Refrigeration

Refrigeration is the process of removing heat from a body or space so that its temperature is lowered below that of the surroundings and maintained there. It needs work (or heat) input and rejects heat to a higher-temperature sink.

Ton of refrigeration (TR)

One TR is the rate of heat removal needed to freeze 1 ton (2000 lb, the US short ton) of water at 0 °C into ice at 0 °C in 24 hours.

$$1\ \text{TR} = \frac{2000 \times 144\ \text{Btu}}{24\ \text{h}} = 12000\ \text{Btu/h} = 211\ \text{kJ/min} = 3.517\ \text{kW}$$

(With a metric tonne of ice per day the value is about 3.87 kW.)

Coefficient of performance

$$\text{COP} = \frac{\text{refrigerating effect}}{\text{work input}} = \frac{Q_L}{W}$$

For a refrigerator the COP can exceed 1. The reversed Carnot (maximum) value is $T_L/(T_H - T_L)$.

Applications

1. Domestic refrigerators and freezers.
2. Cold storage of food, fruits, vegetables, meat and milk; dairy plants.
3. Air-conditioning of buildings and vehicles.
4. Manufacture of ice, and chilling in breweries and chemical plants.
5. Medical use: vaccine and blood storage, and industrial process cooling.

The simple vapour compression cycle uses a refrigerant that evaporates at low pressure (absorbing heat from the cold space) and condenses at high pressure (rejecting heat to the surroundings).

          Q_H (heat rejected)
        +------------------+
   2 -->|    CONDENSER     |--> 3
        +------------------+
   ^                          |
   |                          v
 COMPRESSOR  <-- W        EXPANSION VALVE
   ^                          |
   |                          v
        +------------------+
   1 <--|    EVAPORATOR    |<-- 4
        +------------------+
          Q_L (heat absorbed)

Processes (ideal cycle)

• 1-2 Isentropic compression: dry saturated vapour at $p_e$ is compressed to superheated vapour at $p_c$.
• 2-3 Condensation at constant pressure: the vapour is desuperheated and condensed to saturated liquid, rejecting $Q_H$.
• 3-4 Throttling (isenthalpic): liquid expands in the expansion valve or capillary tube, $h_3 = h_4$; some flashes to vapour; the temperature falls to the evaporator value.
• 4-1 Evaporation at constant pressure: the wet mixture takes heat $Q_L$ from the cold space and becomes dry saturated vapour.

Diagrams

 T-s diagram                  p-h diagram
 T                            p
 |           2                |  3------------2
 |         / |                |  |           /
 |  3-----2' |                |  |         /
 |  |        |                |  |       /
 |  4--------1                |  4------1
 +-------------- s            +----------------- h

In the T-s diagram 3-2' is condensation, 2'-2 is desuperheating and 1-2 is vertical (isentropic). On the p-h diagram, lines 4-1 and 2-3 are horizontal, 3-4 is vertical (constant enthalpy) and 1-2 follows a constant-entropy line sloping upward.

Performance expressions (per kg)

$$\begin{aligned} q_L &= h_1 - h_4 = h_1 - h_3\quad\text{(refrigerating effect)}\\ w &= h_2 - h_1\quad\text{(compressor work)}\\ q_H &= h_2 - h_3\\ \text{COP} &= \frac{h_1 - h_3}{h_2 - h_1} \end{aligned}$$

Mass flow for a capacity $Q$: $\dot m = Q/(h_1 - h_4)$ and power $P = \dot m (h_2 - h_1)$.

The ideal cycle assumes isentropic compression, no pressure drops and exact saturation at the compressor inlet and condenser outlet. A real plant departs from this because of friction, heat transfer across a temperature difference and practical safety requirements.

Causes of deviation in more detail

1. Superheating in the evaporator exit and suction line increases specific volume and compressor work slightly, but ensures dry vapour.
2. Subcooling in the condenser (and liquid line) gives extra refrigerating effect.
3. Pressure drops in the evaporator, condenser and lines require the compressor to start at a lower suction pressure and deliver at a higher pressure.
4. Compressor irreversibility: isentropic efficiency of 70 to 85 %; part of the work appears as extra heat in the vapour.

 T
 |                 2  actual
 |                2s  ideal
 |   3'--3-----2'  |  |
 |    |  |         |  |
 |    4---------1--1'
 +------------------------- s

1-1' superheating; 1'-2s ideal compression; 1'-2 actual compression (entropy rises); 2-2'-3-3' desuperheating, condensation and subcooling; 3'-4 throttling.

(a) Subcooling

Cooling the liquid below the saturation temperature at condenser pressure (3 to 3').

• Enthalpy $h_3$ falls, so the refrigerating effect $h_1 - h_{3'}$ increases.
• Compressor work is unchanged, so COP increases.
• Flash vapour formed in the expansion valve decreases.
• It is obtained with a larger condenser, a liquid-to-suction heat exchanger or a separate subcooler.

(b) Superheating

Heating the vapour above the saturation temperature at evaporator pressure (1 to 1').

• If the heat comes from the cold space (useful superheat), the refrigerating effect rises by $h_{1'} - h_1$; however, the specific volume of the vapour increases and compressor work also increases. The COP may rise or fall, depending on the refrigerant.
• If the heat comes from the surroundings (useless superheat), the refrigerating effect does not increase but the work does, so the COP falls.
• It prevents liquid slugging (wet compression) in the compressor, so a small superheat is always provided.

 p-h diagram (not to scale)
 p
 |   3'--3-----------2
 |   |   |         /
 |   |   |       /
 |   4'  4-----1--1'
 +---------------------- h

Points: 3 to 3' = subcooling, 1 to 1' = superheating.

