Thermodynamic Cycles

Why it matters

Thermodynamic cycles are fundamental to the operation of engines, refrigerators, and power plants. Understanding these cycles helps in designing systems that efficiently convert energy from one form to another, which is crucial for improving performance and reducing energy consumption.

Key ideas

• Thermodynamic Cycle: A series of processes that return a system to its initial state, allowing for continuous operation.
• Carnot Cycle: A reversible cycle with two isothermal and two reversible adiabatic processes; it attains the maximum efficiency between the same two reservoir temperatures.
• Rankine Cycle: Used in steam power plants, involving the conversion of water to steam and back.
• Otto Cycle: Describes the functioning of a typical spark-ignition piston engine, like those in cars.
• Diesel Cycle: Similar to the Otto cycle but used for diesel engines, with idealized constant-pressure heat addition rather than Otto’s constant-volume heat addition; both use isentropic compression in the ideal model.
• Brayton Cycle: Used in gas turbines, involving constant pressure heat addition and rejection.

Air-standard cycle models idealize real engines. The Otto efficiency expression below further assumes constant specific heats and internally reversible compression/expansion. Rankine net work is turbine output minus pump input, and heat input is boiler energy supplied. The same expression W_net/Q_in defines thermal efficiency for any heat-engine cycle.

Formulas

• Carnot Efficiency: η = 1 - (T_c / T_h)

  • η: Efficiency (dimensionless)
  • T_c: Temperature of the cold reservoir (Kelvin)
  • T_h: Temperature of the hot reservoir (Kelvin)

• Otto Cycle Efficiency: η = 1 - (1 / r^(γ-1))

  • η: Efficiency (dimensionless)
  • r: Compression ratio (dimensionless)
  • γ: Specific heat ratio (dimensionless)

• Rankine Cycle Efficiency: η = (W_net / Q_in)

  • η: Efficiency (dimensionless)
  • W_net: Net work output (Joules)
  • Q_in: Heat input (Joules)

Worked example

Problem: Calculate the efficiency of a Carnot engine operating between a hot reservoir at 500 K and a cold reservoir at 300 K.

1. Identify the given temperatures:

   • T_h = 500 K
   • T_c = 300 K

2. Use the Carnot efficiency formula: η = 1 - (T_c / T_h)

3. Substitute the values: η = 1 - (300 / 500)

4. Calculate: η = 1 - 0.6 = 0.4

5. Convert to percentage: η = 0.4 * 100 = 40%

Final Answer: 40%

Common mistakes

• Confusing the temperature units: Use kelvins in absolute-temperature ratios and equations of state; Celsius and kelvin temperature differences have the same numerical value.
• Misapplying the efficiency formulas to the wrong cycle.
• Forgetting to convert efficiency from a decimal to a percentage.

For GATE ME

Questions often involve calculating the efficiency of different cycles, understanding the processes involved in each cycle, and comparing the efficiencies of different cycles. Practice problems on identifying cycle components and calculating work and heat transfer.

Quick check

1. What is the most efficient thermodynamic cycle?
2. Which cycle is used in steam power plants?
3. What is the main difference between the Otto and Diesel cycles?

Answers: 1. A reversible Carnot cycle attains the maximum between specified hot and cold reservoirs. 2. Rankine cycle. 3. Heat addition is at constant volume for ideal Otto and constant pressure for ideal Diesel.