Answer:
For this ideal Rankine cycle:
A) The rate of heat addition into the cycle Qh is 3214.59KJ/Kg.
B) The thermal efficiency of the system εt=0.4354.
C) The rate of heat rejection from the cycle Qc is 1833.32KJ/Kg.
D) The back work ratio is 0.013
Explanation:
To solve the ideal Rankine cycle we have to determinate the thermodynamic information of each point of the cycle. We will use a water thermodynamic properties table. In an ideal Rankine cycle, the process in the turbine and the pump must be isentropic. Therefore S₃=S₄ and S₁=S₂.
We will start with the point 3:
P₃=18000KPa T₃=550ºc ⇒ h₃=3416.12 KJ/Kg S₃=6.40690 KJ/Kg
Point 4:
P₄=9KPa S₄=S₃ ⇒ h₄=2016.60 (x₄=0.7645 is wet steam)
Point 1:
P₁=P₄ in the endpoint of the steam curve. ⇒ h₁= 193.28KJ/Kg S₁=0.62235KJ/Kg x₁=0
Point 2:
P₂=P₃ and S₂=S₁ ⇒ h₂=201.53KJ/Kg
With this information we can obtain the heat rates, the turbine, and the pump work:
[tex]Q_h=Q_{2-3}=h_3-h_2=3214.59KJ/Kg[/tex]
[tex]Q_c=Q_{4-1}=h_4-h_1=1833.32KJ/Kg[/tex]
[tex]W_{turb}=W_{3-4}=h_3-h_4=1399.52KJ/Kg[/tex]
[tex]W_{pump}=W_{1-2}=h_2-h_1=18.25KJ/Kg[/tex]
We can answer the questions with this data:
A) [tex]Q_h=Q_{2-3}=h_3-h_2=3214.59KJ/Kg[/tex]
B) [tex]\epsilon_{ther}=\displaystyle\frac{W_{turb}}{Q_h}=0.4354[/tex]
C)[tex]Q_c=Q_{4-1}=h_4-h_1=1833.32KJ/Kg[/tex]
D) [tex]r_{bW}=\displaystyle\frac{W_{pump}}{W_{turb}}=0.013[/tex]
