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値。
- Use quality for two-phase states and the ideal-gas tables for air.對兩相狀態使用乾度，對空氣使用理想氣體表。
- Look up the gas constant $R = R_u/M$ for common gases.查得常見氣體的氣體常數 $R = R_u/M$。
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。
Two-phase states: quality兩相狀態：乾度
In the two-phase region $T$ and $p$ are not independent, so you need the quality $x$. Any property is the lever-rule blend:在兩相區中，$T$ 與 $p$ 並非獨立，因此需要乾度 $x$。任何性質的桿桿法則展開：
(See Properties of Pure Substances for the full treatment.)(詳見純物質性質。)
Linear interpolation線性插値
When your state falls between two rows, estimate by linear interpolation:當狀態介於兩行之間時，以線性插値估算性質：
The same idea extends to double interpolation (both T and p).同樣思路可延伸至雙重插値（同時對 T 與 p）。
Interpolation trainer插値練習器
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，插値過程展開。嘗試 55 °C。
Representative steam-table values, abridged for practice. Real tables have finer spacing and more columns.
Ideal-gas tables理想氣體表
For ideal gases, $u$ and $h$ depend on temperature only, so a table indexed by $T$ gives $u(T)$, $h(T)$, and the entropy function $s^\circ(T)$. A short air excerpt:對理想氣體，$u$ 與 $h$ 僅與溫度有關，以 $T$ 為索引的表即可給出 $u(T)$、$h(T)$ 與 $s^\circ(T)$。空氣節選：
| T (K) | h (kJ/kg) | u (kJ/kg) | s° (kJ/kg·K) |
|---|---|---|---|
| 300 | 300.19 | 214.07 | 1.70203 |
| 400 | 400.98 | 286.16 | 1.99194 |
| 500 | 503.02 | 359.49 | 2.21952 |
| 1000 | 1046.04 | 758.94 | 2.96770 |
| 1500 | 1635.97 | 1205.41 | 3.44516 |
Use $\Delta h = h(T_2) - h(T_1)$ directly. Entropy: $s_2 - s_1 = s^\circ_2 - s^\circ_1 - R\ln(p_2/p_1)$.直接使用 $\Delta h = h(T_2) - h(T_1)$。熵變化：$s_2 - s_1 = s^\circ_2 - s^\circ_1 - R\ln(p_2/p_1)$。
Gas constants氣體常數
Each gas has its own $R = R_u/M$, with $R_u = 8.314$ kJ/kmol·K:每一氣體各有其常數 $R = R_u/M$，$R_u = 8.314$ kJ/kmol·K：
| Gas | M (kg/kmol) | R (kJ/kg·K) | c_p (kJ/kg·K) | k |
|---|---|---|---|---|
| Air | 28.97 | 0.2870 | 1.005 | 1.40 |
| Nitrogen (N₂) | 28.01 | 0.2968 | 1.039 | 1.40 |
| Oxygen (O₂) | 32.00 | 0.2598 | 0.918 | 1.40 |
| Carbon dioxide (CO₂) | 44.01 | 0.1889 | 0.846 | 1.29 |
| Water vapor (H₂O) | 18.02 | 0.4615 | 1.864 | 1.33 |
| Helium (He) | 4.003 | 2.0769 | 5.193 | 1.67 |
Putting it together綜合應用
Example範例 Interpolating the steam table蒸汽表插値 ›
Given: saturated water vapor at 55 °C. $h_g(50°\mathrm{C})=2592.1$ and $h_g(60°\mathrm{C})=2609.6$ kJ/kg.已知：55 °C 的飽和水蒸氣。$h_g(50°\mathrm{C})=2592.1$，$h_g(60°\mathrm{C})=2609.6$ kJ/kg。
Find: $h_g$ at 55 °C.求：55 °C 的 $h_g$。
Solution. 55 °C is halfway between the rows: $$h_g = 2592.1 + (2609.6-2592.1)\frac{55-50}{60-50} = 2600.9\ \tfrac{\text{kJ}}{\text{kg}}$$解：55 °C 在兩行正中間：$$h_g = 2592.1 + (2609.6-2592.1)\frac{55-50}{60-50} = 2600.9\ \tfrac{\text{kJ}}{\text{kg}}$$
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。
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-5 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-4): $v \approx v_f = 0.001043$ m³/kg, $u \approx u_f = 419.1$ kJ/kg.答：A-5 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-4 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-5: $T_{sat}(100\ \text{kPa}) = 99.6$ °C; $200 > 99.6$ → superheated vapor. Go to the superheated table (A-6) 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-5 表：$T_{sat}(100\ \text{kPa}) = 99.6$ °C；$200 > 99.6$ → 過熱蒸氣。查過熱表（A-6）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-4 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-4 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-5 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-5 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-6 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-6 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-6 at 1 MPa, 300 °C: $v_1 = 0.25799$ m³/kg. Condensation begins when $v_g(T_2) = 0.25799$. Table A-4: $v_g(165\ °\text{C}) = 0.27278$, $v_g(170\ °\text{C}) = 0.24260$. 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-6 表 1 MPa、300 °C：$v_1 = 0.25799$ m³/kg。當 $v_g(T_2) = 0.25799$ 時開始凝結。A-4 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-4: $p = p_{sat} = 101.42$ 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-4 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。