LESSON - 31 POWER CYCLE, VAPOUR POWER CYCLES, CARNOT VAPOUR CYCLE, RANKINE CYCLE

 31.1 POWER CYCLE

In previous lessons we have discussed that the heat engines operate in a cycle and produce net work output. The cycles on which these heat engines operate to produce useful power is called power cycles.

The power cycles are categorised as vapor power cycle and gas power cycle, depending upon the working fluids used in the cycle. In vapour power cycle, the working fluid exists in vapor phase during one part of the cycle and in liquid phase during another part whereas in a gas power cycle, the working fluid remains in the gaseous phase throughout the entire cycle.

In this lesson, we will restrict our study to the analysis of heat engines which operate on vapour power cycle.

31.2. VAPOR POWER CYCLE

Steam power plant as shown in Fig. 31.1 is the example of heat engine that operates on the vapour power cycle.

In a steam power plant, which we have also considered in chapter 12, the following processes take place:

1-2: The heat energy released by combustion of fuel in furnace is utilized to vaporize water into steam in a boiler.

2-3: The steam produced in the boiler is expanded in a steam turbine or steam engine to obtain useful work.

3-4: The vapor leaving the steam turbine or steam engine is condensed in a condenser.

4-1: The condensed steam is pumped back into the boiler to its initial state constituting a cycle.

Thus the steam power plant operates on vapor power cycle in which working fluid is alternatively vaporized and condensed. Such steam power plant is sometimes said to operate on the closed vapor power cycle as working fluid finally returns to the initial state after undergoing a series of processes.

 

Fig. 31.1. Schematic diagram of a steam power plant

31.3. WHY WE NEED IDEAL CYCLE IN THE ANALYSIS OF VAPOR POWER CYCLE

To study the performance of steam power plants it is important to analysis actual vapor power cycle encountered in steam power plants. However, it is difficult to analyse the actual vapor power cycle because the actual cycle consists of irreversible and non-quasi-equilibrium processes whereas we can apply thermodynamics principles only to reversible and quasi-equilibrium processes. So, to make an analytical study of an actual power cycle feasible, it is advantageous to device an idealized vapor power cycle which has reversible and quasi-equilibrium processes along with general characteristics of the actual vapor power cycle and then analyzes the performance of this idealized cycle.

31.4. CARNOT VAPOR CYCLE AS THE IDEAL CYCLE FOR STEAM POWER PLANT

We have mentioned in lesson 14 that the steam power plant operating on Carnot cycle is the most efficient cycle operating between two specified temperature limits. Thus it is natural to look at the Carnot cycle first as a prospective ideal cycle for vapor power plant.

Figure 31.2 shows a steam turbine power plant operating on Carnot cycle. The Carnot vapor cycle is represented on p-v and T-s diagrams as shown in Figure 31.3.

Fig. 31.2. A Steam Turbine Power Plant Operating on Carnot Cycle

 

Fig. 31.3. Carnot vapour cycle on p-v and T-s diagram

Consider 1 kg of working fluid for analysis:-

Process 4-1: Constant pressure, constant temperature heat addition in boiler (boiling) 

In this process, the saturated liquid at state ‘4’ (Saturated liquid) is converted into dry and saturated steam in the boiler by adding heat ‘q_A’ from heat source at the temperature T_H.

Heat added is given by

      q_A = ₄q₁ = h_g,1 –h_f,4  since ₄w₁  = 0 (from 1^st law of thermodynamics for flow process)

Process 1-2: Isentropic expansion of steam in turbine (steam turbine work, w_t )

In this process, dry saturated steam at temperature T₁ (=T_H) enters the steam turbine at state ‘1’ (dry saturated vapor)and expands isentropically to state ‘2’(wet steam) at sink temperature T₂ (=T_L)thus doing the useful work.

     ₁q₂ = 0 (as the process is adiabatic)

Steam turbine work is given by,

      w_t = ₁w₂ = h_g,1 – h₂                         (from 1^st law of thermodynamics for flow process)

Process 2-3: Constant pressure, constant temperature heat rejection in condenser (Condensation)

In this process, the exhaust steam from the turbine at state ‘2’ (wet steam) is condensed and cooled to state ‘3’ (wet steam)in the condenser by rejecting heat ‘q_R’ to heat sink at the temperature T_L.

Heat rejected is given by

        q_R = ₂q₃  = h₂ – h₃      since ₂w₃  = 0     (from 1^st law of thermodynamics for flow process)

Process 3-4: Isentropic compression of steam in compressor (Pump work, w_p)

In this process, the water and vapor from condenser at state ‘3’ (wet steam) are pumped by feed pump to state ‘4’(saturated liquid).

        ₃q₄ = 0 (as the process is adiabatic)

Pump work (w_p) is given by,

         w_p = ₃w₄ = h_f,4 – h₃   (from 1^st law of thermodynamics for flow process)

Net work done during cycle, w_net   = Heat added (q_A) – heat rejected (q_R)

                                                           = (h_g,1 –h_f,4) – (h₂ – h₃)       

 or

= Turbine work (w_t) – Pump work (w_p)

= (h_g,1 – h₂) – (h_f,4 – h₃)= (h_g,1 –h_f,4) – (h₂ – h₃)  

Thermal efficiency   

Even though we know that the steam power plant operating on Carnot cycle is more efficient but it is not possible in practice to operate power vapor cycles on Carnot cycle because of the problem of pumping of wet steam at state `3’ and deliver it as saturated water only at state `4’.

