Entropy and the Second Law of Thermodynamics

Transcription

1 1 Entropy and the Second Law of Thermodynamics

2 Introduction The Second Law of Thermodynamics helps us to understand such things as why: heat flows from hot to cold, a gas flows from high pressure to low pressure, why we cannot have perpetual motion machines, and why a refrigerator only functions plugged in! 2

3 3 Introduction

4 Introduction More specifically, The Second Law of Thermodynamics assists in understanding aspects of physical phenomena such as: Predicting the direction of processes Establishing conditions for equilibrium Determining theoretical performance of cycles Evaluating quantitatively factors that prevent the attainment of maximum performance Defining a thermodynamic temperature scale Developing a means for evaluating properties 4

5 5 Statements of the Second Law The Second Law is frequently stated in two forms which can be shown to be equivalent: Clausius Statement: It is impossible for any system to operate in such a way that the sole result would be energy transfer by heat from a cooler body to a warmer body. Kelvin-Planck Statement: It is impossible for any system to operate in a thermodynamic cycle and deliver a net amount of energy by work to its surroundings while receiving energy by heat from a single thermal reservoir.

6 6 Statements of the Second Law Entropy Statement of 2 nd Law: It is impossible for any system to operate in a way such that entropy is destroyed. change in the amount net amount of entropy amount of entropy of entropy within the transferred in across = + produced within the system during some the systemboundary system during the time time period during the time period period Entropy is a property (we will discuss this in more detail shortly).

7 7 Statements of the Second Law What do these mean? Graphically, the Clausius statement provides for the following:

8 8 Statements of the Second Law What do these mean? Graphically, the Kelvin-Planck statement prohibits the following:

9 9 Statements of the Second Law The Clausius Statement basically implies that it is impossible to construct a refrigeration cycle that operates without a work input! Mathematically, the Kelvin-Planck Statement implies that for a single thermal reservoir: W cycle 0 In simpler terms for a cycle to produce a net work output, we must have at least two thermal reservoirs.

10 Irreversibility The Second Law is also associated with the concept of irreversibility! An irreversible process is one in which the system and all parts of its surroundings cannot be restored to their respective initial state after the process has occurred. A reversible process is one in which both the system and surroundings can be restored to their initial state 10

11 Irreversibility Some common irreversible processes are: Heat transfer across a finite temperature difference Unrestrained expansion of a gas or liquid to lower pressure Spontaneous chemical reaction Spontaneous mixing of matter of different states Sliding friction and fluid friction Inelastic deformation Electric current flow through a resistance 11

12 Irreversibility We also categorize irreversibility as being either internal (within the system) or external (within the surroundings). Engineers should be able to recognize irreversibilities, evaluate their influence, and develop a practical means for reducing them. Some systems rely on irreversibilities for their successful operation! 12

13 Irreversibility Irreversibilities are associated with the concept of entropy production In practice all processes are irreversible, however some processes are approximately reversible, while in others the ideal reversible process forms a limiting case as irreversibilities are reduced, e.g. we have already considered the adiabatic compression/expansion process in a piston 13

14 Irreversibility An Internally Reversible Process is one in which there are no irreversibilities within the system, but irreversibilities can be present in the surroundings, due to heat transfer at boundaries. Such a process can be viewed as a series of equilibrium states. In thermodynamics, this is much like idealizations in mechanics of the point mass, frictionless pulleys, rigid beams, and massless ropes, etc. 14

15 15 Second Law and Cycles The Carnot Engine: Thermal efficiency: η T = W cycle Q H =1 Q C Q H When combined with the Thermodynamic temperature scale we obtain: η max =1 T C T H This is the Carnot efficiency! *** It is a measure of maximum possible thermal performance!

16 16 Second Law and Cycles The Carnot Refrigerator: Coefficient of performance: COP = Q C W cycle = Q C Q H Q C When combined with the Thermodynamic temperature scale we obtain: T C COP max = T H T C *** It is a measure of maximum possible thermal performance!

17 17 Carnot Power Cycles

18 18 Carnot Power Cycles

19 19 Example Consider what the maximum possible thermal efficiency an engineering system could achieve in a heat engine cycle? Considering that steel melts around 1370 C and the coldest ocean is around 0 C. Thus with a thermal sink at 273 K and a maximum thermal source around 1273 K before thermal design issues become a problem, we have a Carnot efficiency of approximately ~78.5%.

