Stoichiometric Reaction: A stoichiometric mixture of ethanol (C_2H_6O), octane (
ID: 898542 • Letter: S
Question
Stoichiometric Reaction: A stoichiometric mixture of ethanol (C_2H_6O), octane (C_8H_18), and AIR initially at T_1 = 300 K, P_1 = 100 kPa, P_2 =4500 kPa, and V_1 = 0.005 m^3 is compressed in a piston-cylinder assembly in a polytropic process. The final volume V_2 = 2.1 Times 10-4(m^3). The work is -2.225kJ. Determine the initial total volume of moles (kmol), the final temperature and the heat transfer (Q) in kJ. Use the equation for heat transfer: Q = W+ n*Cv(T2-T1) When calculating the number of moles for air, the value must include the O_2 and N_2 combined values as shown below:Explanation / Answer
from P1V1/T1= P2V2/T2
P1=100Kpa
V1= 0.005m3
T1=300K
P2=4500 Kpa
V2=2.1X10-4m3
T2=?
T2= P2V2*T1/P1V1= 4500*2.1X10-4*300/(100*0.05)=567K
Basis : 1 mole ethanol
it requires 9 moles of octane and 115.5 moles of oxygen
Air contains 21% oxygen and 79% N2 hence N2 in air =(79/21)*115.5 =3.76*115.5 moles of Nitrogen
Initial toal moles = 115.5(1+3.76)+1+9= 559.78 moles intially
For calculation of Cv average Cv based on all the three reactants is considered
mole fractions
Ethanol 1/559.78= 0.001786
Octane 9/559.78=0.016078
Air 549.78/559.78 =0.9821
Mixture Cv= 0.001786*134.9 + 0.016078*265.3+ 0.9821*21.24 =25.366 Kj/Mol.k
Q= heat transfer= -2.225+559.78*25.366*(567-300)=3791378 KJ