Tuesday, September 30, 2008

TE 303 - HW #4, P2 - Work and Heat Transfer for a Closed, 3-Step Cycle - 15 pts

A closed system undergoes a thermodynamic cycle consisting of the following processes:

Process 1-2: Adiabatic compression from P1 = 50 psia, V1 = 3.0 ft3 to V2 = 1 ft3 along a path described by :



Process 2-3: Constant volume.
Process 3-1: Constant pressure with U1 - U3 = 46.7 Btu.

There are no significant changes in kinetic or potential energies in any of the processes.
a.) Sketch this cycle on a PV Diagram.
b.) Calculate the net work for the cycle in Btu.
c.) Calculate the heat transfer for process 2-3.

6 Old Comments

1 comment:

Anonymous said...

ANSWERS:

Wcycle = -19.79 Btu
Q23 = -84.995 Btu

HINTS:
b) To solve, do similarly to what we did in class -- calculate the work for each step, and then add them up.

c) For a cycle, deltaE = 0because they are properties, hence why we can write Wcycle = Qcycle. Therefore, the sum of deltaUs for the cyle will be zero. Apply the first law for each step. (So, for instance, 1->2 will give -W12 = U2 -U1 since Q12 = 0 (adiabatic).) Plug in what you know from part b (such as W12) and combine equations to find U3 - U2.