∮ Thermodynamics熱力學
Chapters章節  /  03 Laws03 定律

Steady-Flow Devices穩態流動裝置

Nozzles, diffusers, pumps, compressors, turbines, throttles, mixers and heat exchangers: the same mass and energy balances, simplified differently for each. For every device, see what it does, where it is used and how its performance is rated.噴嘴、擴散器、泵、壓縮機、渦輪機、節流裝置、混合室與熱交換器:相同的質量與能量平衡,各自以不同方式化簡。逐一了解每種裝置的功能、應用場合與性能評估方式。

7 devices7 種裝置Typical efficiencies典型效率7 worked examples · 14 exercises7 個範例 · 14 題練習
Overview總覽

One balance, seven devices一個平衡式,七種裝置

Every device below is a steady-state control volume. Start from the steady-flow balances from the previous chapter:以下每種裝置皆為穩態控制體積,從上一章的穩態流動平衡式出發:

$$ \sum_i \dot m_i = \sum_e \dot m_e \qquad 0 = \dot Q_{cv} - \dot W_{cv} + \sum_i \dot m_i\!\left(h_i + \tfrac{\mathcal{V}_i^2}{2} + gz_i\right) - \sum_e \dot m_e\!\left(h_e + \tfrac{\mathcal{V}_e^2}{2} + gz_e\right) $$
Steady flow

For each device, the same three steps apply:每種裝置皆依循相同三步驟:

  • Mass: count inlets and exits; write the mass balance.質量:計算入口與出口數目,寫出質量平衡。
  • Energy: ask what the device is for. Keep that term, and drop the terms that are negligible for this hardware.能量:先問此裝置的目的為何。保留該項,並刪去對此硬體可忽略的項。
  • Performance: compare the real device with an ideal one.性能:將實際裝置與理想裝置比較。
A preview: the isentropic state $2s$預告:等熵狀態 $2s$

Several efficiencies below compare the actual exit state 2 with an ideal exit state $2s$: the state the fluid would reach at the same exit pressure if the device were adiabatic and frictionless. That ideal process has constant entropy, which is defined in Entropy. For now, treat $h_{2s}$ as a given value.以下數種效率將實際出口狀態 2 與理想出口狀態 $2s$ 比較:$2s$ 為裝置若絕熱且無摩擦時,在相同出口壓力下流體會達到的狀態。此理想過程的熵不變,熵的定義見熵一章。目前先將 $h_{2s}$ 視為已知值。

Rating performance性能評估

First-law (energy) efficiency第一定律(能量)效率

Before looking at individual devices, define what "efficiency" means from the first law. A device takes in energy we pay for and delivers energy we want. The rest is lost to the surroundings.在逐一探討裝置前,先由第一定律定義「效率」。裝置吸收我們付出的能量,產出我們想要的能量,其餘則散失至外界。

Device裝置(CV)(CV) Energy input能量輸入paid for: fuel, electricity, heat須付出的:燃料、電力、熱 Useful output有用輸出what we want: work, heated stream想要的:功、被加熱的流體 Losses: stack gas, casing heat, friction損失:排煙、機殼散熱、摩擦 η = useful output / energy inputη = 有用輸出 / 能量輸入
Energy is conserved, so input = useful output + losses. The first-law efficiency is the fraction of the input that ends up as the useful output (green band).能量守恆,故輸入 = 有用輸出 + 損失。第一定律效率即輸入中成為有用輸出(綠色帶)的比例。

Definition定義

$$ \eta_I = \frac{\text{desired energy output}}{\text{required energy input}} = 1 - \frac{\text{energy losses}}{\text{required energy input}} $$
First-law efficiency
  • Built on the energy balance: input = useful output + losses, so $0 \le \eta_I \le 1$.以能量平衡為基礎:輸入 = 有用輸出 + 損失,因此 $0 \le \eta_I \le 1$。
  • "Useful" and "input" are your choice: always state them. The same device can have different efficiencies depending on what you count.「有用」與「輸入」由分析者界定:務必寫明。同一裝置依計入項目不同,效率可能不同。
  • Use rates (kW) for steady devices; the ratio is the same as for energies over a period.穩態裝置使用速率(kW);其比值與一段時間內的能量比相同。

Examples for flow devices流動裝置的例子

$$ \eta_{boiler} = \frac{\dot m_{steam}(h_{out} - h_{in})}{\dot m_{fuel}\,\mathrm{HV}} \qquad \eta_{motor} = \frac{\dot W_{shaft}}{\dot W_{elec}} \qquad \eta_{heater} = \frac{\dot m\,c_p\,\Delta T}{\dot W_{elec}} $$
  • Boiler or furnace: heat absorbed by the stream divided by the fuel energy (mass rate × heating value HV). Losses are hot stack gas and casing heat. Typical: 0.85–0.95.鍋爐或加熱爐:流體吸收的熱除以燃料能量(質量流率 × 熱值 HV)。損失為高溫排煙與機殼散熱。典型值:0.85–0.95。
  • Electric motor or generator: shaft power vs electrical power. Losses are winding resistance and bearing friction. Typical: 0.90–0.98.電動機或發電機:軸功率與電功率之比。損失為繞組電阻與軸承摩擦。典型值:0.90–0.98。
  • Electric water heater: nearly 1, because almost all electricity becomes heat in the water.電熱水器:接近 1,因為幾乎所有電力都成為水中的熱。
  • Chained devices multiply: $\eta_{overall} = \eta_1\,\eta_2\,\eta_3 \cdots$, e.g. motor × compressor.串接裝置效率相乘:$\eta_{overall} = \eta_1\,\eta_2\,\eta_3 \cdots$,例如馬達 × 壓縮機。

Quick example簡例

A gas-fired boiler burns 0.25 kg/s of natural gas (HV = 50 000 kJ/kg). Feedwater at $h_{in} = 640$ kJ/kg leaves as steam at $h_{out} = 3230$ kJ/kg with $\dot m = 4.2$ kg/s.燃氣鍋爐燃燒 0.25 kg/s 天然氣(HV = 50 000 kJ/kg)。給水以 $h_{in} = 640$ kJ/kg 進入,以 $h_{out} = 3230$ kJ/kg 的蒸汽流出,$\dot m = 4.2$ kg/s。

$$ \eta_{boiler} = \frac{4.2\,(3230 - 640)}{0.25\,(50\,000)} = \frac{10\,878}{12\,500} = 0.87 $$

So 13%, about 1.6 MW, leaves mostly with the stack gas.因此 13%(約 1.6 MW)主要隨排煙散失。

Why the first law is not enough for every device為何第一定律不足以評估所有裝置

  • Take an adiabatic turbine. The energy balance says $\dot W = \dot m(h_1 - h_2)$ exactly: every bit of the enthalpy drop becomes work. Measured this way, $\eta_I = 1$ for every adiabatic turbine, good or bad.以絕熱渦輪機為例。能量平衡給出 $\dot W = \dot m(h_1 - h_2)$:焓降全部成為功。以此衡量,任何絕熱渦輪機,不論好壞,$\eta_I = 1$。
  • A poor turbine is not wasting energy. It produces a smaller enthalpy drop between the same pressures, because friction keeps the exit hotter ($h_2 > h_{2s}$).差的渦輪機並非浪費能量,而是在相同壓力間產生較小的焓降,因摩擦使出口較熱($h_2 > h_{2s}$)。
  • So nozzles, diffusers, compressors, pumps and turbines are rated by comparing the real device with an ideal device between the same pressures: the isentropic efficiencies used below.因此噴嘴、擴散器、壓縮機、泵與渦輪機,改以實際裝置與相同壓力間的理想裝置比較來評估,即下文的等熵效率。
  • First-law efficiencies remain the right measure for energy conversion: fuel or electricity to heat or work (boilers, heaters, motors). For heat exchangers, the effectiveness $\varepsilon$ plays a similar role.第一定律效率仍是評估能量轉換(燃料或電力轉為熱或功,如鍋爐、加熱器、馬達)的正確指標。熱交換器則以有效度 $\varepsilon$ 扮演類似角色。
  • For complete cycles, the same idea becomes the thermal efficiency $\eta_{th} = \dot W_{net}/\dot Q_{in}$ and the coefficient of performance COP, developed in the Second Law and cycle chapters.對於完整循環,同一概念成為熱效率 $\eta_{th} = \dot W_{net}/\dot Q_{in}$ 與性能係數 COP,將於第二定律與循環章節中說明。
Device 1裝置 1

