What you'll be able to do本章學習成果
- Describe the physics of phase change and locate every region of the p–v–T surface.說明相變化的物理機制，並在 p–v–T 曲面上定位每個區域。
- Read the p–v, T–v, and p–T diagrams as projections of that surface — isotherms, isobars, the saturation dome, the critical point, and the triple point.將 p–v、T–v 與 p–T 圖視為該曲面的投影並加以解讀——等溫線、等壓線、飽和圓頂、臨界點與三相點。
- 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。
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——尚未沸騰。
- Saturated liquid — on the verge of vaporizing.飽和液體——即將汽化。
- 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——最後一滴剛完全汽化。
- 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. The five states above are numbered 1–5 on the T–v diagram below.汽化過程中吸收的能量為汽化潛熱 $h_{fg}$ ——水在 1 大氣壓下約為 2257 kJ/kg。上述五個狀態在下方 T–v 圖中標為 1–5。
The p–v–T surfacep–v–T 曲面
By the state postulate, two independent intensive properties fix the state of a simple compressible substance. Pick $v$ and $T$ as the two, and pressure follows: $p = p(v, T)$. Plotted in three dimensions this relation is a surface — the p–v–T surface. Every equilibrium state of the substance is a point on it; nothing off the surface can exist at equilibrium.依狀態假設，兩個獨立的強度性質即可決定簡單可壓縮物質的狀態。取 $v$ 與 $T$ 為此二者，壓力隨之確定：$p = p(v, T)$。將此關係繪於三維空間即成一曲面——p–v–T 曲面。物質的每一個平衡狀態都是曲面上的一點；曲面之外的點在平衡下不存在。
The surface above is drawn for a substance that contracts on freezing — the solid is denser than the liquid, so the solid region sits at smaller $v$ than the liquid, the solid–liquid region rises to the left, and on p–T the fusion line leans right (pressure raises the melting point). This is the normal case: CO₂, most metals, paraffin, nearly all organic liquids. Water is the famous exception (with bismuth, gallium, silicon, germanium): ice is less dense than liquid water ($v_{ice} = 0.00109 > v_f = 0.00100$ m³/kg at 0 °C), so on water's surface the solid region lies to the right of the liquid, the solid–liquid region dips the other way, and the fusion line on p–T leans left — pressure lowers the melting point. Consequences: ice floats, pipes burst, an ice-skate blade melts a film of water under it, and glaciers creep on a pressure-melted base.上圖是凝固時收縮的物質：固體比液體密，故固體區位於比液體更小的 $v$ 處，固–液區向左上方升起，p–T 圖上的熔化線向右傾（壓力使熔點升高）。這是常態：CO₂、大多數金屬、石蠟、幾乎所有有機液體。水是著名的例外（還有鉍、鎵、矽、鍺）：冰比液態水疏（0 °C 時 $v_{ice} = 0.00109 > v_f = 0.00100$ m³/kg），故水的曲面上固體區位於液體右側、固–液區反向傾斜，p–T 圖上的熔化線向左傾——壓力使熔點降低。後果：冰會浮、水管會凍裂、冰刀下方會壓出一層水膜、冰川在壓力融化的底層上滑動。
A 3-D surface is awkward to work with, so we project it onto the three coordinate planes. Each projection keeps part of the information and hides part — which is why all three are used side by side.三維曲面不便使用，因此將其投影到三個坐標平面上。每種投影保留部分資訊也隱藏部分資訊——這正是三種圖並用的原因。
The p–v diagramp–v 圖
Project the surface onto the p–v plane and the two-phase regions open up into areas while lines of constant temperature — isotherms — become the family of curves that carry the shape of the surface. Follow one isotherm below $T_{cr}$ from left to right: in the liquid, pressure drops steeply because liquids are nearly incompressible; across the dome the isotherm is horizontal ($p = p_{sat}(T)$); in the vapor it falls roughly hyperbolically.將曲面投影到 p–v 平面，兩相區展開成面積，而定溫線——等溫線——成為承載曲面形狀的一族曲線。沿一條 $T_{cr}$ 以下的等溫線由左向右：在液體區，因液體幾乎不可壓縮，壓力陡降；穿越圓頂時等溫線水平（$p = p_{sat}(T)$）；在蒸氣區則近似雙曲線下降。
