Talk:Spark spread

Multiplication / Division

The text says: "The spark spread value is therefore the power price minus the gas price multiplied by 0.4913, i.e. Spark Spread = Power Price – (Gas price/0.4913)." So it says "multiplied" but then it is divided by the efficiency! —Preceding unsigned comment added by 147.114.226.194 (talk) 18:58, 14 April 2010 (UTC)[reply]

Hi, just a note about this: The above user said there was written "multiplied" and then the formula divided, so someone "corrected" it making a division in both, text and written formula. But that's wrong! It should be a multiplication (and it should be the gas price x 0.50); if it's left as a division, it would mean the plant can generate more MWh than it pays for in the commodity "gas", and no plan can get more energy from gas than the energy gas itself has, it's just the opposite. — Preceding unsigned comment added by 195.144.99.90 (talk) 09:58, 26 November 2018 (UTC)[reply]

Doubts about the chapter "Spark spread as cost of replacement power for intermittent renewables"

What is written here does not apply to any country that I know. In most European countries renewables either are treated like must-run plants so that conventional plants have to be switched off or they crowd out conventional plants in the electricity market due to their low running cost. And I don't know of any country that these conventional plants get a refund for being switched off, which is a well-known challenge for building new conventional plants (e. g. in the UK or Germany). So I wonder the statements in this chapter are rather an exception than a rule. So I suggest to correct or delete this chapter as long as there are no references to original information. — Preceding unsigned comment added by 85.178.146.82 (talk) 10:56, 30 November 2012 (UTC)[reply]

Text reformulated - tag to be removed

I have substantially reformulated the article. I hope that Wiki standard quality can be achieved from this point by further incremental improvemnets. MGTom 00:41, 16 September 2006 (UTC)[reply]

Transwiki tag removed. MGTom 08:24, 10 October 2006 (UTC)[reply]

UK standard spark spread 49.13% efficiency

I have previously deleted a mention of a standard efficiency used for spark spread calcularion: 49.13% . Now I have found confirmation of the figure as used by Argus consultants. The origin of this odd figure remains to be documented. The Dutch standard efficiency is 50%, which seems reasonable - why use odd numbers if they do not bring any real content? (Occams razor) MGTom 08:11, 10 October 2006 (UTC)[reply]

I wonder if it is due to the fact the Dutch use the LCV for gas, whereas the Brits use HCV = there is a 10 % difference...Engineman 22:03, 9 November 2006 (UTC) @100% efficiency, 1 th ≈ 29.30710701583 kWh; and @49.1348395193765% efficiency, 100k th is 1440 MWh, which is 60MW on Adrian Peterson. I don’t think he is Dutch though. — Preceding unsigned comment added by 72.198.106.110 (talk) 05:25, 18 April 2012 (UTC)[reply]

"In reality, each gas-fired plant has a different fuel efficiency, but 49.13% is used as a standard in the UK market because it provides an easy conversion between gas and power volumes (25,000 therms of gas = 15 MWh of energy). The spark spread value is therefore the power price minus the gas cost divided by 0.4913, i.e. Spark Spread = Power Price – (Gas cost/0.4913)." - I dont get this explanation either. If a therm is 29.3 kWh, then 25000 therm of natural gas as fuel input equals 732.500 kWh. Multiplied with an efficiency of 49.13% that results in 359.877 kWh, i.e. 360 MWh and not 15 MWh. Can somebode give a better hint where the 49.13% are coming from. "15 MWh of energy" - is this electricity output or fuel input? --Gunnar (talk) 14:45, 25 February 2018 (UTC)[reply]

edit

I've taken the below sentence out since I don;t see the point being made, or if I do, I think it is erroneous.

Whether transmission and distribution costs are higher or lower depends on the location to which backup power has to be delivered.

The above is surely not true, or misses the point. The point trying to be made is, what does the National Grid itself cost - answer - unequivocally about 0.2p p / kWh on all units, so if you are forced to double it for whatever reason, you might expect to incur costs of about 0.2p p / kWh on all units.

Wind farms always have to quote and allow for their costs of connecting to the grid so the sentence seems to me to be irrelleavnt or missing the point....

I think without the explicit calculation of the cost the whole point is lost.

Happy to be proved wrong.....Engineman 14:42, 19 November 2006 (UTC)[reply]

Changes of Dec. 2, 2023

Greetings Wikipedians! I found an equation that was not accompanied by a citation to a reliable source. So I added a simpler equation that I could support (in text, since I'm not proficient in math notation). In researching this, I found that it's more complex than the earlier text indicated (e.g. hedge ratios), so I addressed that with additional text and a citation. Comments are welcome. Cordially, BuzzWeiser196 (talk) 19:36, 2 December 2023 (UTC)[reply]