The Carnot Vapour Cycle

The Carnot Vapour Cycle

It has been proved earlier that Carnot cycle is the most efficient cycle operating between the specified temperature limits.

Therefore, first we shall consider the Carnot cycle as the prospective ideal cycle However, subsequently it will be shown that Carnot vapour cycle is not suitable for practical development of steam power plant due to inherent difficulties.

Schematic diagram of a Carnot vapour power cycle is shown in Figure A (a) and on (T-S) diagram in Figure (b).

A Carnot cycle consists of two isothermals and two adiabatics.

Various processes of the Carnot cycle (1-2-3-4) as applied to vapour power plant are as follows : 

(1) Process (1-2) :

Wet vapour at pressure P_b and dryness X₁ is compressed in vapour compressor isentropically (reversible adiabatically upto boiler pressure, P₁ such that it becomes a saturated liquid). The work supplied is w_p and temperature of steam increases from T₂ to T₁.

(II) Process (2-3) :

Saturated liquid enters the boiler where heat q₁is absorbed isothermally at temperature T₁ steam becomes dry-saturated represented by state 3. Steam being in vapour state, its temperature, T₁ and pressure, P₁ remain constant.

(III) Process (3-4) :

Dry-saturated steam at P_I enters the steam turbine where it expands isentropically upto condenser pressure P_b and produces work w_T. Temperature of steam drops from T₁ to T₂

(lV) Process (4-1) :

The wet steam then enters the condenser where it rejects heat q_r isothermally (also a constant pressure) to the cooling water circulated in the condenser. The steam returns to original state-1. Thus, it completes a thermodynamic cycle.

Analysis of Carnot vapour cycle :

Assume 1 kg of steam flow through the cycle. We can apply the staedy flow of energy equation (S. F. E. E.) to various process for until mass flow as follow :

q – w = [ (△h) + △(K.E.) + (P.E.) ]  …(1)     q₁

If we neglect the changes in K.E. and P.E.’s the equation (1) reduced to :

q – w = △h  …(2)

(a) Heat supplied, q₁ to steam in bolier in process (2-3) :

Workdone, w = 0. Therefore, from Equation (2)

q₁ = (h₃ – h₂)   ….(i)

(b) Turbine Work, w_T in process (3-4) :

Since the expansion process is turbine is reversible adiabatic process, therefore, the heat transfer is zero. From equation (3), we get 

0 – w_T  = (h₄ – h₃); w_T  = (h₃ – (h₄)  …(ii)

(c) Heat rejected, q_r to cooling water by steam in condenser during the process (4-1) : 

Again, w = 0 , therefore from Equation (2) we get, 

q_r = (h₁ –  h₄);  q_r = (h₄ – h₁) …(iii) 

Negative sign indicates that the heat is transferred from steam to the cooling water by an amount equal to (h₄ – h₁). 

(d) Pump work, w_P in the process (1-2) : 

q = 0, since the process is adiabatic, from Equation (2), we get, 

0 – w_P = ( h2 – h₁) ;  w_P = – (h₂ –  h₁) … (iv) 

Negative sign indicates that the work is supplied to the pump from the surroundings.

(e) Shaft work : 

Net work output (shaft work), w_S is the difference of turbine and pump work. Therefore, 

Shaft work, w_s = w_T – w_P

or, w_s = (h₃ –  h₄) – (h₂ –  h₁) …(3) 

(f) Thermal efficiency : 

The thermal efficiency of the power and is defined as the ratio of shaft work output to the heat supplied. Therefore,

Thermal efficiency, η₁ = Shaft work (w_S) / heat supplied (q_i ) …(4) 

or η₁ = (h₃ –  h₄) – (h₂ –  h₁) / (h₂ –  h₁) … (5) 

(g) Steam rate : 

Steam rate is defined as the amount of steam required to develop 1 kW – hr of work.

Therefore, 

Steam rate (S.R.) = 3600 / w_s  ….(6)

Units of S.R.is kg / kW – hr. 

(h) Work ratio w_r :

Out of turbine work, W_T the pump (vapour compressor) is supplied the work W_T the shaft work, w_s  =W_T – W_P.

The ratio of net work transfer (shaft work, w_s) to the positive work transfers i.e. the turbine work wr is called work ratio Shaft work w , 

Therefore, Work ratio, w_r  = Shaft work w_s / Turbine work , w_r… (7) 

 w_r = (h₃ –  h₄) – (h₂ –  h₁) / (h₃ –  h₄) 

Importance of work ratio : 

A large work ratio means a small negative work transfer and small work ratio means a large negative work transfer. Therefore, on comparing two cycles with the same ideal efficiency the cycle with smaller work ratio would have less actual efficiency since the cycle with high work ratio, has the losses i.e. irreversibility mainly in positive work transfer while with smaller work ratio the irreversibility will be both in positive and negative work transfers.

(i) Carnot efficiency : 

Carnot efficiency, η_C = 1 – (T₂ – T₁) …(9) 

It also equals to thermal efficiency of Carnot cycle given by the Equation (5)