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alberto23
Nov2-05, 10:28 PM
problem says:

6.25 employing the ideal gas model, determine the change in specific entropy between the indicated states, in kJ/kg K. Solve 2 ways: use the appropriate ideal gas table, and a constant specific heat from Table A-20.

(a) air, p1 = 100 kPa, T1 = 20°C → p2 = 100 kPa, T2 = 100°C
(b) air, p1 = 1 bar, T1 = 27°C → p2 = 3 bar, T2 = 377°C
(c) carbon dioxide, p1 = 150 kPa, T1 = 30°C, p2 = 300 kPa, T2 = 300°C
(d) carbon monoxide, T1 = 300K, v1 = 1.1 m3/kg → T2 = 500K, v2 = 0.75 m3/kg
(e) nitrogen, p1 = 2 Mpa, T1 = 800K → p2 = 1 Mpa, T2 = 300K

answers are:
A) 0.24289, 0.2431
B) 0.47632, 0.47684
C) 0.4769, 0.4862
D) 0.2701, 0.2696
E) -0.8373, -0.8389

i did part A fine and got the right asnwer. solving using the first method (appropriate ideal gas table)--which is table A22 in my book (fundamentals of thermodynamics edition 5). you just have to convert temp to kelvin and match it up with the corresponding "s" value. the exact values aren't on table so i had to interpolate. then i did s2-s1 and got the right answer. solving using the second method (Table A20--"ideal gas specific heats of some common gasses"), i used the temperatures to find the cp values. then i averaged the 2 cp values together and used the formula (delta)s=cp*ln(T2/T1). this gave me the right answer.


but...for part B, it is not constant pressure and it doesn't say if its constant volume or not, so i don't know how to use the specific heat table for this.

if you can help me.......thanks.

Astronuc
Nov3-05, 02:26 AM
try

\Delta s\, =\, c_p\, ln\, \frac{T_2}{T_1}\, -\, R\, ln\, \frac{P_2}{P_1}

remembering R = c_p - c_v

alberto23
Nov3-05, 11:47 PM
astronuc, thanks for the help. i found that formula in my book burried in a page of derivations. i'm pretty sure its the right one because my answers were very very close to the ones given. thanks!