EGR 334 Thermodynamics Chapter 5: Sections 10-11 Lecture 22: Carnot Cycle Quiz Today?

Today’s main concepts: State what processes make up a Carnot Cycle. Be able to calculate the efficiency of a Carnot Cycle Be able to give the Classius Inequality Be able to apply the Classisus Inequality to determine if a cycle is reversible, irreversible, or impossible as predicted by the 2nd Law. Reading Assignment: Read Chapter 6, Sections 1-5 Homework Assignment: Problems from Chap 5: 64, 79, 81,86

Recall from last time: Energy Balance: Energy Rate Balance: Entropy Balance: Entropy Rate Balance:

Carnot Cycle The Carnot cycle provides a specific example of a reversible cycle that operates between two thermal reservoirs. Other examples covered in Chapter 9 are the Ericsson and Stirling cycles. In a Carnot cycle, the system executing the cycle undergoes a series of four internally reversible processes: two adiabatic processes (Q = 0) alternated with two isothermal processes ( T = constant) p v 2 3 1 4 T v 2 3 1 4

Carnot Power Cycles The p-v diagram and schematic of a gas in a piston-cylinder assembly executing a Carnot cycle are shown below:

In each of these cases the thermal efficiency is given by Carnot Power Cycles The p-v diagram and schematic of water executing a Carnot cycle through four interconnected components are shown below: In each of these cases the thermal efficiency is given by

The Carnot cycle: QH Gas only cycle Area = Work QH QC QC Sec 5.10 : The Carnot Cycle T v 2 3 1 4 The Carnot cycle: QH Gas only cycle Area = Work QH QC Process 1-2 : Adiabatic Compression. Process 2 -3 : Isothermal Expansion receiving QH. Process 3 – 4 : Adiabatic Expansion. Process 4 – 1 : Isothermal Compression, rejecting QC. QC

Analyzing the Carnot cycle: Sec 5.10 : The Carnot Cycle Analyzing the Carnot cycle: Energy Balance: First look at the two isothermal processes Process 2 -3 : Isothermal Expansion receiving QH. Process 4 – 1 : Isothermal Compression, rejecting QC. and

Analyzing the Carnot cycle: Sec 5.10 : The Carnot Cycle Analyzing the Carnot cycle: Energy Balance: Then look at the two adiabatic processes (Q = 0) Process 1-2 : Adiabatic Compression. The term Thus, Process 3 – 4 : Adiabatic Expansion.

Analyzing the Carnot cycle: Sec 5.10 : The Carnot Cycle Analyzing the Carnot cycle: With and Therefore, We have now proven

Added heat causes further rising. The Carnot Model of a Hurricane Added heat causes further rising. As T to dew point, vapor condenses, releasing hfg and warming air Cools & Expands as P Adiabatic cooling http://www.pepperridgenorthvalley.com/monsoon_basics.php Warm air rises Warm moist air

The Carnot Model of a Hurricane Isothermal Compression Adiabatic vcore>>v outer A Adiabatic B http://www.sciencecases.org/hurricane/hurricane_notes.asp D Isothermal Expansion C

Sketch the cycle on a P-v diagram. Sec 5.10 : The Carnot Cycle Example: (5.76) One-half pound of water executes a Carnot power cycle. During the isothermal expansion, the water is heated at 600°F from a saturated liquid to a saturated vapor. The vapor then expands adiabatically to a temperature of 90°F and a quality of 64.3% Sketch the cycle on a P-v diagram. Evaluate the heat and work for each process in BTU Evaluate the thermal efficiency. p v

Sketch the cycle on a P-v diagram. Sec 5.10 : The Carnot Cycle Example: (5.76) One-half pound of water executes a Carnot power cycle. During the isothermal expansion, the water is heated at 600°F from a saturated liquid to a saturated vapor. The vapor then expands adiabatically to a temperature of 90°F and a quality of 64.3% Sketch the cycle on a P-v diagram. Evaluate the heat and work for each process in BTU Evaluate the thermal efficiency.

