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(Q)₁= 5052.482 kW • Heat rejection (Q)R T3=1400 K (turbine inlet temperature) Ta turbine exit temperature = (6) 1400 (10) T₁

(Q)₁= 5052.482 kW • Heat rejection (Q)R T3=1400 K (turbine inlet temperature) Ta turbine exit temperature = (6) 1400 (10) T₁

Elements Of Electromagnetics
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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A,B and C were answered, proceed to D,E and F. Air enters the compressor of an ideal air standard brayton cycle at 100 kPa, 300 K, with a volumetric flow rate of 5 cu.m/s. The compressor pressure ratio is 10. The turbine inlet temperature is 1400 K. Compute the (a) P, V, and T at each point of the cycle; (b) QA, QR in kW; (c) WC, WT in kW; (d) rp, TBW; (e) e; (f) mep.
Hence the heat adan |(Q)A = 5052.482 kW • Heat rejection (QR T3-1400 K (turbine inlet temprerature) T₁ turbine exit temperature 2=(6) 1400 (10) T₁ T4=725, 126 K (QR=mx C₂ x (T4-T₁) (Q)R 6. 125 x 1.005 x (725. 126-300) (QR-2616.916 kW Hence the heat rejection (Q)R= 2616.916 kW Step 3 (c). Calculation of work consumed by the compressor (Wc) and work done by the turbine (WT): • The work consumed by the compressor (WC) T₂=579.209 K T₁=300 K m=6. 125 kg/s Cp=1.005 kJ/kg K (Wc)=mx cp x (T₂-T₁) (We) 6. 125 x 1.005 × (579.209 300) (We) 1718.706 kW Hence the work consumed by the compressor (We) = 1718.706 kW • The work done by the turbine (WT) (WT) turbine work T₁ turbine inlet temperature T4 turbine exit temperature T;=1400 K T₁=725. 126 K m=6.125 kg/s Cp=1.005 kJ/kg K (WT)=mx c₂ x (T3-T4) (WT)=6. 125 x 1.005 x (1400-725. 126) (WT)=4154.271 kW Hence the work done by the turbine (WT) = 4154.271 kW Step 4 Results: For the given problem the following results are obtained, The heat addtion |(Q), = 5052.482 kW The heat rejection (Q) = 2616.916 kW The work consumed by the compressor (We) = 1718. 706 kW The work done by the turbine (WT) = 4154.271 kW
Transcribed Image Text:Hence the heat adan |(Q)A = 5052.482 kW • Heat rejection (QR T3-1400 K (turbine inlet temprerature) T₁ turbine exit temperature 2=(6) 1400 (10) T₁ T4=725, 126 K (QR=mx C₂ x (T4-T₁) (Q)R 6. 125 x 1.005 x (725. 126-300) (QR-2616.916 kW Hence the heat rejection (Q)R= 2616.916 kW Step 3 (c). Calculation of work consumed by the compressor (Wc) and work done by the turbine (WT): • The work consumed by the compressor (WC) T₂=579.209 K T₁=300 K m=6. 125 kg/s Cp=1.005 kJ/kg K (Wc)=mx cp x (T₂-T₁) (We) 6. 125 x 1.005 × (579.209 300) (We) 1718.706 kW Hence the work consumed by the compressor (We) = 1718.706 kW • The work done by the turbine (WT) (WT) turbine work T₁ turbine inlet temperature T4 turbine exit temperature T;=1400 K T₁=725. 126 K m=6.125 kg/s Cp=1.005 kJ/kg K (WT)=mx c₂ x (T3-T4) (WT)=6. 125 x 1.005 x (1400-725. 126) (WT)=4154.271 kW Hence the work done by the turbine (WT) = 4154.271 kW Step 4 Results: For the given problem the following results are obtained, The heat addtion |(Q), = 5052.482 kW The heat rejection (Q) = 2616.916 kW The work consumed by the compressor (We) = 1718. 706 kW The work done by the turbine (WT) = 4154.271 kW
Step 2 (a). Pressure (P)-volume (V) and temperature (T)-entropy (S) diagram of air standard Brayton cycle: The following P-V and T-S diagram for each process of the Brayton cycle, V₂ T P-P, S₁-S₂ Volume →→→→ Figa: Pressure-Volume diagram of Brayton cycle. P₁-P₁ (10) T₂=579.209 K T3=1400 K V=5 m³/s S-S V₁ Entropy Fig b Temperature-entropy diagram of Brayton cycle. 4 (b). Calculation of heat addition (Q) and heat rejection (Q)K: • Heat addition (Q)A T₂=temperature after compression T₁=300 K (temperature before compress r=10 (compression ratio) y=1.4 (adiabatic indx) p=1.225 kg/m³ (air density) m=p x v Hence the heat addition (Q)A = 5052.482 kW m=1.225 x 5 m=6. 125 kg/s Cp=1.005 kJ/kg K (specific heat) (Q)₁=mx C₂ x (T3-T₂) (Q)₁ 6. 125 x 1.005 x (1400 - 579. 209) (Q)₁=5052.482 kW
Transcribed Image Text:Step 2 (a). Pressure (P)-volume (V) and temperature (T)-entropy (S) diagram of air standard Brayton cycle: The following P-V and T-S diagram for each process of the Brayton cycle, V₂ T P-P, S₁-S₂ Volume →→→→ Figa: Pressure-Volume diagram of Brayton cycle. P₁-P₁ (10) T₂=579.209 K T3=1400 K V=5 m³/s S-S V₁ Entropy Fig b Temperature-entropy diagram of Brayton cycle. 4 (b). Calculation of heat addition (Q) and heat rejection (Q)K: • Heat addition (Q)A T₂=temperature after compression T₁=300 K (temperature before compress r=10 (compression ratio) y=1.4 (adiabatic indx) p=1.225 kg/m³ (air density) m=p x v Hence the heat addition (Q)A = 5052.482 kW m=1.225 x 5 m=6. 125 kg/s Cp=1.005 kJ/kg K (specific heat) (Q)₁=mx C₂ x (T3-T₂) (Q)₁ 6. 125 x 1.005 x (1400 - 579. 209) (Q)₁=5052.482 kW
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