REGENERATION

In the Rankine cycle and reheat, the condensate which is at a fairly low temperature is pumped to the boiler. Thus, there is an irreversible mixing of the cold condensate with hot boiler water. This results in loss of cycle efficiency. Regeneration is a method to heat the feed water from the hot well of the condenser reversibly by interchange of heat within the system, thus improving the cycle efficiency. The cycle is called regenerative cycle.

Figure 4.13 Regenerative heating

In Fig. 4.13, steam is drawn from the boiler at state point 1 and passes through the turbine and goes to the condenser at state point 2. If no heat is transferred from this steam to the surrounding, which include the casing, and the expansion is isentropic. This process is shown by line 1−2′ in the T-s diagram (Fig. 4.14). However, in the regenerative cycle, it is assumed that the condensate after being pumped to the boiler pressure at state point 4 passed through the hollow casing surrounding the turbine rotor to the boiler. Therefore, the steam losses heat to the surrounding water which gets heated. The temperature of steam entering the turbine and temperature of water leaving the turbine are same.

Let T₁ = saturation temperature of steam at the commencement of expansion at state point 1.

T₃ = temperature of saturated water condensate entering the hollow casing of turbine at state point 3. Thus, 1 kg of water is gradually heated along the path 3−4 and 1 kg of steam gradually loses the same amount of heat during expansion process 1−2. Such a heat exchange within the system is called regenerative heating. Therefore, the net heat supplied to the system is in the boiler during process 4−1 and the net heat rejected from the system is in the condenser during process 2−3 (Fig. 4.14).

Heat added, q_a = T₁ (s₁ − s₄)

Heat rejected, q_r = T₂ (s₂ − s₃)

Net work done,

w_net = q_a − q_r

= T₁ (s₁ − s₄) − T₂ (s₂ − s₃)

But s₁ − s₄ = s₂ − s₃

∴ w_net = (T₁ − T₂) (s₁ − s₄)

Figure 4.14 T-s diagram for regenerative Rankine cycle

 

This method of regeneration is not practicable because the steam becomes extremely wet at the later stages of expansion.

1 Regenerative Cycle with Open Heaters

The regeneration effect can be achieved by bleeding or extracting small quantities of steam at different points during the expansion and exploiting the energy of bled steam rather than whole steam. Thus, that part of the steam which continues to expand and to do work, does not condense excessively. A regenerative cycle with three bleeding points and therefore, three open type heaters, where feed water and steam mix, is shown in Fig. 4.15. The corresponding T-s diagram is shown in Fig. 4.16.

Let 1 kg of steam enter the turbine at state point 1 and expand to the state point 2 where m₁ kg of steam is extracted from the turbine and passed on to the heater 3. The remainder (1 − m₁) kg of steam continues to expand isentropically to state points 3 where m₂ kg of steam is extracted and passed on the heater 2. The remainder (1 − m₁ − m₂) kg of steam further continues to expand isentropically to state point 4 where m₃ kg of steam is bled and passed on to heater 1. Thus, only (1 − m₁ − m₂ − m₃) kgof steam does the remaining isentropic expansion to point 5 before entering the condenser, where it is condensed to point 6. The condensate (1 − m₁ − m₂ − m₃) kg at enthalpy b₆ is mixed in heater 1. m₃ kg of steam bled at state point 4 condenses to state point 7 and (1 − m₁ − m₂ −m₃) kg of water gets heated from b₆ to point 7. (1 − m₁ − m₂) kg of feed water at state point 7 is pumped to heater 2 at state point by. In the heater (1 − m₁ − m₂) kg of feed water is heated from state point 7 to state point 8 and m₂ kg of bled steam from point 3 condenses to point 8. Thus, heater 2 has (1 − m₁) kg of feed water at point 8 which is pumped to heater 3 raising the enthalpy of pumped feed water to b₈. This feed water in heater 3 gets heated to point 9 by condensation of m₁ kg of steam bled from point 2, leaving 1kg of feed water in heater 3 to be pumped to boiler pressure by raising its enthalpy to b₉. Heating of 1 kg of feed water from b₉ to point 10 is done in the boiler where it is further heated to steam at point 1.

Figure 4.15 Regenerative feed water heating with open heaters

Figure 4.16 T-s diagram for feed heating with open heaters

For simplified analysis, points 6 and b₆, 7 and b₇, 8, and b₈, 9 and b₉ are treated as coincident points, thus ignoring the very small quantity of pump work. Thus, we have

Heat gained by feed water = Heat lost by bled steam

Heater 1: (1 − m₁ − m₂ − m₃) (h₇ − h₆) = m₃ (h₄ − h₇)

Heater 2: (1 − m₁ − m₂) (h₈ − h₇) = m₂ (h₃ − h₈)

Heater 3: (1 − m₁) (h₉ − h₈) = m₁ (h₂ − h₉)

Net work output, w_out = (h₁ − h₂) + (1 − m₁) (h₂ − h₃) + (1 − m₁ − m₂) (h₃ − h₄) + (1 − m₁ − m₂ − m₃)(h₄ − h₅)

w_net = w_out − ∑w_p

Heat added, q_a = h₁ − h_b9 = h₁ − h₉ − w_p4

Now ∑w_p = (1 − m₁ − m₂ − m₃) (h_b6 − h₆) + (1 − m₁ − m₂) (h_b7 − h₇) + (1 − m₁) (h_b8 − h₈) + (h_b9 − h₉)

= v_f6(p₁ − p₆)

where p₁ = throttle pressure, p₆ = condenser pressure and v_f6 = specific volume of water at condenser pressure.

