Thermodynamics · Chapter 6
Thermodynamics · Chapter 6

The Second Law of Thermodynamics

The first law says energy is conserved, but it never says which direction a process will go. The second law supplies that direction, sets the ceiling on how well engines and refrigerators can perform, and introduces the Carnot cycle.

Learning Objectives

  • State the Kelvin–Planck and Clausius statements of the second law.
  • Define thermal-energy reservoirs, heat engines, refrigerators, and heat pumps.
  • Compute thermal efficiency and coefficient of performance.
  • Distinguish reversible from irreversible processes and list the irreversibilities.
  • Describe the Carnot cycle and apply the Carnot efficiency limit.

Physical meaning

The first law counts energy; the second law limits direction and usefulness. A cyclic engine cannot turn heat from one reservoir entirely into work, and moving heat from cold to hot requires work. Entropy generation measures irreversibility and can be zero for an ideal reversible process but never negative.

Problem-solving strategy

  1. Draw the cyclic device boundary and identify the desired effect: net work, cooling load, or heating delivery.
  2. Use positive magnitudes for reservoir heat transfers and work, while keeping their physical arrows explicit.
  3. Apply cycle energy closure first, then calculate thermal efficiency or the appropriate COP without confusing their definitions.
  4. Convert all reservoir temperatures to kelvin before applying a reversible Carnot limit.
  5. Compare actual performance with the reversible ceiling and audit energy closure, feasibility, uncertainty, and nonnegative entropy generation.
6–1

Introduction to the Second Law

Direction of processes

Processes proceed in a preferred direction: heat flows hot-to-cold, not the reverse, on its own. Satisfying the first law is necessary but not sufficient — the second law decides feasibility.

6–2

Thermal Energy Reservoirs

Thermal reservoir

A body large enough to absorb or supply heat without any change in its temperature — oceans, the atmosphere, or a large furnace. A heat source supplies heat; a heat sink receives it.

6–3

Heat Engines

Heat engine

A device that converts part of the heat it receives into work while operating in a cycle. It takes heat from a high-temperature source, produces work, and rejects the remainder to a low-temperature sink.

Thermal efficiency
\[\eta_{\text{th}} = \frac{W_{\text{net,out}}}{Q_H} = 1 - \frac{Q_L}{Q_H}\]
Kelvin–Planck Statement

No heat engine can convert all the heat it receives into work; some heat must always be rejected to a low-temperature sink.

Equivalently, no cyclic device can produce net work while exchanging heat with only a single reservoir.

T_H (hot)T_L (cold)HEQ_HQ_LW
Figure 6–1  A heat engine receives Q_H from the hot reservoir, produces work W, and rejects Q_L to the cold reservoir.
6–4

Refrigerators and Heat Pumps

Refrigerator and heat pump

Both move heat from cold to hot using work input. A refrigerator's goal is removing heat from the cold space; a heat pump's goal is delivering heat to the warm space.

COP of a refrigerator
\[\text{COP}_R = \frac{Q_L}{W_{\text{net,in}}} = \frac{Q_L}{Q_H - Q_L}\]
COP of a heat pump
\[\text{COP}_{HP} = \frac{Q_H}{W_{\text{net,in}}} = \frac{Q_H}{Q_H - Q_L} = \text{COP}_R + 1\]
Clausius Statement

No device can transfer heat from a colder body to a hotter body without some work input.

The Kelvin–Planck and Clausius statements are equivalent: violating one violates the other.

T_H (hot)T_L (cold)RQ_HQ_LW
Figure 6–2  A refrigerator/heat pump uses work W to move Q_L from the cold reservoir and deliver Q_H to the hot reservoir.
6–5

Perpetual-Motion Machines

Why they fail

A machine that would violate the first law (creating energy) is a PMM1; one that violates the second law (e.g. fully converting heat to work from one reservoir) is a PMM2. Neither can exist.

6–6

Reversible and Irreversible Processes

Reversible process

An idealized process that can be reversed leaving no trace on system or surroundings. It is the best a process can possibly do and sets the performance limit.

Irreversibilities

Real effects that prevent reversal: friction, unrestrained expansion, mixing, heat transfer across a finite temperature difference, and chemical reactions.

