2. An ideal dual cycle, shown below on Figure 2 (P-v diagram) operates with 1 kg of air. The initial pressure (Pi) is 100 kPa and temperature (Ti) is 300K. (Note: Processes 1- 2 and 3-4 are adiabatic) The specific heats are constant and for their processes: C_v (2- x) = 0.933 kJ/kg-K, and C_V(4-n) = 0.815 kJ/kg-K. C_p(x-3) = 1.222 kJ/kg-K, with a compression ratio of 10 (r = 10). T3 is known and is 1700 K. (T3 = 1700 K). Specific heat ratio (k) is known to be 1.4. The constant pressure heat transfer process (x-3) is not known, but total heat transfer from 2-x (constant volume process) is measured as 250 kJ (Q_2-x = 250 kJ) Using the given information, the adiabatic relationships, and calculating changes in enthalpy and internal energy from specific heats, solve the complete cycle starting from Pt. 1 and making your way back to Pt. 1 to find (a) heat rejected from 4-1 (-431.2 kJ) (b) the thermal efficiency of this cycle (60%) (c) the cutoff ratio (r_c)[1.66]
2. An ideal dual cycle, shown below on Figure 2 (P-v diagram) operates with 1 kg of air. The initial pressure (Pi) is 100 kPa and temperature (Ti) is 300K. (Note: Processes 1- 2 and 3-4 are adiabatic) The specific heats are constant and for their processes: C_v (2- x) = 0.933 kJ/kg-K, and C_V(4-n) = 0.815 kJ/kg-K. C_p(x-3) = 1.222 kJ/kg-K, with a compression ratio of 10 (r = 10). T3 is known and is 1700 K. (T3 = 1700 K). Specific heat ratio (k) is known to be 1.4. The constant pressure heat transfer process (x-3) is not known, but total heat transfer from 2-x (constant volume process) is measured as 250 kJ (Q_2-x = 250 kJ) Using the given information, the adiabatic relationships, and calculating changes in enthalpy and internal energy from specific heats, solve the complete cycle starting from Pt. 1 and making your way back to Pt. 1 to find (a) heat rejected from 4-1 (-431.2 kJ) (b) the thermal efficiency of this cycle (60%) (c) the cutoff ratio (r_c)[1.66]
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
Related questions
Question
![2. An ideal dual cycle, shown below on Figure 2 (P-v diagram) operates with 1 kg of
air. The initial pressure (Pi) is 100 kPa and temperature (Ti) is 300K. (Note: Processes 1-
2 and 3-4 are adiabatic) The specific heats are constant and for their processes: C_v (2-
x) = 0.933 kJ/kg-K, and C_V(4-n) = 0.815 kJ/kg-K. C_p(x-3) = 1.222 kJ/kg-K, with a
compression ratio of 10 (r = 10). T3 is known and is 1700 K. (T3 = 1700 K). Specific heat
ratio (k) is known to be 1.4. The constant pressure heat transfer process (x-3) is not
known, but total heat transfer from 2-x (constant volume process) is measured as 250
kJ (Q_2-x = 250 kJ) Using the given information, the adiabatic relationships, and
calculating changes in enthalpy and internal energy from specific heats, solve the
complete cycle starting from Pt. 1 and making your way back to Pt. 1 to find
(a) heat rejected from 4-1 (-431.2 kJ)
(b) the thermal efficiency of this cycle (60%)
(c) the cutoff ratio (r_c)[1.66]
P
qin
2
qin
3
لیا
4
qout](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F58a48ed1-dae5-4a61-bca0-e2398c8b9b2b%2F19fbcc9f-76a5-4122-8338-7ec5ab57b4db%2F6cpzuv_processed.png&w=3840&q=75)
Transcribed Image Text:2. An ideal dual cycle, shown below on Figure 2 (P-v diagram) operates with 1 kg of
air. The initial pressure (Pi) is 100 kPa and temperature (Ti) is 300K. (Note: Processes 1-
2 and 3-4 are adiabatic) The specific heats are constant and for their processes: C_v (2-
x) = 0.933 kJ/kg-K, and C_V(4-n) = 0.815 kJ/kg-K. C_p(x-3) = 1.222 kJ/kg-K, with a
compression ratio of 10 (r = 10). T3 is known and is 1700 K. (T3 = 1700 K). Specific heat
ratio (k) is known to be 1.4. The constant pressure heat transfer process (x-3) is not
known, but total heat transfer from 2-x (constant volume process) is measured as 250
kJ (Q_2-x = 250 kJ) Using the given information, the adiabatic relationships, and
calculating changes in enthalpy and internal energy from specific heats, solve the
complete cycle starting from Pt. 1 and making your way back to Pt. 1 to find
(a) heat rejected from 4-1 (-431.2 kJ)
(b) the thermal efficiency of this cycle (60%)
(c) the cutoff ratio (r_c)[1.66]
P
qin
2
qin
3
لیا
4
qout
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