An Organic Rankine Cycle (ORC) with R-410A has the boiler at 3 MPa superheating to 180°C, and the condenser operates at 800 kPa.
Find all four energy transfers and the cycle efficiency.

An Organic Rankine Cycle (ORC) with R-410A has the boiler at 3 MPa superheating to 180°C, and the condenser operates at 800 kPa.

Find all four energy transfers and the cycle efficiency.

The question is solved for steam please plug in the properties for R-410A using the same procedure as below to solve for R-410A

The answers are as follows

(i) At the boiler

Q₁ = 40.73 kJ·kg⁻¹

(ii) At the turbine

 = 2.908 kJ·kg⁻¹

(iii) At the condenser

Q₂   = 40.272 kJ·kg⁻¹

(iv)  At the feed pump

   = 2.453 kJ·kg⁻¹

Cycle efficiency is 4.4 %

Explanation:

Boiler Pressure p₁ = 3 MPa @ 180 °C

Condenser pressure p₂ = 800 kPa

From the thermodynamics tables we have for steam  

At 3 MPa (p₁), 180 °C: h₁ = 764.198 kJ·kg⁻¹

and                              s₁ = 2.1368 kJ· kg⁻¹· K⁻¹

At 800 kPa (p₂)           h₃ = = 721.018 kJ·kg⁻¹

                                    s₃ =   = 2.0460 kJ· kg⁻¹· K⁻¹

                           = 2047.28 kJ·kg⁻¹       = 6.6615 kJ· kg⁻¹· K⁻¹

                         = 0.00111479 m³·kg⁻¹       = 4.6156 kJ· kg⁻¹· K⁻¹

                       Since s₁ = s₂

                                    2.1368 =   + x₂·  = 2.0460 + x₂·4.6156

From where x₂ = 0.01967

Therefore h₂=  + x₂· = 721.018+0.01967×2047.28 = 761.29 kJ·kg⁻¹

                             h₂ = 761.29 kJ·kg⁻¹

                                =  0.00111479 m³·kg⁻¹×(3000-800) = 2.453 kJ·kg⁻¹

and = 2.453 + 721.018  = 723.471 kJ·kg⁻¹

Therefore for 1 kg of fluid we have

Through the implementation of of the steady flow energy equation to the

i) Boiler

ii) Turbine

iii) Condenser

iv) Pump

(i) For the boiler, we have

                + Q₁ = h₁

                        Q₁ = h₁ -   = 764.198 -723.471  = 40.73 kJ·kg⁻¹

(ii) For turbine

                          h₁ =    Where = Work done by turbine

                             = h₁ -h₂ = 764.198 - 761.29 = 2.908 kJ·kg⁻¹

(iii) For the condenser we have

                               h₂ = Q₂ +   where    =   =  721.018 kJ·kg⁻¹

                          Q₂  = h₂ -   = 761.29 - 721.018 = 40.272 kJ·kg⁻¹

(iv)  The feed pump gives

                              +   =   Where,   = Work done by pump

                                         =   -   = 723.471 - 721.018 = 2.453 kJ·kg⁻¹

Cycle efficiency  is given by

=  4.4 %

Efficiency of the Rankine cycle is =

               = 0.0112 or 1.12 %

the clapeyron equation (also called the clausius-clapeyron equation) relates the slope of a reaction line on a phase diagram to fundamental thermodynamic properties.

explanation:

answer:

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