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**Chapter 19: The Second Law of Thermodynamics**

Directions of thermodynamic processes Irreversible and reversible processes Thermodynamic processes that occur in nature are all irreversible processes which proceed spontaneously in one direction but not the other. In a reversible process the system must be capable of being returned to its original state with no other change in the surroundings. A reversible process proceeds slowly through equilibrium states.

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**Heat engines Heat engine**

Any device that transforms heat partly into work or mechanical energy is called a heat engine. A quantity of matter inside the engine undergoes inflow and outflow of heat, expansion and compression, and sometimes change of phase. This matter inside the engine is called the working substance of the engine. The simplest kind of engine to analyze is one in which the working substance undergoes a cyclic process. All heat engines absorb heat from a source at a relatively high temperature, perform some mechanical work, and discard or reject some heat at a lower temperature. For a cyclic process

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**Heat engines Energy flow diagram for a heat engine**

Temperature of hot reservoir Heat flow from hot reservoir per cycle Temperature of cold reservoir Heat flow from cold reservoir per cycle waste! Net heat Q absorbed per cycle: Net work done W per cycle:

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**Heat engines (cont’d) Energy flow diagram for a heat engine (cont’d)**

Net heat Q absorbed per cycle: Net work done W per cycle: Thermal efficiency e:

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Heat engines (cont’d) Steam engines (external combustion)

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**Internal-combustion engines**

Gasoline engine (a heat engine) First a mixture of air and gasoline vapor flows into a cylinder through an open intake valve while the piston descends, increasing the volume of the cylinder from a minimum of V to a maximum of rV (r: compression ratio 8-10). At the end of intake stroke, the intake valve closes and the mixture is compressed, approximately adiabatically, back to volume V during the compression stroke. Then the mixture is ignited by the spark plug, and the heated gas expands, approximately adiabatically, back to volume rV, pushing on the piston-power stroke.

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**Internal-combustion engines (cont’d)**

The Otto cycle compression stroke (adiabatic compression) ignite fuel (heating at constant volume) power stroke (adiabatic expansion) thermal efficiency in Otto cycle Reject heat to environment (cooling at constant volume) r=8,g=1.4->e=56%

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**Internal-combustion engines (cont’d)**

The Otto cycle

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**Internal-combustion engines (cont’d)**

The Diesel cycle compression stroke (adiabatic compression) ignite fuel (heating at constant pressure) power stroke (adiabatic expansion) Reject heat to environment (cooling at constant volume)

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**Internal-combustion engines (cont’d)**

The Diesel cycle

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**Refrigerators Refrigerator**

A refrigerator works as a reversed heat engine.

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**Refrigerators Energy flow of a refrigerator Coefficient of performance**

A refrigerator takes heat from low temperature reservoir and gives it off to high temperature reservoir. From the first law of thermodynamics Coefficient of performance Carnot refrigerator)

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**Refrigerators Air conditioner A air conditioner works on exactly**

the same principle as a refrigerator. For air conditioners, the quantities of greatest practical importance are the rate of heat removal (the heat current H from the region being cooled) and the power input P=W/t. H in Btu/h, P in watts H/P (energy efficiency rating EER) in (Btu/h)/W. Note 1 W=3.413 Btu/h. EER~7-10.

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**The second law of thermodynamics**

It is impossible for any system to undergo a process in which it absorbs heat from a reservoir at a single temperature and converts the heat completely into mechanical work, with the system ending in the same state in which it began (Kelvin-Planck statement) The kinetic and potential (due to interactions between molecules) energies associated with random motion constitute the internal energy. When a body sliding on a surface comes to rest as a result of friction, the organized motion of the body is converted to random motion of molecules in the body and in the surface. Since we can’t control the motions of individual molecules, we can’t convert this random motion COMPLTELY back to organized motion. It is impossible for any process to have as its sole result the transfer of heat from a cooler to a hotter body (Clausius statement).

