - saving current model) / (saving old model) (4)

In our calculations, where we needed η-values at all kinds of power levels of CCGTs, we have interpolated the above figures using two straight lines. One was used in the range 100% - 60% and the other between 60% and 40%.

When wind electricity is fed into the grid, the conventional generators are ramped down. Once they are running stationary again, the fuel use can be calculated with the pertaining η-values. This produces saving figures as in table 3 of the article.
In the remainder of the article we have interpolated these η-values when computing the fuel use in a stationary mode.

back to article

Quasi stationary cycling.

In quasi stationary cycling, a theoretical construct that does not include the special non-equilibrium phenomena of cyling, we assume that during the actual ramping, which takes a certain time, the fuel consumption is at any power level to be calculated according to (1), using the appropriate η-value. We split the fuel use during a half hour into a ramping part and a stationary part. The ramping part lasts tr,i = Δi / 12 min. The other, stationary part: ts,i = 30 - Δi / 12 min.

Δi = | PGT,i-1 - PGT,i  | (5)

in which PGT,i stands for the power level of the CCGT+ during each half hour. Therefor the quasi stationary fuel use of the CCGT+ each half hour, Fqs,i, is:

Fqs,i = (tr,i / 60) x (PGT,i-1 + PGT,i) / (ηi-1 + ηi) + (ts,i /60) x PGT,i / (ηi)    [MWhtherm] (6)

The fuel use that half hour has to be compared with the consumption of the CCGT without wind. This would be: x 500 / η100% [MWhtherm].
The quasi stationary model thus shows a fuel saving Sqs,i:

Sqs,i = x 500 / η100% - Fqs,i [MWhtherm] (7)

back to article

Cycling.

During a full cycle in one hour from 100% power back to 80% and up again to 100% the CCGT uses 1,01 times the fuel it would consume during one hour at 100%. For our CCGT this means: 1,01 x 500 / η100% = 855,9 MWhtherm, ( 4a of the article). If the unit would have run stationary at full capacity it would have used 847,5 MWhtherm. The difference, 8,44 MWhtherm is equivalent to 950 m3 gas.
Following the quasi stationary model the fuel use would have been Fqs,80% (6) with
tr,80% = 2 x 100 / 12 = 16,7 min
ts,80% = 60 - tr,80% = 43,3 min
η100% = 0,59
η80% = 0,57
PGT,100% = 500 MW
PGT,80% = 400 MW
Resulting in: Fqs,80% = 722,4 MWhtherm.
This means that the nett cycle loss in this example is: 855,9 - 722,4 = 133,5 MWhtherm.

The same cycle loss would apply if it had taken place during a half hour, as long as the ramping speed stays 12 MW/min.
For half a cycle we assume that it would be half as much, i.e. 66,75 MWhtherm. Following the assumptions in 5, 1, 2 & 3 for each half hour we can now calculate the real fuell saving by correcting the quasi stationary result (Sqs,i) with the nett cycle loss Ci = (Δi / 100) x 66,75 MWh.
Thus we obtain for the fuel saved every half hour:

Si = Sqs,i - Ci   [MWhtherm] (8)

The sum of these,
Σi Si   [ MWhtherm] (9)

gives the total savings for the day accounting for the cycling effect.

back to article

Energy costs of building installing and grid adaptation of wind turbines.

We did this for two scenarios, an average lifetime of the installations of 15 year and one of 30 year. We combined the adaptation with the energy costs of turbine construction and installation. In the last case each one requires to subtract 5% of the electricity produced by the wind turbines, together 10%.
In the first case this is 10% each, or 20% both together. This subtracted electric energy must be delivered by the gas turbine, which results in extra fuel consumption. The latter we derived using η100%.

windcapacity 100 MW 200 MW 300 MW
wind el. production [MWh] 448,7 897,4 1346,1
life time 30 yr [MWh] 44,9 89,7 134,6
life time 15 yr [MWh] 89,7 179,5 269.2
fuel costs (30 yr) [MWhtherm] 76,1 152,2 228,2
fuel costs (15 yr) [MWhtherm] 152,1 304,2 456,3

The yellow marked figures have to be deducted from the savings found in the previous to pay back the energy investment in the wind electricity set-up.

back to article

More OCGTs.

Following 8 we suppose that because of requirements about grid stabillity, i.e. to cope with rapid wind variations, a fraction of the high efficiency CCGT-generators has to be replaced by OCGTs. They can be regulated faster. But their efficiency is about half that of a CCGT. We assume that with a windpark of 100 MW 3% of the CCGT capacity has to be replaced by OCGTs. With 200 MW wind: 6% and with 300 MW 10%. We find a new effective ηeff through the following:
First we calculate the average η during the 21,5 hours of CCGT operation. They differ slightly for the different wind parks:
ηav,100 = 0,5858
ηav,200 = 0,5817
ηav,300 = 0,5771
These we use together with the electric energy production of the gas turbine without cycling or life time correction to obtain the fuel used during the day if no OCGTs had been necessary. (Resp. 10301,3 MWh; 9852,6 MWh & 9403,9 MWh) Next we calculate the effective η of the combination of CCGT & OCGT:
ηeff,100 = 0,97 x 0,5858 + 0,03 x 0,32 = 0,5778
ηeff,200 = 0,94 x 0,5817 + 0,06 x 0,32 = 0,5660
ηeff,300 = 0,90 x 0,5771 + 0,10 x 0,32 = 0,5514
Using these efficiency values we calculate anew the fuel used by the gas turbines. The difference between these and the previous values is the fuel loss due to the partial replacement. They amount to: 243,7 MWhtherm; 469,9 MWhtherm & 759,8 MWhtherm.

These fuel quantities have to be subtracted from the savings calculated before. The result is listed in table 4 of the article. It is negative in all configurations.

back to article

Results and conclusions

A loss of fuel of 2,3% of the total amount of fuel which a 100% running 500 MW generator would have used during the 21,5 h period is:

0,023 x 21,5 x 500 / 0,59 = 419,1 MWhtherm (10)
This is eaquivalent to 419,1 x 106 x 3600 = 1508,8 x 109 J.
Which would require 1508,8 x 109 / 32 x 106 = 47150 m3 gas.
That would have released 2,5 x 47150 = 117,9 ton CO2 into the atmosphere.

back to article