As $T$ rises toward $T_{cr}$, $v_f$ and $v_g$ approach each other and the flat segment shrinks to a point — the critical point. The critical isotherm therefore has zero slope and zero curvature there:當 $T$ 趨近 $T_{cr}$，$v_f$ 與 $v_g$ 彼此逼近，水平段縮為一點——即臨界點。因此臨界等溫線在該處斜率與曲率皆為零：
Far to the lower right — low pressure, large specific volume, well away from the dome — the isotherms become exact hyperbolas $pv = RT$. That corner of the surface is the ideal-gas region; how far it extends, and what to do when it doesn't, is the subject of Ideal Gas, Mixtures & Compressibility.在右下方遠處——低壓、大比體積、遠離圓頂——等溫線成為精確的雙曲線 $pv = RT$。曲面的這一角即理想氣體區；其範圍多大、不適用時如何處理，見理想氣體、混合物與壓縮因子。
Numbers worth knowing: steam temperature is set by pressure值得記住的數字：蒸氣溫度由壓力決定
Inside the dome an isotherm is flat — so fixing the pressure fixes the temperature. Wherever water boils or steam condenses, its temperature is not a free choice: it is $T_{sat}(p)$, read off the vaporization line. A few anchor points make the tables feel familiar:圓頂內的等溫線是水平的——固定壓力即固定溫度。凡是水沸騰或蒸氣凝結之處，其溫度不能任意選擇：它就是 $T_{sat}(p)$，由汽化線讀出。記住幾個錨點，查表時便有熟悉感：
| Where場合 | p | Tsat | Note備註 |
|---|---|---|---|
| Sea level, open pot海平面，開放鍋 | 1 atm = 101.3 kPa ≈ 1 bar | 100 °C = 373.15 K | the definition-era anchor; at a round 100 kPa the tables give 99.6 °C早期定義的錨點；在整數 100 kPa 下查表得 99.6 °C |
| 3000 m mountain hut3000 m 高山小屋 | ≈ 70 kPa | ≈ 90 °C | roughly −1 °C per 300 m; rice and eggs cook slowly約每 300 m 降 1 °C；米飯與蛋煮得慢 |
| Everest summit, 8849 m聖母峰頂，8849 m | ≈ 34 kPa | ≈ 72 °C | tea, not boiling water只有溫茶，沒有滾水 |
| Pressure cooker壓力鍋 | ≈ 200 kPa (1 bar g) | ≈ 120 °C | +20 °C cuts cooking time by half or more; household units fall under CNS 3151 / UL 136 — below the boiler codes多 20 °C，烹調時間減半以上；家用品屬 CNS 3151 / UL 136，不在鍋爐法規範圍 |
| "Small boiler" (小型鍋爐), Taiwan小型鍋爐（台灣） | ≤ 1 kgf/cm² g = 0.1 MPa g ≈ 0.2 MPa abs | ≤ 120 °C | 鍋爐及壓力容器安全規則 §3: steam boilers below this (and ≤ 1 m² heating surface) escape full inspection; hot-water boilers likewise if head ≤ 10 m (≈ 0.1 MPa); once-through boilers up to 10 kgf/cm² = 1 MPa g (≤ 10 m²). US counterpart: ASME BPVC Section IV "Heating Boilers" — steam ≤ 15 psig (0.1 MPa g); hot water ≤ 160 psig (1.1 MPa g) and ≤ 250 °F (121 °C)《鍋爐及壓力容器安全規則》第 3 條：蒸汽鍋爐低於此壓力（且傳熱面積 ≤ 1 m²）免受完整檢查；熱水鍋爐水頭 ≤ 10 m（≈ 0.1 MPa）同；貫流鍋爐可達 10 kgf/cm² = 1 MPa g（≤ 10 m²）。美國對應：ASME BPVC Section IV「加熱鍋爐」——蒸汽 ≤ 15 psig（0.1 MPa g），熱水 ≤ 160 psig（1.1 MPa g）、≤ 250 °F（121 °C） |
| Industrial fire-tube boiler工業火管鍋爐 | 7 – 10 kgf/cm² g ≈ 0.7 – 1.0 MPa g (0.8 – 1.1 MPa abs) | 170 – 185 °C | typical Taiwan rating for process steam (food, textiles, laundries); fire-tube construction rarely exceeds ≈ 1.6 – 2 MPa (≈ 200 – 215 °C) — higher pressure needs water-tube designs. Design code: ASME BPVC Section I "Power Boilers" (all steam boilers above the Section IV limits); Taiwan: CNS 2139 boiler construction standard, inspected under 危險性機械及設備安全檢查規則台灣製程蒸氣（食品、紡織、洗衣）的常見額定；火管構造很少超過約 1.6 – 2 MPa（約 200 – 215 °C）——更高壓力須改用水管式。設計規範：ASME BPVC Section I「動力鍋爐」（超過 Section IV 限值之所有蒸汽鍋爐）；台灣：CNS 2139 鍋爐構造標準，依《危險性機械及設備安全檢查規則》檢查 |
