Efficiency of Rankine cycle

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Derive an expression for efficiency of the Rankine cycle? How this cycle is different from the Carnot cycle?

The Rankine cycle is the ideal cycle for vapor power plants. The ideal Rankine cycle does not have any internal irreversibility.

Thermodynamic analysis of cycle :

Process (4‐1): Constant pressure heat addition in a boiler.

Process (1‐2): Isentropic expansion in a turbine.

Process (2‐3): Constant pressure heat rejection in a condenser.

Process (3‐4): Isentropic compression in the pump.

Referring T‐s diagram figure (b).

h₁  represents the enthalpy of steam at turbine inlet in kJ/kg.

h₂  represents the enthalpy of steam at condenser inlet in kJ/kg.

h₃  represents the enthalpy of steam at pump inlet in kJ/kg.

h₄  represents the enthalpy of steam at boiler inlet in kJ/kg.

W_T  is work output from a turbine in kJ/kg.

W_P  is work input to the pump in kJ/kg.

^Q a d d is heat added to the boiler in kJ/kg.

^Q r e j is heat rejected from the condenser in kJ/kg.

Fig.(a) Block diagram of a simple Rankine cycle

Fig.(b) T‐s diagram of simple Rankine cycle.

Heat added in the cycle :

The SFEE (steady flow energy equation) for the boiler,

h₄ + Q_add  = h₁Q_add  = h₁ – h₄

Work-done by turbine :
The SFEE for the turbine,
h₁ = W_T + h₂

W_T  = h₁ – h₂

Heat rejected to condenser :

The SFEE for the condenser,

h₂  = Q _rej + h₃

Q _rej = h₂ – h₃

Work done on pump :

The SFEE for pump,

h₃ + W _P  = h₄

W _P = h₄ – h₃

The thermal efficiency of Rankine cycle is given as:

The pump handles water which is incompressible, i.e. density or specific volume undergoes little change with an increase in pressure. For reversible adiabatic compression, the use of general property relation :

Tds = dh – vdp                       [ ds = 0 ]

dh = vdp

Since change in specific volume is negligible hence assuming it as constant.

So Dh = v Dp.

h₄ – h₃ = v₃ ( p ₄ – p₃ )

W _p  = h₄ – h₃  = v ₃ ( p₁ – p₂ )

Usually, pump work is very small compared to the turbine output and is sometimes neglected. Then h₄ – h₃ , and the cycle efficiency approximately becomes

cycle efficiency ^{h=  h}1 ^{– h}2 /h₁ – h₄