[thermodynamics] Is it possible problem.

  • Thread starter Thread starter constfang
  • Start date Start date
  • Tags Tags
    Thermodynamics
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
3 replies · 2K views
constfang
Messages
10
Reaction score
0
[thermodynamics] "Is it possible" problem.

Air in a piston-cylinder device, heat input
Initially: T1 = 21Celsius, P1 = 2 atm
Finally: T2 = 38Celsius, P2 = 2 atm = P1 (isobaric)
Heat source: TH = 260Celsius, qH = 23 kJ/kg (heat input)
Find:
a) specific work output, w =?
b) whether the process is possible?

Illustration figure can be found here:
hxxp://img247.imageshack.us/img247/4540/tempca1.jpg

The process have a positive total entropy change but the professor said it is impossible, positive total entropy change is true but still something missing, anyone know what is it?
 
Physics news on Phys.org
Have you calculated the temperature increase when 23 kJ is added to one kilogram of air?
 
Mapes said:
Have you calculated the temperature increase when 23 kJ is added to one kilogram of air?

Genius guy! we actually only need a qH of 17.073kJ/kg to raise the temperture of air to 38Celsius, but I'm not sure if it's the answer that the professor wanted, because Cp was not given, if Cp by some reason becomes large then it's still possible.
 
Here's a little detail about my solution:
from 1st law:
delta(u) = q - w

q = 23kJ/kg
delta(u) = u2 - u1 = 221.25 - 210.40 = 10.85 kJ/kg (u obtained from table)
=> w = 23kJ - 10.85 kJ/kg = 12.15kJ/kg

Entropy change of the system: (treated as ideal gas):
delta(s)system = Cp*ln(T2/T1) = 1.005*ln(311K/294K)=0.00565kJ/kgK

Entropy change of surrounding:
delta(s)surround = q/T = 23/533K = 0.04315kJ/kgK

delta(s)total = delta(s)surround - delta(s)system = 0.04315kJ/kgK - 0.00565kJ/kgK = 0.0375kJ/kgK (>0)

Total Entropy change is larger than zero, the process seems to be possible...
what's missing ?