20 20 Steam Plant Efficiency Consider the history of thermal efficiency improvements to the simple steam engine / steam plant through the ages: First plot is heat engine efficiency Second plot is second law efficiency, actual work output divided by reversible work output

21 21 Steam Plant Efficiency

22 22 Steam Plant Efficiency

23 23 Example 1 A residential heat pump is used to provide heating during the winter season. A house is to be maintained at 21 C and on a typical day the heat transfer rate from the house is kj/hr when the outdoor temperature is -4 C. The heat pump has a COP of 3.7 under these conditions. Determine the power input required for the heat pump and the heat transfer rate from the cold air to the home. If the heat pump is replaced with a Carnot heat pump, what is the COP and the required work.

24 24 Example 2 A heat engine with a thermal efficiency of 35% produces 750 kj of work. Heat transfer to the engine is from a reservoir at 550 K, and the heat transfer from the engine is to the surrounding air which is at 300 K. Determine the heat transfer to the engine and from the engine to the air. If the heat engine is replaced with a Carnot heat engine that produces the same work, determine the efficiency and the heat transfer to the engine and from the engine to the air.

25 Entropy What is Entropy? Entropy is an extensive property of matter. In order to gain an appreciation for entropy you must understand how it is used and what it is used for. The change in a system s Entropy is defined by the integral: S 2 S 1 = In an adiabatic, internally reversible process, the entropy remains constant. 2 1 δq T in rev 25

26 Entropy Just as you learned how to use internal energy and enthalpy, we will now consider specific entropy data as a property of matter. Specific entropy is tabulated for saturated states, superheated vapor states, and compressed liquid states. Specific entropy is also tabulated for ideal gases. The units of entropy in SI system are kj/k and kj/ kgk or kj/kmolk for specific entropy. 26

27 27 T-dS Equations T-dS equations are used for property analysis and other thermodynamic manipulations. See text for derivation. First T-dS Equation: TdS = du + pdv Tds = du + pdv Second T-dS Equation: TdS = dh Vdp Tds = dh vdp

28 28 Saturated Data Saturation data are used in a manner similar to v,u, and h for mixtures, i.e. s = (1 x)s f + xs g s = s f + x(s g s f ) We also deduce states based on entropy in a similar manner using the pressure and temperature tables: s < s f compressed s f < s < s g mixture s > s g superheated

29 29 Incompressible Substances We may develop simple expressions for the entropy change in incompressible substances using the 1 st T-dS equation: ds = du T + pdv T T 2 c s 2 s 1 = v (T)dT T T 1 s 2 s 1 = c v ln T 2 T 1 dv =0 for incompressible If c v is constant then we simply integrate

30 30 Ideal Gases For ideal gases we can combine the T- ds equations with the ideal gas law and obtain: T 2 s(t 2,v 2 ) s(t 1,v 1 ) = c v (T) dt T + Rln v 2 T1 v 1 T 2 s(t 2, p 2 ) s(t 1, p 1 ) = c p (T) dt T Rln p 2 T1 p 1 The second of these is combined with tabulated data to obtain: s(t 2, p 2 ) s(t 1, p 1 ) = s (T 2 ) s (T 1 ) Rln p 2 tabulated p 1

31 31 Ideal Gases The two terms given in the last equation are tabulated properties in the gas tables Assuming constant specific heats (over small temperature changes) we can use: s(t 2,v 2 ) s(t 1,v 1 ) = c v ln T 2 T 1 + Rln v 2 v 1 s(t 2, p 2 ) s(t 1, p 1 ) = c p ln T 2 T 1 Rln p 2 p 1 Finally, we also define entropy data on a per mol basis much the same as enthalpy.

32 32 Closed System Balance For a closed system the entropy balance is obtained from: change in the amount net amount of entropy amount of entropy of entropy within the transferred in across = + produced within the system during some the systemboundary system during the time time period during the time period period Mathematically, this becomes: S 2 S 1 entropy change = 2 δq 1 T entropy transfer + σ entropy production

33 33 Closed System Balance The entropy production term σ is always greater or equal to zero. If it is zero, the process is reversible and if it is greater than zero then the process is irreversible The entropy transfer term can be viewed as entropy transfer accompanying heat transfer and in the same direction of the heat transfer. The entropy change can be either positive, negative or zero.

