Different states of a Carnot Cycle

In summary, the engine operates with a thermal efficiency of .75 and a mechanical power output of 720 W.
  • #1
cheme2019
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0

Homework Statement


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?

Homework Equations


Steam Tables, TS diagram.

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!
 
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  • #2
cheme2019 said:

Homework Statement


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?
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.
 

1. What is a Carnot Cycle?

A Carnot Cycle is a theoretical thermodynamic cycle that describes the most efficient way to convert heat into work. It consists of four reversible processes: isothermal expansion, adiabatic expansion, isothermal compression, and adiabatic compression.

2. What states does a Carnot Cycle go through?

A Carnot Cycle goes through four states: the initial state (state 1), the first isothermal expansion state (state 2), the second adiabatic expansion state (state 3), and the final isothermal compression state (state 4).

3. How does a Carnot Cycle differ from other thermodynamic cycles?

A Carnot Cycle is unique in that it is reversible and operates at the maximum theoretical efficiency, known as the Carnot efficiency. Other thermodynamic cycles, such as the Otto and Diesel cycles, are irreversible and operate at lower efficiencies.

4. What is the purpose of a Carnot Cycle?

The purpose of a Carnot Cycle is to serve as a theoretical benchmark for the maximum possible efficiency of a heat engine. It also helps in understanding the principles of thermodynamics and the relationship between heat, work, and temperature.

5. How is the efficiency of a Carnot Cycle calculated?

The efficiency of a Carnot Cycle is calculated by dividing the work output by the heat input. In other words, it is the ratio of the temperature difference between the hot and cold reservoirs to the temperature of the hot reservoir (η = 1 - Tc/Th), where Tc is the temperature of the cold reservoir and Th is the temperature of the hot reservoir.

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