| Steam condenser (power plant)發電廠凝結器 | 5 – 10 kPa | 33 – 46 °C | a deep vacuum — set by the cooling-water temperature; performance tested to ASME PTC 12.2, tubes to HEI Standards深真空——由冷卻水溫決定；性能試驗依 ASME PTC 12.2，管束依 HEI 標準 |
| Critical point臨界點 | 22.06 MPa | 373.95 °C | above this there is no $T_{sat}$ at all超過此點便不存在 $T_{sat}$ |
Units note. Taiwan boiler codes still quote gauge pressure in kgf/cm²: 1 kgf/cm² = 98.07 kPa ≈ 1 bar ≈ 1 atm. Tables A-4/A-5 use absolute pressure, so add ≈ 100 kPa before looking up $T_{sat}$. The steam tables themselves follow the IAPWS-IF97 formulation — the same international standard adopted by ASME and by power-plant performance test codes (ASME PTC).單位提醒。台灣鍋爐法規仍以 kgf/cm² 標示表壓力：1 kgf/cm² = 98.07 kPa ≈ 1 bar ≈ 1 atm。表 A-4/A-5 使用絕對壓力，查 $T_{sat}$ 前須先加約 100 kPa。本章蒸氣表本身依據 IAPWS-IF97 公式——ASME 與電廠性能試驗（ASME PTC）所採用的同一套國際標準。
Power-station boilers climb the dome. Once the steam leaves the dome it can be superheated to any temperature the tube metal allows, so utility boilers are classed by pressure — how far up the vaporization line they operate — and by the final steam temperature. All are built to ASME BPVC Section I (or EN 12952 in Europe); the pressure ceiling of each class is really a materials limit set by the allowable stresses in ASME Section II-D:發電鍋爐沿圓頂往上爬。蒸氣一離開圓頂，便可過熱到管材所容許的任何溫度，因此電廠鍋爐依壓力——即它們在汽化線上運轉的高度——與最終蒸氣溫度分類。全部依 ASME BPVC Section I（歐洲為 EN 12952）建造；各等級的壓力上限實際上是材料限制，由 ASME Section II-D 的容許應力決定：
| Class等級 | Main steam p主蒸氣壓力 | Main steam T主蒸氣溫度 | Note備註 |
|---|---|---|---|
| Subcritical (drum boiler)次臨界（汽包鍋爐） | 10 – 18 MPa | 540 – 565 °C | boiling at $T_{sat}$ = 310–360 °C in the drum, then superheated; plant efficiency ≈ 36–39 %汽包內於 $T_{sat}$ = 310–360 °C 沸騰後再過熱；電廠效率約 36–39 % |
| Supercritical (SC)超臨界（SC） | 24 – 25 MPa | 540 – 580 °C | $p > p_{cr}$: no boiling, no drum — water turns to steam continuously in a once-through boiler; ≈ 40–42 %$p > p_{cr}$：無沸騰、無汽包——水在直流鍋爐中連續轉為蒸氣；約 40–42 % |
| Ultra-supercritical (USC)超超臨界（USC） | 25 – 30 MPa | 600 – 620 °C | today's state of the art; limited by creep strength of 9–12 % Cr ferritic steels (ASME SA-213 T91/T92) and austenitic tubing (Super304H); ≈ 43–47 %當今主流最先進；受 9–12 % Cr 鐵素體鋼（ASME SA-213 T91/T92）與奧氏體管材（Super304H）蠕變強度限制；約 43–47 % |
| Advanced USC (700 °C class)先進超超臨界（700 °C 級） | 30 – 35 MPa | 700 – 760 °C | demonstration stage; needs nickel-base alloys (Inconel 740H — ASME Code Case 2702 — and alloy 617); efficiency target ≈ 50 %示範階段；須用鎳基合金（Inconel 740H——ASME Code Case 2702——及 617 合金）；效率目標約 50 % |
Read the two tables against the p–v diagram: the first walks up the vaporization line from the condenser vacuum to the critical point; the second leaves the dome entirely and heads for the upper right, where higher $T$ at higher $p$ buys efficiency — the reason, once we reach the second law, that supercritical plants exist.對照 p–v 圖讀這兩張表：第一張沿汽化線從凝結器真空一路爬到臨界點；第二張則完全離開圓頂、朝右上方前進——在更高 $p$ 下取得更高 $T$ 以換取效率，這正是我們學到第二定律時會明白的、超臨界電廠存在的理由。
The T–v diagramT–v 圖
