What you'll be able to do本章學習成果
- Understand entropy transfer, entropy production, and the increase-of-entropy principle.
- Evaluate entropy and entropy change, and analyze isentropic processes with property data.
- Represent heat transfer as an area on a T–s diagram.
- Apply entropy balances to closed systems and control volumes, and evaluate isentropic efficiencies of turbines, nozzles, compressors, and pumps.
Key equations重要公式
Defining entropy change
Starting from the Clausius inequality, consider two internally reversible processes linking the same two states. The integral $\int \delta Q/T$ comes out identical for both — it depends only on the end states. A quantity whose change is path-independent is a property. We call it entropy, $S$:
You compute it by imagining any internally reversible path between the states — but because entropy is a property, the result holds for the actual process too, reversible or not. Entropy is extensive; specific entropy $s$ has units kJ/kg·K. For two-phase mixtures, $s = s_f + x\,s_{fg}$, just like $v$, $u$, and $h$.
Entropy & heat transfer熵與熱傳
In the Second Law chapter we met Boltzmann's $S = k_B\ln\Omega$. But nobody can count $\Omega$ for a real tank of steam, so how does it turn into something we can measure? Follow what heat does to the molecules.在第二定律章節中我們見過波茲曼的 $S = k_B\ln\Omega$。但沒有人能實際數出一槽蒸汽的 $\Omega$，那麼它如何變成可量測的量？看看熱對分子做了什麼。
- Heat adds energy to molecular motion. More energy gives the molecules more ways to share it among themselves — more speeds, more vibration and rotation. So $\Omega$ grows, and $S$ goes up. Work done by a piston does not have to do this; heat always does.熱增加分子運動的能量。能量越多，分子之間分配能量的方式就越多——更多種速度、振動與轉動。因此 $\Omega$ 增加，$S$ 上升。活塞作功不一定如此；熱則必然如此。
- The same heat matters more when the system is cold. A cold system has little energy and few arrangements, so a small amount of heat multiplies $\Omega$ by a large factor. A hot system already has an enormous $\Omega$; the same heat changes it proportionally much less. Like giving 1000 dollars to someone with nothing versus someone with a million, the effect scales as $1/T$.同樣的熱量，對冷的系統影響更大。冷的系統能量少、排列方式少，少量的熱就能使 $\Omega$ 成倍增加。熱的系統 $\Omega$ 已極為龐大，同樣的熱量造成的相對變化小得多。就像把 1000 元給一無所有的人與給百萬富翁，效果與 $1/T$ 成正比。
- This is what temperature means. Statistical mechanics defines temperature by exactly this sensitivity — how fast entropy rises as energy is added, with volume held fixed so no work is done:這正是溫度的意義。統計力學正是以此敏感度定義溫度——在體積固定（無作功）下，加入能量時熵上升的速率：
With no work, the energy added is the heat, $dU = \delta Q$. The result is Clausius's definition from the previous section, now derived from molecules. Both describe the same property: $\Omega$ explains what entropy is, and $\delta Q/T$ is how we measure it with a thermometer and a calorimeter.無作功時，加入的能量就是熱，$dU = \delta Q$。所得結果即為前一節克勞修斯的定義，如今由分子層次推導出來。兩者描述同一性質：$\Omega$ 說明熵是什麼，$\delta Q/T$ 則是用溫度計與量熱計量測它的方法。
On a differential basis $dS = (\delta Q/T)_{int\,rev}$. So heat into a system raises its entropy and heat out lowers it — entropy transfer accompanies heat transfer, in the same direction. An internally reversible, adiabatic process has no heat transfer and therefore constant entropy — an isentropic process.
Rearranging, $\delta Q_{int\,rev} = T\,dS$, or per unit mass $\delta q_{int\,rev} = T\,ds$. The heat transferred in an internally reversible process is the area under the process curve on a temperature–entropy diagram — the T–s counterpart to "work is area under the p–v curve."