2 Regenerative Cycle with Closed Heaters

In this type of heater, the feed water is contained in the tubes about which the bled steam passes and condenses. One pump can be used for two or more heaters placed in series. The heaters can be placed in various combinations. One such arrangement is shown in Fig. 4.17. The corresponding T-s diagram is shown in Fig. 4.18.

Work done per kg of steam supplied to turbine,

w_t = (h₁ − h₂) + (1 − m₁) (h₂ − h₃) + (1 − m₁ − m₂) (h₃ − h₄)

Total pump work, ∑w_p = w_p1 + w_p2 + w_p3

= (1 − m₁ − m₂) (h₆ − h₅) + m₂ (h₁₀ − h₉) + m₁ (h₁₂ − h₁₁)

where h₆ − h₅ = v_f5 (p₁ − p₄) × 10² kJ/kg

h₁₀ − h₉ = v_f9 (p₁ − p₃) ×10² kJ/kg

h₁₂ − h₁₁ = v_f11 (p₁ − p₂) × 10² kJ/kg

where v_f is in m³/kg and p is in bar.

Now, v_f5 = v_f9 = v_f11 = saturated volume of water

Figure 4.17 Regenerative feed water heating with closed heaters

Figure 4.18 T-s diagram for regenerative Rankine cycle with closed heaters

Net work done, w_net = w_t − ∑w_p

Heat added in the boiler, q_a = h₁ − h_f13

Neglecting heat losses and considering heat balance at feed heaters, we have

m₂ (h₃ − h_f9) = (1 − m₁ − m₂) (h_f9 − h_f 5)

and m₁ (h₂ − h_f11) = (1 − m₁ − m₂) (h_f11 − h_f9)

Neglecting enthalpy increase due to pump,

h_f6 = h_f5

also h_f7 = h_f9

and h_f8 = h_f11

The condition of feed water h₁₃ entering the boiler is given by,

(1 − m₁ − m₂) h_f11 + m₂h_f9 + m₁h_f11 = 1 × h₁₃

or h₁₃ = (1 − m₂) h_f11 + m₂h_f 9

Example 4.7

A boiler feeds a turbine at 56 bar and 600°C. Before being passed on to the condenser at 30°C, the steam is bled for regenerative feed heating at 6.5 bar. For an ideal regenerative cycle and 1kg/s of throttle steam, determine (a) the amount of bled steam, (b) net work done, and (c) the ideal efficiency of the cycle.

Solution

The T-s diagram is shown in Fig. 4.19.

At p₁ = 56 bar, 600°C from steam tables,

h₁ = 3659.5 kJ/kg, s₁ = 7.2011 kJ/kg.K

At 6.5 bar, h₂ = 2987 kJ/kg

and t₂ = 265°C

s₃ = s₁ = s_f3 + x₃ s_gh3

At 30°C, s_f3 = 0.4369 kJ/kg.K, s_fg3 = 8.01674 kJ/kg.K

or 7.2011 = 0.436 + x₃ × 8.0164

or x₃ = 0.8438

h₃ = h_f3 + x₃h_gh3 = 125.79 + 0.8438 × 2430.5 = 2176.6 kJ/kg

At 6.5 bar, h_f6 = 684.26 kJ/kg, v_f6 = 0.001104 m³/kg

h₇ = h_f6 + v_f6 (p₁ − p₂) × 10² = 684.26 + 0.001104 (56 − 6.5) × 10²

= 684.26 + 5.47 = 689.72 kJ/kg

w_p2 = 5.47 kJ/kg

v_f4 = 0.001004 m³/kg

p₄ corresponding to 30°C = 0.042461 bar

w_p1 = v_f 4 (p₂ − p₄) = 0.001004(6.5 − 0.042461) × 10²

= 0.648 kJ/kg

∑w_p = w_p1 + w_p2 = 0.648 + 5.47 = 6.118 kJ/kg

Figure 4.19 Regenerative Rankine cycle with a single heater

1. Neglecting pump work, for heat balance of heater,m₁ (h₂ − h_f6) = (1 − m₁) (h_f7 − h_f4)
2. w_net = w_t − ∑w_p = (h₁ − h₂) + (1 − m₁) (h₂ − h₃) −∑w_p    = (3659.5 − 2987) + (1 − 0.195) (2987 − 2176.6) − 6.118    = 1318.754 kJ/kg
3. Heat added, q_a = (h₁ − h₇) = 3659.5 − 689.72 = 2969.78 kJ/kgCycle efficiency, 