Project onto the T–v plane and the curves become isobars. This is the natural picture for the constant-pressure heating of §Phases: state 1 is compressed liquid; at 2 the first bubble forms (saturated liquid, $v_f$); from 2 to 4 temperature holds at $T_{sat}$ while $v$ grows through the mixture; at 4 the last drop vanishes (saturated vapor, $v_g$); beyond 4 the vapor superheats.投影到 T–v 平面，曲線成為等壓線。這是「相變化」一節中定壓加熱過程的自然圖像：狀態 1 為壓縮液體；到 2 時出現第一個氣泡（飽和液體，$v_f$）；2 到 4 之間溫度維持 $T_{sat}$，$v$ 隨混合物增長；到 4 最後一滴液體消失（飽和蒸氣，$v_g$）；4 之後蒸氣過熱。
The saturation temperature $T_{sat}$ and saturation pressure $p_{sat}$ are one-to-one: fix either and the other follows. The locus of saturated-liquid states ($x=0$) is the saturated liquid line; the locus of saturated-vapor states ($x=1$) is the saturated vapor line. They meet at the critical point and enclose the saturation dome — the same dome as on the p–v diagram, viewed from a different side of the surface.飽和溫度 $T_{sat}$ 與飽和壓力 $p_{sat}$ 一一對應：固定其一，另一隨之確定。飽和液態（$x=0$）的軌跡為飽和液線；飽和氣態（$x=1$）的軌跡為飽和氣線。兩線在臨界點相會，圍成飽和圓頂——與 p–v 圖上的圓頂相同，只是從曲面的另一側觀看。
The p–T phase diagramp–T 相圖
Project onto the p–T plane and something dramatic happens: because $p$ and $T$ are locked together in every two-phase region, each ruled surface collapses edge-on into a line. Each line is named for the phase change crossing it in one direction; crossing the other way is the same equilibrium, just reversed:投影到 p–T 平面時發生了戲劇性的變化：因為每個兩相區內 $p$ 與 $T$ 相互鎖定，每個直紋面側視時塌縮成一條線。每條線以單向穿越它的相變化命名；反向穿越是同一平衡，只是方向相反：
- Sublimation line (solid ⇌ vapor) — solid → vapor is sublimation; vapor → solid is deposition (frost forming).昇華線（固 ⇌ 氣）——固 → 氣為昇華；氣 → 固為凝華（結霜）。
- Vaporization line (liquid ⇌ vapor) — liquid → vapor is vaporization (boiling / evaporation); vapor → liquid is condensation.汽化線（液 ⇌ 氣）——液 → 氣為汽化（沸騰／蒸發）；氣 → 液為凝結。
- Fusion line (solid ⇌ liquid) — solid → liquid is fusion (melting); liquid → solid is freezing (solidification).熔化線（固 ⇌ 液）——固 → 液為熔化；液 → 固為凝固。
The three lines meet at the triple point, the single $(p, T)$ where all three phases coexist. The vaporization line ends at the critical point; the fusion line, as far as is known, does not end.三條線交於三相點，即三相共存的唯一 $(p, T)$。汽化線終止於臨界點；熔化線據目前所知並無終點。
p–v for work ($\int p\,dv$ is an area) and for the shape of isotherms. T–v for heating and cooling processes and for reading quality. p–T to identify the phase from two measured properties — but never to locate a state inside a two-phase region, since $p$ and $T$ alone cannot.p–v 用於功（$\int p\,dv$ 為面積）與等溫線形狀。T–v 用於加熱與冷卻過程及讀取乾度。p–T 用於由兩個量測性質判別相——但絕不能用來定位兩相區內部的狀態，因為僅憑 $p$ 與 $T$ 無法辨別。
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. Raise the temperature and watch $h_{fg} = h_g - h_f$ shrink with the width of the dome: 2257 kJ/kg at 100 °C, about 1400 kJ/kg at 300 °C, zero at the critical point — where liquid and vapor have become the same thing and there is no longer anything to vaporize.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，比體積與焓遵循槓桿法則。提高溫度，可見 $h_{fg} = h_g - h_f$ 隨圓頂寬度一同縮小：100 °C 時 2257 kJ/kg，300 °C 約 1400 kJ/kg，臨界點為零——此時液體與蒸氣已無區別，再無可汽化之物。
