Нeat 3-3 A “real" Brayton cycle is shown in Figure 3-3. The cycle starts at 300K (T1) and a pressure of 100 kPa. The compressor has an isentropic efficiency of 85% and the turbine has an isentropic efficiency given in Table 2, along with other process temperatures and the net power output. Assume there is no pressure loss in the piping P2=P3=P4 and Ps =P6 =P1 and the working fluid is air (use actual properties.) The heat recovery HX has a given effectiveness (ɛ) as defined by hz – hz h5 – h2 6. tombustor www Recovery Compressor Turbine Figure 3. Brayton with heat recovery E = Additionally, you are given Nturb T4 (K)|Heat Exchanger Effectiveness (8) Pressure ratio Net power (kW) 0.91 1300 0.75 10 6000 With this information, solve (a) mass flow rate of air [kg/s], (b) rate of heat added at the combustor (kW), (c) thermal efficiency of the cycle, and (d) Back work ratio (BWR)
Нeat 3-3 A “real" Brayton cycle is shown in Figure 3-3. The cycle starts at 300K (T1) and a pressure of 100 kPa. The compressor has an isentropic efficiency of 85% and the turbine has an isentropic efficiency given in Table 2, along with other process temperatures and the net power output. Assume there is no pressure loss in the piping P2=P3=P4 and Ps =P6 =P1 and the working fluid is air (use actual properties.) The heat recovery HX has a given effectiveness (ɛ) as defined by hz – hz h5 – h2 6. tombustor www Recovery Compressor Turbine Figure 3. Brayton with heat recovery E = Additionally, you are given Nturb T4 (K)|Heat Exchanger Effectiveness (8) Pressure ratio Net power (kW) 0.91 1300 0.75 10 6000 With this information, solve (a) mass flow rate of air [kg/s], (b) rate of heat added at the combustor (kW), (c) thermal efficiency of the cycle, and (d) Back work ratio (BWR)
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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![Нeat
3-3 A “real" Brayton cycle is shown in Figure
3-3. The cycle starts at 300K (T1) and a
pressure of 100 kPa. The compressor has an
isentropic efficiency of 85% and the turbine
has an isentropic efficiency given in Table 2,
along with other process temperatures and the
net power output. Assume there is no
pressure loss in the piping P2=P3=P4 and P5
=P6=P1 and the working fluid is air (use
actual properties.) The heat recovery HX has
a given effectiveness (ɛ) as defined by
hz – h2
h5 – h2
6.
tombustor
3
Recovery
Compressor
Turbine
Figure 3. Brayton with heat recovery
E =
Additionally, you are given
Nturb T4 (K) Heat Exchanger Effectiveness (8) Pressure ratio Net power (kW)
0.91 1300
0.75
10
6000
With this information, solve (a) mass flow rate of air [kg/s], (b) rate of heat added at the combustor
(kW), (c) thermal efficiency of the cycle, and (d) Back work ratio (BWR)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F523684c4-df8f-40dd-8c0d-907308e61cee%2F5db33a6b-16a2-4015-83fd-4be9afe6dd5b%2Fdk94ddkp_processed.png&w=3840&q=75)
Transcribed Image Text:Нeat
3-3 A “real" Brayton cycle is shown in Figure
3-3. The cycle starts at 300K (T1) and a
pressure of 100 kPa. The compressor has an
isentropic efficiency of 85% and the turbine
has an isentropic efficiency given in Table 2,
along with other process temperatures and the
net power output. Assume there is no
pressure loss in the piping P2=P3=P4 and P5
=P6=P1 and the working fluid is air (use
actual properties.) The heat recovery HX has
a given effectiveness (ɛ) as defined by
hz – h2
h5 – h2
6.
tombustor
3
Recovery
Compressor
Turbine
Figure 3. Brayton with heat recovery
E =
Additionally, you are given
Nturb T4 (K) Heat Exchanger Effectiveness (8) Pressure ratio Net power (kW)
0.91 1300
0.75
10
6000
With this information, solve (a) mass flow rate of air [kg/s], (b) rate of heat added at the combustor
(kW), (c) thermal efficiency of the cycle, and (d) Back work ratio (BWR)
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