Nozzle噴嘴

A nozzle accelerates a flow by converting enthalpy (pressure) into kinetic energy. Its whole purpose is a high exit velocity $\mathcal{V}_2$.噴嘴將焓(壓力)轉換為動能,使流體加速。其目的就是得到高出口速度 $\mathcal{V}_2$。

1p1 high · V1 low2p2 low · V2 highh ↓ p ↓ V ↑Ẇ = 0 · Q̇ ≈ 0
A subsonic nozzle: the duct narrows, so the flow speeds up (longer blue arrows) while pressure and enthalpy fall.次音速噴嘴:管道收縮,流速增加(藍色箭頭變長),壓力與焓下降。

Mass balance質量平衡

$$ \dot m_1 = \dot m_2 \;\Rightarrow\; \frac{A_1\mathcal{V}_1}{v_1} = \frac{A_2\mathcal{V}_2}{v_2} $$

The mass balance sizes the exit area once $\mathcal{V}_2$ is known.求得 $\mathcal{V}_2$ 後,由質量平衡決定出口面積。

Energy balance能量平衡

  • $\dot W_{cv} = 0$: no shaft; the walls are rigid.$\dot W_{cv} = 0$:無軸,管壁為剛性。
  • $\dot Q_{cv} \approx 0$: the flow passes too quickly to exchange much heat.$\dot Q_{cv} \approx 0$:流體通過太快,來不及大量換熱。
  • $\Delta PE \approx 0$: inlet and exit are at nearly the same height.$\Delta PE \approx 0$:入口與出口高度幾乎相同。
  • Keep $\Delta KE$: it is the whole point of the device.保留 $\Delta KE$:這正是此裝置的目的。
$$ 0 = (h_1 - h_2) + \frac{\mathcal{V}_1^2 - \mathcal{V}_2^2}{2} \quad\Rightarrow\quad \mathcal{V}_2 = \sqrt{\mathcal{V}_1^2 + 2(h_1 - h_2)} $$
Nozzle
  • Subsonic flow: the nozzle converges. Supersonic flow: it must diverge (see Compressible Flow).次音速流:噴嘴為漸縮;超音速流:必須漸擴(見可壓縮流)。
  • Remember the units: $h$ in kJ/kg must be multiplied by 1000 to give $\mathcal{V}^2$ in m²/s².注意單位:kJ/kg 的 $h$ 須乘以 1000 才得到 m²/s² 的 $\mathcal{V}^2$。

Where you find it實際應用

  • Jet and rocket engines: the exhaust nozzle produces thrust.噴射與火箭引擎:排氣噴嘴產生推力。
  • Steam and gas turbines: stationary blade rows are nozzles that accelerate the flow onto the moving blades.蒸汽與燃氣渦輪機:靜葉列即噴嘴,將流體加速沖向動葉。
  • Fuel injectors and spray nozzles: atomize liquids in engines, burners and painting.燃油噴射器與噴霧噴嘴:在引擎、燃燒器與噴漆中霧化液體。
  • Ejectors and jet pumps: a high-speed jet entrains and pumps a second fluid.引射器與噴射泵:高速噴流夾帶並抽送第二股流體。
  • Water-jet cutting and fire hoses.水刀切割與消防水帶。

How performance is rated性能評估指標

Real nozzles lose some kinetic energy to friction. Compare the actual exit KE with the ideal (frictionless, adiabatic) value at the same exit pressure, state $2s$:實際噴嘴因摩擦損失部分動能。將實際出口動能與相同出口壓力下的理想值(無摩擦、絕熱,狀態 $2s$)比較:

$$ \eta_N = \frac{\mathcal{V}_2^2/2}{\mathcal{V}_{2s}^2/2} = \frac{h_1 - h_2}{h_1 - h_{2s}} \qquad C_{\mathcal V} = \frac{\mathcal{V}_2}{\mathcal{V}_{2s}} = \sqrt{\eta_N} $$
Nozzle efficiency
  • Typical $\eta_N$: 0.90–0.99 for well-designed nozzles.典型值 $\eta_N$:設計良好的噴嘴為 0.90–0.99。
  • $C_\mathcal{V}$ is the velocity coefficient, often quoted for liquid jets (≈ 0.95–0.99).$C_\mathcal{V}$ 為速度係數,常用於液體噴流(約 0.95–0.99)。

Worked example範例

Example範例 Exit velocity and exit area of an air nozzle空氣噴嘴的出口速度與出口面積 ›

Given: air enters an adiabatic nozzle at $p_1 = 300$ kPa, $T_1 = 500$ K, $\mathcal{V}_1 = 30$ m/s with $\dot m = 0.8$ kg/s. It leaves at $p_2 = 100$ kPa, $T_2 = 380$ K. The ideal (isentropic) exit temperature is $T_{2s} = 365$ K. Ideal gas, $R = 0.287$, $c_p = 1.005$ kJ/kg·K.已知:空氣以 $p_1 = 300$ kPa、$T_1 = 500$ K、$\mathcal{V}_1 = 30$ m/s、$\dot m = 0.8$ kg/s 流入絕熱噴嘴,以 $p_2 = 100$ kPa、$T_2 = 380$ K 流出。理想(等熵)出口溫度 $T_{2s} = 365$ K。理想氣體,$R = 0.287$、$c_p = 1.005$ kJ/kg·K。

Find: $\mathcal{V}_2$, the inlet and exit areas $A_1$, $A_2$, and the nozzle efficiency $\eta_N$.求:$\mathcal{V}_2$、入口與出口面積 $A_1$、$A_2$,以及噴嘴效率 $\eta_N$。

Step 1 — Assumptions. Steady state; $\dot W = 0$; adiabatic, $\dot Q = 0$; $\Delta PE = 0$. The SFEE reduces to $$0 = (h_1 - h_2) + \frac{\mathcal{V}_1^2 - \mathcal{V}_2^2}{2}$$步驟 1 — 假設。穩態;$\dot W = 0$;絕熱 $\dot Q = 0$;$\Delta PE = 0$。SFEE 化簡為 $$0 = (h_1 - h_2) + \frac{\mathcal{V}_1^2 - \mathcal{V}_2^2}{2}$$

Step 2 — Enthalpy drop. $$h_1 - h_2 = c_p(T_1 - T_2) = 1.005(500 - 380) = 120.6\ \text{kJ/kg} = 120\,600\ \text{J/kg}$$步驟 2 — 焓降。$$h_1 - h_2 = c_p(T_1 - T_2) = 1.005(500 - 380) = 120.6\ \text{kJ/kg} = 120\,600\ \text{J/kg}$$

Step 3 — Exit velocity. $$\mathcal{V}_2 = \sqrt{\mathcal{V}_1^2 + 2(h_1 - h_2)} = \sqrt{30^2 + 2(120\,600)} = 492\ \text{m/s}$$ The inlet KE ($450$ J/kg) is tiny next to the enthalpy drop.步驟 3 — 出口速度。$$\mathcal{V}_2 = \sqrt{\mathcal{V}_1^2 + 2(h_1 - h_2)} = \sqrt{30^2 + 2(120\,600)} = 492\ \text{m/s}$$ 入口動能(450 J/kg)遠小於焓降。

