Answer:
1. The diagram T-s or H-s is attached to this answer.
2. The fraction of the steam extracted is 4.088Kg/s
3. The net Power produced per kg of steam exiting the boiler is 1089.5KJ/Kg.
4. The mass flow rate of steam supplied by the boiler is 16.352Kg/s
5. the net power produced by the plant is 11016.2KJ/s.
6. The utilization factor is 0.218.
Explanation:
To analyze this problem we need to find all the thermodynamic coordinates of the system. In the second image attached to this answer, we can see the entire ideal cogeneration steam plant system.
From a water thermodynamic properties chart, we can obtain the information for each point.
+ Steam enters the turbine from the boiler at 7 Mpa and 500 degrees C:
h₆=3410.56KJ/Kg
s₆=6.7993 KJ/Kg
This is an ideal cogeneration steam system, therefore: s₆=s₇=s₈
+One-fourth of the steam is extracted from the turbine at 600-kPa:
h₇(s₇) = 2773.74 KJ/Kg (overheated steam)
+The remainder of the steam continues to expand and exhausts to the condenser at 10 kPa.
h₈(s₈)=2153.58 KJ/Kg (this is wet steam with title X=0.8198)
h₁(P=10Kpa)= 191.83 KJ/Kg (condensed water) s₁=0.64925KJ/Kg
-This flow is pumped to 600KPa, so:
s₂=s₁
h₂(s₂)=192.585KJ/Kg
+The steam extracted for the process heater is condensed in the heater:
h₃(P=600KPa)=670.42KJ/Kg (condensed water)
+The steam extracted for the process heater is condensed in the heater and mixed with the feed-water at 600 kPa:
The mixing process of the flow of point 2 and 3 is an adiabatic process, therefore:
[tex]\dot{Q}_4=\dot{Q}_2+\dot{Q}_3=\dot{m}_2 h_2+\dot{m}_3h_3\\\dot{m}_4 h_4=\dot{m}_2 h_2+\dot{m}_3h_3=\dot{m}_2 0.75h_4+\dot{m}_30.25h_4\\h_4=0.75h_2+0.25h_3=312.043KJ/Kg[/tex]
s₄=1.02252
+the mixture is pumped to the boiler pressure of 7 Mpa:
s₅=s₄
h₅(s₅)=323.685KJ/Kg
1)Now we have all the thermodynamic coordinates and we can draw the diagram of the system.
2) To determine the fraction of steam, the mass flow that is extracted from the turbine at state 7, we use the information that this flow is used to generate 8600KJ/s in a process of heat. Therefore:
[tex]P=8600KJ/s=\dot{m}_{3-7}(h_7-h_3)\\\dot{m}_{3-7}=8600KJ/s/(h_7-h_3)=4.088Kg/s[/tex]
3)The net power produced per kg of steam exiting the boiler can be obtained as the rest between all the power obtained in the turbine less the power used in the pumps:
[tex]P_{turb}/Kg=(h_6-h_7)+0.75(h_7-h_8)=1101.94KJ/Kg\\P_{pump1}/Kg=h_2-h_1=0.755KJ/kg\\P_{pump2}/kg=h_5-h_4=11.642KJ/Kg\\P_{net}/kg=P_{turb}-P_{pump1}-P_{pump2}=1089.543KJ/Kg[/tex]
4) To determine the mass flow rate of steam that must be supplied by the boiler, we only have to remember that the flow used in point 2) is a fourth of the total flow. therefore:
[tex]0.25\dot{m}_{tot}=\dot{m}_{3-7}\\\dot{m}_{tot}=4\dot{m}_{3-7}=16.352Kg/s[/tex]
5)The net power supplied by the plant is the net power calculated in point 3) less the power used in the heat process 7-3:
[tex]P_{net sys}=P_{net}/kg \cdot \dot{m}_t-6800Kj/s=11016.2KJ/s[/tex]
6) The utilization factor is obtained as the division between net power supplied by the plant and the power used to heat the steam. In this case:
[tex]UF=P_{net sys}/P_{boil}=P_{net sys}/[\dot{m}_t(h_6-h_5)]=0.21[/tex]
