What you'll be able to do本章學習成果
- Apply conservation of mass to steady- and unsteady-flow control volumes.
- Identify the energy a stream carries — internal, flow work, kinetic, potential — and fold the first two into enthalpy.
- Solve steady-flow energy balances for nozzles, turbines, compressors, throttles, mixers, and heat exchangers.
- Analyze unsteady (uniform-flow) charging and discharging processes.
Key equations重要公式
Conservation of mass質量守恆
Mass entering or leaving a control volume is tracked by the mass rate balance:
The mass flow rate through an area is $\dot m = \rho A V = AV/v$. At steady state the inventory is constant, so total inflow equals total outflow: $\sum \dot m_i = \sum \dot m_e$. For a single stream this is just $\dot m_1 = \dot m_2$, and for incompressible liquids the volume flow rate is conserved too.
Flow work & enthalpy流動功與焓
Pushing a unit of mass across the boundary takes flow work equal to $pv$. So each stream carries internal energy plus flow work, plus kinetic and potential energy. The first two combine into the property we already met:
This is the real reason enthalpy is so useful: for any flowing stream, the combination of internal energy and the flow work needed to move it is automatically accounted for by $h$. Control-volume analysis is "the enthalpy version" of the energy balance.
Control-volume energy rate balance控制體積能量速率平衡
Substituting enthalpy into the energy balance gives the general one-dimensional-flow form:
Here $\dot W_{cv}$ is everything except flow work — shaft, boundary, and electrical work — because flow work is already inside the enthalpy terms.
Steady-flow form (SFEE)穩態流動形式(SFEE)
At steady state $dE_{cv}/dt = 0$. For one inlet and one exit, dividing by $\dot m$ gives the steady-flow energy equation (SFEE):
Analyzing any device is then a matter of deciding which terms are negligible.
From the balance to real equipment從平衡式到實際裝置
The power of the SFEE is that each device just keeps the terms that matter and drops the rest — a turbine keeps $\dot W$, a nozzle keeps the kinetic term, a throttle keeps almost nothing ($h_2 = h_1$). That device-by-device treatment, together with the isentropic efficiencies that rate how real hardware compares with the ideal, is developed in Flow Devices & Isentropic Efficiency — including the interactive SFEE device explorer.
Unsteady (uniform-flow) processes非穩態(均勻流動)過程
Charging or discharging a tank changes the inventory with time. Integrating the rate balances over the process gives the mass and energy accounting:
In the uniform-flow model, inlet and exit properties are taken constant, so $U_{cv}(t) - U_{cv}(0) = Q_{cv} - W_{cv} + \sum m_i h_i - \sum m_e h_e$. With all ports closed, this collapses back to the closed-system energy balance — the two views are consistent.
Try it — filling a tank動手試試——儲罐充氣
Charge a rigid, insulated tank from a high-pressure supply line. The uniform-flow energy balance gives a striking result: the gas already in the tank does compression work on the incoming gas, so the final temperature ends up above the line temperature. Fill an evacuated tank and $T_2 = k\,T_{line}$ — about 40% hotter for air.
Rigid adiabatic tank, air as ideal gas. Filled until tank pressure reaches the supply-line pressure.
Steady-flow energy balance穩態流動能量平衡
Example範例 Exit velocity from a nozzle噴嘴出口速度 ›
Given: air enters an adiabatic nozzle at 420 K with 10 m/s and exits at 300 K. $c_p=1.005$ kJ/kg·K.
Find: the exit velocity.
Solution. With no work or heat, $(h_1-h_2)=\tfrac{1}{2}(V_2^2-V_1^2)$, so $$V_2=\sqrt{V_1^2+2c_p(T_1-T_2)\cdot10^3}=\sqrt{10^2+2(1005)(120)}=491\ \tfrac{\text{m}}{\text{s}}.$$ The enthalpy drop became kinetic energy.