Second Law of Thermodynamics and Carnot Engine Study Guide

The Second Law of Thermodynamics and Engine Principles

  • The First Law of Thermodynamics (FIOTFIOT) focuses on the conservation of energy through the equation ΔQ=ΔU+ΔW\Delta Q = \Delta U + \Delta W.
  • The Second Law of Thermodynamics (SIOTSIOT) is necessary to determine the direction of heat flow and whether a specific process is possible.
  • Statement #1 (Kelvin): It is impossible to construct an engine operating continuously in a cycle that takes heat from a single source and converts it completely into work. This implies that an engine can never be 100%100\% efficient. While all work can be converted into heat, all heat can never be converted completely into work.
  • Statement #2 (Clausius): It is impossible to cause heat to flow from a cold body to a hot body without the expenditure of work. Heat flow from a hot body to a cold body is a spontaneous or natural process, but the reverse requires external work.

Heat Engines and Efficiency

  • A heat engine is a device that converts heat energy into mechanical work.
  • Components of a Heat Engine:
    • Source (High-Temperature Reservoir - HTR): Provides heat at temperature T1T_1.
    • Sink (Low-Temperature Reservoir - LTR): Accepts rejected heat at temperature T2T_2, where T1>T2T_1 > T_2.
    • Working Substance: The material (e.g., gas or steam) that undergoes thermodynamic changes to perform work.
  • Efficiency ($\eta$): Defined as the ratio of work output to heat input.
    • η=outputinput=WQ1\eta = \frac{\text{output}}{\text{input}} = \frac{W}{Q_1}
    • Since W=Q1−Q2W = Q_1 - Q_2, efficiency can be written as: η=Q1−Q2Q1=1−Q2Q1\eta = \frac{Q_1 - Q_2}{Q_1} = 1 - \frac{Q_2}{Q_1}
    • Percentage Efficiency: η=(1−Q2Q1)×100\eta = (1 - \frac{Q_2}{Q_1}) \times 100
    • In terms of temperature: η=(1−T2T1)×100\eta = (1 - \frac{T_2}{T_1}) \times 100

Numerical Examples of Efficiency

  • Example 1: A heat engine takes 1200 J1200\,J of heat (Q1Q_1) and performs 800 J800\,J of work (WW).
    • η=WQ1×100=8001200×100\eta = \frac{W}{Q_1} \times 100 = \frac{800}{1200} \times 100
    • η=0.67×100=67%\eta = 0.67 \times 100 = 67\%
  • Example 2: An engine operates with a source at 327∘C327^{\circ}C and a sink at 100∘C100^{\circ}C.
    • Convert to Kelvin: T1=327+273=600 KT_1 = 327 + 273 = 600\,K; T2=100+273=373 KT_2 = 100 + 273 = 373\,K
    • η=(1−373600)×100=(1−0.62)×100\eta = (1 - \frac{373}{600}) \times 100 = (1 - 0.62) \times 100
    • η=0.38×100=38%\eta = 0.38 \times 100 = 38\%

Internal and External Combustion Engines

  • Internal Combustion (IC) Engine: Fuel is burnt inside the main cylinder of the engine.
    • Examples: Car engine, bike engine.
    • Components typically include valves, cylinder, and a spark plug.
  • External Combustion (EC) Engine: Fuel is burnt outside the main cylinder of the engine.
    • Examples: Train engine, steam engine.
    • The fuel (like coal or wood) is used to heat a working fluid (like water into steam) outside the cylinder.

Thermodynamic Processes and Cycles

  • Cyclic Process: A sequence of processes that leaves the system back at its initial state (P1,V1,T1P_1, V_1, T_1) after going through various states like (P2,V2,T2P_2, V_2, T_2) and (P3,V3,T3P_3, V_3, T_3).
  • Reversible Process: A process that can be returned to its initial condition without causing any change to its surroundings.
  • Irreversible Process: A process that cannot be returned to its initial state. Most natural processes are inherently irreversible.

Refrigerators and Coefficient of Performance

  • A refrigerator is essentially a reverse heat engine. It is a device that takes heat (QcQ_c) from a cold body and, by doing some external work (WW), expels heat (QHQ_H) to a hot body.
  • Coefficient of Performance (K or COP): The ratio of heat extracted from the cold body to the work input.
    • K=heat extractedwork input=QcWK = \frac{\text{heat extracted}}{\text{work input}} = \frac{Q_c}{W}
    • Since QH=Qc+WQ_H = Q_c + W, then W=QH−QcW = Q_H - Q_c
    • K=QcQH−QcK = \frac{Q_c}{Q_H - Q_c}
  • Example: If Qc=20 JQ_c = 20\,J and QH=30 JQ_H = 30\,J, then W=10 JW = 10\,J.
    • K=2010=2K = \frac{20}{10} = 2
  • Note: KK is always greater than 11 (100%100\%) conceptually in terms of heat transfer ratio.