Sketch the cycle on a P-v diagram. Sec 5.10 : The Carnot Cycle Example: (5.76) One-half pound of water executes a Carnot power cycle. During the isothermal expansion, the water is heated at 600°F from a saturated liquid to a saturated vapor. The vapor then expands adiabatically to a temperature of 90°F and a quality of 64.3% Sketch the cycle on a P-v diagram. Evaluate the heat and work for each process in BTU Evaluate the thermal efficiency. Isothermal Adiabatic Isothermal Adiabatic state 1 2 3 4 T (°F/R) 600 90 p (psi) x 0.643 v (ft3/lb) u (Btu/lb) state 1 2 3 4 T (°F/R) 600 90 p (psi) 1541 0.6988 x 0.643 v (ft3/lb) 0.02363 0.2677 300.74 u (Btu/lb) 609.9 1090.0 687.8 Using Table A-2

Evaluate the heat and work for each process in BTU Sec 5.10 : The Carnot Cycle Example: (5.76) Evaluate the heat and work for each process in BTU Isothermal Adiabatic Isothermal Adiabatic state 1 2 3 4 T (°F/R) 600 90 p (psi) 1541 0.6988 x 0.643 v (ft3/lb) 0.02363 0.2677 300.74 u (Btu/lb) 609.9 1090.0 687.8 Process U Q W 1 - 2 240 274.8 34.81 2 - 3 3 - 4 4 - 1 Process U Q W 1 - 2 2 - 3 3 - 4 4 - 1 Process 1-2

Evaluate the heat and work for each process in BTU Sec 5.10 : The Carnot Cycle Example: (5.76) Evaluate the heat and work for each process in BTU Isothermal Adiabatic Isothermal Adiabatic state 1 2 3 4 T (°F/R) 600 90 p (psi) 1541 0.6988 x 0.643 v (ft3/lb) 0.02363 0.2677 300.74 u (Btu/lb) 609.9 1090.0 687.8 Process U Q W 1 - 2 240 274.8 34.81 2 - 3 3 - 4 4 - 1 Process U Q W 1 - 2 240 274.8 34.81 2 - 3 -201 201 3 - 4 4 - 1 Process 2-3 (adiabatic process)

Evaluate the heat and work for each process in Btu Sec 5.10 : The Carnot Cycle Example: (5.76) Evaluate the heat and work for each process in Btu Isothermal Adiabatic Isothermal Adiabatic state 1 2 3 4 T (°F/R) 600 90 p (psi) 1541 0.6988 x 0.643 v (ft3/lb) 0.02363 0.2677 300.74 u (Btu/lb) 609.9 1090.0 687.8 Process U Q W 1 - 2 240 274.85 34.81 2 - 3 -201 201 3 - 4 -142.6 4 - 1 For Process 3 – 4: for the Carnot cycle: where QH = Q12 = 274.8 Btu

Evaluate the heat and work for each process in Btu Sec 5.10 : The Carnot Cycle Example: (5.76) Evaluate the heat and work for each process in Btu Isothermal Adiabatic Isothermal Adiabatic state 1 2 3 4 T (°F/R) 600 90 p (psi) 1541 0.6988 x 0.643 v (ft3/lb) 0.02363 0.2677 300.74 u (Btu/lb) 609.9 1090.0 687.8 Process U Q W 1 - 2 240 274.85 34.81 2 - 3 -201 201 3 - 4 -142.6 4 - 1 continuing for Process 3 – 4: recalling h = u + pv

Evaluate the heat and work for each process in Btu Sec 5.10 : The Carnot Cycle Example: (5.76) Evaluate the heat and work for each process in Btu Isothermal Adiabatic Isothermal Adiabatic state 1 2 3 4 T (°F/R) 600 90 p (psi) 1541 0.6988 x 0.643 0.368 v (ft3/lb) 0.02363 0.2677 300.74 172.1 u (Btu/lb) 609.9 1090.0 687.8 419.5 state 1 2 3 4 T (°F/R) 600 90 p (psi) 1541 0.6988 x 0.643 v (ft3/lb) 0.02363 0.2677 300.74 u (Btu/lb) 609.9 1090.0 687.8 Process U Q W 1 - 2 240 274.85 34.81 2 - 3 -201 201 3 - 4 -142.6 4 - 1 continuing for Process 3 – 4: Then using Table A2 at h4 = 441.5 Btu/lbm and T4 = 90 deg. which let the state 4 intensive properties be found:

Evaluate the heat and work for each process in BTU Sec 5.10 : The Carnot Cycle Example: (5.76) Evaluate the heat and work for each process in BTU Isothermal Adiabatic Isothermal Adiabatic state 1 2 3 4 T (°F/R) 600 90 p (psi) 1541 0.6988 x 0.643 0.368 v (ft3/lb) 0.02363 0.2677 300.74 172.1 u (Btu/lb) 609.9 1090.0 687.8 419.5 Process U Q W 1 - 2 240 274.85 34.81 2 - 3 -201 201 3 - 4 -142.6 4 - 1 Process U Q W 1 - 2 240 274.85 34.81 2 - 3 -201 201 3 - 4 -134.2 -142.6 -8.32 4 - 1 and for Process 4-1:

