Chapter 4.
Entropy and the Second Law
Soong Ho Um
Sungkyunkwan University
Chemical Engineering
1
stLaw to 2
ndLaw
• The first law, discussed in the previous chapter, expresses a fundamental principle, “energy conservation”.
• There is another principle at work that determines whether a process that satisfies the first law is indeed possible or not. This principle is known as the second law of thermodynamics and entropy, the most fundamental concept in thermodynamics. (Show some examples explaining 2nd law from 1st law.)
Some examples for 2
ndlaw verification
• If we throw a piece of ice in warm water and leave the system undisturbed, the ice will melt and the water will become
colder. What if it occurs in reverse?
• Oxygen and nitrogen placed into contact with each other will mix without further interference from the outside but air does not spontaneously unmix into oxygen and nitrogne.
• A bouncing ball must eventually come to rest by converting its kinetic energy into internal energy, but a ball at rest will not spontaneously begin to bounce by converting some of its internal energy into kinetic energy
What You will Learn in This Chapter
1. Calculate entropy changes of real fluids along special paths.
2. Determine, after the fact, whether a process was conducted reversibly or not.
3. Apply the first and second law to determine if a proposed process is thermodynamically feasible.
4. Apply the notion of a cycle to determine thermodynamic efficiency.
The Second Law in a Closed System
This process is spontaneous and adiabatic. This tendency defines a preferred direction:
towards the equilibrium state, never away from it.
Formulation of Second Law:
A spontaneous adiabatic process in a closed system proceeds in the direction that satisfies the inequality,
[dS]adiabatic > 0, where S is a new extensive state function, which we call entropy. At equilibrium, the inequality reduces to an exact equality:
[dS]equilibrium = 0
To a reversible adiabatic process, [dS]adiabatic ≥ 0
Relating Entropy to Measurable Quantities
• Since entropy is a state function, we can express it as a mathematical function of two intensive properties. S = S(U, V)
Entropy
expressed now as measurable quantities (heat, temperature)
0
Calculation of Entropy in
Constant-Pressure Path, No Phase Change
This situation gives the entropy difference between two states at the same pressure but at different temperatures.
Constant-Volume Path, No Phase Change
Const. V
Example 4.1. Entropy change at constant pressure
Toluene is heated at constant pressure from 12 C, 1 bar, to 55 C. Determine the entropy change of toluene.
Example 4.2. Entropy of solid
Calculate the entropy change for copper, from initial state at 120 C, 1 bar, to 20 C, 5 bar. The heat capacity of copper is Cp = 0.38 kJ/ kG K.
Tip: Entropy of solids and liquids
The entropy of incompressible phases (solids, liquids far from the critical point) is nearly independent of pressure. Entropy in this case depends only on temperature and may be calculated using the
equation in slide 7, even if pressure is not the same in the two states.
Entropy and Phase Change
The entropy of vaporization
Q = ∆Hvap while T is constant in a tie line of PV diagram
Example 4.3. Energy balances with phase change: using Cp and ∆Hvap
Toluene vapor at 1.013 bar, 150 C, is to be cooled under constant pressure to final temperature of 22 C. (a) Determine the entropy change. (b) Repeat the calculation if the final state is a vapor-liquid mixture that contains 47 % vapor by mass.
Tip: Cp/R = a0 + a1T + a2T2 + a3T3 + a4T4.
Entropy Change of Bat
A heat bath is a system so large that its temperature does not change in any appreciable amount when it absorbs or rejects heat.
Since temperature is constant, the integral reduces to the beside.
Q = -Qbath is the heat defined with respect of the system .
The + sign if the bath receive the heat from the system, and the – sign if it transfers it to the system.
When the bath absorbs heat its entropy increases, and when it rejects heat its entropy decreases.
Example 4.4. Entropy change of bath
A piece of copper (Cp = 0.38 kJ/ kg K), initially at 250 C, is left to cool in open air at 1 bar, 22 C.
