Optimizing Energy Conversion: Exploring the Efficiency of Rankine Cycle

Derive an expression for efficiency of 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).

*h*_{1}* *represents the enthalpy of steam at turbine inlet in kJ/kg.

*h*_{2}* *represents the enthalpy of steam at condenser inlet in kJ/kg.

*h*_{3}* *represents the enthalpy of steam at pump inlet in kJ/kg.

*h*_{4}* *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,

*h*_{4} +* Q*_{add}* *=* h*_{1}*Q*_{add}* *=* h*_{1} –* h*_{4}

Work-done by turbine :

The SFEE for the turbine,*h*_{1} =* W** _{T}* +

*h*

_{2}

*W*_{T}* *=* h*_{1} –* h*_{2}

**Heat rejected to condenser :**

The SFEE for the condenser,

*h*_{2}* *=* Q ** _{rej}* +

*h*

_{3}

*Q ** _{rej}* =

*h*

_{2}–

*h*

_{3}

**Work done on pump :**

The SFEE for pump,

*h*_{3} +* W *_{P}* *=* h*_{4}

*W ** _{P}* =

*h*

_{4}–

*h*

_{3}

*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 D*h *=* v *D*p*.

*h*_{4} –* h*_{3} =* v*_{3} (* p *_{4} –* p*_{3} )

*W *_{p}* *=* h*_{4} –* h*_{3}* *=* v *_{3} (* p*_{1} –* p*_{2} )

##### Usually, pump work is very small compared to the turbine output and is sometimes neglected. Then *h*_{4} –* h*_{3} , and the cycle efficiency approximately becomes

**cycle efficiency** ^{h= h}**1 **^{–}^{h}**2 /***h*_{1}**–*** h*_{4}