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: saturated liquid at 0.01 °C, $u = 0$; R-134a: saturated liquid at −40 °C, $h = 0$) — 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，使用線性插值。表格建立於參考狀態之上（水：0.01 °C 飽和液體，$u = 0$；R-134a：−40 °C 飽和液體，$h = 0$）——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 $v(T,p) \approx v_f(T)$、$u(T,p) \approx u_f(T)$。
Locating states and reading processes定位狀態與解讀過程
Example範例 Quality and the lever rule乾度與槓桿法則 ›
Given: liquid–vapor mixture of water at 100 °C, $x=0.60$. $v_f=0.001043$, $v_g=1.672$ m³/kg, $h_f=419$, $h_g=2676$ kJ/kg.已知：100 °C 的水的液氣混合物，$x=0.60$。$v_f=0.001043$、$v_g=1.672$ m³/kg，$h_f=419$、$h_g=2676$ kJ/kg。
Find: specific volume and enthalpy.求：比體積與比焓。
Solution. $$v=v_f+x(v_g-v_f)=0.001043+0.60(1.671)=1.004\ \tfrac{\text{m}^3}{\text{kg}}$$ $$h=h_f+x\,h_{fg}=419+0.60(2257)=1773\ \tfrac{\text{kJ}}{\text{kg}}$$解： $$v=0.001043+0.60(1.671)=1.004\ \tfrac{\text{m}^3}{\text{kg}}$$ $$h=419+0.60(2257)=1773\ \tfrac{\text{kJ}}{\text{kg}}$$
Example範例 Locate the state — which region?定位狀態——在哪個區域？ ›
Given: water at (a) 300 kPa, 200 °C; (b) 150 °C, $v = 0.20$ m³/kg; (c) 5 MPa, 100 °C. Saturation values (supplied here; in the next module you will look them up): $T_{sat}(300\ \text{kPa}) = 133.5$ °C; at 150 °C, $v_f = 0.001091$, $v_g = 0.3928$ m³/kg, $p_{sat} = 476$ kPa; $T_{sat}(5\ \text{MPa}) = 263.9$ °C; $v_f(100\ °\text{C}) = 0.001043$ m³/kg.已知：水在 (a) 300 kPa、200 °C；(b) 150 °C、$v = 0.20$ m³/kg；(c) 5 MPa、100 °C。飽和值（此處直接給出；下一模組將學習查表）：$T_{sat}(300\ \text{kPa}) = 133.5$ °C；150 °C 時 $v_f = 0.001091$、$v_g = 0.3928$ m³/kg、$p_{sat} = 476$ kPa；$T_{sat}(5\ \text{MPa}) = 263.9$ °C；$v_f(100\ °\text{C}) = 0.001043$ m³/kg。
Find: the phase region of each, and where it sits on the T–v and p–v diagrams.求：各狀態所在的相區，及其在 T–v 與 p–v 圖上的位置。
Method. Compare the given property with its saturation value. Given $p$ and $T$: compare $T$ with $T_{sat}(p)$. Given $T$ and $v$: compare $v$ with $v_f(T)$ and $v_g(T)$.方法。將已知性質與其飽和值比較。已知 $p$、$T$：比較 $T$ 與 $T_{sat}(p)$。已知 $T$、$v$：比較 $v$ 與 $v_f(T)$、$v_g(T)$。
(a) $T_{sat}(300\ \text{kPa}) = 133.5$ °C. $T = 200 > T_{sat}$ → superheated vapor. On T–v: on the 300 kPa isobar, to the right of the dome, above the flat segment.(a) $T_{sat}(300\ \text{kPa}) = 133.5$ °C。$T = 200 > T_{sat}$ → 過熱蒸氣。T–v 圖：在 300 kPa 等壓線上、圓頂右側、水平段之上。
(b) At 150 °C: $v_f = 0.001091$, $v_g = 0.3928$ m³/kg. $v_f < 0.20 < v_g$ → saturated mixture, $x = (0.20 - 0.0011)/(0.3928 - 0.0011) = 0.51$. On T–v: on the 150 °C horizontal, about halfway between the lines. $p = p_{sat}(150\ °\text{C}) = 476$ kPa.(b) 150 °C 時：$v_f = 0.001091$、$v_g = 0.3928$ m³/kg。$v_f < 0.20 < v_g$ → 飽和混合物，$x = (0.20 - 0.0011)/(0.3928 - 0.0011) = 0.51$。T–v 圖：在 150 °C 水平線上，約在兩線中點。$p = p_{sat}(150\ °\text{C}) = 476$ kPa。
(c) $T_{sat}(5\ \text{MPa}) = 263.9$ °C. $T = 100 < T_{sat}$ → compressed liquid. On p–v: far left, on the 100 °C isotherm's steep liquid leg; $v \approx v_f(100\ °\text{C}) = 0.001043$ m³/kg.(c) $T_{sat}(5\ \text{MPa}) = 263.9$ °C。$T = 100 < T_{sat}$ → 壓縮液體。p–v 圖：最左側，在 100 °C 等溫線的陡峭液體段上；$v \approx v_f(100\ °\text{C}) = 0.001043$ m³/kg。
Example範例 Read a process — cooling a rigid tank解讀過程——剛性容器冷卻 ›
