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Regenerative scheme with open feed water heater

Short answer

Steam bled from the turbine heats condensate in a direct-contact open feedwater heater, raising mean heat addition temperature. The saturated mixture is pumped to boiler pressure; less heat is rejected in the condenser, improving efficiency.

List the common methods used for performance of a steam turbine power plant. Sketch layout for a regenerative scheme with an open feed water heater. Explain its working with corresponding processes on T-S diagram.

Common methods of improving performance

The efficiency of a Rankine (steam) cycle is η=1−TlowTm,in\eta = 1 - \dfrac{T_{\text{low}}}{T_{m,\text{in}}} roughly, so it improves by raising the mean temperature of heat addition TmT_{m} or lowering the temperature of heat rejection. Common methods:

  1. Increasing boiler (steam) pressure - raises the saturation temperature at which heat is added.
  2. Superheating the steam (higher turbine inlet temperature) - raises mean temperature of heat addition and dryness at exhaust.
  3. Reducing condenser pressure (better vacuum) - lowers heat rejection temperature.
  4. Reheating - steam partly expanded in HP turbine is reheated in the boiler and expanded in LP turbine.
  5. Regenerative feed water heating - steam bled from the turbine heats the feed water in feed water heaters.
  6. Using economiser and air preheater - recover heat from flue gases.
  7. Supercritical / binary vapour and combined cycles - use very high pressure, or a topping cycle (gas turbine) whose waste heat runs the steam cycle.
  8. Reducing losses: better turbine blading, insulation, reduced throttling.

Regenerative scheme with an open feed water heater

An open (direct contact) feed water heater is a mixing chamber in which bled steam and feed water mix directly; the outlet is saturated water at heater pressure.

              1
  +--------+------>+---------------+   +-----+
  | Boiler |       |    Turbine    |===| Gen |
  |  + S/H |       +---------------+   +-----+
  +--------+          | m kg (2)  | (1-m) kg (3)
      ^ 7             v           v
      |         +-----------+ +-----------+
  +--------+  6 |  Open FWH | | Condenser |
  | Pump 2 |<---| (mixing)  | +-----------+
  +--------+    +-----------+       | 4
                      ^ 5           |
                 +--------+         |
                 | Pump 1 |<--------+
                 +--------+

Working

  1. Superheated steam from the boiler enters the turbine at state 1.
  2. A fraction mm kg/kg is extracted at the intermediate pressure (state 2) and sent to the open heater.
  3. The remaining (1−m)(1-m) kg expands to condenser pressure (3) and is condensed to saturated water (4).
  4. Pump 1 raises the condensate to heater pressure (5).
  5. In the open heater the bled steam condenses and gives its latent heat to the feed water; the mixture leaves as saturated water at heater pressure (6). Energy balance: mh2+(1−m)h5=h6m h_2 + (1-m) h_5 = h_6.
  6. Pump 2 raises it to boiler pressure (7); the boiler heats it from state 7 instead of from cold condensate.

Because feed water enters the boiler hot, the mean temperature of heat addition rises and less heat is rejected in the condenser, so the efficiency increases. A separate pump is needed after each open heater; the open heater also removes dissolved gases (deaerator).

T-s diagram

  T
  ^                    1
  |                   /|
  |      f----------g  |
  |     /              |
  |    7               |
  |    6 . . . . . . . 2  bleed (m kg)
  |   /                |
  |  5                 |
  |  4-----------------3  (1-m) kg
  +-------------------------> s
  4-5 pump 1, 5 and 2 mix in FWH -> 6,
  6-7 pump 2, 7-f-g-1 boiler,
  1-2-3 turbine, 3-4 condenser
η=(h1−h2)+(1−m)(h2−h3)−[(1−m)Wp1+Wp2]h1−h7\eta = \frac{(h_1 - h_2) + (1-m)(h_2 - h_3) - [(1-m)W_{p1} + W_{p2}]}{h_1 - h_7}

✓ Answer checked by Claude Opus 5.5 (high effort)

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