Step 4 — Areas from the mass balance. $v_1 = RT_1/p_1 = 0.287(500)/300 = 0.478$ m³/kg and $v_2 = 0.287(380)/100 = 1.091$ m³/kg. $$A_1 = \frac{\dot m v_1}{\mathcal{V}_1} = \frac{0.8(0.478)}{30} = 0.0128\ \text{m}^2, \qquad A_2 = \frac{0.8(1.091)}{492} = 0.00177\ \text{m}^2$$步驟 4 — 由質量平衡求面積。$v_1 = RT_1/p_1 = 0.287(500)/300 = 0.478$ m³/kg,$v_2 = 0.287(380)/100 = 1.091$ m³/kg。$$A_1 = \frac{\dot m v_1}{\mathcal{V}_1} = \frac{0.8(0.478)}{30} = 0.0128\ \text{m}^2, \qquad A_2 = \frac{0.8(1.091)}{492} = 0.00177\ \text{m}^2$$

Step 5 — Efficiency. $$\eta_N = \frac{h_1 - h_2}{h_1 - h_{2s}} = \frac{T_1 - T_2}{T_1 - T_{2s}} = \frac{500 - 380}{500 - 365} = 0.89$$步驟 5 — 效率。$$\eta_N = \frac{h_1 - h_2}{h_1 - h_{2s}} = \frac{T_1 - T_2}{T_1 - T_{2s}} = \frac{500 - 380}{500 - 365} = 0.89$$

Step 6 — Check. The area shrinks about 7× while the speed rises 16×, because the gas also expands ($v$ more than doubles). A converging nozzle is consistent with subsonic flow at the inlet.步驟 6 — 檢驗。面積縮小約 7 倍而流速增加 16 倍,因為氣體同時膨脹($v$ 增加一倍以上)。入口為次音速流,與漸縮噴嘴一致。

Answer: $\mathcal{V}_2 \approx 492$ m/s, $A_1 \approx 0.0128$ m², $A_2 \approx 17.7$ cm², $\eta_N \approx 0.89$.答:$\mathcal{V}_2 \approx 492$ m/s,$A_1 \approx 0.0128$ m²,$A_2 \approx 17.7$ cm²,$\eta_N \approx 0.89$。

Exercises練習題

Device 2裝置 2

Diffuser擴散器

A diffuser does the reverse: it slows a flow down and recovers its kinetic energy as a rise in pressure.擴散器作用相反:使流體減速,並將其動能回收為壓力上升。

1V1 high · p1 low2V2 low · p2 highV ↓ p ↑ h ↑Ẇ = 0 · Q̇ ≈ 0
A subsonic diffuser: the reverse of a nozzle. The duct widens, the flow slows down and kinetic energy is recovered as pressure.次音速擴散器:噴嘴的反向。管道擴張,流速降低,動能回收為壓力。

Mass balance質量平衡

$$ \dot m_1 = \dot m_2 \;\Rightarrow\; \frac{A_1\mathcal{V}_1}{v_1} = \frac{A_2\mathcal{V}_2}{v_2} $$

Energy balance能量平衡

Same assumptions as the nozzle ($\dot W = 0$, $\dot Q \approx 0$, $\Delta PE \approx 0$), read the other way:假設與噴嘴相同($\dot W = 0$、$\dot Q \approx 0$、$\Delta PE \approx 0$),只是方向相反:

$$ h_2 - h_1 = \frac{\mathcal{V}_1^2 - \mathcal{V}_2^2}{2} \quad\Rightarrow\quad T_2 = T_1 + \frac{\mathcal{V}_1^2 - \mathcal{V}_2^2}{2c_p}\ \ (\text{ideal gas}) $$
Diffuser
  • Subsonic flow: the diffuser diverges. Supersonic flow: it converges first.次音速流:擴散器為漸擴;超音速流:須先漸縮。
  • If $\mathcal{V}_2 \ll \mathcal{V}_1$, the exit is close to the stagnation state, $h_0 = h_1 + \mathcal{V}_1^2/2$.若 $\mathcal{V}_2 \ll \mathcal{V}_1$,出口接近滯止狀態,$h_0 = h_1 + \mathcal{V}_1^2/2$。

Where you find it實際應用

  • Jet-engine inlets: slow the incoming air before the compressor (ram compression).噴射引擎進氣道:在壓縮機前使來流減速(衝壓壓縮)。
  • Centrifugal compressors and pumps: the vaned diffuser and volute after the impeller turn speed into pressure.離心式壓縮機與泵:葉輪後的葉片擴散器與蝸殼將速度轉為壓力。
  • Hydro-turbine draft tubes: recover exit kinetic energy.水輪機尾水管:回收出口動能。
  • Wind tunnels and HVAC ducts: reduce velocity and pressure loss downstream.風洞與空調風管:降低下游流速與壓損。

How performance is rated性能評估指標

$$ \eta_D = \frac{h_{2s} - h_1}{h_2 - h_1} \qquad C_{pr} = \frac{p_2 - p_1}{\tfrac12\rho_1\mathcal{V}_1^2} $$
Diffuser efficiency
  • $\eta_D$ compares the ideal enthalpy rise needed to reach $p_2$ with the actual one. Typical: 0.80–0.95.$\eta_D$ 比較達到 $p_2$ 所需的理想焓升與實際焓升。典型值:0.80–0.95。
  • $C_{pr}$, the pressure-recovery coefficient, is the fraction of inlet dynamic pressure turned into static pressure. Typical: 0.6–0.85. Diffusers are harder to make efficient than nozzles because the slowing flow tends to separate from the wall.$C_{pr}$ 為壓力回收係數,即入口動壓轉為靜壓的比例。典型值:0.6–0.85。擴散器比噴嘴難做到高效率,因為減速流容易自壁面分離。

Worked example範例

Example範例 Jet-engine inlet diffuser噴射引擎進氣擴散器 ›

Given: air enters an aircraft-engine diffuser at $p_1 = 80$ kPa, $T_1 = 250$ K and $\mathcal{V}_1 = 240$ m/s through $A_1 = 0.5$ m². It leaves at $\mathcal{V}_2 = 60$ m/s and $p_2 = 105$ kPa. Adiabatic; ideal gas, $R = 0.287$, $c_p = 1.005$ kJ/kg·K.已知:空氣以 $p_1 = 80$ kPa、$T_1 = 250$ K、$\mathcal{V}_1 = 240$ m/s 經 $A_1 = 0.5$ m² 流入飛機引擎擴散器,以 $\mathcal{V}_2 = 60$ m/s、$p_2 = 105$ kPa 流出。絕熱;理想氣體,$R = 0.287$、$c_p = 1.005$ kJ/kg·K。

Find: the mass flow rate, the exit temperature and the exit area.求:質量流率、出口溫度與出口面積。

Step 1 — Mass flow at the inlet. $v_1 = RT_1/p_1 = 0.287(250)/80 = 0.897$ m³/kg. $$\dot m = \frac{A_1\mathcal{V}_1}{v_1} = \frac{0.5(240)}{0.897} = 133.8\ \text{kg/s}$$步驟 1 — 入口質量流率。$v_1 = RT_1/p_1 = 0.287(250)/80 = 0.897$ m³/kg。$$\dot m = \frac{A_1\mathcal{V}_1}{v_1} = \frac{0.5(240)}{0.897} = 133.8\ \text{kg/s}$$

Step 2 — Reduce the SFEE. $\dot W = 0$, $\dot Q = 0$, $\Delta PE = 0$: $$h_2 - h_1 = \frac{\mathcal{V}_1^2 - \mathcal{V}_2^2}{2}$$步驟 2 — 化簡 SFEE。$\dot W = 0$、$\dot Q = 0$、$\Delta PE = 0$:$$h_2 - h_1 = \frac{\mathcal{V}_1^2 - \mathcal{V}_2^2}{2}$$