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**The second law of thermodynamics (cont’d)**

Proof : Kelvin-Planck statement = Clausius statement impossible impossible A workless refrigerator violates 2nd law (K-P statement). If it existed, it could be used to make a 100% efficient engine, which violate K-P statement. (b) An impossible 100% efficient engine violates C statement. If it existed, it could be used to make a workless refrigerator, which violates C state- ment. K-P statement C-statement

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**The second law of thermodynamics (cont’d)**

The conversion of work to heat, as in friction or viscous fluid flow, and heat flow from hot to cold across a finite temperature gradient, are irreversible processes. Two equivalent statements of the 2nd law state that these processes can be only partially reversed Gases, for example, always seep spontaneously through an opening from a region of high pressure to a region of low pressure. The 2nd law is an expression of the inherent one-way aspect of this and many other irreversible processes. To have the maximum efficiency of a heat engine, we must avoid all irreversible processes Answer: Carnot cycle In any process in which the temperature of the working substance of the engine is intermediate between TH and TC , there must be NO heat transfer between the engine and either reservoir because such heat transfer could not be reversible. Any process in which the temperature of the working substance changes must be adiabatic. As heat flow through a finite temperature drop is an irreversible, during heat transfer, there must be NO finite temperature difference.

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**The Carnot cycle The Carnot cycle reversible expansion**

adiabatic compression adiabatic expansion reversible compression

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**The Carnot cycle The Carnot cycle : ideal gas approximation isothermal**

expansion isothermal expansion Adiabatic processes: thermal efficiency of a Carnot engine

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**The Carnot cycle (cont’d)**

The Carnot refrigerator A reversed Carnot engine = a Carnot refrigerator coefficient of performance of a Carnot refrigerator When the temperature difference is small, K is much larger than unity, in which case a lot of heat can be pumped from the lower to the higher temperature with only a little expenditure of work.

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**The Carnot cycle (cont’d)**

The Carnot cycle and the second law More efficient than a Carnot engine impossible A Carnot refrigerator Since each step in the Carnot cycle is reversible, the entire cycle may be reversed. If you run the entire backward, the engine becomes a refrigerator.

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**The Carnot cycle (cont’d)**

The Carnot cycle and the second law (cont’d) The refrigerator does work W=-|W|, takes in heat QC from the cold reservoir, and expels hear |QH| to the hot reservoir. The superefficient heat engine expels heat |QC|, but to do so, it takes in a greater amount of heat QH+D. Its work output is then W+D. The net effect of the two machines together is to take a quantity of heat D and convert it completely into work, which violates the 2nd law. No engine can be more efficient than a Carnot engine. All Carnot engines operating between the same two temperatures have the same efficiency, irrespective of the nature of the working substance.

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**Entropy Entropy and disorder**

Entropy provides a quantitative measure of disorder. Define the infinitesimal entropy change dS during an infinitesimal reversible process at absolute temperature T as: If a total amount of heat Q is added during a reversible isothermal process at absolute temperature T, the total entropy change is given by: Higher temperature means greater randomness of motion Adding heat Q causes a substantial fractional increase in molecular motion and randomness

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**Entropy (cont’d) Entropy and disorder (cont’d)**

Generalization of definition of entropy change is to include ANY reversible process leading from one state to another, whether it is isothermal or not: Let us represent the process as a series of infinitesimal reversible steps. During a step, an infinitesimal quantity of heat dQ is added to the system at absolute temperature T. Then the change of entropy for the entire process is: entropy change in a reversible process The change in entropy does not depend on the path leading from the initial to the final state but is the same for all possible processes. Since entropy is a function only of the current state of the system, we can compute entropy change in irreversible processes using the above formula. ( we need to invent a path connecting the given initial and final state that consist entirely of reversible, equilibrium processes and compute the total entropy change as for the actual path).

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**Entropy (cont’d) Entropy in cyclic processes**

For any reversible process (e.g. Carnot cycle): The entropy of a system in a given state is independent of the path taken to get there, and is thus a state variable.