34 34 Closed System Balance We may also write the entropy balance for a closed system as: Or in rate form: S 2 S 1 = j Q j T j +σ ds dt = j Q j T j + σ

35 35 Open System Balance We can extend the entropy balance to an open system with multiple inlets and exits and account for entropy transfer from streams: ds CV dt = j Q j + T j m i s i m e s e i e entropy transfer from flows In a steady state we have: 0 = j + m T i s i m e s e j i e Q j entropy transfer from flows + σ CV + σ CV

36 36 Example 3 Consider a rigid adiabatic container with a moveable piston which separates the tank into two regions. Each region contains the same gas at different temperature, pressure, and mass. Over time, the piston moves slowly (with friction) to a new position such that the pressures and temperatures in each region are now equal. Find the final pressure, temperature, and entropy production for the process. Initially, the mass in region A is 2.5 kg, T= 250 C, and p = 500 kpa. The mass in region B is 0.5 kg, T = 70 C, and p = 50 kpa.

37 37 Example 4 CO 2 gas undergoes a change of state from an initial temperature and pressure of 45 C and 190 kpa to a a final state of 80 C and 375 kpa. Determine the entropy change of the gas by using a) tables of ideal gas properties and b) average specific heats.

38 38 Example 5 Nitrogen at 500 kpa and 400 K is contained in a closed piston assembly that has an initial volume of 750 cm 3. The nitrogen is heated isothermally and it expands until its pressure is 100 kpa. During this process it is found that the work done during expansion is 0.55 kj. Determine if the process is internally reversible, irreversible, and calculate the entropy change and entropy production.

39 39 Example 6 Water enters a heat exchanger at a rate of 60 kg/s as a saturated liquid at 50 kpa and leaves at 250 C. The heating of the water is accomplished by heat transfer from a hot stream of air that enters the heat exchanger at 1000 C and leaves at 450 C. Assuming ideal gas behaviour for the air, negligible pressure drops for the two streams, and negligible kinetic and potential energy changes, find the rate of entropy production associated with this process.

40 40 Isentropic Processes Isentropic processes are those in which the entropy of the initial state is the same as the entropy of the final state. Isentropic processes are approximated in many compression and expansion processes. If a process is adiabatic and internally reversible, the net entropy change is zero, i.e. if Q and σ are ~ 0. S 2 S 1 entropy change = 2 δq 1 T entropy transfer + σ entropy production

41 41 Isentropic Processes For an ideal gas, we may develop expressions for simple isentropic processes (compression/expansion) s(t 2, p 2 ) s(t 1, p 1 ) = s (T 2 ) s (T 1 ) Rln p 2 tabulated 0 = s (T 2 ) s (T 1 ) Rln p 2 tabulated s (T p 2 = exp s (T exp 2 ) 2) s (T 1 ) p 1 R = R exp s (T 1) R v 2 = v r2 v 1 v r1 p 1 p 1 = p r2 p r1

42 42 Isentropic Processes In applications of turbines, compressors, pumps, and nozzles we make use of the isentropic process and define isentropic efficiencies. For example, for a turbine: 2 nd law gives: W cv = m (h 1 h 2 ) m (s 2 s 1 ) = For reversible processes we define: j Q j T j + σ CV > 0 W cv,rev = m (h 1 h 2s )

43 43 Isentropic Processes Where h 2s is found when s 1 =s 2. We can now define a 2 nd law efficiency as: η ts = W t W = (h 1 h 2 ) t,rev (h 1 h 2s ) In a turbine entropy changes are due to heat loss to surroundings, and entropy production due to friction.

44 44 Isentropic Processes For a compressor or pump we define the efficiency in a similar manner: η cs = W c,rev W = (h h ) 2s 1 c (h 2 h 1 ) η ps = W p,rev W = (h 2s h 1 ) p (h 2 h 1 )

45 45 Example 7 Steam enters an adiabatic turbine at 1 MPa and 300 C and leaves at a pressure of 15 kpa. The work output of the turbine is measured to be 600 kj/kg of flowing steam. Determine the second law efficiency of the turbine and the exit state.