Step 3 — Exit temperature. $$\frac{240^2 - 60^2}{2} = 27\,000\ \text{J/kg} = 27.0\ \text{kJ/kg}, \qquad T_2 = 250 + \frac{27.0}{1.005} = 276.9\ \text{K}$$步驟 3 — 出口溫度。$$\frac{240^2 - 60^2}{2} = 27\,000\ \text{J/kg} = 27.0\ \text{kJ/kg}, \qquad T_2 = 250 + \frac{27.0}{1.005} = 276.9\ \text{K}$$

Step 4 — Exit area. $v_2 = 0.287(276.9)/105 = 0.757$ m³/kg. $$A_2 = \frac{\dot m v_2}{\mathcal{V}_2} = \frac{133.8(0.757)}{60} = 1.69\ \text{m}^2$$步驟 4 — 出口面積。$v_2 = 0.287(276.9)/105 = 0.757$ m³/kg。$$A_2 = \frac{\dot m v_2}{\mathcal{V}_2} = \frac{133.8(0.757)}{60} = 1.69\ \text{m}^2$$

Step 5 — Check. The air is warmed by about 27 K and compressed before reaching the compressor, at no shaft cost: this is "ram" compression. The area increases, consistent with a subsonic diffuser.步驟 5 — 檢驗。空氣在進入壓縮機前升溫約 27 K 並被壓縮,無須消耗軸功,即「衝壓」壓縮。面積增加,與次音速擴散器一致。

Answer: $\dot m \approx 134$ kg/s, $T_2 \approx 277$ K, $A_2 \approx 1.69$ m².答:$\dot m \approx 134$ kg/s,$T_2 \approx 277$ K,$A_2 \approx 1.69$ m²。

Exercises練習題

Device 3裝置 3

Pump & compressor泵與壓縮機

A compressor (gas) or pump (liquid) takes in shaft work to raise the fluid pressure.壓縮機(氣體)或泵(液體)輸入軸功以提高流體壓力。

ẆinQ̇ (small, out)compressor壓縮機or pump或泵1 · low p2 · high pp ↑ h ↑ Ẇ in
Compressor (gas) or pump (liquid). Shaft work enters the CV and raises the fluid pressure. The trapezoid narrows because the compressed gas occupies less volume.壓縮機(氣體)或泵(液體)。軸功輸入 CV 並提高流體壓力。梯形逐漸變窄,代表氣體受壓後體積變小。

Mass balance質量平衡

$$ \dot m_1 = \dot m_2 = \dot m $$

Energy balance能量平衡

  • Keep $\dot W_{cv}$: negative (work in).保留 $\dot W_{cv}$:為負(功輸入)。
  • $\dot Q_{cv}$: usually small and outward; zero if adiabatic. Intercooled compressors reject heat on purpose.$\dot Q_{cv}$:通常很小且向外;絕熱時為零。具中間冷卻的壓縮機則刻意排熱。
  • $\Delta KE$, $\Delta PE$: usually negligible compared with $\Delta h$.$\Delta KE$、$\Delta PE$:通常遠小於 $\Delta h$,可忽略。
$$ \dot W_{cv} = \dot Q_{cv} + \dot m\,(h_1 - h_2) \quad\xrightarrow{\ \dot Q = 0\ }\quad \dot W_{in} = \dot m\,(h_2 - h_1) $$
Compressor

For a pump moving an incompressible liquid at nearly constant temperature, $u_2 \approx u_1$, so $h_2 - h_1 = v(p_2 - p_1)$. Keeping KE and PE, which can matter for pumps:對於在近乎定溫下輸送不可壓縮液體的泵,$u_2 \approx u_1$,故 $h_2 - h_1 = v(p_2 - p_1)$。保留對泵可能重要的動能與位能:

$$ \frac{\dot W_{in}}{\dot m} = v(p_2 - p_1) + \frac{\mathcal{V}_2^2 - \mathcal{V}_1^2}{2} + g(z_2 - z_1) $$
Pump
  • Because $v$ of a liquid is about 1000× smaller than that of a gas, pumping a liquid takes far less work than compressing a gas through the same $\Delta p$.液體的 $v$ 約為氣體的千分之一,因此在相同 $\Delta p$ 下,泵送液體所需的功遠小於壓縮氣體。

Where you find it實際應用

  • Gas turbines and jet engines: axial compressors deliver air to the combustor.燃氣渦輪機與噴射引擎:軸流式壓縮機將空氣送入燃燒室。
  • Refrigerators and heat pumps: compress refrigerant vapor.冰箱與熱泵:壓縮冷媒蒸氣。
  • Turbochargers: boost engine intake air.渦輪增壓器:提高引擎進氣壓力。
  • Boiler feed pumps in power plants; water supply, irrigation and chemical process pumps.電廠的鍋爐給水泵;供水、灌溉與化工製程泵。
  • Industrial compressed air and gas pipelines.工業壓縮空氣與天然氣管線。

How performance is rated性能評估指標

Compare the ideal (isentropic) work to reach the same exit pressure with the actual work. The ideal work is smaller, so it goes on top:比較達到相同出口壓力的理想(等熵)功與實際功。理想功較小,故置於分子:

$$ \eta_c = \frac{h_{2s} - h_1}{h_2 - h_1} \qquad \eta_p = \frac{v(p_2 - p_1)}{h_2 - h_1} = \frac{w_s}{w_a} $$
Isentropic efficiency
  • Typical $\eta_c$: centrifugal 0.75–0.85; axial 0.85–0.92.典型值 $\eta_c$:離心式 0.75–0.85;軸流式 0.85–0.92。
  • Typical $\eta_p$: 0.60–0.90, higher for large pumps.典型值 $\eta_p$:0.60–0.90,大型泵較高。
  • Overall efficiency from electricity to fluid, combining isentropic and first-law efficiencies: $\eta_{overall} = \eta_c\,\eta_{mech}\,\eta_{motor}$; electric motors are typically 0.90–0.97.從電力到流體的總效率,結合等熵效率與第一定律效率:$\eta_{overall} = \eta_c\,\eta_{mech}\,\eta_{motor}$;電動機通常為 0.90–0.97。
  • Cooled compressors are sometimes rated by an isothermal efficiency, $w_T/w_a$, since isothermal compression needs the least work.具冷卻的壓縮機有時以等溫效率 $w_T/w_a$ 評估,因為等溫壓縮所需功最少。

Worked example範例

Example範例 Air compressor with an isentropic efficiency具等熵效率的空氣壓縮機 ›

Given: air at 100 kPa, 300 K is compressed to 800 kPa at $\dot m = 1.5$ kg/s. The compressor is adiabatic with $\eta_c = 0.85$; the isentropic exit temperature is $T_{2s} = 543$ K. The motor efficiency is 0.95. $c_p = 1.005$ kJ/kg·K; neglect KE and PE.已知:100 kPa、300 K 的空氣以 $\dot m = 1.5$ kg/s 壓縮至 800 kPa。壓縮機為絕熱,$\eta_c = 0.85$;等熵出口溫度 $T_{2s} = 543$ K。馬達效率 0.95。$c_p = 1.005$ kJ/kg·K;忽略動能與位能。

Find: the actual exit temperature, the compressor power and the electric power.求:實際出口溫度、壓縮機功率與電功率。

Step 1 — Reduce the SFEE. Steady, adiabatic, KE and PE neglected: $$\dot W_{in} = \dot m\,(h_2 - h_1)$$步驟 1 — 化簡 SFEE。穩態、絕熱,忽略動能與位能:$$\dot W_{in} = \dot m\,(h_2 - h_1)$$

