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Refrigeration cycles – vapour compression

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Introduction to Refrigeration and Vapour Compression Cycles

Refrigeration is the process of removing heat from a space or substance to lower its temperature below the ambient level. This process is essential in many engineering applications such as food preservation, air conditioning, and industrial cooling. A refrigeration cycle is a sequence of thermodynamic processes that accomplish this heat removal continuously and efficiently.

Among various refrigeration methods, the vapour compression refrigeration cycle is the most widely used due to its efficiency and practicality. It uses a working fluid called a refrigerant that circulates through a closed loop, absorbing heat at low temperature and rejecting it at a higher temperature.

Thermodynamics plays a crucial role in understanding and analyzing refrigeration cycles. It helps us quantify energy transfers, work input, and the effectiveness of the cycle. In this chapter, we will explore the vapour compression cycle in detail, understand its components, analyze its performance, and solve practical problems related to it.

Vapour Compression Refrigeration Cycle

The vapour compression refrigeration cycle consists of four main components connected in a loop:

  • Compressor
  • Condenser
  • Expansion Valve
  • Evaporator

The working fluid (refrigerant) undergoes four key thermodynamic processes as it moves through these components:

Compressor Condenser Expansion Valve Evaporator 1 2 3 4

Let us understand each process in the cycle with reference to the numbered points in the diagram:

  1. Compression (1 -> 2): The refrigerant enters the compressor as a low-pressure, low-temperature vapour. The compressor does work on the refrigerant, increasing its pressure and temperature. This process requires energy input (work).
  2. Condensation (2 -> 3): The high-pressure, high-temperature vapour flows into the condenser, where it releases heat to the surroundings (usually air or water). The refrigerant condenses into a high-pressure liquid at nearly constant pressure.
  3. Expansion (3 -> 4): The liquid refrigerant passes through the expansion valve (throttling device), where its pressure and temperature drop suddenly. This process is isenthalpic (constant enthalpy), and the refrigerant becomes a low-pressure mixture of liquid and vapour.
  4. Evaporation (4 -> 1): The low-pressure refrigerant enters the evaporator, where it absorbs heat from the space or substance to be cooled. It evaporates completely into a low-pressure vapour, completing the cycle.

This continuous cycle extracts heat from a low-temperature reservoir (inside the evaporator) and rejects it to a high-temperature reservoir (outside the condenser), thus providing cooling.

Thermodynamic Analysis Using P-h and T-s Diagrams

Thermodynamic diagrams help visualize the refrigeration cycle and calculate important parameters like work input and heat transfer. Two common diagrams are:

  • Pressure-Enthalpy (P-h) Diagram: Shows pressure vs. enthalpy, useful for calculating energy changes.
  • Temperature-Entropy (T-s) Diagram: Shows temperature vs. entropy, useful for understanding irreversibility and efficiency.
P-h Diagram Enthalpy (h) Pressure (P) 1 2 3 4 T-s Diagram Entropy (s) Temperature (T) 1 2 3 4

In these diagrams:

  • Points 1 to 4 correspond to the same states as in the cycle schematic.
  • The P-h diagram helps calculate the enthalpy changes during compression and expansion, which relate directly to work and heat transfer.
  • The T-s diagram shows entropy changes, helping identify irreversibilities and efficiency losses.

For example, the work done by the compressor per unit mass is given by the enthalpy difference between states 2 and 1:

Compressor Work

\[W = m (h_2 - h_1)\]

Work input required to compress the refrigerant

m = Mass flow rate (kg/s)
\(h_1\) = Enthalpy at compressor inlet (kJ/kg)
\(h_2\) = Enthalpy at compressor outlet (kJ/kg)

Coefficient of Performance (COP)

The Coefficient of Performance (COP) is a measure of the efficiency of refrigeration and heat pump systems. It compares the useful heat transfer to the work input required.

