THERMODYNAMICSReferring to the T–s diagram of Fig. 9.10, which of the two cycles does not allow use of a regenerator: 1–2-3-4-1 or 1–2 -3-4-1? Explain.T3aS=Cp =c941p = cFIG. 9.10 Air-standard ideal Brayton cyclea

Given: Temperature entropy (T-S) diagram of two cycles are given,

Answered: Referring to the T-s diagram of Fig.…

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

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An ideal air-standard Brayton cycle operating at steady state produces 10 MVWof power. Operating data at principal states in the cycle are given in the table below. The states are numbered as in the figure below. State p(kPa) T (K) h (kJ/kg) 1 100 300 300 1000 574.1 579.7 3 1000 1800 2002 4 100 1034 1085
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Consider a cold air-standard Diesel cycle. Operating data at principal states in the cycle are given in the table below. The states are numbered as in Fig. 9.5 with p1 = 100 kPa, T1 = 340 K, and T3 = 4850.3 K, r = 12 and rc = 2. For n = 1.3, cv = 0.718 kJ/(kg-K), and cp = 1.005 kJ/(kg-K), determine p2 in kPa. (A) 3380.02 B) 4431.27 (C) 2528.92 D) 3675.83
In an Otto cycle, the volume at the beginning of the isentropic compression is 0.74 L. If the displaced volume of the cycle is 0.68 L, the specific heat ratio is y=1.3 and the heat rejected is 405 kJ/kg, the heat added in kJ per kg of air is: Select one: O a. 860.6 O b. 624.5 O c. 1005.2 O d. 421.4
Consider a cold air-standard Diesel cycle. Operating data at principal states in the cycle are given in the table below. The states are numbered as in Fig. 9.5 with p1 = 100 kPa, T1 = 340 K, and T3 = 4850.3 K, r = 12 and rc = 2. For n = 1.3, cv = 0.718 kJ/(kg-K), and cp = 1.005 kJ/(kg-K), determine p2 in kPa.
Consider the operation of the ideal Brayton cycle with a regenerator. The maximum temperature in the cycle is set. What happens with increasing pressure ratio? Group of answer choices The amount of heat added in the cycle up to a point will decrease with net work output remaining the same and thermal efficiency will increase. Thermal efficiency will increase Compressor exit temperature will increase with turbine exit temperature remaining the same. When the two temperatures are equal, the benefit of the regenerator will be zero. Thermal efficiency will decrease
Consider a cold air-standard Diesel cycle. Operating data at principal states in the cycle are given in the table below. The states are numbered as in Fig. 9.5 with p1 = 100 kPa, T1 = 340 K, and T3 = 4850.3 K, r = 12 and rc = 2. For n = 1.3, cv = 0.718 kJ/(kg-K), and cp = 1.005 kJ/(kg-K), determine the heat transfer during heat addition process per unit mass in kJ/kg. A) 3865.1 B) 4104.58 C) 3838.71 (D) 4154.44
Thermodynamics 2
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Referring to the T-s diagram of Fig. 9.10, which of the two cycles does not allow use of a regenerator: 1-2-3-4-1 or 1-2-3-4-1? Explain. a b. p =c p =c FIG. 9.10 Air-standard ideal Brayton cycle a