The first law for open systems: enthalpy carries the ledger
도입됨| English | 의미 |
|---|---|
| control volume | control volume |
| steady flow | steady flow |
| enthalpy/enˈθælpi/ | A state function equal to internal energy plus the product of pressure and volume. |
| flow work | flow work |
| isentropic | isentropic |
| nozzle | nozzle |
Pipes, not pistons
- Real machines are open: mass crosses a control volume 控制体积 boundary and carries energy with it. At steady flow 稳定流动, inflow equals outflow and nothing inside accumulates - the per-port ledger is $q - w = \Delta h + \Delta ke + \Delta pe$.
- The steady-flow energy equation is the first law rewritten for flow: shaft work and heat at the boundary, enthalpy 焓 $h = u + pv$ at the ports, and velocity changes where they matter.
The steady-flow energy equation runs on internal energy u at the ports.
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It runs on enthalpy - u plus the flow work pv.
Where the pv goes
- Pushing 1 kg of fluid into the control volume costs $pv$ of flow work 流动功; the same term reappears pushing it out. Absorbing that transport term into the internal energy is exactly what enthalpy is - which is why the open-system ledger runs on $h$, not $u$.
A turbine stage, priced
- Steam enters at $3\ \mathrm{MPa}$, $400^\circ\mathrm{C}$ ($h_1 = 3230.9\ \mathrm{kJ/kg}$, $s_1 = 6.9212\ \mathrm{kJ/(kg\,K)}$) and expands adiabatically to $50\ \mathrm{kPa}$. The ideal leg is isentropic 等熵过程: $s_2 = s_1$ gives $x_{2s} = (6.9212 - 1.0910)/6.5029 = \mathbf{0.89655}$, so $h_{2s} = 340.49 + 0.89655 \times 2305.4 = \mathbf{2407.4}$ and $w_s = 3230.9 - 2407.4 = \mathbf{823.5}\ \mathrm{kJ/kg}$.
- The real leg prices the imperfection with an isentropic efficiency $\eta_t = 0.85$: $w = 0.85 \times 823.5 = \mathbf{700.0}$, back-solving $h_2 = 3230.9 - 700.0 = \mathbf{2530.9}\ \mathrm{kJ/kg}$. Isentropic, then efficiency - the two-step turbine method.
Which device, which term?
Route each duty to the device that owns it.
Isentropic turbine exit quality at 50 kPa (four decimals)?
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x = (6.9212 - 1.0910)/6.5029.
Isentropic exit enthalpy (one decimal, kJ/kg)?
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340.49 + 0.89655 x 2305.4.
Isentropic work (one decimal, kJ/kg)?
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3230.9 - 2407.4.
Actual work with eta_t = 0.85 (one decimal, kJ/kg)?
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0.85 x 823.5.
A pump and a nozzle
- A pump lifts liquid water $0.5 \to 3\ \mathrm{MPa}$: liquid $v$ is tiny, so $w_p \approx v\,\Delta p = 0.001005 \times 2500 = \mathbf{2.5}\ \mathrm{kJ/kg}$.
- A nozzle 喷嘴 accelerates air $10 \to 300\ \mathrm{m/s}$ adiabatically with no shaft work: $\Delta h = -\Delta ke = -(300^2 - 10^2)/2 = \mathbf{-44.95}\ \mathrm{kJ/kg}$. Three devices, one equation - and the kinetic term is kept exactly where velocity is the product.
Pump work 0.5 to 3 MPa liquid water (one decimal, kJ/kg)?
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v dP = 0.001005 x 2500.
Nozzle 10 to 300 m/s: enthalpy drop (two decimals, kJ/kg)?
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(300^2 - 10^2)/2 in J/g.
Across an adiabatic nozzle with no shaft work:
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q - w = 0, so dh = -d(ke).
Three port slips
- Dropping $ke$ for a nozzle. It is the point of the device; keep it for pumps too only when velocities are huge (they are not).
- Using $c_p \Delta T$ for a two-phase expansion exit. Inside the dome, $h$ is not $c_p \Delta T$ from a superheated start.
- Answering power when asked for work per kilogram. $\dot W = \dot m\,w$ is the last step, not the first.
Recap
- $q - w = \Delta h + \Delta ke + \Delta pe$; enthalpy absorbs the flow work.
- Turbine: $s_{2s} = s_1$, then $\eta_t$: $0.89655 \to 2407.4 \to 823.5 \to 700.0\ \mathrm{kJ/kg}$.
- Pump $v\,\Delta p$; nozzle $\Delta h = -\Delta ke$.