∮ Thermodynamics熱力學
Chapters章節  /  05 Cycle05 循環

Refrigeration & Heat Pump Systems製冷與熱泵系統

Run a power cycle backwards and you can move heat uphill — from cold to hot — at the cost of work. The vapor-compression cycle in your fridge, AC, and heat pump is exactly that. We analyze it, define the coefficient of performance, and see what raises or lowers it.

COP explorer R-134a T–s
Overview總覽

What you'll be able to do本章學習成果

  • Describe the four-component vapor-compression refrigeration and heat-pump cycles and sketch their T–s diagrams.
  • Apply mass and energy balances to evaluate capacity, compressor power, and COP.
  • Account for irreversibilities via the isentropic compressor efficiency.
  • Explain how varying evaporator and condenser temperatures changes performance, and outline absorption and gas (Brayton) refrigeration.

Key equations重要公式

Refrigerator COP製冷機性能係數
$\beta = (h_1-h_4)/(h_2-h_1)$
Heat-pump COP熱泵性能係數
$\gamma = (h_2-h_3)/(h_2-h_1) = \beta + 1$
Carnot limits卡諾極限
$\beta_{max}=\tfrac{T_C}{T_H-T_C},\;\; \gamma_{max}=\tfrac{T_H}{T_H-T_C}$
Throttling valve節流閥
$h_4 = h_3$
The cycle循環

The vapor-compression cycle蒸汽壓縮循環

The most common refrigeration cycle in use today has four components and four processes:

  • 4–1 Evaporator: a low-pressure two-phase mixture evaporates, absorbing heat from the refrigerated space.
  • 1–2 Compressor: vapor is compressed to high temperature and pressure (work input).
  • 2–3 Condenser: vapor condenses to liquid, rejecting heat to the warmer surroundings.
  • 3–4 Expansion valve: liquid throttles back to evaporator pressure.
Engineering model工程模型

Each component is a steady-state control volume. The compressor is adiabatic; the valve is a throttling process ($h_4 = h_3$, constant enthalpy); kinetic and potential energy changes are ignored. "Dry compression" means the refrigerant entering the compressor is vapor.

Analysis分析

Component balances各組件平衡式

Per unit mass of refrigerant, mass and energy balances give:

$$ \frac{\dot Q_{in}}{\dot m}=h_1-h_4 \quad \frac{\dot W_c}{\dot m}=h_2-h_1 \quad \frac{\dot Q_{out}}{\dot m}=h_2-h_3 \quad h_4=h_3 $$
Eq. 10.3–10.6

The evaporator term $\dot Q_{in}$ is the refrigeration capacity, in kW. A traditional unit is the ton of refrigeration ≈ 211 kJ/min ≈ 3.52 kW.

Performance性能

Coefficient of performance性能係數

Because the "benefit" of a refrigerator is heat removed and the "cost" is compressor work, performance is a coefficient of performance — which can exceed 1:

$$ \beta_{ref} = \frac{\dot Q_{in}/\dot m}{\dot W_c/\dot m} = \frac{h_1-h_4}{h_2-h_1} $$
Eq. 10.7

The maximum theoretical value for any cycle operating between cold and hot regions at $T_C$ and $T_H$ is the Carnot COP:

$$ \beta_{max} = \frac{T_C}{T_H-T_C} $$
Carnot
Interactive互動

COP explorerCOP 探索器

Set the evaporator and condenser temperatures and the compressor's isentropic efficiency. The cycle traces on the R-134a T–s diagram while the COP, capacity, and second-law efficiency (vs. Carnot) update. Try narrowing the temperature gap — the COP climbs sharply, exactly as the Carnot limit predicts.

Teaching model on a compact R-134a saturation table; superheated compression approximated with $c_p \approx 0.95$ kJ/kg·K.

Reality現實情況

Actual vs. ideal cycle實際循環與理想循環比較

Real cycles deviate from the ideal in two ways. Heat transfer is irreversible: the refrigerant must be colder than $T_C$ in the evaporator and hotter than $T_H$ in the condenser, which widens the effective temperature span and lowers COP. And compression is irreversible: entropy rises across the compressor, so the actual work exceeds the isentropic ideal. The isentropic compressor efficiency captures this:

$$ \eta_c = \frac{(\dot W_c/\dot m)_s}{\dot W_c/\dot m} = \frac{h_{2s}-h_1}{h_2-h_1} $$
Eq. 6.48

Since the refrigeration effect $h_1-h_4$ is unchanged but the work grows, irreversible compression always lowers the COP. Drag the efficiency slider in the explorer to see it directly.

Reversing the goal目標反向

Heat pumps熱泵

A heat pump uses the identical hardware but its objective is the warm side — keeping a space above ambient. The benefit is now the heat delivered by the condenser:

$$ \gamma = \frac{\dot Q_{out}/\dot m}{\dot W_c/\dot m} = \frac{h_2-h_3}{h_2-h_1}, \qquad \gamma_{max} = \frac{T_H}{T_H-T_C} $$
Eq. 10.9, 10.10

Note $\gamma = \beta + 1$: every unit of work shows up in the delivered heat plus the heat pulled from outdoors, which is why heat pumps can deliver several units of heat per unit of electricity. Toggle the explorer to "Heat pump" mode to compare.

Beyond vapor-compression超越蒸汽壓縮

Other refrigeration systems其他製冷系統

  • Absorption refrigeration (e.g. ammonia–water) replaces the compressor with an absorber, pump, and generator. Pumping a liquid solution takes far less work than compressing vapor, and the generator can run on waste heat or solar — attractive where cheap heat is available.
  • Brayton (gas) refrigeration keeps the working fluid a gas throughout — the reversed Brayton cycle. The turbine helps drive the compressor; used in aircraft and cryogenics.
  • Refrigerant selection weighs performance, safety (toxicity, flammability), and environmental impact — ozone depletion (CFCs/HCFCs are phased out) and global warming potential (many HFCs are high-GWP; natural refrigerants like CO₂ and ammonia are low-GWP).
Worked example範例

Vapor-compression performance蒸汽壓縮性能

Example範例 Capacity and COP with R-134aR-134a 的製冷量與 COP ›

Given: R-134a cycle with $h_1=241.4$, $h_2=280.2$, $h_3=h_4=91.5$ kJ/kg and $\dot m=0.08$ kg/s.

Find: compressor power, refrigeration capacity, and COP.

Solution. $$\dot W_c=\dot m(h_2-h_1)=0.08(38.8)=3.1\text{ kW},\quad \dot Q_{in}=\dot m(h_1-h_4)=0.08(149.9)=12.0\text{ kW }(3.4\text{ tons}).$$ $$\beta=\frac{h_1-h_4}{h_2-h_1}=\frac{149.9}{38.8}=3.86.$$