Expansion is an isentropic process. if we are providing pressure gauges we can measure the pressure at state point 3 and 4. (T3/T4)= (P3/p4)^(1.4/.4) where T4 is the temperate at the end of expansion.

rahul if you can't understand clearly sent reply i'll try give a correct answer after discussing with my professors

But T4 and P4 both are unknown

I think T3 has something to do with calorific value of fuel

Equating H(calorific)=Cp*(T3-T2) should yield something but i am nt sure about its validity

It wud be gr8 if u can confirm it with ur profs :)

We have to know the mass flow rate of the fuel entering and its calorific value and by assuming combustion efficiency we can calculate the change in enthalpy of the working substance.Then using mf.CV . eff of combustion + ma.cpa.T2 = (ma+mf)T3......By mass balance and energy balance we can get that....

hmmm... this looks good but the equation requires some Constant multiplied to RHS->(ma+mf)T3 in order to make it dimension-ally correct ...... wat is that constant ?

see...start with energy balance....in this case.....as u want to theoretically calculate how much T0 you require....taking all consideratiions try solving the energy equations with the help of other individual processes...if u still arenot getting the answer..it means u need more data......and also please read the following BIG essay of mine about practically measuring pressure using gauges..its not true....

may be expansion is an approximated isentropic process....but dont forget the T-s diagram of the engine cycles go like around a few 100 times each second....so within this second your combustion process occurs by DETONATION/EXPLOSION/DEFLAGRATION phenomena....so using normal pressure gauges will never give us the pressure in combustion....the combustion /explosion pressure is different.....and moreover if u use a wall mounted pressure gauge all it gives is the wall instantaneous static pressure that too for 1millisecond or less.....there are shock tube techniques to compute such stuff.....So the summary is that COMBUSTION PRESSURE IS NEVER MEASURED by normal gauges. please if u are not sure dont post wrong explanations....:)

Rahul singh sir actually in the RHS it shuld be mutliplied with Cp of the mixture.. Then the equation becomes ...

mf.CV . eff of combustion + ma.cpa.T2 = Cp(mix).(ma+mf)T3 and u can determine the values required

We can Calculate the temperature in the following way

if T1,T2,T3 and T4 are the temperature at cardinal points and T3 is the temperature to be calculated after combustion

then T3=(T2/T1)*T4

One can do some theoretical calculations by knowing the inlet temperature, inlet pressure, pressure ratio or compression ratio (be careful this second one is a volume ratio), appproximate pressure drop through the intake system, the flow rate of the air, and the flow rate of the fuel. Calculate the temperature after compression using isentropic compression but with a correction factor for efficiency of the compression process and heat exchange with the engine coolant. Write a combustion equation using the air flow and fuel flow then do a heat balance utilizing the heating value of the fuel and then iterate until you get a combustion temperature that gives you a heat balance. This will not give you the actual temperature because there are heat losses to the cylinder walls, head, and eventually cooling water. Use the enthalpies of the individual components of the combustion gases from a set of gas tables (or look up tables if you are doing this by computer).

The heat addition process in an Otto cycle is assumed to occur at constant volume. Given the thermodynamic state of the non-reacted mixture before combustion, the post-combustion state can be found by solving the conservation of energy of the gas kept at constant volume. Since the volume is assumed adiabatic, there is no heat transfer, and because th volume is kept constant, there is no work done. This means that the internal energy of the gas remains constant. The end state is assumed to be in chmical equilibrium. Given the constraints of fixed volume and internal energy, the equilibrium composition and state can found by maximizing the entropy of the gas.

Dedicated chemical eqilibrium codes can be used for this calculation, such as NASA's Chemical Equilibrium with Applications (CEA) code, available at

http://www.grc.nasa.gov/WWW/CEAWeb/

There is also an online applet for this calculation.

i guess by this tym u must've found your answer but in case u haven't

mf* LCV= (ma + mf) Cp* (T3-T2)

where Cp is the mean Cp of air and the fuel mixture.... even if u use Cp air that would give fairly accurate results. mf is the mass of fuel and LCV the lower calorific value, T2 you can find out using isentropic compression relation if u know the pressure ratio