Evaluate the heat and work for each process in BTU Sec 5.10 : The Carnot Cycle Example: (5.76) Evaluate the heat and work for each process in BTU Isothermal Adiabatic Isothermal Adiabatic state 1 2 3 4 T (°F/R) 600 90 p (psi) 1541 0.6988 x 0.643 0.368 v (ft3/lb) 0.02363 0.2677 300.74 172.1 u (Btu/lb) 609.9 1090.0 687.8 419.5 Process U Q W 1 - 2 240 274.85 34.81 2 - 3 -201 201 3 - 4 -134.2 -142.6 -8.32 4 - 1 Process U Q W 1 - 2 240 274.85 34.81 2 - 3 -201 201 3 - 4 -134.2 -142.6 -8.32 4 - 1 95.2 -95.2 Finally, process 4 – 1:

(c) Evaluate the thermal efficiency. Sec 5.10 : The Carnot Cycle Example: (5.76) (c) Evaluate the thermal efficiency. Isothermal Adiabatic Isothermal Adiabatic state 1 2 3 4 T (°F/R) 600 90 p (psi) 1541 0.6988 x 0.643 0.368 v (ft3/lb) 0.02363 0.2677 300.74 172.1 u (Btu/lb) 609.9 1090.0 687.8 419.5 Process U Q W 1 - 2 240 274.85 34.81 2 - 3 -201 201 3 - 4 -134.2 -142.6 -8.32 4 - 1 95.2 -95.2 Thermal efficiency of the cycle:

scycle = 0 no irreversibilities present within the system Clausius Inequality The Clausius inequality is developed from the Kelvin- Planck statement of the second law and can be expressed as: The nature of the cycle executed is indicated by the value of scycle: scycle = 0 no irreversibilities present within the system scycle > 0 irreversibilities present within the system scycle < 0 impossible

For an ideal/reversible process Sec 5.11 : The Clausius Inequality We have shown that: and thus Therefore: For an ideal/reversible process Now consider a general process Each part of the cycle is divided into an infinitesimally small process p v Then sum (integrate) the entire process dQ = heat transfer at boundary T = absolute T at that part of the cycle.

For a real process, Qreal > Qreversible Sec 5.11 : The Clausius Inequality For a real process, Qreal > Qreversible Therefore: We can then define σ, where and σcycle = 0  reversible process σcycle > 0  irreversible process σcycle < 0  impossible process P v We now also have the mathematical definition of enthalpy. More on this in Chapter 6

Sec 5.11 : The Clausius Inequality Example: (5.81) A system executes a power cycle while receiving heat transfer at a temperature of 500 K and discharging 1000 kJ by heat transfer at a temperature of 300 K. There are no other heat transfers. Appling the Clausius Inequality, determine σcycle if the thermal efficiency is (a) 60% (b) 40% (c) 20 %. In each case is the cycle reversible, or impossible? W=? TH= 500 K TH= 300 K Q = 50 kJ

Since σcycle is negative, the cycle is impossible. Sec 5.11 : The Clausius Inequality Example: (5.81) A system executes a power cycle while receiving heat transfer at a temperature of 1000 K and discharging 1000 kJ by heat transfer at a temperature of 300 K. There are no other heat transfers. Appling the Clausius Inequality, determine σcycle if the thermal efficiency is (a) 60% (b) 40% (c) 20 %. In each case is the cycle reversible, or impossible? W=? TH= 500 K TH= 300 K and Therefore: Q = 50 kJ Since σcycle is negative, the cycle is impossible.

Since σcycle is zero, the cycle is internally reversible. Sec 5.11 : The Clausius Inequality Example: (5.81) A system executes a power cycle while receiving heat transfer at a temperature of 1000 K and discharging 1000 kJ by heat transfer at a temperature of 300 K. There are no other heat transfers. Appling the Clausius Inequality, determine σcycle if the thermal efficiency is (a) 60% (b) 40% (c) 20 %. In each case is the cycle reversible, or impossible? W=? TH= 500 K TH= 300 K Q = 50 kJ Since σcycle is zero, the cycle is internally reversible.

Since σcycle is positive, the cycle is irreversible. Sec 5.11 : The Clausius Inequality Example: (5.81) A system executes a power cycle while receiving heat transfer at a temperature of 1000 K and discharging 1000 kJ by heat transfer at a temperature of 300 K. There are no other heat transfers. Appling the Clausius Inequality, determine σcycle if the thermal efficiency is (a) 60% (b) 40% (c) 20 %. In each case is the cycle reversible, or impossible? W=? TH= 500 K TH= 300 K Q = 50 kJ Since σcycle is positive, the cycle is irreversible.

End of Lecture 21 Slides