# Different states of a Carnot Cycle

1. Jan 31, 2017

### cheme2019

1. The problem statement, all variables and given/known data
A Carnot engine with water as the working fluid operates with a water recirculation rate of 1 kg/s. For TH = 475 K and TC = 300 K, determine:
a. The pressure of each state
b. The quality of each state
c. The rate of heat addition
d. The rate of heat rejection
e. The mechanical power for each of the four steps
f. The thermal efficiency of the cycle
g. Is the cycle feasible? Why or why not?

2. Relevant equations
Steam Tables, TS diagram.

3. The attempt at a solution
I just need to get over the first hurdle, then I can finish the problem..
I assumed at the 1st point, the quality of the mixture is 0 (sat'D liquid). Using T= 202 deg C and the steam tables, I get
P = 1.55 MPa
V= .1159 m^3/kg
U = 851 kJ/kg
H = 853 kJ/kg
S = 2.34 kJ/kgK
I know at the 2nd state, the temperature is the same, and that the quality should be 1 (sat'D vapor), however this would imply that the pressure didn't change, which according to a PV diagram isn't the case. Can someone help guide me in the right direction? Thanks!

2. Feb 2, 2017

### rude man

Until someone wiser responds:
I agree, you have to make an assumption about the thermal coordinates of one of the 4 corners on the p-V diagram. And I see no reason not to go with your choice. I also believe the cycle has to operate within the 0-100% quality section of the p-V diagram, i.e. the water is never saturated liquid nor superheated vapor. Every reversible cycle need not observe these limits but as I see it a Carnot cycle, by definition, does.

So the problem is can we configure a closed cycle remaining within the confines of the vaporization lines (the saturation curves).

The area of the closed path is the work done per kg of water per cycle. Since there is 1 cycle/sec. the area is also the power output of the engine.Since we have exactly 1 kg of water we can graph specific volume on the x axis.

I will look at this further if I can but perhaps you can take up the problem yourself at this point if you agree with the foregoing, absent other and better posts.