Given: a rigid tank holds superheated steam at 1 MPa, 300 °C. It is cooled slowly until condensation just begins, and then a little further.已知：剛性容器內有 1 MPa、300 °C 的過熱蒸氣。緩慢冷卻至剛開始凝結，然後再稍微冷卻。
Find: which property stays fixed, which line the state meets, and the path on the T–v, p–v, and p–T diagrams. (No numbers needed — the quantitative version is Stage 2 in Reading Property Tables.)求：哪個性質保持不變、狀態碰到哪條線，以及過程在 T–v、p–v、p–T 圖上的路徑。（無需數值——量化版本為讀取性質表中的第二階段。）
Key observation. Rigid tank, fixed mass → $v$ is constant. On T–v and p–v the path is a vertical line at the initial $v$. Condensation begins where this line meets the saturated-vapor line — the state at which $v_g(T) $ equals the tank's $v$. Since the initial state is superheated and to the right of the critical point, the vertical line must hit the vapor line, not the liquid line.關鍵觀察。剛性容器、質量固定 → $v$ 不變。在 T–v 與 p–v 圖上，路徑是初始 $v$ 處的垂直線。當此線碰到飽和氣線時開始凝結——即 $v_g(T)$ 等於容器 $v$ 的狀態。由於初始狀態為過熱且在臨界點右側，垂直線必碰到飽和氣線而非飽和液線。
What happens to p and T. Both fall while the steam is superheated. At the vapor line the state becomes saturated: from here on $T = T_{sat}(p)$, so further cooling lowers $p$ and $T$ together along the saturation relation, and the state moves down inside the dome with $x$ decreasing from 1.p 與 T 如何變化。蒸氣過熱時兩者皆下降。到達飽和氣線後狀態成為飽和：此後 $T = T_{sat}(p)$，繼續冷卻使 $p$、$T$ 一同沿飽和關係下降，狀態在圓頂內向下移動，$x$ 由 1 遞減。
On p–T. In the vapor region $p$ falls with $T$ along a nearly straight line aimed at the origin ($p \propto T$ at fixed $v$ for a near-ideal gas) until it reaches the vaporization line. Any further cooling follows the vaporization line downward — $p$ and $T$ are now locked, and the state moves into the dome on the other two diagrams.p–T 圖上。在蒸氣區，$p$ 隨 $T$ 沿一條近乎指向原點的直線下降（近理想氣體在定 $v$ 下 $p \propto T$），直到抵達汽化線。繼續冷卻則沿汽化線下行——此時 $p$ 與 $T$ 相互鎖定，狀態在另兩張圖上進入圓頂。
Practice & board work練習與板書活動
Every saturation value you need is given in the problem. The only skill tested here is placing a state on the diagrams and naming its region. Stage 2 — the same problems, but you find the values in the steam tables yourself — is in Reading Property Tables.所需的每個飽和值都已在題目中給出。此處只考驗將狀態放到圖上並說出其區域。第二階段——相同題目，但需自行查蒸汽表取值——見讀取性質表。
Try it first. Type any two properties and watch one state appear on all three projections; or switch to Pin it yourself, pick a practice item, click where you think it lies, then reveal. Superheated and compressed-liquid points are placed with $v \approx v_g(T_{sat})\,T/T_{sat}$ and $v \approx v_f(T)$ — right to within a few percent, indistinguishable on a log axis.先動手試試。輸入任意兩個性質，同一狀態即出現在三張投影圖上；或切換到「自己釘點」，選一道練習題，點擊你認為的位置，再揭曉答案。過熱與壓縮液體點以 $v \approx v_g(T_{sat})\,T/T_{sat}$ 與 $v \approx v_f(T)$ 定位——誤差僅數個百分點，在對數軸上無法分辨。
Part A — classify. Each item states the saturation value it needs. Decide the region first, then open the answer.A 部分——判別相區。每題皆已給出所需的飽和值。先決定區域，再開啟答案。
A1 Water at 500 kPa, 100 °C · given $T_{sat}$(500 kPa) = 151.8 °C水，500 kPa、100 °C ・ 已知 $T_{sat}$(500 kPa) = 151.8 °C ›
Answer. $T = 100 < T_{sat}(500\ \text{kPa}) = 151.8$ °C → compressed liquid. T–v: left of the dome on the 500 kPa isobar, below its flat segment.答：$T = 100 < T_{sat}(500\ \text{kPa}) = 151.8$ °C → 壓縮液體。T–v：500 kPa 等壓線上、圓頂左側、水平段之下。