The entropy balance熵的平衡
Entropy is accounted for like mass and energy — but with a production term. For a closed system:
The first term is entropy transfer with heat (evaluated at the boundary); $\sigma$ is entropy production from internal irreversibilities. Its sign is decisive:
- $\sigma = 0$ — no internal irreversibilities
- $\sigma > 0$ — irreversibilities present
- $\sigma < 0$ — impossible
For an adiabatic process the transfer term vanishes, so $\sigma = S_2 - S_1$: a negative entropy change would be impossible. This is the increase-of-entropy principle in action — the test the explorer below applies.
Control-volume entropy balance控制體積的熵平衡
Streams of matter carry entropy across a control surface. At steady state, for one inlet and one exit:
A throttling valve is a clean example: adiabatic with $h_2 = h_1$, so $\dot\sigma_{cv}/\dot m = s_2 - s_1 > 0$ — the entropy production traces directly to the unrestrained expansion to lower pressure.
Calculating entropy change熵變化的計算
The property tables are built from the T ds equations, $T\,ds = du + p\,dv = dh - v\,dp$. Two models give closed forms. For an incompressible substance with constant $c$: $\Delta s = c\,\ln(T_2/T_1)$. For an ideal gas with constant specific heats:
Setting $s_2 = s_1$ recovers the familiar isentropic relation $T_2/T_1 = (p_2/p_1)^{(k-1)/k}$.
When specific heats vary appreciably, the temperature integral is tabulated instead. The ideal-gas table lists the entropy function $s^\circ(T) = \int_{T_\text{ref}}^{T} c_p(T)\,dT/T$ alongside $h$ and $u$, and the pressure term is added separately:當比熱變化明顯時，改以查表處理溫度積分。理想氣體表在 $h$、$u$ 之外另列熵函數 $s^\circ(T) = \int_{T_\text{ref}}^{T} c_p(T)\,dT/T$，壓力項則另外加上：
| T K | s° kJ/kg·K |
|---|---|
| 300 | 1.70203 |
| 400 | 1.99194 |
| 500 | 2.21952 |
| 1000 | 2.96770 |
| 1500 | 3.44516 |
Air, ideal-gas table (Table A-22), excerpt. $h$ and $u$ for the same rows are in Equations of State. The relative-pressure $p_r$ column in the full table handles isentropic processes the same way.空氣理想氣體表（表 A-22）節選。同列的 $h$ 與 $u$ 見狀態方程式。完整表格中的相對壓力 $p_r$ 欄亦以同樣方式處理等熵過程。
Isentropic efficiency等熵效率
Because adiabatic devices can only reach exit states with $s_2 \ge s_1$, the isentropic (constant-entropy) endpoint is the best case. For a turbine, the most work is the isentropic work; the efficiency is the actual fraction of it:
For a compressor or pump the ideal is the minimum work, so the ratio inverts: $\eta_c = (h_{2s}-h_1)/(h_2-h_1)$. In every case the actual process lands to the right of the isentropic one on a T–s or Mollier diagram — that horizontal shift is the entropy generated.
Isentropic-efficiency lab等熵效率實驗室
Expand or compress air between two pressures at a chosen isentropic efficiency. The ideal path (1→2s) is vertical — constant entropy; the actual path (1→2) leans right by exactly the entropy generated. Watch the work and $\sigma$ respond as you lower the efficiency.
Air as ideal gas, constant $c_p = 1.005$ kJ/kg·K, $k = 1.4$. Adiabatic device, so actual Δs equals the entropy generated.
Entropy change of an ideal gas理想氣體的熵變化
Example範例 Air heated and compressed空氣加熱並壓縮 ›
Given: air goes from 300 K, 100 kPa to 600 K, 300 kPa. Use $c_p=1.005$, $R=0.287$ kJ/kg·K.
Find: the specific entropy change.
Solution. $$\Delta s=c_p\ln\frac{T_2}{T_1}-R\ln\frac{p_2}{p_1}=1.005\ln 2-0.287\ln 3=0.696-0.315=0.381\ \tfrac{\text{kJ}}{\text{kg·K}}.$$ Heating raises $s$; the pressure rise pulls it back down.