Data that have to be used for solving the practical examples and the problems: Atmospheric pressure for all problems and practical examples is B = 946 mbar. For all practical examples and problems consider air as ideal and perfect gas. Universal gas constant R₁ = 8314.472 J/(kmol.K). Air ideal gas constant R = 287 J/(kg.K). Atomic weights in kg/kmol: Carbon - 12.0107; Nitrogen - 14.0067; Oxygen - 15.9994. Isentropic exponent: k=1.67 for monatomic gases; k=1.40 for diatomic gases; k=1.29 for polyatomic gases. Wien's displacement constant: b=0.0028977685 m.K. Stefan-Boltzmann constant: σ = 5.6704-108 W/(m²K) Standard state conditions: P=101325 Pa, T.-273.15 K. A closed system contains an ideal gas, which molecular weight is W=63.47 kg/kmol, and its standard state entropy is so-0. The system undergoes the following cycle: at state 1 the temperature is 294.15 K, the pressure is 95 kPa (absolute), and the entropy is 46.450 J/(kg.K). The gas is compressed polytropically at n=1.46 until the specific volume is 10 times lower than that at state 1 (state 2). Then 85651.7 J/kg of heat is added at constant specific volume (state 3). After that heat is added at constant pressure until entropy is 333.333 J/(kg.K) (state 4). In the next process the system undergoes isentropic expansion (and reaches state 5). Finally there is a constant volume rejection of heat (until state 1). Determine a) the values of p, v, T and s, at each cycle point; b) the p-V and T-s diagrams of the cycle; c) the heat added to the system; d) the heat rejected by the system; e) the cycle net work; f) the mean effective pressure, g) the cycle thermal efficiency.
Data that have to be used for solving the practical examples and the problems: Atmospheric pressure for all problems and practical examples is B = 946 mbar. For all practical examples and problems consider air as ideal and perfect gas. Universal gas constant R₁ = 8314.472 J/(kmol.K). Air ideal gas constant R = 287 J/(kg.K). Atomic weights in kg/kmol: Carbon - 12.0107; Nitrogen - 14.0067; Oxygen - 15.9994. Isentropic exponent: k=1.67 for monatomic gases; k=1.40 for diatomic gases; k=1.29 for polyatomic gases. Wien's displacement constant: b=0.0028977685 m.K. Stefan-Boltzmann constant: σ = 5.6704-108 W/(m²K) Standard state conditions: P=101325 Pa, T.-273.15 K. A closed system contains an ideal gas, which molecular weight is W=63.47 kg/kmol, and its standard state entropy is so-0. The system undergoes the following cycle: at state 1 the temperature is 294.15 K, the pressure is 95 kPa (absolute), and the entropy is 46.450 J/(kg.K). The gas is compressed polytropically at n=1.46 until the specific volume is 10 times lower than that at state 1 (state 2). Then 85651.7 J/kg of heat is added at constant specific volume (state 3). After that heat is added at constant pressure until entropy is 333.333 J/(kg.K) (state 4). In the next process the system undergoes isentropic expansion (and reaches state 5). Finally there is a constant volume rejection of heat (until state 1). Determine a) the values of p, v, T and s, at each cycle point; b) the p-V and T-s diagrams of the cycle; c) the heat added to the system; d) the heat rejected by the system; e) the cycle net work; f) the mean effective pressure, g) the cycle thermal efficiency.
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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Solve the problem with the data below
![Data that have to be used for solving the practical examples and the problems:
Atmospheric pressure for all problems and practical examples is B = 946 mbar.
For all practical examples and problems consider air as ideal and perfect gas.
Universal gas constant R₁ = 8314.472 J/(kmol.K).
Air ideal gas constant R = 287 J/(kg.K).
Atomic weights in kg/kmol: Carbon - 12.0107; Nitrogen - 14.0067; Oxygen - 15.9994.
Isentropic exponent: k=1.67 for monatomic gases; k=1.40 for diatomic gases; k=1.29 for
polyatomic gases.
Wien's displacement constant: b=0.0028977685 m.K.
Stefan-Boltzmann constant: σ = 5.6704-108 W/(m²K)
Standard state conditions: P=101325 Pa, T.-273.15 K.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa737805f-cf95-4086-92d5-3bea761ebae4%2F69ceaa93-b155-4bc6-8ac5-2c4e95d13872%2Frok6p3_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Data that have to be used for solving the practical examples and the problems:
Atmospheric pressure for all problems and practical examples is B = 946 mbar.
For all practical examples and problems consider air as ideal and perfect gas.
Universal gas constant R₁ = 8314.472 J/(kmol.K).
Air ideal gas constant R = 287 J/(kg.K).
Atomic weights in kg/kmol: Carbon - 12.0107; Nitrogen - 14.0067; Oxygen - 15.9994.
Isentropic exponent: k=1.67 for monatomic gases; k=1.40 for diatomic gases; k=1.29 for
polyatomic gases.
Wien's displacement constant: b=0.0028977685 m.K.
Stefan-Boltzmann constant: σ = 5.6704-108 W/(m²K)
Standard state conditions: P=101325 Pa, T.-273.15 K.
![A closed system contains an ideal gas, which molecular weight is W=63.47 kg/kmol, and its
standard state entropy is so-0. The system undergoes the following cycle: at state 1 the
temperature is 294.15 K, the pressure is 95 kPa (absolute), and the entropy is 46.450 J/(kg.K).
The gas is compressed polytropically at n=1.46 until the specific volume is 10 times lower than
that at state 1 (state 2). Then 85651.7 J/kg of heat is added at constant specific volume (state 3).
After that heat is added at constant pressure until entropy is 333.333 J/(kg.K) (state 4). In the
next process the system undergoes isentropic expansion (and reaches state 5). Finally there is a
constant volume rejection of heat (until state 1). Determine a) the values of p, v, T and s, at each
cycle point; b) the p-V and T-s diagrams of the cycle; c) the heat added to the system; d) the heat
rejected by the system; e) the cycle net work; f) the mean effective pressure, g) the cycle thermal
efficiency.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa737805f-cf95-4086-92d5-3bea761ebae4%2F69ceaa93-b155-4bc6-8ac5-2c4e95d13872%2Fpcc3mw7_processed.jpeg&w=3840&q=75)
Transcribed Image Text:A closed system contains an ideal gas, which molecular weight is W=63.47 kg/kmol, and its
standard state entropy is so-0. The system undergoes the following cycle: at state 1 the
temperature is 294.15 K, the pressure is 95 kPa (absolute), and the entropy is 46.450 J/(kg.K).
The gas is compressed polytropically at n=1.46 until the specific volume is 10 times lower than
that at state 1 (state 2). Then 85651.7 J/kg of heat is added at constant specific volume (state 3).
After that heat is added at constant pressure until entropy is 333.333 J/(kg.K) (state 4). In the
next process the system undergoes isentropic expansion (and reaches state 5). Finally there is a
constant volume rejection of heat (until state 1). Determine a) the values of p, v, T and s, at each
cycle point; b) the p-V and T-s diagrams of the cycle; c) the heat added to the system; d) the heat
rejected by the system; e) the cycle net work; f) the mean effective pressure, g) the cycle thermal
efficiency.
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