The reversed Carnot cycle (isothermal heat absorption, isentropic compression, isothermal heat rejection, isentropic expansion) has the maximum COP between two temperatures, $\text{COP} = T_L/(T_H - T_L)$. It is used only as a standard of comparison, for these reasons:

1. Wet compression: the compression starts in the two-phase region. Liquid droplets damage the valves and cylinder wall; it is hard to compress a liquid-vapour mixture isentropically.
2. Expansion engine: the isentropic expansion would need an expander handling a wet mixture. Its work output is very small compared to the cost and friction, so it is replaced by a throttle valve.
3. Practical isothermal processes need an infinitely large heat exchanger or very slow heat transfer.
4. Slow, bulky plant for a given capacity, and the cycle is sensitive to control.
5. Operating close to the critical point is difficult.

Comparison with the vapour compression cycle

Cycle states

• State 1: dry saturated vapour, $h_1 = 239.16$ kJ/kg.
• State 2: after isentropic compression, $h_2 = 275.39$ kJ/kg.
• State 3: saturated liquid, $h_3 = 95.47$ kJ/kg.
• State 4: after throttling, $h_4 = h_3 = 95.47$ kJ/kg.

Capacity: $Q_L = 5 \times 3.517 = 17.585$ kW.

(a) Refrigerating effect

$$q_L = h_1 - h_4 = 239.16 - 95.47 = 143.69\ \text{kJ/kg}$$

(b) Mass flow rate

$$\dot m = \frac{Q_L}{q_L} = \frac{17.585}{143.69} = 0.1224\ \text{kg/s}$$

(c) Compressor power

$$w = h_2 - h_1 = 275.39 - 239.16 = 36.23\ \text{kJ/kg}$$ $$P = \dot m\,w = 0.1224 \times 36.23 = 4.43\ \text{kW}$$

(d) COP

$$\text{COP} = \frac{q_L}{w} = \frac{143.69}{36.23} = 3.97$$

(e) Heat rejected in condenser

$$Q_H = \dot m (h_2 - h_3) = 0.1224 \times (275.39 - 95.47) = 22.02\ \text{kW}$$

Check: $Q_L + P = 17.585 + 4.43 = 22.02$ kW, equal to $Q_H$.

(f) Volume flow rate at compressor inlet

$$\dot V_1 = \dot m\,v_1 = 0.1224 \times 0.1402 = 0.0172\ \text{m}^3/\text{s}$$

Answer: (a) 143.7 kJ/kg; (b) 0.1224 kg/s; (c) 4.43 kW; (d) 3.97; (e) 22.0 kW; (f) 0.0172 m³/s.

Capacity: $Q_L = 3 \times 3.517 = 10.551$ kW. Throttling gives $h_4 = h_3 = 88.5$ kJ/kg.

(a) Refrigerating effect

$$q_L = h_1 - h_4 = 243.5 - 88.5 = 155.0\ \text{kJ/kg}$$

(b) Mass flow rate

$$\dot m = \frac{10.551}{155.0} = 0.0681\ \text{kg/s}$$

(c) Compressor power

$$w = h_2 - h_1 = 280.5 - 243.5 = 37.0\ \text{kJ/kg}$$ $$P = 0.0681 \times 37.0 = 2.52\ \text{kW}$$

(d) COP

$$\text{COP} = \frac{155.0}{37.0} = 4.19$$

Comment

The COP rises from 3.97 to 4.19, about 5.6 %, because subcooling increased the refrigerating effect (from 143.7 to 155.0 kJ/kg) while the extra compressor work due to superheat was small (36.2 to 37.0 kJ/kg).

Answer: (a) 155 kJ/kg; (b) 0.0681 kg/s; (c) 2.52 kW; (d) 4.19.

Refrigerating effect

The cooling is obtained in the evaporator, so (suction-line heat gain is a loss from outside the cold space):

$$q_L = h_{evap\ exit} - h_4 = 239.16 - 95.47 = 143.69\ \text{kJ/kg}$$

(a) Compressor work per kg

$$\begin{aligned} w_s &= h_{2s} - h_1 = 278.6 - 242.2 = 36.4\ \text{kJ/kg}\\ w_a &= \frac{w_s}{\eta_s} = \frac{36.4}{0.75} = 48.53\ \text{kJ/kg} \end{aligned}$$

Actual discharge enthalpy: $h_2 = 242.2 + 48.53 = 290.7$ kJ/kg.

(b) Mass flow rate

$$\dot m = \frac{10}{143.69} = 0.0696\ \text{kg/s}$$

(c) Power

$$\begin{aligned} P_{comp} &= \dot m\,w_a = 0.0696 \times 48.53 = 3.38\ \text{kW}\\ P_{elec} &= \frac{P_{comp}}{\eta_{motor}} = \frac{3.38}{0.9} = 3.75\ \text{kW} \end{aligned}$$

(d) Actual COP

Based on shaft power: $\text{COP} = 10/3.38 = 2.96$. Based on electrical input: $\text{COP} = 10/3.75 = 2.66$.

(For comparison, the ideal COP with no losses is $143.69/36.23 = 3.97$.)

(e) Condenser heat rejection

$$Q_H = \dot m (h_2 - h_3) = 0.0696 \times (290.7 - 95.47) = 13.59\ \text{kW}$$

(Energy check: $10 + 0.21$ suction-line gain $+ 3.38 = 13.59$ kW.)

Answer: (a) 48.5 kJ/kg; (b) 0.0696 kg/s; (c) 3.38 kW at the shaft, 3.75 kW electrical; (d) 2.96 (shaft) or 2.66 (electrical); (e) 13.6 kW.