Question

Answers are given.show me how to get them

4 This problem is modified from Matsoukas (original notes) Example 4.7. First, we will look at the ideal gas case. In a flowing system at steady-state, two components l = CO2 (17% mol) and 2-n-pentane are compressed adiabatically and slowly enough to be considered reversible. The conditions start at T 398 K and 1.3 bar, and the outlet T is measured at 423 K. What is the total change in entropy? What is the outlet pressure, and how much work was done per mol? CPD -1 . 1 57h405 0 Species Tc (K) Pc (bar) CPA CPB CPC 0 ? CO2 304.2 73.8 0.225 5.457 1.045E-03 nCs 469.6 33.7 0.251 2.464 4.535E-02 -1.411E-05 CPD,1 answer: AS 0, P 3.56 bar, W 3430 J/mol

Answer 1

There's slight difference in asnwer. Method is correct.

There's slight difference in answer. Method is correct. first and second law for a closed reversible process Q = delta u - w Rightarrow 1 delta s = Q/T + s_gen Rightarrow 2 Q = 0 for an adiabatic process therefore from equation 2 delta s = 0 for a reversible adiabatic process Rightarrow 3 from equation 1 delta u = w Rightarrow 4 for an ideal gas system delta s_ig = integral cp/T dT - R d ln p Rightarrow 5 R ln p_2/p_1 = integral_T_1^T_2 sum x_i cpi/T dT = integral (x_1 cp_1/T + x_2 cp_2/T) d T Rightarrow 6 Here x_1 = 0.17, x_2 = 0.83 cp_1 = R[5.457 + 1.045 times 10^-3 T - 1.157 times 10^5/T^2] cp_2 = R[2.464 + 4.535 times 10^-2 T - 1.411 times 10^-5 T^2] substituting above in equation 6 R ln p_2/p_1 = R integral[(5.457/T + 1.045 times 10^-3 T/T - 1.157 times 10^5/T T^2) times 0.17 + (2.464/T + 4.535 times 10^-2 T/T