What you'll be able to do本章學習成果
- Describe the physics of phase-change and read the T–v, P–v, and P–T diagrams.說明相變化的物理機制並讀取 T–v、P–v 與 P–T 圖。
- Determine properties from tables, including quality for two-phase mixtures.由性質表查得性質,包括兩相混合物的乾度。
- Use linear interpolation and the saturated-liquid approximation for compressed liquids.使用線性插値與壓縮液體的飽和液近似。
Key equations重要公式
Phases & phase change相與相變化
A pure substance has a fixed chemical composition throughout. Heat a compressed liquid at constant pressure and it marches through:純物質全體具有固定的化學組成。在定壓下加熱壓縮液體,將依序經歷:
- Compressed (subcooled) liquid — not about to vaporize.壓縮(未飽和)液體——尚未达到永魔。
- Saturated liquid — on the verge of vaporizing.飽和液體——即將永魔。
- Saturated liquid–vapor mixture — both phases coexist; temperature holds constant while latent heat is absorbed.飽和液氣混合物——兩相共存;溫度保持不變而潛熱被吸收。
- Saturated vapor — the last drop has just vaporized.飽和蒸氣——最後一滴剛完全永魔。
- Superheated vapor — heated beyond saturation.過熱蒸氣——已加熱至超過飽和狀態。
The energy absorbed during vaporization is the latent heat of vaporization $h_{fg}$ — for water at 1 atm, about 2257 kJ/kg.永魔過程中吸收的能量為永魔潛熱 $h_{fg}$ ——水在 1 大氣壓下約為 2257 kJ/kg。
Saturation & property diagrams飽和與性質圖
The saturation temperature $T_{sat}$ and saturation pressure $P_{sat}$ trace the saturation curve. On a T–v diagram, the saturated-liquid and saturated-vapor lines meet at the critical point, enclosing the two-phase dome. Above the critical pressure there is no distinct boiling. The triple point (water: 0.01 °C, 0.6117 kPa) is where solid, liquid, and vapor coexist.飽和溫度 $T_{sat}$ 與飽和壓力 $P_{sat}$ 描繪出飽和曲線。在 T–v 圖上,飽和液線與飽和氣線在臨界點交汇,圍成兩相圓頂。在臨界壓力以上則無明顯永魔。三相點(水:0.01 °C、0.6117 kPa)是化固、液體與蒸氣三相共存之處。
Quality of a mixture混合物的乾度
Inside the dome, temperature and pressure are not independent — so we need another property to fix the state: the quality $x$, the vapor mass fraction:在圓頂內,溫度與壓力並非獨立——因此需另一個性質來固定狀態:乾度 $x$,即蒸氣質量分率:
The same lever-rule form gives $u = u_f + x\,u_{fg}$ and $h = h_f + x\,h_{fg}$. Geometrically, the mixture state sits a fraction $x$ of the way from the liquid line to the vapor line.同樣形式給出 $u = u_f + x\,u_{fg}$ 與 $h = h_f + x\,h_{fg}$。在圓頂圖上,混合狀態點位於液線到氣線之間 $x$ 的位置。
Saturation dome & quality lab飽和圓頂與乾度實驗
Pick a saturation temperature, then slide the quality from 0 (saturated liquid, point f) to 1 (saturated vapor, point g). The state point glides along the tie-line, and specific volume and enthalpy follow the lever rule.選擇飽和溫度,再將乾度從 0(飽和液體, f 點)滑至 1(飽和蒸氣, g 點)。狀態點沿結線滑動,比體積與焓遵循桿桿法則。
Saturated-water data (teaching subset). The specific-volume axis is logarithmic to show the full liquid-to-vapor span.
Property tables性質表
Most substances are too complex for simple equations, so properties are tabulated. The steam tables are the canonical example: saturation tables (by T and by p) list $v_f, v_g, u, h, s$; superheated and compressed-liquid tables cover single-phase regions where $T$ and $p$ are independent. When a state falls between entries, use linear interpolation. Tables are built on a reference state (water: 0.01 °C; R-134a: −40 °C) — only changes matter.大多數物質太複雜而難以簡單方程式描述,因此性質均被表格化。蒸汽表是典型範例:飽和表(按 T 與按 p)列出 $v_f, v_g, u, h, s$;過熱蒸氣表與壓縮液體表涵蓋單相區。狀態介於兩項間時,使用線性插値。表格不同性質的參考狀態各不相同(水:0.01 °C;R-134a:−40 °C)——實際上只有變化量才有意義。
Compressed-liquid properties depend far more on temperature than pressure, so $v(T,p) \approx v_f(T)$ and $u(T,p) \approx u_f(T)$.壓縮液體性質對溫度的依賴遠大於壓力,因此 $v(T,p) \approx v_f(T)$、$u(T,p) \approx u_f(T)$。
Once a substance is well into the vapor region at low density, the tables give way to the ideal-gas model and the compressibility factor — covered in Ideal Gas & Compressibility.當物質在低密度蒸氣區時,表格讓位於理想氣體模型與壓縮因子——見理想氣體與壓縮因子。
Properties of a wet mixture濕混合物的性質
Example範例 Quality and the lever rule乾度與桿桿法則 ›
Given: liquid–vapor mixture of water at 100 °C, $x=0.60$. $v_f=0.001$, $v_g=1.673$ m³/kg, $h_f=419$, $h_g=2676$ kJ/kg.已知:100 °C 的水的液氣混合物,$x=0.60$。$v_f=0.001$、$v_g=1.673$ m³/kg,$h_f=419$、$h_g=2676$ kJ/kg。
Find: specific volume and enthalpy.求:比體積與比焓。
Solution. $$v=v_f+x(v_g-v_f)=0.001+0.60(1.672)=1.00\ \tfrac{\text{m}^3}{\text{kg}}$$ $$h=h_f+x\,h_{fg}=419+0.60(2257)=1773\ \tfrac{\text{kJ}}{\text{kg}}$$解: $$v=0.001+0.60(1.672)=1.00\ \tfrac{\text{m}^3}{\text{kg}}$$ $$h=419+0.60(2257)=1773\ \tfrac{\text{kJ}}{\text{kg}}$$