Heat Engines and The Carnot Cycle

First Statement of the Second Law of Thermodynamics The first statement of the second law is a statement from common experience. 1.When two objects, one hot and the other cold, come into contact, heat energy will be transferred from the system at higher temperature to the system at lower temperature, but not vice versa. 2.The 1 st law of thermodynamics does not clarify this, that is why we have the 2 nd Law

Second Statement of the Second Law: Heat Engines A heat engine is a device that is capable of changing thermal energy (Q H ), also known as the input heat or heat of combustion of the fuel, into useful work (W).

Heat Engines The diagram shows that heat Q H flows from the high temp reservoir and part fo the heat is transformed to work and part of the thermal energy is discarded as heat Q L to a low temperature reservoir (large area of constant temperature) Simplified Heat Engine Process

Heat engines cannot be made to be 100% efficient and while part of the heat energy is converted to useful work, the remaining heat energy will be rejected to the environment or surroundings as waste heat (Q c ) like the exhaust from a car engine for example. Therefore, W = Q H - Q c

Efficiency of Engines e - efficiency e = W/Q H x 100 or e = 1- Q L /Q H

Example I An internal combustion engine operates with an efficiency of 18%. The engine produces 2100W of power, on average, while operating at peak performance. A) What is the rate of energy flow from the high temperature reservoir to the engine? B) What is the rate of energy flow from the engine to the low temperature reservoir?

Answer 12000J/s, 9900J/s

Carnot Cycle Sadi Carnot (1824) published that the maximum efficiency for a heat engine depends on the ratio of the absolute operating temperature T H and T L.

CARNOT ENGINE Carnot imagined the “ideal engine” known as a Carnot engine, energy losses due to internal friction or turbulence present in the fuel after ignition, etc. are not considered. The most efficient heat engine cycle is the Carnot cycle, consisting of two isothermals and two adiabatic processes. All 4 processes are said to be reversible or in equilibrium states.

Carnot Explained The Carnot cycle can be thought of as the most efficient heat engine cycle allowed by physical laws. When the second law of thermodynamics states that not all the supplied heat in a heat engine can be used to do work, the Carnot efficiency sets the limiting value on the fraction of the heat which can be so used.

Carnot Efficiency e = W/Q H –Can be rewritten as e = (Q H – Q c )/Q H Only Valid for Ideal Engines T c /T H = Q c /Q H

Carnot determined that the maximum efficiency (e) that could be realized from a heat engine depends on the temperature of the input heat (T H ) and the exhaust heat (T C ) where T H and T C are expressed in Kelvin.

Carnot Explained By comparing the work done by the gas in the first two processes to the work done on the gas in the final two processes we see that the difference in the area bounded by the cycle 1,2,3,4. The gas moved through a complete cycle, so the net change in internal energy is zero ΔU = 0

The maximum efficiency (e) or Carnot efficiency of a heat engine is determined by: Using the input and waste heat temperatures: e = (Q H – Q L )/Q H e = (T H – T c )/T H Is it possible to be mathematically 100% efficient???????

Carnot Impossibility If e = 100% –Then ΔU = 0 If ΔU = 0 then ΔS = 0 Why is it impossible for ΔS = 0

Carnot Impossibility ΔS = 0 only occurs at absolute zero which is impossible to achieve. No device can transform a given amount of heat energy into work. –Automobiles have efficiencies in the 20 to 35%.

Example II What is the maximum efficiency possible of an engine that operates between 500°C and 250°C?

Answer 32%

Example III The exhaust temperature of a heat engine is 240°C. The engine produces 500J of work and has a Carnot efficiency of 45%. A) What is the temperature of the heat source? B) What heat is extracted from the heat source? C) What heat is delivered to the cold reservoir?

Answer 930K, 1100J, 610J