What you'll be able to do本章學習成果
- Navigate saturation, superheated, and compressed-liquid tables.查閱飽和表、過熱蒸氣表與壓縮液體表。
- Interpolate linearly between table entries.在表格項目之間進行線性插值。
- Use quality to find properties of two-phase states.以乾度求兩相狀態的性質。
Key relations關鍵公式
Why tables?為何查表？
For most real substances, the relationships among properties are too complex for a single equation. Measured values are compiled into tables — the steam tables for water, separate tables for each refrigerant, and ideal-gas tables for air. Correct table reading is a core engineering skill.對於大多數實際物質，各性質間的關係難以單一方程式描述。因此量測值被編成表格——水的蒸汽表、各冷媒專用表、空氣的理想氣體表。正確查表是工程师的核心學能。
Anatomy of the tables表格結構詳解
A typical substance comes with several tables:一般物質有多種表格：
- Saturation tables — indexed by T and by p. List saturated-liquid (f) and saturated-vapor (g) properties plus vaporization change (fg). Use inside the dome.飽和表——分別以 T 與 p 為索引。列出飽和液 (f)、飽和蒸氣 (g) 與永魔變化量 (fg)。用於圓頂內部。
- Superheated-vapor tables — right of the dome, where T and p are independent. List $v, u, h, s$.過熱蒸氣表——圓頂右側，T 與 p 獨立。列出 $v, u, h, s$。
- Compressed-liquid tables — often approximated by $y(T,p) \approx y_f(T)$.壓縮液體表——常以 $y(T,p) \approx y_f(T)$ 近似。
Each table uses a reference state — only changes in $u$, $h$, $s$ matter.每個表以一個參考狀態為基準——只有 $u$、$h$、$s$ 的變化量才有意義。
Linear interpolation線性插值
Tables list properties only at discrete rows. When your state falls between two rows, assume the property varies along a straight line between them. Let $a$ be the variable you know (e.g. $T$ or $p$) and $y$ the property you want. The straight line through $(a_1, y_1)$ and $(a_2, y_2)$ has slope $(y_2 - y_1)/(a_2 - a_1)$, so性質表只在離散的列上列出數值。當狀態落在兩列之間時，假設性質在兩列之間呈直線變化。令 $a$ 為已知變數（如 $T$ 或 $p$），$y$ 為所求性質。通過 $(a_1, y_1)$ 與 $(a_2, y_2)$ 的直線斜率為 $(y_2 - y_1)/(a_2 - a_1)$，因此
Read the fraction $\dfrac{a - a_1}{a_2 - a_1}$ as "how far along" the state is from row 1 to row 2: it is 0 at row 1, 1 at row 2, and 0.5 halfway. The same idea extends to double interpolation when both $T$ and $p$ fall between rows.將分式 $\dfrac{a - a_1}{a_2 - a_1}$ 理解為狀態由第 1 列走向第 2 列的「進度」：在第 1 列為 0、第 2 列為 1、正中間為 0.5。當 $T$ 與 $p$ 皆落在兩列之間時，同樣思路可延伸為雙重插值。
Outside the dome this straight line is an approximation — the true curve bends, and finer table spacing makes the error smaller. Inside the dome, as the next section shows, the same formula becomes exact.在圓頂外，這條直線只是近似——真實曲線是彎曲的，表格間距越細，誤差越小。在圓頂內，如下一節所示，同一公式卻是精確的。
Two-phase states: quality兩相狀態：乾度
In the two-phase region $T$ and $p$ are not independent — fixing one fixes the other — so they cannot locate the state along the horizontal line of the dome. The saturation table gives only its two ends: saturated liquid ($y_f$) and saturated vapor ($y_g$). The missing coordinate is the quality, the mass fraction of vapor:在兩相區中，$T$ 與 $p$ 並非獨立——固定其一即固定另一——因此無法用來定出狀態在圓頂水平線上的位置。飽和表只給出兩端：飽和液體（$y_f$）與飽和蒸氣（$y_g$）。缺少的座標就是乾度，即蒸氣的質量分率：