6–7

The Carnot Cycle

The Carnot cycle

The most efficient cycle operating between two temperatures, built from two reversible isothermal and two reversible adiabatic (isentropic) processes.

sTT_HT_L1234
Figure 6–3  The Carnot cycle on a T–s diagram: two isothermal legs (1–2, 3–4) and two isentropic legs (2–3, 4–1) form a rectangle.
6–8

The Carnot Principles

Carnot Principles

(1) No real engine between two reservoirs can be more efficient than a reversible engine between the same reservoirs. (2) All reversible engines operating between the same two reservoirs have the same efficiency.

Efficiency therefore depends only on the reservoir temperatures, not on the working fluid or design.

6–9

The Thermodynamic Temperature Scale

Absolute temperature scale

Because reversible-engine efficiency depends only on temperature, heat ratios can define a temperature scale independent of any substance.

Kelvin relation for reversible engines
\[\frac{Q_H}{Q_L} = \frac{T_H}{T_L}\]
6–10

The Carnot Heat Engine

Carnot (maximum) efficiency
\[\eta_{\text{th,rev}} = 1 - \frac{T_L}{T_H}\quad\text{(T in K)}\]
6–11

The Carnot Refrigerator and Heat Pump

Carnot COP (refrigerator)
\[\text{COP}_{R,\text{rev}} = \frac{1}{T_H/T_L - 1} = \frac{T_L}{T_H - T_L}\]
Carnot COP (heat pump)
\[\text{COP}_{HP,\text{rev}} = \frac{1}{1 - T_L/T_H} = \frac{T_H}{T_H - T_L}\]
Key IdeaThese are the highest COPs physically attainable between the two temperatures; real refrigerators and heat pumps fall short of them.
EXAMPLE 1Waste-heat engine balance
A cyclic waste-heat engine receives 2.40 MJ from a source and rejects 1.56 MJ to cooling water each cycle. Determine net work output and thermal efficiency, and state whether energy conservation alone proves feasibility.

Recognition strategy

The word cyclic makes the working-fluid energy change zero. One hot input and one cold rejection identify a heat engine, for which desired output is net work.

Given information

  • Per cycle \(Q_H=2.40\ \text{MJ}\), \(Q_L=1.56\ \text{MJ}\); reservoir temperatures and property tables are not supplied or needed for the first-law result.

Model and interpretation

The boundary encloses the cyclic engine, not the reservoirs. Heat and work magnitudes follow their arrows, temperatures would require kelvin and pressures absolute. The basis is one cycle, kinetic and potential changes close to zero, and no ideal-gas, phase, quality, or component-efficiency model is invoked.

Step-by-step solution
1. Use cycle closure \(\Delta E_{cycle}=0\), so energy conservation requires \(W_{net,out}=Q_H-Q_L\).
2. Calculate \(W_{net,out}=2.40-1.56=0.840\ \text{MJ/cycle}\).
3. Define rather than guess thermal efficiency: \(\eta_{th}=W_{net,out}/Q_H\).
4. Evaluate \(\eta_{th}=0.840/2.40=0.350=35.0\%\).

Verification

Units: Every cycle quantity is in MJ and their ratio is dimensionless; work output retains the stated MJ-per-cycle basis.

Signs and direction: Heat enters from the source, rejection leaves to cooling water, and their positive difference correctly gives outward work.

Plausibility: The independent identity \(2.40=1.56+0.840\ \text{MJ}\) closes energy, but feasibility still requires reservoir temperatures and \(S_{gen}\ge0\).

Final result: \(W_{net,out}=0.840\ \text{MJ/cycle}\) and \(\eta_{th}=35.0\%\).

Engineering interpretation

Energy conservation is necessary but not sufficient. A professional review would measure reservoir boundary temperatures, compare with the Carnot ceiling, quantify uncertainty, and confirm that calculated entropy generation is nonnegative.

EXAMPLE 2Cold-room refrigeration demand
A refrigerator removes 18.0 kW from a cold room while drawing 6.00 kW of electrical power. Determine its COP, heat rejection rate, and the corresponding heat-pump COP for the same cyclic device.

Recognition strategy

Cooling load divided by work identifies refrigerator COP; the same hardware viewed for heating uses heat delivered divided by work and therefore has a different COP.

Given information

  • \(\dot Q_L=18.0\ \text{kW}\), \(\dot W_{in}=6.00\ \text{kW}\); steady cyclic operation is given, while temperatures and refrigerant properties are not required.