For refrigeration mode, the COP is defined as:

Coefficient of Performance (Refrigeration)

\[\mathrm{COP}_{\mathrm{R}} = \frac{Q_{L}}{W_{net}}\]

Ratio of heat absorbed in evaporator to work input

\(Q_L\) = Heat absorbed in evaporator (kJ)
\(W_{net}\) = Net work input to compressor (kJ)

For heat pump mode, where the system is used for heating, the COP is:

Coefficient of Performance (Heat Pump)

\[\mathrm{COP}_{\mathrm{HP}} = \frac{Q_{H}}{W_{net}}\]

Ratio of heat rejected in condenser to work input

\(Q_H\) = Heat rejected in condenser (kJ)
\(W_{net}\) = Net work input (kJ)

Since the heat rejected is always greater than the work input, COP values are typically greater than 1, indicating the system transfers more heat than the energy it consumes.

Worked Examples

Example 1: Calculating COP for a Simple Vapour Compression Cycle Easy
A vapour compression refrigeration system uses refrigerant R134a. The enthalpy at the compressor inlet (state 1) is 250 kJ/kg and at the compressor outlet (state 2) is 280 kJ/kg. The enthalpy at the evaporator outlet (state 1) is 250 kJ/kg and at the evaporator inlet (state 4) is 100 kJ/kg. Calculate the COP of the refrigeration cycle.

Step 1: Calculate the refrigeration effect \(Q_L\) per unit mass:

\[ Q_L = h_1 - h_4 = 250 - 100 = 150 \text{ kJ/kg} \]

Step 2: Calculate the work input \(W\) per unit mass:

\[ W = h_2 - h_1 = 280 - 250 = 30 \text{ kJ/kg} \]

Step 3: Calculate the COP:

\[ \mathrm{COP}_R = \frac{Q_L}{W} = \frac{150}{30} = 5 \]

Answer: The COP of the refrigeration cycle is 5.

Example 2: Determining Refrigeration Effect and Compressor Work Medium
In a vapour compression system using refrigerant R22, the enthalpy at the evaporator outlet is 400 kJ/kg and at the evaporator inlet is 250 kJ/kg. The enthalpy at the compressor outlet is 430 kJ/kg. Calculate the refrigeration effect and the work done by the compressor per kg of refrigerant.

Step 1: Refrigeration effect \(Q_L\):

\[ Q_L = h_1 - h_4 = 400 - 250 = 150 \text{ kJ/kg} \]

Step 2: Work done by compressor \(W\):

\[ W = h_2 - h_1 = 430 - 400 = 30 \text{ kJ/kg} \]

Answer: Refrigeration effect is 150 kJ/kg and compressor work is 30 kJ/kg.

Example 3: Effect of Condenser Pressure on COP Medium
A vapour compression cycle operates with the following enthalpy values (in kJ/kg): \(h_1 = 250\), \(h_2 = 280\), \(h_3 = 120\), \(h_4 = 100\). If the condenser pressure is increased causing \(h_2\) to rise to 300 kJ/kg, how does the COP change?

Step 1: Calculate initial COP:

\[ Q_L = h_1 - h_4 = 250 - 100 = 150 \text{ kJ/kg} \]

\[ W = h_2 - h_1 = 280 - 250 = 30 \text{ kJ/kg} \]

\[ \mathrm{COP}_1 = \frac{150}{30} = 5 \]

Step 2: Calculate new COP with increased \(h_2\):

\[ W_{new} = 300 - 250 = 50 \text{ kJ/kg} \]

\[ \mathrm{COP}_2 = \frac{150}{50} = 3 \]

Answer: Increasing condenser pressure reduces COP from 5 to 3, indicating lower efficiency.

Example 4: Comparing Performance of R134a and R22 Refrigerants Hard
Two vapour compression systems operate under identical conditions. For R134a, \(h_1=250\), \(h_2=280\), \(h_4=100\). For R22, \(h_1=270\), \(h_2=310\), \(h_4=110\). Calculate and compare the COP of both systems.

Step 1: Calculate COP for R134a:

\[ Q_{L,R134a} = 250 - 100 = 150 \text{ kJ/kg} \]

\[ W_{R134a} = 280 - 250 = 30 \text{ kJ/kg} \]

\[ \mathrm{COP}_{R134a} = \frac{150}{30} = 5 \]

Step 2: Calculate COP for R22:

\[ Q_{L,R22} = 270 - 110 = 160 \text{ kJ/kg} \]

\[ W_{R22} = 310 - 270 = 40 \text{ kJ/kg} \]

\[ \mathrm{COP}_{R22} = \frac{160}{40} = 4 \]

Answer: R134a has a higher COP (5) compared to R22 (4), indicating better efficiency under the given conditions.