A2 Water at 500 kPa, x = 0 · given $T_{sat}$(500 kPa) = 151.8 °C水，500 kPa、x = 0 ・ 已知 $T_{sat}$(500 kPa) = 151.8 °C ›
Answer. $x = 0$ → saturated liquid, on the saturated liquid line at $T = 151.8$ °C. Two-phase boundary: $p$ and $T$ are not independent here, so $x$ was needed.答：$x = 0$ → 飽和液體，位於飽和液線上，$T = 151.8$ °C。此為兩相邊界：$p$ 與 $T$ 不獨立，故需 $x$。
A3 Water at 100 kPa, 200 °C · given $T_{sat}$(100 kPa) = 99.6 °C水，100 kPa、200 °C ・ 已知 $T_{sat}$(100 kPa) = 99.6 °C ›
Answer. $T = 200 > T_{sat}(100\ \text{kPa}) = 99.6$ °C → superheated vapor. p–T: below the vaporization line.答：$T = 200 > T_{sat}(100\ \text{kPa}) = 99.6$ °C → 過熱蒸氣。p–T：汽化線下方。
A4 Water at 250 °C, v = 0.030 m³/kg · given $v_f$ = 0.001252, $v_g$ = 0.05013 m³/kg, $p_{sat}$ = 3.98 MPa水，250 °C、v = 0.030 m³/kg ・ 已知 $v_f$ = 0.001252、$v_g$ = 0.05013 m³/kg、$p_{sat}$ = 3.98 MPa ›
Answer. $v_f < 0.030 < v_g$ → saturated mixture, $x = (0.030 - 0.00125)/(0.05013 - 0.00125) = 0.59$. $p = p_{sat}(250\ °\text{C}) = 3.98$ MPa — not free to choose.答：$v_f < 0.030 < v_g$ → 飽和混合物，$x = (0.030 - 0.00125)/(0.05013 - 0.00125) = 0.59$。$p = p_{sat}(250\ °\text{C}) = 3.98$ MPa——非自由選擇。
A5 Water at 25 MPa, 400 °C · given $p_{cr}$ = 22.06 MPa水，25 MPa、400 °C ・ 已知 $p_{cr}$ = 22.06 MPa ›
Answer. $p > p_{cr}$ → supercritical. No saturation temperature exists at this pressure; the isobar passes above the dome on T–v. Neither “liquid” nor “vapor” is a meaningful label.答：$p > p_{cr}$ → 超臨界。此壓力下不存在飽和溫度；T–v 圖上等壓線越過圓頂上方。「液體」與「蒸氣」皆非有意義的標籤。
A6 Water at 20 °C, 0.5 kPa · given $p_{sat}$(20 °C) = 2.34 kPa, $p_{tp}$ = 0.61 kPa水，20 °C、0.5 kPa ・ 已知 $p_{sat}$(20 °C) = 2.34 kPa、$p_{tp}$ = 0.61 kPa ›
Answer. $p = 0.5 < p_{sat}(20\ °\text{C}) = 2.34$ kPa → superheated vapor (same test as A3, phrased in pressure). Bonus: 0.5 kPa is below the triple-point pressure (0.61 kPa), so cooling at this pressure would meet the sublimation line — the vapor would deposit directly to ice.答：$p = 0.5 < p_{sat}(20\ °\text{C}) = 2.34$ kPa → 過熱蒸氣（與 A3 同一檢驗，改以壓力表述）。延伸：0.5 kPa 低於三相點壓力（0.61 kPa），故在此壓力下冷卻將碰到昇華線——蒸氣直接凝華為冰。
A7 Dry ice (solid CO₂) on a bench at 1 atm, 25 °C room · given CO₂ triple point 518 kPa, −56.6 °C; sublimation at 101 kPa: −78.5 °C乾冰（固態 CO₂）置於 1 atm、25 °C 的實驗桌上 ・ 已知 CO₂ 三相點 518 kPa、−56.6 °C；101 kPa 下的昇華溫度 −78.5 °C ›
Answer. $p = 101 < p_{tp} = 518$ kPa → the horizontal $p = 1$ atm on the p–T diagram passes below the triple point and can only cross the sublimation line (at −78.5 °C). Liquid CO₂ cannot exist at room pressure; the block goes solid → vapor directly, which is why it is "dry". On the p–v–T surface the state moves along the solid–vapor ruled surface, never touching the liquid. To see liquid CO₂ you need $p > 518$ kPa — a fire extinguisher (≈ 6 MPa at 25 °C) holds it as a saturated liquid–vapor mixture on the vaporization line. Same test as A6, other substance.答：$p = 101 < p_{tp} = 518$ kPa → p–T 圖上水平線 $p = 1$ atm 通過三相點下方，只可能穿越昇華線（−78.5 °C）。室壓下不存在液態 CO₂；乾冰由固態直接變為蒸氣，故稱「乾」。在 p–v–T 曲面上，狀態沿固–氣直紋面移動，不會碰到液體。要看到液態 CO₂ 需 $p > 518$ kPa——滅火器（25 °C 約 6 MPa）內即以飽和液氣混合物形式儲存於汽化線上。與 A6 相同檢驗，不同物質。