Derivation from the mixture. A specific property of the mixture is its total divided by the total mass. The liquid and vapor each carry their own saturated value:由混合物推導。混合物的比性質為總量除以總質量；液體與蒸氣各自帶有其飽和值：
The same result from interpolation. Now apply the interpolation formula along the horizontal line, using quality as the known variable $a$. Row 1 is saturated liquid ($a_1 = x = 0$, $y_1 = y_f$); row 2 is saturated vapor ($a_2 = x = 1$, $y_2 = y_g$):由插值得到同一結果。現在沿水平線套用插值公式，以乾度作為已知變數 $a$。第 1 列為飽和液體（$a_1 = x = 0$，$y_1 = y_f$）；第 2 列為飽和蒸氣（$a_2 = x = 1$，$y_2 = y_g$）：
Expanding $y_f + x(y_g - y_f)$ gives back $(1-x)\,y_f + x\,y_g$ — the two routes agree. So the quality relation is linear interpolation between the saturated-liquid and saturated-vapor states, with $x$ playing the role of "how far along." And here it is exact, not approximate: mass-averaging is genuinely linear in $x$. $y_{fg}$ is tabulated directly for $u$, $h$ and $s$ (for $h$ it is the latent heat $h_{fg}$).展開 $y_f + x(y_g - y_f)$ 即回到 $(1-x)\,y_f + x\,y_g$——兩種途徑結果一致。因此乾度關係式就是飽和液體與飽和蒸氣兩狀態之間的線性插值，$x$ 正扮演「進度」的角色。而此處是精確而非近似：質量平均對 $x$ 本來就是線性的。$u$、$h$、$s$ 的 $y_{fg}$ 直接列於表中（對 $h$ 而言即潛熱 $h_{fg}$）。
Run it backwards when a property is known instead of $x$ — e.g. given $v$ at a saturation pressure:若已知的是某性質而非 $x$，則反向求解——例如已知飽和壓力下的 $v$：
This is the same "how far along" fraction as in the interpolation formula. (See Properties of Pure Substances for the dome and the quality lab.)這正是插值公式中的同一個「進度」分式。（圓頂與乾度實驗見純物質性質。）
Interpolation trainer插值練習器
A real saturated-water excerpt. Slide the target temperature: bracketing rows highlight and interpolation is worked out. Try 55 °C.實際飽和水表節選。滑動目標溫度：對應行將被標示，插值過程展開。嘗試 55 °C。
Saturated water, temperature table (IAPWS-95 values, as in the standard textbook Table A-2), abridged to 10 °C steps. Full tables use finer spacing near the ends and list $u_{fg}$, $s_{fg}$ as well.飽和水溫度表（IAPWS-95 數值，與標準教科書表 A-2 相同），以 10 °C 間距節錄。完整表格在兩端間距更細，並另列 $u_{fg}$、$s_{fg}$。
What about gases?那氣體呢？
Everything above is for substances that change phase — water, refrigerants — where you need two properties and several tables to fix a state. Gases far from their dome behave as ideal gases, and their tables are much simpler: $u$ and $h$ depend on $T$ alone, so there is only one column to look up. Those tables, and the gas constants $R$ for common gases, are in Equations of State.以上內容適用於會發生相變化的物質——水、冷媒——需要兩個性質與多張表格才能定出狀態。遠離圓頂的氣體則表現為理想氣體，其性質表簡單得多：$u$ 與 $h$ 僅與 $T$ 有關，只需查一欄。這些表格以及常見氣體的氣體常數 $R$，請見狀態方程式。
Putting it together綜合應用
Example範例 Interpolating the steam table蒸汽表插值 ›
Given: saturated water vapor at 55 °C. $h_g(50°\mathrm{C})=2591.3$ and $h_g(60°\mathrm{C})=2608.8$ kJ/kg.已知：55 °C 的飽和水蒸氣。$h_g(50°\mathrm{C})=2591.3$，$h_g(60°\mathrm{C})=2608.8$ kJ/kg。
Find: $h_g$ at 55 °C.求：55 °C 的 $h_g$。
Solution. 55 °C is halfway between the rows: $$h_g = 2591.3 + (2608.8-2591.3)\frac{55-50}{60-50} = 2600.05\ \tfrac{\text{kJ}}{\text{kg}}$$解：55 °C 在兩行正中間：$$h_g = 2591.3 + (2608.8-2591.3)\frac{55-50}{60-50} = 2600.05\ \tfrac{\text{kJ}}{\text{kg}}$$
