Physics Help Forum Finding temperature before combustion

 Thermodynamics and Fluid Mechanics Thermodynamics and Fluid Mechanics Physics Help Forum

 Mar 18th 2017, 11:23 AM #1 Junior Member   Join Date: Mar 2017 Posts: 3 Finding temperature before combustion Hi All, I'm completely stuck I've been trying to find the temperature of air in an otto cycle before the fuel is ignited. I know exhaust air is 733^C, Cp is 1005, air flow mass rate is 0.00460 fuel mass flow rate is 0.000393. The cal val for fuel is 46.5 Mj/kg, and im using the equation Q=m x cv (t3-t2) but...i just get a ridiculous answer every time. 46.58106 x 0.000393=18 kj/kg 1.225 * 0.00460=0.005635 cv=718 Q=m x cv (t3-t2)=>(Q/(m x Cv))-t2=-t1 (18 x 103 / (718 x 0.005635))-1006=3442k for the lower temperature, I'm not sure whats going wrong. If anyone could shed some light that would be great Many thanks
 Mar 18th 2017, 02:29 PM #2 Senior Member   Join Date: Apr 2015 Location: Somerset, England Posts: 926 First and most important. You should be using Cv for the Otto cycle - it is a constant volume process. The Diesel cycle is the constant pressure process. For the Otto cycle T2/T1 = T3/T4 Edit Sorry you started by stating Cp, I'm not quite sure why? But I now see that you are using Cv in your calcs. You also need the compression ratio. topsquark likes this. Last edited by studiot; Mar 18th 2017 at 03:03 PM.
 Mar 19th 2017, 07:36 AM #3 Senior Member   Join Date: Apr 2015 Location: Somerset, England Posts: 926 OK after much head scratching I have finally identified your problem. This could have been done much more easily if you had laid your work out clearly. You have correctly identified the exhaust temperature this T3 in the cycle and converted 733 degrees celsius to 1006 degrees Kelvin and also that the pre ignition state is state 2 on the diagram. Then you have said heat gained by air = heat supplied by fuel air mass per cycle x Cv x (T3-T2) = cal value x fuel mass per cycle Both of these are measured in Joules, not J/kg as you have written. However the above equation is not quite true which accounts for the enormous difference between the left hand side and right hand side. This is because the heat energy from the fuel goes to increase the total internal energy of the air when it goes from point 2 to point 3 in the PV diagram. Both the temperature and the pressure change (increase) at constant volume and both changes contribute to the increase in internal energy. So in going from state 2 to state 3 {P2, V2 and T2} go to {P3, V3 and T3} Of course this is a constant volume change so V2 = V3. You will need to use the gas laws to calculate back to P2 and T2 from the internal energy change as supplied by the fuel (which you calculated correctly except for the units as already noted) I will leave you to have a go before posting more. Incidentally, you have quoted the mass flow rate but not supplied units. These will be per cycle, not per second. topsquark likes this.
 Mar 19th 2017, 10:20 AM #4 Junior Member   Join Date: Mar 2017 Posts: 3 Thanks to you both for your input, very much appreciated. Really sorry for the presentation I have approached it from a different angle using the efficiency equation for Otto cycles and found the temperature. My original attempt was to try to find an unknown compression ratio in an test engine where I had the thermal efficiency, V1,T1,T3 η=1-1/(rv^(γ-1) )=59.26%=>0.5926-1=-1/(rv^(γ-1) ) Heat capacity ratio, γ=1.4 for air 0.5926-1* r_v^0.4=-1 r_v^0.4=(-1)/(-0.4074) 0.4 In rv=In2.4546 In rv=In2.4546/0.4 In rv=((In 2.4546))/0.4 In rv=2.245 e^2.245=9.44=rv T_2=T_1 r_v^(γ-1) T_2=286*〖9.44〗^0.4=702.02K This has lead me to another query, my T2 is around 702.02K, yet the auto ignition of petrol is much lower. I'm guessing this is probably something to do with pressure and octane values but the internet is not giving any clear answers. Again any feedback would be appreciated Kindest regards

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