Step 2 — Ideal work. $$w_s = h_{2s} - h_1 = c_p(T_{2s} - T_1) = 1.005(543 - 300) = 244.2\ \text{kJ/kg}$$步驟 2 — 理想功。$$w_s = h_{2s} - h_1 = c_p(T_{2s} - T_1) = 1.005(543 - 300) = 244.2\ \text{kJ/kg}$$

Step 3 — Actual work from the efficiency. $$w_a = \frac{w_s}{\eta_c} = \frac{244.2}{0.85} = 287.3\ \text{kJ/kg}$$步驟 3 — 由效率求實際功。$$w_a = \frac{w_s}{\eta_c} = \frac{244.2}{0.85} = 287.3\ \text{kJ/kg}$$

Step 4 — Actual exit temperature. $$T_2 = T_1 + \frac{w_a}{c_p} = 300 + \frac{287.3}{1.005} = 586\ \text{K}$$ Hotter than $T_{2s}$: friction ends up as extra internal energy.步驟 4 — 實際出口溫度。$$T_2 = T_1 + \frac{w_a}{c_p} = 300 + \frac{287.3}{1.005} = 586\ \text{K}$$ 高於 $T_{2s}$:摩擦最終成為額外的內能。

Step 5 — Power. $$\dot W_{in} = 1.5(287.3) = 431\ \text{kW}, \qquad \dot W_{elec} = \frac{431}{0.95} = 454\ \text{kW}$$步驟 5 — 功率。$$\dot W_{in} = 1.5(287.3) = 431\ \text{kW}, \qquad \dot W_{elec} = \frac{431}{0.95} = 454\ \text{kW}$$

Step 6 — Check. The real compressor needs about 18% more work than the ideal one, and the motor adds another 5%.步驟 6 — 檢驗。實際壓縮機比理想壓縮機多需約 18% 的功,馬達再增加 5%。

Answer: $T_2 \approx 586$ K, $\dot W_{in} \approx 431$ kW, electric input $\approx 454$ kW.答:$T_2 \approx 586$ K,$\dot W_{in} \approx 431$ kW,電力輸入 $\approx 454$ kW。

Exercises練習題

Device 4裝置 4

Turbine & expander渦輪機與膨脹機

A turbine extracts shaft work as a high-pressure fluid expands through rows of blades. An expander is the general name, including piston and scroll machines.渦輪機使高壓流體通過多列葉片膨脹而輸出軸功。膨脹機為統稱,亦包括活塞式與渦卷式機械。

ẆoutQ̇ (small, out)turbine渦輪機or expander或膨脹機1 · high p2 · low pp ↓ h ↓ Ẇ out
Turbine or expander. The fluid expands through the blades, so the trapezoid widens; the enthalpy drop leaves the CV as shaft work.渦輪機或膨脹機。流體通過葉片膨脹,梯形因此變寬;焓降以軸功的形式離開 CV。

Mass balance質量平衡

$$ \dot m_1 = \dot m_2 = \dot m $$

Large steam turbines often have extraction ports that bleed steam part-way through. Then $\dot m_1 = \dot m_2 + \dot m_3 + \dots$大型蒸汽渦輪機常有抽汽口,在中途抽出部分蒸汽,此時 $\dot m_1 = \dot m_2 + \dot m_3 + \dots$

Energy balance能量平衡

  • Keep $\dot W_{cv}$: positive (work out).保留 $\dot W_{cv}$:為正(功輸出)。
  • $\dot Q_{cv}$: small heat loss through the casing; often taken as adiabatic.$\dot Q_{cv}$:經機殼的少量熱損失;常視為絕熱。
  • $\Delta KE$, $\Delta PE$: usually negligible.$\Delta KE$、$\Delta PE$:通常可忽略。
$$ \dot W_{cv} = \dot Q_{cv} + \dot m\,(h_1 - h_2) \quad\xrightarrow{\ \dot Q = 0\ }\quad \dot W_{out} = \dot m\,(h_1 - h_2) $$
Turbine

For a hydraulic turbine, the available energy is the water head $H$: $\dot W_{ideal} = \dot m\,gH$.對於水輪機,可用能量為水頭 $H$:$\dot W_{ideal} = \dot m\,gH$。

Where you find it實際應用

  • Steam power plants: coal, nuclear, geothermal, solar-thermal.蒸汽動力廠:燃煤、核能、地熱、太陽熱能。
  • Gas turbines: power generation, aircraft engines, ships.燃氣渦輪機:發電、航空引擎、船舶。
  • Turbochargers: the exhaust turbine drives the intake compressor.渦輪增壓器:排氣渦輪驅動進氣壓縮機。
  • Hydroelectric turbines; organic Rankine cycle expanders for waste heat.水力發電水輪機;用於廢熱回收的有機朗肯循環膨脹機。
  • Cryogenic expanders in air separation and gas liquefaction.空氣分離與氣體液化中的低溫膨脹機。

How performance is rated性能評估指標

Now the actual work is the smaller one, so it goes on top:此時實際功較小,故置於分子:

$$ \eta_t = \frac{h_1 - h_2}{h_1 - h_{2s}} = \frac{w_a}{w_s} \qquad \eta_{hydro} = \frac{\dot W_{out}}{\dot m\,gH} $$
Isentropic efficiency
  • Typical $\eta_t$: large steam turbines 0.85–0.92; gas turbines 0.88–0.92; small turbochargers and expanders 0.65–0.85.典型值 $\eta_t$:大型蒸汽渦輪機 0.85–0.92;燃氣渦輪機 0.88–0.92;小型渦輪增壓器與膨脹機 0.65–0.85。
  • Typical $\eta_{hydro}$: 0.90–0.95, among the highest of any energy converter.典型值 $\eta_{hydro}$:0.90–0.95,為能量轉換機械中最高者之一。
  • Generator efficiency (≈ 0.98) multiplies in for electrical output.計算電力輸出時,還須乘上發電機效率(約 0.98)。

Worked example範例

Example範例 Steam turbine with heat loss and kinetic energy具熱損失與動能的蒸汽渦輪機 ›

Given: steam enters a turbine at 4 MPa, 500 °C ($h_1 = 3445.3$ kJ/kg) with $\mathcal{V}_1 = 50$ m/s and leaves at 10 kPa with $h_2 = 2350$ kJ/kg, $\mathcal{V}_2 = 120$ m/s. The isentropic exit enthalpy is $h_{2s} = 2160$ kJ/kg. $\dot m = 20$ kg/s; heat loss 100 kW; $\Delta z = 0$.已知:蒸汽以 4 MPa、500 °C($h_1 = 3445.3$ kJ/kg)、$\mathcal{V}_1 = 50$ m/s 流入渦輪機,以 10 kPa、$h_2 = 2350$ kJ/kg、$\mathcal{V}_2 = 120$ m/s 流出。等熵出口焓 $h_{2s} = 2160$ kJ/kg。$\dot m = 20$ kg/s;熱損失 100 kW;$\Delta z = 0$。

Find: the power output and the isentropic efficiency.求:輸出功率與等熵效率。

Step 1 — Reduce the SFEE. Steady, $\Delta z = 0$: $$\dot W_{cv} = \dot Q_{cv} + \dot m\left[(h_1 - h_2) + \frac{\mathcal{V}_1^2 - \mathcal{V}_2^2}{2}\right]$$ with $\dot Q_{cv} = -100$ kW (heat leaves).步驟 1 — 化簡 SFEE。穩態、$\Delta z = 0$:$$\dot W_{cv} = \dot Q_{cv} + \dot m\left[(h_1 - h_2) + \frac{\mathcal{V}_1^2 - \mathcal{V}_2^2}{2}\right]$$ 其中 $\dot Q_{cv} = -100$ kW(熱流出)。

