## T-s Diagram Of Regenerative Rankine Cycle

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or h4. = h3 - ^turbine(h3 - h4) = 3151 - 0.8(3151 - 2787) = 2860 kJ/kg.

At point 5, P5 = 13 bar and T5 = 390°C. From superheated steam tables, h5 = 3238 kJ/kg and s5 = 7.212 kJ/kg-K.

At point 6, s6 = s5 = 7.212 kJ/kg-K. From the problem definition, P6 = 0.16 bar. From steam tables, s6f = 0.772 kJ/kg-K and s6g = 7.985 kJ/kg-K. Therefore, at this point, we have a wet vapor and its dryness fraction is s - sf _ 7.212 - G.772 7.9S5 - G.772

G.S93

At a pressure of 0.16 bar, hf = 232 kJ/kg and hfg = 2369 kJ/kg; therefore, h6 = hf + xhfg = 232 + 0.893 X 2369 = 2348 kJ/kg.

To find h6', we need to use Eq. (10.1) for turbine efficiency:

h6, = h5 - nturbine(h5 - h6) = 3238 - 0.8(3238 - 2348) = 2526 kJ/kg

At point 1, the pressure is also 0.16 bar. Therefore, from steam tables at saturated liquid state, we have v1 = 0.001015 m3/kg and h1 = 232 kJ/kg. From Eq. (10.5), h2'

n pump

7.592 kJ/kg

Therefore, hr = 232 + 7.592 = 239.6 kJ/kg. Finally, the cycle efficiency is given by Eq. (10.7):

(h - h4.) + h - h6.) - (hy - ht) (h3 - ft2.) + (h5 - h4 ) (3151 - 2S6G) + (323S - 2526) - (239.6 - 232) (3151 - 239.6) + (323S - 2S6G)

The efficiency of the simple Rankine cycle is much less than the Carnot efficiency, because some of the heat supplied is transferred while the temperature of the working fluid varies from T3 to Tj. If some means could be found to transfer this heat reversibly from the working fluid in another part of the cycle, then all the heat supplied from an external source would be transferred at the upper temperature and efficiencies close to the Carnot cycle efficiency could be achieved. The cycle where this technique is used is called a regenerative cycle.

In a regenerative cycle, expended steam is extracted at various points in the turbine and mixed with the condensed water to pre-heat it in the feedwater heaters. This process, with just one bleed point, is shown in Figure 10.11, in which the total steam flow rate is expanded to an intermediate point 6, where a fraction, f, is bled off and taken to a feedwater heater; the remaining (1 - f) is expanded to the condenser pressure and leaves the turbine at point 7. After condensation to state 1, the (1 - f) kg of water is compressed in the first feed pump to the bleeding pressure, P6. It is then mixed in the feedwater heater with f kg of bled steam in state 6 and the total flow rate of the mixture leaves the heater in state 3 and is pumped to the boiler, 4.

Although one feedwater heater is shown in Figure 10.11, in practice, a number of them can be used; the exact number depends on the steam conditions. Because this is associated with additional cost, however, the number of heaters and the proper choice of bleed pressures is a matter of lengthy optimization calculations. It should be noted that if x number of heaters are used, x + 1 number of feed pumps are required.

Heater

 Boiler 5 1 * A) Pump Heater Condenser Turbine FIGURE 10.11 Rankine power plant cycle with regeneration. (a) Regenerative Rankine cycle schematic. (b) T-s diagram. FIGURE 10.11 Rankine power plant cycle with regeneration. (a) Regenerative Rankine cycle schematic. (b) T-s diagram. The reheat cycle efficiency is given by = h - h6) + (1 - f )(h - hy) - (1 - f)(h - hi) - h - h) where f = fraction of steam in the turbine bled at state 6 to mix with the feedwater. In this cycle, the enthalpy at state 3 can be found by an energy balance as from which m h3 = fmh6 + (1 - f )rhh2 Example 1G.2 In a regenerative cycle, steam leaves the boiler to enter a turbine at a pressure of 60 bar and a temperature of 500°C. In the turbine, it expands to 5 bar, then a part of this steam is extracted to pre-heat the feedwater in a heater that produces saturated liquid, also at 5 bar. The rest of the steam is further expanded in the turbine to a pressure of 0.2 bar. Assuming a pump and turbine efficiency of 100%, determine the fraction of steam used in the feedwater heater and the cycle efficiency. Solution At point 5, P5 = 60 bar and T5 = 500°C. From superheated steam tables, s5 = 6.879 kJ/kg-K and h5 = 3421 kJ/kg. At point 6, s6 = s5 = 6.879 kJ/kg-K and P6 = 5 bar. Again from superheated steam tables by interpolation, h6 = 2775 kJ/kg. At point 7, P7 = 0.2 bar and s7 is also equal to s5 = 6.879 kJ/kg-K. At this pressure, Sf = 0.832 kJ/kg-K and sg = 7.907 kJ/kg-K. Therefore, the dryness fraction is s - sf _ 6.879 - G.832 7.9G7 - G.832 G.855 s fg At the same pressure, hf = 251 kJ/kg and hfg = 2358 kJ/kg. Therefore, h7 = hf + xhfg = 251 + 0.855 X 2358 = 2267 kJ/kg. At point 1, the pressure is 0.2 bar, and because we have saturated liquid, hi = hf = 251 kJ/kg and vi = 0.001017 m3/kg. At point 2, P2 = 5 bar and as h2 - h1 = y1(P2 - P1) h2 = 251 + 0.001017 0.2) X 102 = 251.5 kJ/kg. At point 3, P3 = 5 bar. From the problem definition, the water at this point is a saturated liquid. So, v3 = 0.001093 m3/kg and h3 = 640 kJ/kg. Using Eq. (10.9), h3 = fh6 + (1 - f)h2, or f ha, - h2 _ 64G - 251.5 G.154 At point 4, P4 = 60 bar. Therefore, h4 - h3 = v3(P4 - P3) or h4 = h3 + v3 (P4 - P3) = 640 + 0.001093(60 - 5) X 102 = 646 kJ/kg. Finally, the cycle efficiency is obtained from Eq. (10.8): = (h - h6) + (1 - f)(h6 - h,) - (1 - f )(h2 - h) - (h - ft,) (h5 - h4) = (3421 - 2775) + (1 - 0.154)(2775 - 2267) - (1 - 0.154)(251.5 - 251) - (646 - 640)

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### Responses

• Bisrat
How to find h6 on a steam table?
8 years ago