Locate the state — from the tables定位狀態——由性質表
In Properties of Pure Substances the saturation values were handed to you. Now find them yourself: open the saturated-water table (by T or by p), decide the region, then compute what is asked. Keep the T–v and p–v diagrams in front of you and mark every state.在純物質性質中，飽和值直接給出。現在請自行查找：翻開飽和水表（按 T 或按 p），判定區域，再計算所求。將 T–v 與 p–v 圖放在面前，標出每個狀態。
Worked example範例 Water at 800 kPa, 250 °C. Region, and v, u, h?水，800 kPa、250 °C。區域及 v、u、h？ ›
Every "locate the state" problem follows the same three steps. Work through this one before trying A1–A5.所有「定位狀態」題目都依循相同三步驟。先完成此例，再做 A1–A5。
Step 1 — Look up the saturation table at the given pressure. $p$ is given, so open the pressure table (A-3) and read the saturation temperature:步驟 1——依已知壓力查飽和表。已知 $p$，故翻開壓力表（A-3），讀取飽和溫度：
| p kPa | Tsat °C | vf m³/kg | vg m³/kg | uf kJ/kg | ug kJ/kg |
|---|---|---|---|---|---|
| 700 | 164.95 | 0.001108 | 0.27278 | 696.37 | 2571.8 |
| 750 | 167.75 | 0.001111 | 0.25552 | 708.64 | 2573.9 |
| 800 | 170.41 | 0.001115 | 0.24035 | 720.02 | 2576.0 |
| 850 | 172.94 | 0.001118 | 0.22690 | 731.00 | 2577.9 |
Saturated water — pressure table (A-3), excerpt.飽和水——壓力表（A-3）節錄。
Step 2 — Compare the given $T$ with $T_{sat}$ to identify the region.步驟 2——比較已知 $T$ 與 $T_{sat}$，判定區域。
| If…若… | Region區域 | Go to查表 |
|---|---|---|
| $T < T_{sat}$ | compressed liquid壓縮液體 | A-5, or $y \approx y_f(T)$ from A-2A-5，或由 A-2 取 $y \approx y_f(T)$ |
| $T = T_{sat}$ | saturated mixture飽和混合物 | A-3 — need one more property ($x$ or $v$)A-3——需再一個性質（$x$ 或 $v$） |
| $T > T_{sat}$ | superheated vapor過熱蒸氣 | A-4 at the given $p$A-4，於已知 $p$ |
Here $T = 250\ °\text{C} > T_{sat} = 170.41\ °\text{C}$ → superheated vapor. The saturation values $v_f$, $v_g$ are no longer needed — they only told us which side of the dome we are on.此處 $T = 250\ °\text{C} > T_{sat} = 170.41\ °\text{C}$ → 過熱蒸氣。飽和值 $v_f$、$v_g$ 已不再需要——它們只用來告訴我們位於圓頂的哪一側。
Step 3 — Read the values from the table for that region. Open the superheated table (A-4) to the block for $p = 0.80$ MPa and find the 250 °C row:步驟 3——由該區域的表格讀值。翻開過熱表（A-4）中 $p = 0.80$ MPa 的區塊，找到 250 °C 那一列：
| p = 0.80 MPa (Tsat = 170.41 °C) | ||||
|---|---|---|---|---|
| T °C | v m³/kg | u kJ/kg | h kJ/kg | s kJ/kg·K |
| Sat. | 0.24035 | 2576.0 | 2768.3 | 6.6616 |
| 200 | 0.26088 | 2631.1 | 2839.8 | 6.8177 |
| 250 | 0.29321 | 2715.9 | 2950.4 | 7.0402 |
| 300 | 0.32416 | 2797.5 | 3056.9 | 7.2345 |
Superheated water (A-4), excerpt. The first row repeats the saturated-vapor values from A-3 — a built-in check that you are in the right block.過熱水（A-4）節錄。第一列重複 A-3 的飽和蒸氣值——可用來確認翻到正確的區塊。
250 °C is a listed row, so no interpolation is needed:250 °C 為表列行，無需插值：
