What you'll be able to do本章學習成果
- Classify any property as intensive or extensive, and form specific (per-mass) properties.將任何性質分類為強度或廣延性質,並形成比(單位質量)性質。
- Define mass, density, specific volume, pressure, and temperature with their units.定義質量、密度、比體積、壓力與溫度及其單位。
- Define the energy properties — internal energy, enthalpy, and entropy — and the specific heats $c_v$, $c_p$.定義能量性質——內能、焓與熵——以及比熱 $c_v$、$c_p$。
Pairs with Basic Concepts (systems, state, the state postulate).與基本概念(系統、狀態、狀態假設)配對學習。
Key definitions關鍵定義
Intensive vs extensive強度性質與廣延性質
A property is any measurable characteristic of a system. Properties split into two kinds:性質是系統任何可量測的特徵。性質分為兩類:
- Intensive — independent of system size: temperature, pressure, density.強度性質——與系統大小無關:溫度、壓力、密度。
- Extensive — proportional to extent: mass, volume, total energy, entropy.廣延性質——與系統大小成比例:質量、體積、總能量、熵。
Divide an extensive property by mass and you get a specific (intensive) property. Quick test: split the system in two — whatever halves is extensive; whatever stays the same is intensive.廣延性質除以質量就得到「比」(強度)性質。快速測試:將系統一分為二——半掉的是廣延性質;不變的是強度性質。
The test — cut the system in half測試——將系統一分為二
Take a smaller portion of the same gas and watch what changes. Mass, volume, and total energy scale with the portion (extensive); temperature, pressure, density, and specific volume hold fast (intensive).取同一氣體的較小部分,觀察何者改變。質量、體積與總能量隨部分比例縮放(廣延);溫度、壓力、密度與比體積不變(強度)。
Mass, density & specific volume質量、密度與比體積
We treat matter as a continuum. Density is mass per unit volume; specific volume is its reciprocal:我們將物質視為連續體。密度是单位體積的質量;比體積是其倒數:
Specific gravity is the ratio of a substance's density to that of water at 4 °C.比重是物質密度與 4 °C 水的密度之比。
Pressure壓力
Pressure is the normal force a fluid exerts per unit area (SI: $1\,\mathrm{Pa} = 1\,\mathrm{N/m^2}$). Gauges read relative to atmosphere:壓力是流體對各單位面積施加的法向力(SI:$1\,\mathrm{Pa}=1\,\mathrm{N/m^2}$)。壓力表讀數相對於大氣:
Pressures below atmospheric are vacuum pressures. Thermodynamic relations use absolute pressure. In a static fluid, $\Delta p = \rho g h$.低於大氣壓的為真空壓。熱力學關係式一律用絕對壓力。靜止流體中,$\Delta p = \rho g h$。
Temperature溫度
Temperature measures the tendency of a system to exchange heat; equality of temperature is the condition for thermal equilibrium (the zeroth law). The absolute Kelvin scale:溫度表徵系統交換熱量的傾向;溫度相等是熱平衡的條件(第零定律)。絕對克耳文溫標:
Thermodynamic relations involving temperature ratios require the absolute scale.涉及溫度比的熱力學關係式必須使用絕對溫標。
Internal energy內能
Internal energy $U$ is the energy stored in a system's molecular motion and interactions. Specific internal energy $u = U/m$ (kJ/kg). For an ideal gas, $u$ depends on temperature alone.內能 $U$ 是儲存於系統分子運動與互動中的能量。比內能 $u = U/m$(kJ/kg)。對理想氣體,$u$ 僅與溫度有關。
Enthalpy焓
The combination $u + pv$ appears so often it earns its own name, enthalpy:$u + pv$ 的組合極常出現,因此專門命名為焓:
It is the natural energy property for flowing streams and constant-pressure processes.焓是流動氣流與定壓過程的自然能量性質。
Entropy熵
Entropy $S$ is the extensive property that measures the dispersal of energy and the direction of spontaneous change. For an internally reversible process:熵 $S$ 是量度能量分散與自發變化方向的廣延性質。對內部可逆過程:
Unlike mass and energy, entropy can be produced by irreversibilities but never destroyed — the heart of the second law. (See Entropy.)與質量、能量不同,熵可由不可逆性產生,卻永不減少——這是第二定律的核心。(見熵)
Specific heats比熱
The specific heats quantify how much a property changes per degree of temperature — at constant volume or constant pressure:比熱量化性質每度溫度的變化——在定體積或定壓下:
For an ideal gas $c_p - c_v = R$; for an incompressible solid or liquid, $c_p = c_v = c$. Despite the name, they are properties — not heat transfer $Q$.對理想氣體,$c_p - c_v = R$;對不可壓縮圖體或液體,$c_p = c_v = c$。尽管名稱含「熱」,它們是性質——而非熱轉移量 $Q$ 本身。
Specific properties from totals由總量求比性質
Example範例 Density, specific volume, enthalpy密度、比體積、焓 ›
Given: 2 kg of gas in 0.80 m³ at 150 kPa, $u=210$ kJ/kg.已知:2 kg 氣體在 150 kPa 下占據 0.80 m³,$u=210$ kJ/kg。
Find: density, specific volume, and specific enthalpy.求:密度、比體積與比焓。
Solution. $$\rho=2/0.80=2.5\ \tfrac{\text{kg}}{\text{m}^3},\quad v=0.40\ \tfrac{\text{m}^3}{\text{kg}},\quad h=210+150(0.40)=270\ \tfrac{\text{kJ}}{\text{kg}}$$解: $$\rho=2/0.80=2.5\ \tfrac{\text{kg}}{\text{m}^3},\quad v=0.40\ \tfrac{\text{m}^3}{\text{kg}},\quad h=210+150(0.40)=270\ \tfrac{\text{kJ}}{\text{kg}}$$