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**Entropy (cont’d) Entropy in irreversible processes**

Consider a combined system with a Carnot engine and a cyclic system. heat reservoir TR Carnot engine Carnot cycle combined cyclic system T Cyclic system

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**Entropy (cont’d) Entropy in irreversible processes**

Consider a combined system with a Carnot engine and a cyclic system. heat reservoir TR Carnot engine If WC > 0, a cyclic device exchanging heat with a single heat reservoir and producing an equivalent amount of work. - Violation of K-P statement of 2nd law- combined cyclic system T Cyclic system -the 2nd law- The equality is true for a reversible process. Clausius inequality

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**Entropy (cont’d) Entropy in irreversible processes (cont’d)**

Processes: aAb and aBb reversible Process : aCb irreversible T A reversible heat transfer causes changes in entropy of both the system and the reservoir (at least one needed except for adiabatic process) In an irreversible heat transfer process, there is a finite source of energy instead of an energy reservoir, and the temperature difference during the heat transfer process is finite. A C B S For an isolated system dQ=0, so dS>=0. (> for irreversible and = for reversible process)

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**Entropy (cont’d) Entropy and the second law**

When all systems taking part in a process are included, the entropy either remains constant or increases. OR No process is possible in which the total entropy decreases, when all systems taking part in the process are included.

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Exercises Problem 1 A Carnot engine takes 2000 J of heat from a reservoir at 500 K, does some work, and discards some heat to a reservoir at 350 K. How much work does it do, how much heat is discarded, and what is the efficiency? Solution The heat discarded by the engine is: From the first law, the work done by the engine is: The thermal efficiency is:

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Exercises Problem 2 One kilogram of water at 0oC is heated to 100oC. Compute its change in entropy. Solution In practice, the process described would be done irreversibly, perhaps by setting a pan of water on an electric range whose cooking surface is maintained at 100oC. But the entropy change of the water depends only on the initial and final states of the system, and is the same whether the process is reversible or not. Hence we can imagine that the temp. of the water is increased reversibly in a series of infinitesimal steps, in each of which the temp is raised by an infinitesimal amount dT.

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**Example: Refrigerator**

A refrigerator pumps heat from the inside of the freezer (-5°C) to the room (25°C). What is the maximum coefficient of performance? i.e. 8.9 Joules of heat would be pumped from the freezer for every Joule of work done by the compressor. (typical K = 3-5)

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**Exercises Problem 3 What is the entropy change in a free**

expansion process, when the volume is doubled. Solution The work done by n moles of ideal gas in an isothermal expansion from V1 to V2 is: Therefore the entropy change for n=1 is:

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Exercises Problem 4 A physics student immerses one end of a copper rod in boiling water at at 100oC and the other end in an ice-water mixture at 0oC. The sides of the rod are insulated. After steady-state conditions have been achieved in the rod, kg of ice melts in a certain time interval. For this time interval find the entropy change of the boiling water; (b) the entropy change of the ice-water mixture; (c) the entropy change of the copper rod; (d) the total entropy change of the entire system. Solution (a) DS=Q/T=-mLf/T=-(0.160 kg)(334x103 J/kg)/( K)=-143 J/K. (b) DS=Q/T=mLf/T=(0.160 kg)(334x103 J/kg)/( K)= 196 J/K. (c) From the time equilibrium has been reached, there is no net heat exchange between the rod and its surroundings, so the entropy change of the copper rod is zero. (d) 196 J/K-143 J/K=53 J/K.

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Exercises Problem 5 An experimental power plant at the Natural Energy Lab generates electricity from the temperature gradient of the ocean. The surface and deep-water temperatures are 27oC and 6oC, respectively. (a) What is the maximum theoretical efficiency of this power plant? (b) If the power plant is to produce 210 kW of power, at what rate must heat be extracted from the warm water? At what rate must heat be absorbed by the cold water? (c) The cold water that enters the plant leaves it at a temperature of 10oC. What must be the flow rate of cold water through the system? Solution (a) (b) (c)

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Chapter 24 Thermodynamics

Chapter 24 Thermodynamics

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