$$ v = 0.29321\ \tfrac{\text{m}^3}{\text{kg}}, \qquad u = 2715.9\ \tfrac{\text{kJ}}{\text{kg}}, \qquad h = 2950.4\ \tfrac{\text{kJ}}{\text{kg}} $$Check: $v = 0.29321 > v_g = 0.24035$ at 800 kPa — consistent with being to the right of the dome. And $h = u + pv = 2715.9 + 800 \times 0.29321 = 2950.5$ kJ/kg ✓.檢查：$v = 0.29321 > v_g = 0.24035$（800 kPa）——與位於圓頂右側一致。且 $h = u + pv = 2715.9 + 800 \times 0.29321 = 2950.5$ kJ/kg ✓。
What if $T$ had been 150 °C? Step 2 gives $150 < 170.41$ → compressed liquid. Liquid properties depend mainly on $T$, so read the saturated-liquid row at 150 °C from the temperature table (A-2) — not at 800 kPa:若 $T$ 為 150 °C 呢？步驟 2 得 $150 < 170.41$ → 壓縮液體。液體性質主要取決於 $T$，故由溫度表（A-2）讀取 150 °C 的飽和液體列——而非 800 kPa：
| T °C | psat kPa | vf m³/kg | uf kJ/kg | hf kJ/kg |
|---|---|---|---|---|
| 145 | 415.68 | 0.001085 | 610.64 | 611.09 |
| 150 | 476.16 | 0.001091 | 631.66 | 632.18 |
| 155 | 543.49 | 0.001096 | 652.79 | 653.39 |
$v \approx v_f = 0.001091$ m³/kg, $u \approx u_f = 631.66$ kJ/kg. This is exactly the route A1 takes.$v \approx v_f = 0.001091$ m³/kg，$u \approx u_f = 631.66$ kJ/kg。這正是 A1 的解題路線。
Part A — classify, then quantify. Table values quoted in the answers are from the standard saturated-water tables.A 部分——先判別再量化。答案中引用的表值取自標準飽和水表。
A1 Water at 500 kPa, 100 °C. Region, and v, u?水，500 kPa、100 °C。區域及 v、u？ ›
Answer. Table A-3 at 500 kPa: $T_{sat} = 151.8$ °C. $100 < 151.8$ → compressed liquid. No compressed-liquid table reaches 500 kPa, so use the liquid approximation at 100 °C (Table A-2): $v \approx v_f = 0.001043$ m³/kg, $u \approx u_f = 419.1$ kJ/kg.答：A-3 表 500 kPa：$T_{sat} = 151.8$ °C。$100 < 151.8$ → 壓縮液體。壓縮液體表未涵蓋 500 kPa，故用 100 °C 的液體近似（A-2 表）：$v \approx v_f = 0.001043$ m³/kg，$u \approx u_f = 419.1$ kJ/kg。
A2 Water at 100 kPa, 200 °C. Region, and v, h?水，100 kPa、200 °C。區域及 v、h？ ›
Answer. Table A-3: $T_{sat}(100\ \text{kPa}) = 99.6$ °C; $200 > 99.6$ → superheated vapor. Go to the superheated table (A-4) at 0.1 MPa, 200 °C row: $v = 2.1724$ m³/kg, $h = 2875.5$ kJ/kg. No interpolation needed — 200 °C is a listed row.答：A-3 表：$T_{sat}(100\ \text{kPa}) = 99.6$ °C；$200 > 99.6$ → 過熱蒸氣。查過熱表（A-4）0.1 MPa、200 °C 行：$v = 2.1724$ m³/kg，$h = 2875.5$ kJ/kg。無需插值——200 °C 為表列行。
A3 Water at 250 °C, v = 0.030 m³/kg. Region, x, p, h?水，250 °C、v = 0.030 m³/kg。區域、x、p、h？ ›
Answer. Table A-2 at 250 °C: $v_f = 0.001252$, $v_g = 0.05013$ m³/kg, $p_{sat} = 3976$ kPa, $h_f = 1085.8$, $h_{fg} = 1715.3$ kJ/kg. $v_f < 0.030 < v_g$ → saturated mixture. $x = (0.030 - 0.001252)/(0.05013 - 0.001252) = 0.588$. $p = 3.98$ MPa (not free). $h = 1085.8 + 0.588 \times 1715.3 = 2094$ kJ/kg.答：A-2 表 250 °C：$v_f = 0.001252$、$v_g = 0.05013$ m³/kg、$p_{sat} = 3976$ kPa、$h_f = 1085.8$、$h_{fg} = 1715.3$ kJ/kg。$v_f < 0.030 < v_g$ → 飽和混合物。$x = (0.030 - 0.001252)/(0.05013 - 0.001252) = 0.588$。$p = 3.98$ MPa（非自由）。$h = 1085.8 + 0.588 \times 1715.3 = 2094$ kJ/kg。
A4 Water at 300 kPa, x = 0.35. T, v, u?水，300 kPa、x = 0.35。T、v、u？ ›
