Hi again Sorry I did not read your mail very well. You wrote that the pilot fuel should be 10% of the gas energy. That means 1/11 or 9,0909% of the fuel supply to the engine which means 109 kWh = 10,32 l of diesel and you need 1091 kWh of gas which means 739 m3. Regards Bjorn Dahlroth
Hi If you want to get 20 kW of electricity for 21 hours this means 20 x 21 = 420 kWh of electricity. If the total efficiency of your diesel generating set is 35% you divide this by = 0.35 and you get 1200 kWh of fuel supply to the engine. Now it seems that you need a fuel mix of 10% diesel oil and 90% generator gas. Supposedly the diesel oil is required to ignite the gas in the engine. This means that you need 120 kWh of diesel oil which is 120/10.563 = 11.36 l of diesel. You also need 0.9 x 1200/1.476 = 732 m3 of gas. Then the problem is if your efficiency figures are reasonably correct, if the heat value of the gas is reasonably correct and if you really need more or less than 10% pilot fuel. These are questions which I have no reply to and you must find out from equipment suppliers or people running similar plants. Small engines have less efficiency than big ones and if you have load variations with much running at low load the diesel generating set could have a much worse efficiency. Bjorn Dahlroth -----Ursprungligt meddelande----- Från: [email protected] [mailto:[email protected]] För [email protected] Skickat: den 15 december 2010 21:00 Till: [email protected] Ämne: Gasification Digest, Vol 54, Issue 11 Send Gasification mailing list submissions to [email protected] To subscribe or unsubscribe via the World Wide Web, visit http://listserv.repp.org/mailman/listinfo/gasification_listserv.repp.org or, via email, send a message with subject or body 'help' to [email protected] You can reach the person managing the list at [email protected] When replying, please edit your Subject line so it is more specific than "Re: Contents of Gasification digest..." Today's Topics: 1. Re: engine exhaust relative humidity (Robert Kana) ---------------------------------------------------------------------- Message: 1 Date: Wed, 15 Dec 2010 06:50:56 +0700 From: Robert Kana <[email protected]> To: [email protected] Subject: Re: [Gasification] engine exhaust relative humidity Message-ID: <[email protected]> Content-Type: text/plain; charset=ISO-8859-1; format=flowed Good afternoon , Would this be a good argument of showing efficiency of gasification? Any suggestions? Here are my assumptions: -Size of System: 20kW (for first trial) -Daily Hours of Operation: 21 -Thermal Energy of SynGas: 1.476 kW-hr/m3 -Thermal Energy of Diesel: 10,563 kW-hr/m3 -Generator Efficiency: 35% -The energy input to the generator from diesel is 10% of the energy input from SynGas The daily SynGas requirement is therefore 813m3/day based on: (20kW * 21 hours) / (1.476 kW-hr/m3 * 35%) = 813m3. Therefore the total SynGas energy input to the generator is 813m3 x 1.476 kW-hr/m3 = 1200 kW-hr. Thus, the diesel energy input is 120 KW-hr (10% of 1200 kW-hr). To create 120kW-hr from diesel only requires 120 kW-hr/10,563 kW-hr/m3 = .01136m3 diesel. So, about 11.36 liters of diesel are needed daily. Many thanks, Robert Kana, Pt. Biomass Energy ------------------------------ _______________________________________________ The Gasification list has moved to [email protected] - please update your email contacts to reflect the change. Please visit http://info.bioenergylists.org for more news on the list move. Thank you, Gasification Administrator End of Gasification Digest, Vol 54, Issue 11 ******************************************** _______________________________________________ The Gasification list has moved to [email protected] - please update your email contacts to reflect the change. Please visit http://info.bioenergylists.org for more news on the list move. Thank you, Gasification Administrator