Step 2 — Evaluate per unit mass. $$h_1 - h_2 = 1095.3\ \text{kJ/kg}, \qquad \frac{50^2 - 120^2}{2} = -5950\ \text{J/kg} = -5.95\ \text{kJ/kg}$$步驟 2 — 計算單位質量各項。$$h_1 - h_2 = 1095.3\ \text{kJ/kg}, \qquad \frac{50^2 - 120^2}{2} = -5950\ \text{J/kg} = -5.95\ \text{kJ/kg}$$

Step 3 — Power. $$\dot W_{cv} = -100 + 20(1095.3 - 5.95) = -100 + 21\,787 = 21\,687\ \text{kW}$$步驟 3 — 功率。$$\dot W_{cv} = -100 + 20(1095.3 - 5.95) = -100 + 21\,787 = 21\,687\ \text{kW}$$

Step 4 — Isentropic efficiency. $$\eta_t = \frac{h_1 - h_2}{h_1 - h_{2s}} = \frac{1095.3}{3445.3 - 2160} = \frac{1095.3}{1285.3} = 0.852$$步驟 4 — 等熵效率。$$\eta_t = \frac{h_1 - h_2}{h_1 - h_{2s}} = \frac{1095.3}{3445.3 - 2160} = \frac{1095.3}{1285.3} = 0.852$$

Step 5 — Check. KE and heat loss together change the power by about 1%: neglecting them would have been acceptable. $\eta_t$ falls in the typical 0.85–0.92 range.步驟 5 — 檢驗。動能與熱損失合計僅使功率改變約 1%,忽略亦可接受。$\eta_t$ 落在典型範圍 0.85–0.92。

Answer: $\dot W_{cv} \approx 21.7$ MW, $\eta_t \approx 0.85$.答:$\dot W_{cv} \approx 21.7$ MW,$\eta_t \approx 0.85$。

Exercises練習題

Device 5裝置 5

Throttling device節流裝置

A throttle drops pressure by forcing the flow through a restriction: a partly closed valve, a porous plug or a long capillary tube.節流裝置迫使流體通過限縮處而降壓:部分關閉的閥、多孔塞或細長毛細管。

1 · p₁ high2 · p₂ lowh₂ = h₁ (ideal gas: T₂ = T₁)partly closed valve部分關閉的閥
A throttling valve. The flow squeezes through a small opening and loses pressure with no work and almost no heat transfer, so enthalpy is unchanged. A porous plug or capillary tube behaves the same way.節流閥。流體擠過狹小開口而損失壓力,無功、幾乎無熱傳,因此焓不變。多孔塞或毛細管的行為相同。

Mass balance質量平衡

$$ \dot m_1 = \dot m_2 = \dot m $$

Energy balance能量平衡

  • $\dot W_{cv} = 0$: no moving parts deliver work.$\dot W_{cv} = 0$:沒有輸出功的運動部件。
  • $\dot Q_{cv} \approx 0$: small device, fast flow.$\dot Q_{cv} \approx 0$:裝置小、流速快。
  • $\Delta KE$, $\Delta PE \approx 0$: measured far enough downstream, velocities are similar.$\Delta KE$、$\Delta PE \approx 0$:在足夠下游處量測,流速相近。
$$ h_2 = h_1 $$
Throttle
  • Ideal gas: $h = h(T)$, so $T_2 = T_1$.理想氣體:$h = h(T)$,故 $T_2 = T_1$。
  • Real fluids: $T$ usually drops. The rate is the Joule–Thomson coefficient $\mu_{JT} = (\partial T/\partial p)_h$.真實流體:$T$ 通常下降,其變化率為焦耳–湯姆森係數 $\mu_{JT} = (\partial T/\partial p)_h$。
  • A saturated liquid throttled to low pressure partly flashes to vapor and becomes a cold liquid–vapor mixture: the basis of refrigeration.飽和液體節流至低壓時部分閃化為蒸氣,成為低溫的液汽混合物:這是冷凍的基礎。

Where you find it實際應用

  • Expansion valves and capillary tubes in refrigerators, air conditioners and heat pumps.冰箱、冷氣與熱泵中的膨脹閥與毛細管。
  • Pressure regulators on gas cylinders and supply lines.氣瓶與供氣管線上的調壓器。
  • Throttling calorimeter: measures the quality of wet steam.節流卡計:量測濕蒸汽的乾度。
  • Linde gas liquefaction (Joule–Thomson expansion).林德氣體液化法(焦耳–湯姆森膨脹)。
  • Flow-control valves throughout process plants.製程工廠中隨處可見的流量控制閥。

How performance is rated性能評估指標

  • A throttle has no efficiency in the usual sense: it is not meant to produce or absorb work.節流裝置沒有一般意義下的效率:其目的不在產生或吸收功。
  • It is fully irreversible. The same pressure drop through a turbine or expander would deliver work; the throttle throws that work potential away. This loss is quantified later as entropy generation and exergy destruction.它是完全不可逆的。同樣的壓降若經過渦輪機或膨脹機可輸出功,節流則將這份作功潛力白白浪費。此損失日後以熵產生與火用破壞量化。
  • Valves are rated instead by a flow coefficient $C_v$ (flow rate per $\sqrt{\Delta p}$), a fluid-mechanics quantity.閥門改以流量係數 $C_v$(每 $\sqrt{\Delta p}$ 的流量)評估,屬流體力學的量。

Worked example範例

Example範例 Refrigerant expansion valve冷媒膨脹閥 ›

Given: saturated liquid R-134a at 0.8 MPa ($T_{sat} = 31.3$ °C, $h_1 = h_f = 95.47$ kJ/kg) is throttled to $-10$ °C. At $-10$ °C: $h_f = 38.55$, $h_g = 244.51$ kJ/kg (table values).已知:0.8 MPa 的飽和液態 R-134a($T_{sat} = 31.3$ °C,$h_1 = h_f = 95.47$ kJ/kg)節流至 $-10$ °C。$-10$ °C 時:$h_f = 38.55$、$h_g = 244.51$ kJ/kg(查表值)。

Find: the exit quality and the temperature drop.求:出口乾度與溫度降。

Step 1 — Reduce the SFEE. $\dot W = 0$, $\dot Q \approx 0$, $\Delta KE \approx \Delta PE \approx 0$: $$h_2 = h_1 = 95.47\ \text{kJ/kg}$$步驟 1 — 化簡 SFEE。$\dot W = 0$、$\dot Q \approx 0$、$\Delta KE \approx \Delta PE \approx 0$:$$h_2 = h_1 = 95.47\ \text{kJ/kg}$$

Step 2 — Locate the exit state. At $-10$ °C, $h_f < h_2 < h_g$, so the exit is a liquid–vapor mixture.步驟 2 — 確定出口狀態。在 $-10$ °C,$h_f < h_2 < h_g$,故出口為液汽混合物。

Step 3 — Quality. $$x_2 = \frac{h_2 - h_f}{h_g - h_f} = \frac{95.47 - 38.55}{244.51 - 38.55} = \frac{56.92}{205.96} = 0.276$$步驟 3 — 乾度。$$x_2 = \frac{h_2 - h_f}{h_g - h_f} = \frac{95.47 - 38.55}{244.51 - 38.55} = \frac{56.92}{205.96} = 0.276$$

Step 4 — Temperature drop. $\Delta T = 31.3 - (-10) = 41.3$ K. About 28% of the liquid flashes to vapor; that evaporation cools the rest.步驟 4 — 溫度降。$\Delta T = 31.3 - (-10) = 41.3$ K。約 28% 的液體閃化為蒸氣,其蒸發使其餘液體降溫。

Step 5 — Check. Enthalpy is constant but temperature drops sharply: the opposite of an ideal gas, where $T$ would stay fixed. This cold mixture then absorbs heat in the evaporator.步驟 5 — 檢驗。焓不變而溫度驟降,與理想氣體 $T$ 不變的情況相反。此低溫混合物隨後在蒸發器中吸熱。

Answer: $x_2 \approx 0.28$; the temperature falls by about 41 K.答:$x_2 \approx 0.28$;溫度下降約 41 K。