Even with the impossibility, the Carnot cycle, however, permits the designer to arrive at the maximum possible efficiency that can be obtained in simple vapor power plant under given conditions.

31.5. RANKINE CYCLE AS THE IDEAL CYCLE FOR STEAM POWER PLANT

We have seen that Carnot cycle is not the theoretical/ideal cycle for steam turbine power plant because of the difficulty of pumping a mixture of water and steam and delivering it as saturated water only. However, this difficulty is eliminated in Rankin cycle by complete condensation of water vapor in the condenser, and then, pumping the water isentropically to boiler pressure. Rankine cycle is a theoretical/ideal cycle for comparing the performance of steam power plants.

Fig. 31.4 shows a steam turbine power plant working on Rankine cycle. The Rankin cycle is represented on p-v and T-s diagrams as shown in Figure 31.5. 

 

Fig. 31.4: Rankine cycle steam turbine power plant

 

 

 

 

 

 

 

 

 Fig. 31.5. Rankine cycle on p-v and T-s diagrams

(i) Boiler: In boiler the working fluid (water) at state ‘4’in sub cooled condition is converted into dry saturated steam at state ‘1’ by receiving heat ‘q_A’ from combustion of fuel (high temperature heat source) through the following processes.

Process 4-5: As thewater enters the boiler from pump in sub cooled condition state ‘4’ at pressure P_H, it is first heated up to the saturated state 5 at constant pressure (sensible heating). 

Process 5-1: Then water at saturated condition 5 is further heated up at constant pressure P_H and constant saturation temperature to the saturated steam at state 1 (latent heat of vaporization).

Process 4-5-1: Total heat addition in boiler,

              q_A= ₄q₁= h_g,1 –h_sub,4                       since ₄w₁  = 0                                           (from 1^st law of thermodynamics for flow process)

(ii) Steam Turbine: In the steam turbine, the dry saturated steam from the boiler at state ‘1’ at pressure ‘p_H’ expands isentropically to state ‘2’ (wet steam) at pressure ‘p_L’ and thus produce mechanical work, w_t.

Process 1-2: Isentropic expansion of steam in turbine (Steam turbine work, w_t)

₁q₂ =  0                                            (as the process is adiabatic)

Steam turbine work is given by,

       w_t = ₁w₂ =  h_g,1 – h₂                 (from 1^st law of thermodynamics for flow process)

(iii) Condenser: In the condenser, the exhaust wet steam from turbine at state ‘2’ is condensed by rejecting heat ‘q_R’ to the cooling water.

Process 2-3: Constant pressure (back pressure), constant temperature heat rejection in condenser (Condensation)

Total heat rejected in condenser,  

                          q_R = ₂q₃ =  h₂ – h_f,₃                          since  ₂w₃ = 0                                                 (from 1^st law of thermodynamics for flow process)

(iv) Feed pump: The feed pump is used to pump the condensate at state ‘3’ (saturated water) from the hot-well to the boiler at the boiler pressure, p_H.

Process 3-4: Isentropic compression of water in pump (Pump work, wp)

          ₃q₄ = 0                                              (as the process is adiabatic)

Pump work (w_p) is given by,

           w_p = ₃w₄ = h_sub,4 – h_f,3                  (from 1^st law of thermodynamics for flow process)

Pump work, w_p  is also given by

           w_p = v_f,3 (p_{4 –} p₃)

Net work done during cycle, w_net = Heat added (q_A) – heat rejected (q_R)

    = (h_g,1 –h_sub,4) – (h₂ – h_f,3)  

                         or

    = Turbine work (w_t) – Pump work (w_p)

    = (h_g,1 – h₂) – (h_sub,4 – h_f,3) = (h_g,1 –h_sub,4) – (h₂ – h_f,3)

Thermal efficiency, 

                                                                                                or           

Rankine cycle efficiency of a good steam power plant may be in the range of 35% to 45%.

Rankine cycle neglecting pump work 

In a Rankine cycle the pump work may be neglected as it is very small compared with other energy transfers. Hence we have

           w_p= 0    

As shown in Figure 31.6 neglecting the pump work (p₄'-p₃')v_f,3' denoted by area 44'3'3 in the p-v diagram for the Rankine cycle will be presented by 1234.

 

Fig. 31.6. Rankine cycle on p-v and T-s diagrams (neglecting pump work)

Total heat addition in boiler in process 3-1 is given by,  

q_A = ₃q₁  = h_g,1 –h_f,3 since ₃w₁  = 0                       (from 1^st law of thermodynamics for flow process)

Total heat rejected in condenser,  

q_R = ₂q₃ =  h₂ – h_f,₃        since  ₂w₃ = 0                 (from 1^st law of thermodynamics for flow process)

Net work done during cycle, w_net = Heat added (q_A) – heat rejected (q_R)

    = (h_g,1 –h_f,3) - (h₂ – h_f,3) 

                                                 =  (h_g,1 –h₂)

Therefore, Thermal efficiency