Part B — board sketches. One student sketches the process on the two named diagrams at the board, marking the start, end, and every crossing of the dome. Then open the answer and compare.B 部分——板書繪圖。一位學生在黑板上將過程畫在指定的兩張圖上，標出起點、終點與每一次穿越圓頂的位置。然後開啟答案比對。
B1 Isobaric heating: compressed liquid → superheated vapor, on T–v and p–v定壓加熱：壓縮液體 → 過熱蒸氣，T–v 與 p–v ›
Check: on T–v the flat segment must lie inside the dome at $T_{sat}$ of that pressure, and the liquid leg hugs the liquid line. On p–v nothing bends — the isobar is the process.檢查：T–v 圖上水平段必須在圓頂內、位於該壓力的 $T_{sat}$，液體段緊貼飽和液線。p–v 圖上無任何彎折——等壓線即過程線。
B2 Isothermal compression: superheated vapor → saturated liquid, on p–v and T–v等溫壓縮：過熱蒸氣 → 飽和液體，p–v 與 T–v ›
Check: on p–v the curve must flatten exactly at the vapor line and stay flat to the liquid line; the flat level is $p_{sat}(T)$. On T–v the whole process is one horizontal line — the only clue that anything happened is the change in $v$.檢查：p–v 圖上曲線必須恰在飽和氣線處轉為水平，並維持水平至飽和液線；水平高度為 $p_{sat}(T)$。T–v 圖上整個過程只是一條水平線——唯一的線索是 $v$ 的變化。
B3 Constant-volume cooling of a vapor into the dome, on T–v and p–T蒸氣的定容冷卻進入圓頂，T–v 與 p–T ›
Check: on T–v the path is strictly vertical — a common error is to bend it. On p–T the path changes direction at the vaporization line: straight toward the origin before, along the line after. If $v$ were larger than $v_g$ at the triple point, the line would meet the sublimation curve instead.檢查：T–v 圖上路徑必須嚴格垂直——常見錯誤是把它畫彎。p–T 圖上路徑在汽化線處改變方向：之前為指向原點的直線，之後沿汽化線。若 $v$ 大於三相點的 $v_g$，直線將改為碰到昇華線。
Part C — choose the diagram.C 部分——選擇圖表。
C1 A steam condenser runs at 8 kPa (given: $T_{sat}$(8 kPa) = 41.5 °C). Which diagram shows the condensing temperature, and where do you read it?蒸汽冷凝器在 8 kPa 下運轉（已知 $T_{sat}$(8 kPa) = 41.5 °C）。哪張圖能顯示冷凝溫度？在何處讀取？ ›
Answer. The p–T diagram: draw the horizontal $p = 8$ kPa until it meets the vaporization line; the $T$ there is $T_{sat}$ (41.5 °C). The same number is the height of the flat segment on T–v — but p–T gives it as a single point, which is why the saturation table is really a tabulated vaporization line.答：p–T 圖：畫水平線 $p = 8$ kPa 直到與汽化線相交，該處的 $T$ 即 $T_{sat}$（41.5 °C）。同一數值在 T–v 圖上是水平段的高度——但 p–T 圖將其表示為單一點，這正是飽和表本質上就是「表格化的汽化線」的原因。
C2 A piston compresses wet steam isothermally from x = 0.9 to x = 0.1. Which diagrams show the process as a visible line, and which hides it as a point?活塞將濕蒸氣由 x = 0.9 等溫壓縮至 x = 0.1。哪些圖能將此過程顯示為可見的線？哪張圖將其隱藏為一點？ ›
Answer. Visible on T–v and p–v as a horizontal segment inside the dome ($v$ shrinks from near $v_g$ to near $v_f$). On p–T the entire process is one point on the vaporization line — $p$ and $T$ never change. This is the cleanest demonstration that p–T cannot show quality.答：在 T–v 與 p–v 圖上可見，為圓頂內的水平段（$v$ 由接近 $v_g$ 縮至接近 $v_f$）。在 p–T 圖上整個過程是汽化線上的一點——$p$ 與 $T$ 從未改變。這是「p–T 圖無法顯示乾度」最清楚的示範。
Key terms & equations關鍵術語與重要公式
Everything in this chapter reduces to one picture — the p–v–T surface — and the vocabulary for its regions and edges. If you can place a state on the T–v and p–v diagrams from two given properties and name the region, you have the chapter.本章一切歸結為一幅圖——p–v–T 曲面——以及描述其區域與邊界的詞彙。若能由兩個已知性質將狀態定位於 T–v 與 p–v 圖上並說出其區域，本章便已掌握。