Answer:
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 %
Step-by-step 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 %