Example 5: Calculating Power Consumption and Cost Estimation Hard
A refrigeration system with a mass flow rate of 0.05 kg/s uses refrigerant R134a. Enthalpy at compressor inlet is 250 kJ/kg and at outlet is 280 kJ/kg. Calculate the power consumed by the compressor in kW. If the system operates 10 hours daily and electricity costs Rs.8 per kWh, estimate the monthly electricity cost (30 days).

Step 1: Calculate compressor work per kg:

\[ W = h_2 - h_1 = 280 - 250 = 30 \text{ kJ/kg} \]

Step 2: Calculate power consumed (kW):

\[ \text{Power} = m \times W = 0.05 \times 30 = 1.5 \text{ kW} \]

Step 3: Calculate daily energy consumption:

\[ E_{daily} = 1.5 \times 10 = 15 \text{ kWh} \]

Step 4: Calculate monthly energy consumption:

\[ E_{monthly} = 15 \times 30 = 450 \text{ kWh} \]

Step 5: Calculate monthly cost:

\[ \text{Cost} = 450 \times 8 = Rs.3600 \]

Answer: The compressor power is 1.5 kW and the monthly electricity cost is Rs.3600.

Formula Bank

Coefficient of Performance (Refrigeration)
\[\mathrm{COP}_{\mathrm{R}} = \frac{Q_{L}}{W_{net}}\]
where: \(Q_L\) = Refrigeration effect (kJ), \(W_{net}\) = Net work input (kJ)
Coefficient of Performance (Heat Pump)
\[\mathrm{COP}_{\mathrm{HP}} = \frac{Q_{H}}{W_{net}}\]
where: \(Q_H\) = Heat rejected in condenser (kJ), \(W_{net}\) = Net work input (kJ)
Refrigeration Effect
\[ Q_{L} = m (h_1 - h_4) \]
where: \(m\) = mass flow rate (kg/s), \(h_1\) = enthalpy at evaporator outlet (kJ/kg), \(h_4\) = enthalpy at evaporator inlet (kJ/kg)
Work Done by Compressor
\[ W = m (h_2 - h_1) \]
where: \(m\) = mass flow rate (kg/s), \(h_2\) = enthalpy at compressor outlet (kJ/kg), \(h_1\) = enthalpy at compressor inlet (kJ/kg)

Tips & Tricks

Tip: Always label points in the cycle clearly on diagrams.

When to use: When solving problems involving state properties and enthalpy values.

Tip: Use steam tables or refrigerant property charts efficiently by interpolating between values.

When to use: When exact property values are not directly available.

Tip: Remember COP for refrigeration is always greater than 1.

When to use: To quickly check if your answer is reasonable.

Tip: Convert all units to metric system before calculations.

When to use: To avoid unit inconsistency errors.

Tip: Use mnemonic "C-C-E-E" for component order: Compressor, Condenser, Expansion valve, Evaporator.

When to use: To recall cycle sequence quickly during exams.

Common Mistakes to Avoid

❌ Confusing refrigeration effect with work input.
✓ Refrigeration effect is heat absorbed in evaporator; work input is energy supplied to compressor.
Why: Both involve energy terms but represent different physical quantities.
❌ Using incorrect enthalpy values at cycle points.
✓ Always verify state points and use correct pressure and temperature to find enthalpy.
Why: Wrong state assumptions lead to large calculation errors.
❌ Ignoring unit conversions especially for mass flow rate and energy.
✓ Convert all units to SI (kg, kJ, seconds) before calculations.
Why: Mixing units causes incorrect results.
❌ Misinterpreting P-h and T-s diagrams by mixing cycle directions.
✓ Follow the correct direction of cycle processes as per standard conventions.
Why: Incorrect cycle direction leads to wrong process identification.
❌ Assuming COP can be less than 1 for refrigeration.
✓ COP for refrigeration cycle is always >1 under normal operation.
Why: COP <1 indicates calculation or conceptual error.
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