The Petrol Engine (Otto Cycle)

  • A petrol engine is an internal combustion, four-stroke engine that follows the Otto cycle.
  • Construction: Includes inlet valve, outlet valve, spark plug, piston, connecting rod, crankshaft, and flywheel.
  • The Four Strokes:
    1. Intake Stroke: Inlet valve opens, and a petrol-air mixture enters while the piston moves outward.
    2. Compression Stroke: Both valves close. Piston moves inward, causing adiabatic compression. Temperature and pressure increase.
    3. Power Stroke: Fuel is ignited by a spark from the spark plug. Rapid pressure and temperature increase leads to adiabatic expansion, pushing the piston outward.
    4. Exhaust Stroke: Outlet valve opens. Piston moves inward, expelling residual gases.
  • Efficiency: Typically ranges between 25%−30%25\%-30\%.

The Diesel Engine

  • Named after Rudolf Diesel, who created the original design in 18921892.
  • Developed in two classes: two-stroke and four-stroke cycles. Most diesel engines use the four-stroke cycle.
  • Key Difference: There is no spark plug. Ignition occurs due to high compression.
  • Efficiency: Typically higher than petrol engines, ranging between 35%−40%35\%-40\%.

The Carnot Engine and Carnot Cycle

  • The Carnot Engine is a theoretical, ideal engine that possesses maximum efficiency and is perfectly reversible.
  • The Carnot Cycle (4 Steps):
    1. Isothermal Expansion: The system takes heat from a reservoir at constant high temperature T1T_1.
    2. Adiabatic Expansion: The system expands further without heat exchange (Q=0Q = 0), and temperature drops from T1T_1 to T2T_2.
    3. Isothermal Compression: Heat is expelled to a sink at constant low temperature T2T_2.
    4. Adiabatic Compression: The system is compressed back to its initial state without heat exchange, and temperature rises from T2T_2 back to T1T_1.
  • Carnot Efficiency:
    • η=1−Q2Q1=1−T2T1\eta = 1 - \frac{Q_2}{Q_1} = 1 - \frac{T_2}{T_1}
    • Efficiency is directly proportional to the difference between T1T_1 and T2T_2 (Q1Q_1 and Q2Q_2).
    • Efficiency is inversely proportional to the ratio of temperatures (T2T1\frac{T_2}{T_1}).
  • Comparison Example:
    • Engine A: T1=150 K,T2=100 K→η=1−100150=1−0.6=0.4 or 40%T_1 = 150\,K, T_2 = 100\,K \rightarrow \eta = 1 - \frac{100}{150} = 1 - 0.6 = 0.4 \text{ or } 40\%
    • Engine B: T1=200 K,T2=100 K→η=1−100200=1−0.5=0.5 or 50%T_1 = 200\,K, T_2 = 100\,K \rightarrow \eta = 1 - \frac{100}{200} = 1 - 0.5 = 0.5 \text{ or } 50\%

Entropy and Degradation of Energy

  • Entropy (S): A measure of the randomness or disorder of a system.
  • It represents thermal energy per unit temperature that is unavailable for doing "useful work."
  • Formula: ΔS=ΔQT\Delta S = \frac{\Delta Q}{T}
    • Units: J K−1J\,K^{-1}
    • Heat added (QQ is positive) results in +ΔS+\Delta S (increase in entropy).
    • Heat extracted (QQ is negative) results in −ΔS-\Delta S (decrease in entropy).
  • Entropy of the Universe: During any process, the entropy of the universe either remains constant (reversible) or increases (irreversible).
  • Degradation of Energy: Increase in entropy signifies the degradation of energy.
    • Derivation of energy loss: Let work $W_1$ be done at temperature $T_1$ and $W_2$ at $T_2$.
    • W1=Q(1−T0T1)W_1 = Q(1 - \frac{T_0}{T_1}) and W2=Q(1−T0T2W_2 = Q(1 - \frac{T_0}{T_2}
    • Difference in work: W1−W2=Q(1−T0T1)−Q(1−T0T2)W_1 - W_2 = Q(1 - \frac{T_0}{T_1}) - Q(1 - \frac{T_0}{T_2})
    • W1−W2=Q(1−T0T1−1+T0T2)=Q(T0(1T2−1T1))W_1 - W_2 = Q(1 - \frac{T_0}{T_1} - 1 + \frac{T_0}{T_2}) = Q(T_0(\frac{1}{T_2} - \frac{1}{T_1}))
    • W1−W2=T0(QT2−QT1)=T0(S2−S1)W_1 - W_2 = T_0(\frac{Q}{T_2} - \frac{Q}{T_1}) = T_0(S_2 - S_1)
    • Final Result: W1−W2=T0×ΔSW_1 - W_2 = T_0 \times \Delta S