Exercises練習題

Device 6裝置 6

Mixing chamber混合室

A mixing chamber combines two or more streams directly. The exit stream has an intermediate state.混合室將兩股或多股流體直接合併,出口流體的狀態介於其間。

1 · hot2 · cold3 · mixedṁ3 = ṁ1 + ṁ2Ẇ = 0 · Q̇ ≈ 0 (insulated)
A mixing chamber. Two inlet streams combine directly and leave as one. The exit state lies between the two inlet states.混合室。兩股入口流體直接混合後以一股流出,出口狀態介於兩入口狀態之間。

Mass balance質量平衡

$$ \dot m_1 + \dot m_2 = \dot m_3 $$
Two inlets, one exit

Energy balance能量平衡

  • $\dot W_{cv} = 0$, $\dot Q_{cv} \approx 0$ (usually insulated).$\dot W_{cv} = 0$、$\dot Q_{cv} \approx 0$(通常有隔熱)。
  • $\Delta KE$, $\Delta PE \approx 0$.$\Delta KE$、$\Delta PE \approx 0$。
$$ \dot m_1 h_1 + \dot m_2 h_2 = \dot m_3 h_3 $$
Mixer

Eliminate $\dot m_3$ with the mass balance to get the mixing ratio:以質量平衡消去 $\dot m_3$,得到混合比:

$$ \frac{\dot m_1}{\dot m_2} = \frac{h_3 - h_2}{h_1 - h_3} \quad\xrightarrow{\ h = cT\ }\quad \frac{\dot m_1}{\dot m_2} = \frac{T_3 - T_2}{T_1 - T_3} $$
  • The exit enthalpy is the mass-weighted average of the inlets: a "lever rule".出口焓為各入口的質量加權平均,即「槓桿法則」。
  • Inlet pressures must be about equal; otherwise the high-pressure stream would flow back into the low-pressure line.各入口壓力須大致相等,否則高壓流體會倒流入低壓管線。

Where you find it實際應用

  • Open feedwater heaters in steam plants: bleed steam mixes with feedwater.蒸汽動力廠的開放式給水加熱器:抽汽與給水混合。
  • Desuperheaters (attemperators): spray water into superheated steam to control temperature.減溫器:將水噴入過熱蒸汽以控制溫度。
  • HVAC mixing boxes: blend return air with outdoor air.空調混風箱:混合回風與外氣。
  • Showers and faucets; tempering valves for hot-water systems.淋浴與水龍頭;熱水系統的混水閥。

How performance is rated性能評估指標

  • There is no standard efficiency. Energy is conserved exactly; the design goal is a target exit temperature or composition.沒有標準效率。能量完全守恆;設計目標為特定出口溫度或成分。
  • Practical quality is judged by mixing uniformity at the exit.實務上以出口的混合均勻度判斷品質。
  • Mixing streams at different temperatures is irreversible, which shows up later as exergy destruction.不同溫度的流體混合為不可逆過程,日後以火用破壞呈現。

Worked example範例

Example範例 Desuperheater (spray attemperator)減溫器(噴水減溫) ›

Given: superheated steam at 3 MPa, 450 °C ($h_1 = 3344.0$ kJ/kg) flows at $\dot m_1 = 10$ kg/s. Liquid water ($h_2 = 420$ kJ/kg) is sprayed in to bring the exit to $h_3 = 3100$ kJ/kg at the same pressure. Adiabatic; neglect KE and PE.已知:3 MPa、450 °C($h_1 = 3344.0$ kJ/kg)的過熱蒸汽以 $\dot m_1 = 10$ kg/s 流動。噴入液態水($h_2 = 420$ kJ/kg),使出口在相同壓力下達 $h_3 = 3100$ kJ/kg。絕熱;忽略動能與位能。

Find: the spray-water flow rate and the exit flow rate.求:噴水流率與出口流率。

Step 1 — Mass balance. Two inlets, one exit: $$\dot m_3 = \dot m_1 + \dot m_2$$步驟 1 — 質量平衡。兩入口、一出口:$$\dot m_3 = \dot m_1 + \dot m_2$$

Step 2 — Energy balance. $\dot Q = \dot W = 0$: $$\dot m_1 h_1 + \dot m_2 h_2 = \dot m_3 h_3$$步驟 2 — 能量平衡。$\dot Q = \dot W = 0$:$$\dot m_1 h_1 + \dot m_2 h_2 = \dot m_3 h_3$$

Step 3 — Eliminate $\dot m_3$. $$\dot m_1 (h_1 - h_3) = \dot m_2 (h_3 - h_2) \quad\Rightarrow\quad \dot m_2 = \dot m_1 \frac{h_1 - h_3}{h_3 - h_2}$$步驟 3 — 消去 $\dot m_3$。$$\dot m_1 (h_1 - h_3) = \dot m_2 (h_3 - h_2) \quad\Rightarrow\quad \dot m_2 = \dot m_1 \frac{h_1 - h_3}{h_3 - h_2}$$

Step 4 — Numbers. $$\dot m_2 = 10\,\frac{3344.0 - 3100}{3100 - 420} = 10\,\frac{244}{2680} = 0.910\ \text{kg/s}, \qquad \dot m_3 = 10.91\ \text{kg/s}$$步驟 4 — 代入數值。$$\dot m_2 = 10\,\frac{3344.0 - 3100}{3100 - 420} = 10\,\frac{244}{2680} = 0.910\ \text{kg/s}, \qquad \dot m_3 = 10.91\ \text{kg/s}$$

Step 5 — Check. A small amount of water (about 9% of the steam flow) is enough because each kilogram absorbs a large enthalpy change as it evaporates and superheats.步驟 5 — 檢驗。少量的水(約蒸汽流量的 9%)即足夠,因為每公斤水在蒸發與過熱時吸收大量焓變。

Answer: $\dot m_2 \approx 0.91$ kg/s, $\dot m_3 \approx 10.9$ kg/s.答:$\dot m_2 \approx 0.91$ kg/s,$\dot m_3 \approx 10.9$ kg/s。

Exercises練習題

Device 7裝置 7

Heat exchanger熱交換器

A heat exchanger transfers heat from a hot stream to a cold one through a wall, without mixing them.熱交換器經由壁面將熱從熱流體傳給冷流體,兩者不混合。

A in (hot)A 入(熱)A outA 出B in (cold)B 入(冷)B outB 出Overall CV: Q̇cv ≈ 0, Ẇ = 0整體 CV:Q̇cv ≈ 0、Ẇ = 0CV around stream A only: Q̇ = −Q̇(A→B)僅包住流體 A 的 CV:Q̇ = −Q̇(A→B)
A counterflow tube-in-tube heat exchanger. Hot stream A (inner tube) passes heat through the wall to cold stream B (annulus) without mixing. The overall CV has no external heat transfer; a CV around one stream does.逆流式套管熱交換器。熱流體 A(內管)經管壁將熱傳給冷流體 B(環隙),兩者不混合。整體 CV 無對外熱傳;只包住一股流體的 CV 則有熱傳。

Mass balance質量平衡

The streams stay separate, so each has its own mass balance:兩股流體互不混合,因此各自有其質量平衡:

$$ \dot m_{A,in} = \dot m_{A,out} = \dot m_A, \qquad \dot m_{B,in} = \dot m_{B,out} = \dot m_B $$

Energy balance能量平衡

The answer depends on where you draw the CV:結果取決於 CV 的畫法:

  • Overall CV (both streams): $\dot W = 0$ and $\dot Q_{cv} \approx 0$ if the shell is insulated. The heat between streams is internal and does not appear.整體 CV(包含兩股流體):$\dot W = 0$;若外殼隔熱,$\dot Q_{cv} \approx 0$。流體間的熱傳屬內部,不出現在式中。
  • CV around one stream: the exchanged heat now crosses its boundary, $\dot Q = \dot m(h_{out} - h_{in})$.只包住一股流體的 CV:交換的熱穿越其邊界,$\dot Q = \dot m(h_{out} - h_{in})$。
$$ \dot m_A\,(h_{A,in} - h_{A,out}) = \dot m_B\,(h_{B,out} - h_{B,in}) = \dot Q_{A\to B} $$
Heat exchanger

Heat lost by the hot stream equals heat gained by the cold stream.熱流體失去的熱等於冷流體獲得的熱。

Where you find it實際應用

  • Boilers, condensers and evaporators in power and refrigeration cycles.動力與冷凍循環中的鍋爐、冷凝器與蒸發器。
  • Car radiators, oil coolers, charge-air intercoolers.汽車水箱、機油冷卻器、增壓空氣中間冷卻器。
  • Recuperators and regenerators recovering exhaust heat in gas turbines.燃氣渦輪機中回收排氣熱的回熱器與蓄熱器。
  • Closed feedwater heaters, district heating, HVAC coils, electronics cold plates.封閉式給水加熱器、區域供熱、空調盤管、電子散熱冷板。

How performance is rated性能評估指標

The effectiveness compares the actual heat transfer with the maximum possible, set by the stream with the smaller heat-capacity rate $C = \dot m c_p$:有效度將實際熱傳量與最大可能值比較,最大值由熱容率 $C = \dot m c_p$ 較小的流體決定:

$$ \varepsilon = \frac{\dot Q}{\dot Q_{max}}, \qquad \dot Q_{max} = C_{min}\,(T_{h,in} - T_{c,in}) $$
Effectiveness
  • Typical $\varepsilon$: 0.6–0.95. Counterflow reaches higher values than parallel flow for the same size.典型值 $\varepsilon$:0.6–0.95。相同尺寸下,逆流式比平行流式更高。
  • Fired heat exchangers are rated by boiler efficiency, a first-law efficiency: heat to the steam divided by fuel energy (heating value), typically 0.85–0.95.燃燒式熱交換器以鍋爐效率(一種第一定律效率)評估:蒸汽吸收的熱除以燃料能量(熱值),通常為 0.85–0.95。
  • A heat-loss fraction accounts for an imperfectly insulated shell: $\dot Q_{B} = (1 - f)\,\dot Q_{A}$.熱損失比例用以考量外殼隔熱不完美:$\dot Q_{B} = (1 - f)\,\dot Q_{A}$。

Worked example範例

Example範例 Oil cooler: heat duty and effectiveness機油冷卻器:熱負荷與有效度 ›

Given: engine oil ($c_p = 2.1$ kJ/kg·K, $\dot m = 0.5$ kg/s) enters a counterflow cooler at 120 °C. Cooling water ($c = 4.18$ kJ/kg·K, $\dot m = 0.4$ kg/s) enters at 20 °C and leaves at 55 °C. The shell is insulated; neglect KE, PE and pressure effects.已知:機油($c_p = 2.1$ kJ/kg·K,$\dot m = 0.5$ kg/s)以 120 °C 流入逆流式冷卻器。冷卻水($c = 4.18$ kJ/kg·K,$\dot m = 0.4$ kg/s)以 20 °C 流入、55 °C 流出。外殼隔熱;忽略動能、位能與壓力效應。

Find: the heat transfer rate, the oil exit temperature and the effectiveness.求:熱傳率、機油出口溫度與有效度。

Step 1 — Mass balances. Each stream is separate: $\dot m_{oil}$ and $\dot m_w$ are each constant.步驟 1 — 質量平衡。兩股流體分開:$\dot m_{oil}$ 與 $\dot m_w$ 各自不變。

Step 2 — CV around the water. $$\dot Q = \dot m_w c_w (T_{w,out} - T_{w,in}) = 0.4(4.18)(55 - 20) = 58.5\ \text{kW}$$步驟 2 — 包住水的 CV。$$\dot Q = \dot m_w c_w (T_{w,out} - T_{w,in}) = 0.4(4.18)(55 - 20) = 58.5\ \text{kW}$$

Step 3 — Overall CV gives the oil exit. Insulated shell, so heat lost by the oil = heat gained by the water: $$T_{oil,out} = 120 - \frac{58.5}{0.5(2.1)} = 120 - 55.7 = 64.3\ ^\circ\text{C}$$步驟 3 — 由整體 CV 求機油出口。外殼隔熱,故機油失熱 = 水得熱:$$T_{oil,out} = 120 - \frac{58.5}{0.5(2.1)} = 120 - 55.7 = 64.3\ ^\circ\text{C}$$

Step 4 — Maximum possible heat transfer. $C_{oil} = 0.5(2.1) = 1.05$ kW/K, $C_w = 0.4(4.18) = 1.67$ kW/K, so $C_{min} = 1.05$ kW/K. $$\dot Q_{max} = C_{min}(T_{h,in} - T_{c,in}) = 1.05(120 - 20) = 105\ \text{kW}$$步驟 4 — 最大可能熱傳。$C_{oil} = 0.5(2.1) = 1.05$ kW/K,$C_w = 0.4(4.18) = 1.67$ kW/K,故 $C_{min} = 1.05$ kW/K。$$\dot Q_{max} = C_{min}(T_{h,in} - T_{c,in}) = 1.05(120 - 20) = 105\ \text{kW}$$

Step 5 — Effectiveness. $$\varepsilon = \frac{58.5}{105} = 0.56$$步驟 5 — 有效度。$$\varepsilon = \frac{58.5}{105} = 0.56$$

Step 6 — Check. The oil (smaller $C$) changes temperature more than the water: 55.7 K vs 35 K. Only an infinitely long counterflow unit could cool it all the way to 20 °C.步驟 6 — 檢驗。熱容率較小的機油溫度變化較大:55.7 K 對 35 K。只有無限長的逆流式熱交換器才能將它冷卻至 20 °C。

Answer: $\dot Q \approx 58.5$ kW, $T_{oil,out} \approx 64$ °C, $\varepsilon \approx 0.56$.答:$\dot Q \approx 58.5$ kW,$T_{oil,out} \approx 64$ °C,$\varepsilon \approx 0.56$。

Exercises練習題

Summary總整理

Side-by-side comparison並列比較

Device裝置$\dot W$$\dot Q$$\Delta KE$$\Delta PE$Reduced balance化簡後的平衡式Typical efficiency典型效率
Nozzle噴嘴—≈0✓≈0$h_1 - h_2 = (\mathcal{V}_2^2 - \mathcal{V}_1^2)/2$$\eta_N$ 0.90–0.99
Diffuser擴散器—≈0✓≈0$h_2 - h_1 = (\mathcal{V}_1^2 - \mathcal{V}_2^2)/2$$\eta_D$ 0.80–0.95
Pump / compressor泵/壓縮機✓ insmall≈0≈0$\dot W_{in} = \dot m(h_2 - h_1)$$\eta_c$ 0.75–0.92, $\eta_p$ 0.6–0.9
Turbine / expander渦輪機/膨脹機✓ outsmall≈0≈0$\dot W_{out} = \dot m(h_1 - h_2)$$\eta_t$ 0.65–0.92
Throttle節流裝置—≈0≈0≈0$h_2 = h_1$—
Mixer混合室—≈0≈0≈0$\textstyle\sum \dot m h_{in} = \sum \dot m h_{out}$—
Heat exchanger熱交換器—internal≈0≈0$\dot m_A \Delta h_A = \dot m_B \Delta h_B$$\varepsilon$ 0.6–0.95

✓ kept · ≈0 usually negligible · — zero. Typical efficiency ranges are indicative; check manufacturer data for a specific machine.✓ 保留 · ≈0 通常可忽略 · — 為零。典型效率範圍僅供參考,特定機器請查閱製造商資料。