Example 4.8

Determine the improvement of efficiency which would results if a single stage of regenerative feed heating were added to a steam cycle having terminal conditions of 16 bar, 300°C and 0.05 bar. The steam for feed heating is to be extracted at 1.16 bar. The drain from the heat exchanger at saturation temperature of bled steam pressure is returned to the system at a point downstream from the heat exchanger, the feed water getting heated to saturation temperature at pressure of bled steam. Neglect change in enthalpy due to liquid pump work.

Solution

The regenerative cycle and h-s diagram is shown in Fig. 4.20(a) and (b).

From the Mollier diagram,

h₁ = 3035 kJ/kg, h₂ = 2575 kJ/kg, h₃ = 2098 kJ/kg

From steam tables, h_f1 = 856.6 kJ/kg, h_f2 = 475.4 kJ/kg, h_f3 = 137.8 kJ/kg

For heat balance of heater,

m (h₂ − h_f3) = (h_f2 − h_f3)

Figure 4.20 Single stage regenerative heating Rankine cycle: (a) Schematic diagram, (b) Mollier diagram

or 

w_net = (h₁ − h₂) + (1 − m) (h₂ − h₃)

= (3035 − 2575) + (1 − 0.138) (2575 − 2098) = 871 kJ/kg

q_a = h₁ − h_f2 = 3035 − 475.4 = 2559.6 kJ/kg

Thermal efficiency with regeneration, 

Without feed water heating,

w_net = h₁ − h₃ = 3035 − 2098 = 937 kJ/kg

q_a = h₁ − h_f3 = 3035 − 137.8 = 2897.2 kJ/kg

Thermal efficiency without regeneration, 

Improvement in thermal efficiency = 

Example 4.9

A regenerative cycle with three-stage bleed heating works between 30 bar, 450°C, and 0.04 bar. The bleed temperatures are chosen at equal temperature ranges. Determine the efficiency of the cycle. Neglect the pump work.

Solution

The schematic diagram of the cycle, T-s, and h-s diagrams are shown in Fig. 4.21(a) to (c).

At p = 0.04 bar, t_s = 30°C

Temperature range = 450 − 30 = 420°C

Equal temperature difference = 

The temperatures of bled steam are: 450 − 105 = 345°C, 240°C, 135°C

From Mollier chart,

h₁ = 3340 kJ/kg, h₂ = 3150 kJ/kg, h₃ = 2935 kJ/kg,

h₄ = 2720 kJ/kg, h_s = 2120 kJ/kg

The pressure at 2, 3, 4 points are noted from the h–s chart. Then from steam tables,

h_f1 = 1008.35 kJ/kg, h_f2 = 837.45 kJ/kg, h_f3 = 684.0 kJ/kg,

h_f4 = 517.62 kJ/kg, h_f5 = 121.4 kJ/kg

For heat balance of heaters,

m₃ (h₄ − h_f4) = (1 − m₁ − m₂ − m₃) (h_f4 − h_f5)

m₂ (h₃ − h_f3) = (1− m₁ − m₂ − m₃) (h_f3 − h_f4)

m₁ (h₂ − h_f2) = (1 − m₁) (h_f2 − h_f3)

∴ m₃ (2720 − 517.62) = (1 − m₁ − m₂ − m₃) (517.62 − 121.4)

or 2202.38 m₃ = (1 − m₁ − m₂ − m₃) × 396.22

m₂ (2935 − 684) = (1 − m₁ − m₂) (684 − 517.62)

Figure 4.21 Regenerative Rankine cycle with three-stage bleeding: (a) Schematic diagram, (b) T-s diagram, (c) h-s diagram

or 2251 m₂ = (1 − m₁ − m₂) × 166.38

m₁ (3120 − 837.45) = (1 − m₁) (837.45 − 684)

or 2282.55 m₁ = (1 − m₁) × 153.45

Solving the above three equations, we get

m₁ = 0.063 kg, m₂ = 0.0645 kg, m₃ = 0.1332 kg

Net work done, w_net = 1 × (h₁ − h₂) + (1 − m₁) (h₂ − h₃) + (1 − m₁ − m₂) (h₃ − h₄) + (1 − m₁ − m₂ − m₃)(h₄ − h₅)

= (3340 − 3120) + (1 − 0.063) (3120 − 2935) + (1 − 0.063 − 0.0645) (2935 − 2720) + (1 − 0.063 − 0.0645 − 0.13327) (2720 − 2120) = 1024.6 kJ/kg

Heat added, q_a = h₁ − h_f2 = 3340 − 837.45 = 2502.55 kJ/kg

Thermal efficiency of the cycle,