Respuesta :
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
[tex]W_T[/tex] = 2.908 kJ·kg⁻¹
(iii) At the condenser
Q₂ = 40.272 kJ·kg⁻¹
(iv) At the feed pump
[tex]W_p[/tex] = 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 = [tex]h_{f(p_2)}[/tex] = 721.018 kJ·kg⁻¹
s₃ = [tex]s_{f(p_2)}[/tex] = 2.0460 kJ· kg⁻¹· K⁻¹
[tex]h_{fg(p_2)}[/tex] = 2047.28 kJ·kg⁻¹ [tex]s_{g(p_2)}[/tex] = 6.6615 kJ· kg⁻¹· K⁻¹
[tex]v_{f(p_2)}[/tex] = 0.00111479 m³·kg⁻¹ [tex]s_{fg(p_2)}[/tex] = 4.6156 kJ· kg⁻¹· K⁻¹
Since s₁ = s₂
2.1368 = [tex]s_{f(p_2)}[/tex] + x₂· [tex]s_{fg(p_2)}[/tex] = 2.0460 + x₂·4.6156
From where x₂ [tex]=\frac{2.1368-2.0460}{4.6156}[/tex] = 0.01967
Therefore h₂=[tex]h_{f(p_2)}[/tex] + x₂·[tex]h_{fg(p_2)}[/tex] = 721.018+0.01967×2047.28 = 761.29 kJ·kg⁻¹
h₂ = 761.29 kJ·kg⁻¹
[tex]h_{f4}-h_{f(p_2)} = v_{f(p_2)} (p_1-p_2)[/tex]
= 0.00111479 m³·kg⁻¹×(3000-800) = 2.453 kJ·kg⁻¹
and [tex]h_{f4}[/tex] = 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
[tex]h_{f4}[/tex] + Q₁ = h₁
Q₁ = h₁ - [tex]h_{f4}[/tex] = 764.198 -723.471 = 40.73 kJ·kg⁻¹
(ii) For turbine
h₁ = [tex]W_T+h_2[/tex] Where [tex]W_T[/tex] = Work done by turbine
[tex]W_T[/tex] = h₁ -h₂ = 764.198 - 761.29 = 2.908 kJ·kg⁻¹
(iii) For the condenser we have
h₂ = Q₂ + [tex]h_{f3}[/tex] where [tex]h_{f3}[/tex] = [tex]h_{f2}[/tex] = 721.018 kJ·kg⁻¹
Q₂ = h₂ - [tex]h_{f3}[/tex] = 761.29 - 721.018 = 40.272 kJ·kg⁻¹
(iv) The feed pump gives
[tex]h_{f3}[/tex] + [tex]W_p[/tex] = [tex]h_{f4}[/tex] Where, [tex]W_p[/tex] = Work done by pump
[tex]W_p[/tex] = [tex]h_{f4}[/tex] - [tex]h_{f3}[/tex] = 723.471 - 721.018 = 2.453 kJ·kg⁻¹
Cycle efficiency is given by
[tex]\frac{h_1-h_2}{h_1-h_3}[/tex] = [tex]\frac{764.198-761.29}{764.198-721.018} =0.044[/tex] 4.4 %
Efficiency of the Rankine cycle is =
[tex]\eta_{Rankine} = \frac{W_{net}}{Q_1} = \frac{W_T-W_P}{Q_1}[/tex]
[tex]=\frac{2.908-2.453}{40.73}[/tex] = 0.0112 or 1.12 %