Model and interpretation

The boundary encloses the complete refrigerator cycle. Rates are positive magnitudes along their arrows. Pressures would be absolute and temperatures kelvin if states were evaluated. No phase quality, table property, or ideal/real compressor efficiency is invented; cycle kinetic and potential energy changes vanish.

Step-by-step solution
1. Apply cyclic energy conservation: \(\dot Q_H=\dot Q_L+\dot W_{in}\).
2. Calculate heat rejection \(\dot Q_H=18.0+6.00=24.0\ \text{kW}\).
3. Evaluate refrigerator performance: \(COP_R=\dot Q_L/\dot W_{in}=18.0/6.00=3.00\).
4. Evaluate heat-pump performance: \(COP_{HP}=\dot Q_H/\dot W_{in}=24.0/6.00=4.00\).

Verification

Units: All three rates use kW, so their ratios are dimensionless coefficients of performance, not percentages.

Signs and direction: Work enters, heat is removed from the cold room, and the larger heat rate leaves toward the warm sink in the correct directions.

Plausibility: The identity \(COP_{HP}=COP_R+1\) gives \(4.00=3.00+1\), independently confirming energy closure; second-law feasibility still requires \(S_{gen}\ge0\).

Final result: \(COP_R=3.00\), \(\dot Q_H=24.0\ \text{kW}\), and \(COP_{HP}=4.00\).

Engineering interpretation

A COP greater than one is normal because work moves heat rather than becoming the entire useful effect. Equipment selection still needs rated-condition vendor maps, refrigerant phase states, defrost loads, and uncertainty margins.

EXAMPLE 3District heat-pump limit
A proposed district heat pump must deliver 1.20 MW to a loop at 330 K while drawing from groundwater at 280 K. Determine the reversible COP ceiling, minimum work input, and maximum groundwater heat extraction.

Recognition strategy

Two fixed absolute-temperature reservoirs call for a reversible Carnot benchmark, not an assumed real efficiency. The bound tests claims but does not predict actual hardware.

Given information

  • \(T_L=280\ \text{K}\), \(T_H=330\ \text{K}\), and required \(\dot Q_H=1200\ \text{kW}\); temperatures are absolute and the environment reference is the stated groundwater and loop pair.

Model and interpretation

The boundary encloses a cyclic heat pump between two ideal reservoirs. Heat and work are positive magnitudes along arrows. The analysis is a reversible upper bound with \(S_{gen}=0\), not a real component model; no working-fluid table, pressure, phase quality, compressor efficiency, kinetic energy, or potential energy data are required.

Step-by-step solution
1. Compute the reversible heat-pump ceiling \(COP_{HP,rev}=T_H/(T_H-T_L)\).
2. Evaluate \(COP_{HP,rev}=330/(330-280)=6.60\).
3. Use the COP ceiling to obtain minimum work: \(\dot W_{min}=\dot Q_H/COP_{HP,rev}=1200/6.60=181.8\ \text{kW}\).
4. Apply cycle energy closure to find maximum source extraction: \(\dot Q_{L,max}=1200-181.8=1018.2\ \text{kW}\).

Verification

Units: The temperature ratio and COP are dimensionless, while subtracting two heat rates leaves work rate in kW.

Signs and direction: Heat moves from 280 K to 330 K only with positive work input; all arrows therefore satisfy the Clausius statement.

Plausibility: The ratio \(1200/181.8=6.60\) confirms the COP; reversible entropy transfer \(1018.2/280=1200/330\) gives \(S_{gen}=0\), not negative.

Final result: \(COP_{HP,max}=6.60\), \(\dot W_{min}=182\ \text{kW}\), and \(\dot Q_{L,max}=1.02\ \text{MW}\).

Engineering interpretation

Any real installation must require more work for the same source load because compressors, heat exchangers, and pressure drops generate entropy. Release decisions require vendor maps, approach temperatures, seasonal source variation, controls, and uncertainty analysis.

Chapter recap

  • Use the second law to test direction and quality after satisfying the first law.
  • Distinguish thermal efficiency from refrigerator and heat-pump COP.
  • Use reversible Carnot expressions only with absolute reservoir temperatures and never as real component efficiencies.
  • Before moving on, be able to close a cycle, identify the desired effect, compare with a reversible limit, and demonstrate nonnegative entropy generation.