Answer. Given $x$, the state is inside the dome: Table A-3 at 300 kPa: $T = T_{sat} = 133.5$ °C, $v_f = 0.001073$, $v_g = 0.60582$ m³/kg, $u_f = 561.1$, $u_{fg} = 1982.1$ kJ/kg. $v = 0.001073 + 0.35(0.60582 - 0.001073) = 0.2127$ m³/kg; $u = 561.1 + 0.35 \times 1982.1 = 1254.8$ kJ/kg.答：已知 $x$，狀態在圓頂內：A-3 表 300 kPa：$T = T_{sat} = 133.5$ °C、$v_f = 0.001073$、$v_g = 0.60582$ m³/kg、$u_f = 561.1$、$u_{fg} = 1982.1$ kJ/kg。$v = 0.001073 + 0.35(0.60582 - 0.001073) = 0.2127$ m³/kg；$u = 561.1 + 0.35 \times 1982.1 = 1254.8$ kJ/kg。
A5 Water at 1 MPa, 320 °C. Region, and h? (320 °C is not a table row.)水，1 MPa、320 °C。區域及 h？（320 °C 非表列行。） ›
Answer. $T_{sat}(1\ \text{MPa}) = 179.9$ °C → superheated. Table A-4 at 1 MPa: $h(300\ °\text{C}) = 3051.6$, $h(350\ °\text{C}) = 3158.2$ kJ/kg. Interpolate: $h = 3051.6 + (3158.2 - 3051.6)\,\dfrac{320 - 300}{350 - 300} = 3094.2$ kJ/kg.答：$T_{sat}(1\ \text{MPa}) = 179.9$ °C → 過熱。A-4 表 1 MPa：$h(300\ °\text{C}) = 3051.6$、$h(350\ °\text{C}) = 3158.2$ kJ/kg。插值：$h = 3051.6 + (3158.2 - 3051.6)\,\dfrac{320 - 300}{350 - 300} = 3094.2$ kJ/kg。
Part B — a process with numbers. The qualitative version of this problem was sketched at the board in the previous module.B 部分——含數值的過程。此題的定性版本已於前一模組在黑板上繪出。
B1 A rigid tank holds steam at 1 MPa, 300 °C. It is cooled until condensation just begins. Find T and p at that moment.剛性容器內有 1 MPa、300 °C 的蒸氣，冷卻至剛開始凝結。求此時的 T 與 p。 ›
Answer. Rigid → $v$ constant. Table A-4 at 1 MPa, 300 °C: $v_1 = 0.25799$ m³/kg. Condensation begins when $v_g(T_2) = 0.25799$. Table A-2: $v_g(165\ °\text{C}) = 0.27278$, $v_g(170\ °\text{C}) = 0.24260$. Interpolate on $v_g$: $T_2 = 165 + 5\,\dfrac{0.27278 - 0.25799}{0.27278 - 0.24260} = 167.5$ °C; then $p_2 = p_{sat}(167.5\ °\text{C}) \approx 746$ kPa (interpolating 700.9 → 792.2 kPa). On T–v: vertical drop from the 1 MPa isobar to the vapor line; on p–T: straight line toward the origin, meeting the vaporization line at (167.5 °C, 746 kPa).答：剛性 → $v$ 不變。A-4 表 1 MPa、300 °C：$v_1 = 0.25799$ m³/kg。當 $v_g(T_2) = 0.25799$ 時開始凝結。A-2 表：$v_g(165\ °\text{C}) = 0.27278$、$v_g(170\ °\text{C}) = 0.24260$。對 $v_g$ 插值：$T_2 = 165 + 5\,\dfrac{0.27278 - 0.25799}{0.27278 - 0.24260} = 167.5$ °C；則 $p_2 = p_{sat}(167.5\ °\text{C}) \approx 746$ kPa（由 700.9 → 792.2 kPa 插值）。T–v 圖：由 1 MPa 等壓線垂直下降至飽和氣線；p–T 圖：指向原點的直線，於 (167.5 °C, 746 kPa) 碰到汽化線。
B2 Continue cooling the tank of B1 to 100 °C. Find p and x.將 B1 的容器繼續冷卻至 100 °C。求 p 與 x。 ›
Answer. Still $v = 0.25799$ m³/kg, now inside the dome at 100 °C. Table A-2: $p = p_{sat} = 101.42$ kPa, $v_f = 0.001043$, $v_g = 1.6720$ m³/kg. $x = (0.25799 - 0.001043)/(1.6720 - 0.001043) = 0.154$. About 85 % of the mass has condensed, yet the liquid occupies only $x_f v_f / v \approx 0.3$ % of the volume — the tank still looks full of vapor.答：仍為 $v = 0.25799$ m³/kg，現在位於 100 °C 的圓頂內。A-2 表：$p = p_{sat} = 101.42$ kPa、$v_f = 0.001043$、$v_g = 1.6720$ m³/kg。$x = (0.25799 - 0.001043)/(1.6720 - 0.001043) = 0.154$。約 85 % 的質量已凝結，但液體僅佔體積約 0.3 %——容器看起來仍充滿蒸氣。