The Rankine cycle: spending the limit
Введено| English | Значение |
|---|---|
| Rankine cycle | The power plant: pump to boiler to turbine to condenser , and back. |
| boiler | boiler |
| condenser | condenser |
| superheat | superheat |
| exhaust quality | exhaust quality |
| reheat | reheat |
Four components, two dome crossings
- The Rankine cycle 朗肯循环 is the power plant: pump to boiler 锅炉 to turbine to condenser 冷凝器, and back. The working fluid changes phase twice per lap, which is exactly why the cycle hugs the saturation dome on both diagrams.
- Heat enters at the high-pressure isobar, work leaves at the isentropic expansion, heat leaves at the condenser pressure, and the pump pays a tiny $v\,\Delta p$ to close the loop - liquid $v$ is small, so the pump is nearly free.

The heat-in term of the Rankine cycle starts from:
Сложность 1/5 · 100 монет + небольшой бонус за первое завершение сегодня
Heat begins after the pump raises the pressure to the boiler level.
The ideal baseline
- Boiler at $4\ \mathrm{MPa}$, saturated vapour ($h_3 = 2801.4$, $s_3 = 6.0700\ \mathrm{kJ/(kg\,K)}$); condenser at $10\ \mathrm{kPa}$. The turbine leg: $s_4 = s_3$ gives $x_4 = (6.0700 - 0.6492)/7.5009 = \mathbf{0.7227}$, $h_4 = 191.81 + 0.7227 \times 2392.8 = \mathbf{1921.1}\ \mathrm{kJ/kg}$.
- The pump leg: $h_2 = 191.81 + 0.001010 \times 3990 = \mathbf{195.8}$. So work out $2801.4 - 1921.1 = \mathbf{880.3}$, work in $\mathbf{4.0}$, heat in $2801.4 - 195.8 = \mathbf{2605.6}$, and $\eta = 876.3/2605.6 = \mathbf{33.6\%}$ - every number one dome walk or one $v\,\Delta p$.
Which component owns the step?
Route each cycle quantity to its component.
Baseline turbine exit quality at 10 kPa (four decimals)?
Сложность 2/5 · 140 монет + небольшой бонус за первое завершение сегодня
(6.0700 - 0.6492)/7.5009.
Baseline turbine exit enthalpy (one decimal, kJ/kg)?
Сложность 2/5 · 140 монет + небольшой бонус за первое завершение сегодня
191.81 + 0.7227 x 2392.8.
Turbine work (one decimal, kJ/kg)?
Сложность 2/5 · 140 монет + небольшой бонус за первое завершение сегодня
2801.4 - 1921.1.
Heat in per kilogram (one decimal, kJ/kg)?
Сложность 2/5 · 140 монет + небольшой бонус за первое завершение сегодня
2801.4 - 195.8.
Baseline thermal efficiency (one decimal, percent)?
Сложность 2/5 · 140 монет + небольшой бонус за первое завершение сегодня
(880.3 - 4.0)/2605.6.
Two upgrades, priced
- Superheat 过热 to $4\ \mathrm{MPa}$, $480^\circ\mathrm{C}$ ($h_3 = 3398.3$, $s_3 = 6.7677$): $x_4 = \mathbf{0.8157}$, $h_4 = \mathbf{2143.6}$, and $\eta$ rises to $\mathbf{39.1\%}$ - the exhaust quality 排气干度 climbs from 0.72 toward 0.82, still wet.
- Reheat 再热 expands from an intermediate $800\ \mathrm{kPa}$ back up to $480^\circ\mathrm{C}$ before the final expansion: work rises again and the exhaust crosses above 0.9, clearing the 0.88 blade-erosion line. Raising boiler pressure does the same to efficiency - the same cycle at $1$ versus $4\ \mathrm{MPa}$ moves $\eta$ from $25.3\%$ to $32.3\%$ - because the average heat-in temperature rises while the pump cost barely moves.
Efficiency with 480 C superheat (one decimal, percent)?
Сложность 2/5 · 140 монет + небольшой бонус за первое завершение сегодня
x4 = 0.8157, h4 = 2143.6.
A thermodynamically fine cycle can still ruin turbine blades through wet exhaust.
Сложность 1/5 · 100 монет + небольшой бонус за первое завершение сегодня
x4 below about 0.88 erodes blades - superheat and reheat exist for this.
Three cycle slips
- Subtracting $h_1$ instead of $h_2$ in the heat-in term. Heat starts at the pump outlet.
- Setting $s_2 = s_1$ across the pump. Use $h_2 = h_1 + v\,\Delta p$; the pump is not the turbine.
- Ignoring exhaust quality. A thermodynamically fine cycle can still ruin blades.
Recap
- Pump, boiler, turbine, condenser; isentropic expansion, liquid pump.
- Baseline: $x_4 = 0.7227$, $\eta = 33.6\%$; superheat: $39.1\%$ with $x_4 = 0.8157$; reheat keeps the exhaust dry.
- Boiler pressure raises the average heat-in temperature - efficiency climbs, metallurgy caps it.