Calculate the entropy change of the copper and of the air.
Entropy in the Ideal-Gas State
Quite simple in the special case that both states are in the ideal- gas state. The process is isothermal
and mechanically reversible.
Using tabulated values
• As a state function, entropy can be tabulated. The steam
tables in the appendix include the specific entropy of water in the saturated region (liquid, vapor) and in the superheated vapor region.
• If tabulated values are not available, then entropy changes may be calculated from S = Q/T.
• Without the tabulation values, the calculation requires the amount of heat that is exchanged along the path and this is generally obtained by application of the first law. This method can be applied to several specific cases.
Example 4.5. Calculation of entropy of steam
Steam is heated under constant pressure from 15 bar, 300 C to 500 C. Determine the entropy
change.
Ans. Check the steam table at each state.
Energy balances using entropy
The above relates entropy to temperature and heat.
The equation has a simple graphical interpretation on the TS graph: heat in a mechanically reversible process is equal to the area under the path.
Vertical lines represent lines of constant entropy
(isentropic lines, dS = 0, reversible adiabatic process is isentropic).
Reversible adiabatic process
in the ideal-gas state
Entropy generation
• All processes must result in nonzero entropy generation; a process that results in negative entropy generation is not feasible.
Carnot cycle
Continued to the entropy cycle and energy balance with entropy changes
Reversible cycle consisting of two isothermal branches, AB and CD, and two isentropic branches, BC and DA.
TH > TL Reversible isothermal heating
Reversible adiabatic expansion:
WBC Reversible
adiabatic compression:
WAD
Reversible isothermal cooling
Overall energy balance (1st law):
∆U = Q + W = 0 Q = QH + QL
W = WAB + WBC +WCD + WDA
Overall entropy balance (2nd law):
∆SSYS = 0
∆SH = -QH/TH, ∆SL = -QL/TL
Since the carnot cycle is reversible, the entropy generation is zero:
Sgen = ∆Ssys + ∆Ssur = 0 0 – QH/TH – QL/TL = 0 Overall energy balance (1st law):
W = -QH (1-TL/TH)
Net work has a negative sign because of TH>TL and QH<0.
Thermodynamic efficiency:
-W/QH = 1-TL/TH < 1
Since TL < TH, right-hand side is less than 1. the cycle can not convert the entire amount of heat it receives into work.
ηCarnot = 1 – TL/TH
Example 4.6. Carnot cycle using an ideal gas
A fluid undergoes the carnot cycle in the previous Figure with TL = 300 C, TH = 500 C. Calculate the amounts of heat and work for each step, as well as the thermodynamic efficiency of the cycle (net
amount of work produced over amount of heat absorbed) assuming the fluid is in the ideal-gas state.
Ans.
Alternative statements of the second law
• Alternative statement 1: For any change of state, the following inequality holds true, dS ≥ dQ/T. This becomes an exact
equaltiy if the process is reversible.
• Alternative statement 2 (Clausius): it is not possible to
construct a device whose sole effect is to transfer heat from a colder body to a hotter body.
• Alternative statement 3 (Kelvin-Planck): it is not possible to construct a device that operates in a cycle and whose sole effect is to produce work by transferring heat from a single body.
Example 4.7. Clausius statement
Prove that the Clausius statement is a necessary consequence of the second law, as formulated in this chapter.
Ans.
Carnot cycle and the “Work Value” of heat
Example 4.8. Feasibility of Process
Looking for ways to improve energy utilization in a chemical plant, a young intern proposes to design a process that will use steam in a closed system to produce work. The steam is available at 10 bar,
650 C, and will be delivered at 1 bar, 150 C. The intern expects to generate 1200 kJ/kg from this process. A heat source is available at 200 C, should the process need any heat. Determine whether the company should go ahead with this plan.
Ideal and lost work
Example 4.9. Maximum work
How should the specifications of the proposed design in Example 4.8 be modified to make it thermodynamically feasible?
Ans.
