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5. A gas of mass 1.5 kg undergoes a quasi-static expansion which follows a relationship p-a-hV. where a and b are constants. The initial and final pressures are 1000 kPa and 200 kPa respectively and the corresponding volumes are 0.20 m and 1.20 m. The specific internal energy of the gas is given by the relation u=1.5 pv-85 kJ/kg?
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5. A gas of mass 1.5 kg undergoes a quasi-static expansion which follo...
Given parameters:
- Mass of gas = 1.5 kg
- Expansion relationship = p-a-hV
- Initial pressure = 1000 kPa
- Final pressure = 200 kPa
- Initial volume = 0.20 m
- Final volume = 1.20 m
- Specific internal energy = 1.5 pv - 85 kJ/kg
To solve the problem, we need to find the values of a and b in the expansion relationship p-a-hV and then use it to calculate the work done by the gas during the expansion process.
Determine the values of a and b
- Since the expansion is quasi-static, we can assume that the gas is in thermodynamic equilibrium at all times.
- Therefore, we can use the first law of thermodynamics to relate the change in internal energy of the gas to the work done on it and the heat added to it.
- For a quasi-static process, the change in internal energy can be expressed as du = Tds - pdV, where T is the temperature, s is the specific entropy, and V is the volume.
- Since the gas is an ideal gas, we can use the ideal gas law to relate pressure, volume, and temperature: pV = mRT, where m is the mass of the gas, R is the gas constant, and T is the temperature.
- Combining these equations, we get du = CvdT + (p/a)dV, where Cv is the specific heat at constant volume and a = mR.
- Integrating this equation, we get u = Cv(T - T0) + (p/a)(V - V0), where T0 and V0 are the initial temperature and volume, respectively.
- Substituting the given expression for u, we get 1.5 pv - 85 = Cv(T - T0) + (p/a)(V - V0).
- Since the gas is undergoing an isobaric process, Cv(T - T0) = Cp(T - T0), where Cp is the specific heat at constant pressure.
- Therefore, we can rewrite the equation as 1.5 pv - 85 = Cp(T - T0) + (p/a)(V - V0).
- At constant pressure, we have Cp = Cv + R, where R is the gas constant.
- Substituting this in the equation, we get 1.5 pv - 85 = (Cv + R)(T - T0) + (p/a)(V - V0).
- Since pV = mRT, we can rewrite this as 1.5 p(mRT/pV) - 85 = (Cv + R)(T - T0) + (p/a)(V - V0).
- Simplifying this, we get 1.5 RT - 85/pV = (Cv + R)(T - T0) + (mR/a)(V - V0).
- Since T and V are varying during the process, we need to use the initial and final values to solve for a and b.
- At the initial state, we have T0 = p0V0/mR and 1.5 R T0 - 85/p0V0 = Cv T0 + R T0 - (mR/a)V0.
- At the final state, we have Tf = pfVf/m
- Mass of gas = 1.5 kg
- Expansion relationship = p-a-hV
- Initial pressure = 1000 kPa
- Final pressure = 200 kPa
- Initial volume = 0.20 m
- Final volume = 1.20 m
- Specific internal energy = 1.5 pv - 85 kJ/kg
To solve the problem, we need to find the values of a and b in the expansion relationship p-a-hV and then use it to calculate the work done by the gas during the expansion process.
Determine the values of a and b
- Since the expansion is quasi-static, we can assume that the gas is in thermodynamic equilibrium at all times.
- Therefore, we can use the first law of thermodynamics to relate the change in internal energy of the gas to the work done on it and the heat added to it.
- For a quasi-static process, the change in internal energy can be expressed as du = Tds - pdV, where T is the temperature, s is the specific entropy, and V is the volume.
- Since the gas is an ideal gas, we can use the ideal gas law to relate pressure, volume, and temperature: pV = mRT, where m is the mass of the gas, R is the gas constant, and T is the temperature.
- Combining these equations, we get du = CvdT + (p/a)dV, where Cv is the specific heat at constant volume and a = mR.
- Integrating this equation, we get u = Cv(T - T0) + (p/a)(V - V0), where T0 and V0 are the initial temperature and volume, respectively.
- Substituting the given expression for u, we get 1.5 pv - 85 = Cv(T - T0) + (p/a)(V - V0).
- Since the gas is undergoing an isobaric process, Cv(T - T0) = Cp(T - T0), where Cp is the specific heat at constant pressure.
- Therefore, we can rewrite the equation as 1.5 pv - 85 = Cp(T - T0) + (p/a)(V - V0).
- At constant pressure, we have Cp = Cv + R, where R is the gas constant.
- Substituting this in the equation, we get 1.5 pv - 85 = (Cv + R)(T - T0) + (p/a)(V - V0).
- Since pV = mRT, we can rewrite this as 1.5 p(mRT/pV) - 85 = (Cv + R)(T - T0) + (p/a)(V - V0).
- Simplifying this, we get 1.5 RT - 85/pV = (Cv + R)(T - T0) + (mR/a)(V - V0).
- Since T and V are varying during the process, we need to use the initial and final values to solve for a and b.
- At the initial state, we have T0 = p0V0/mR and 1.5 R T0 - 85/p0V0 = Cv T0 + R T0 - (mR/a)V0.
- At the final state, we have Tf = pfVf/m
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5. A gas of mass 1.5 kg undergoes a quasi-static expansion which follo...
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5. A gas of mass 1.5 kg undergoes a quasi-static expansion which follows a relationship p-a-hV. where a and b are constants. The initial and final pressures are 1000 kPa and 200 kPa respectively and the corresponding volumes are 0.20 m and 1.20 m. The specific internal energy of the gas is given by the relation u=1.5 pv-85 kJ/kg?
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5. A gas of mass 1.5 kg undergoes a quasi-static expansion which follows a relationship p-a-hV. where a and b are constants. The initial and final pressures are 1000 kPa and 200 kPa respectively and the corresponding volumes are 0.20 m and 1.20 m. The specific internal energy of the gas is given by the relation u=1.5 pv-85 kJ/kg? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about 5. A gas of mass 1.5 kg undergoes a quasi-static expansion which follows a relationship p-a-hV. where a and b are constants. The initial and final pressures are 1000 kPa and 200 kPa respectively and the corresponding volumes are 0.20 m and 1.20 m. The specific internal energy of the gas is given by the relation u=1.5 pv-85 kJ/kg? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 5. A gas of mass 1.5 kg undergoes a quasi-static expansion which follows a relationship p-a-hV. where a and b are constants. The initial and final pressures are 1000 kPa and 200 kPa respectively and the corresponding volumes are 0.20 m and 1.20 m. The specific internal energy of the gas is given by the relation u=1.5 pv-85 kJ/kg?.
5. A gas of mass 1.5 kg undergoes a quasi-static expansion which follows a relationship p-a-hV. where a and b are constants. The initial and final pressures are 1000 kPa and 200 kPa respectively and the corresponding volumes are 0.20 m and 1.20 m. The specific internal energy of the gas is given by the relation u=1.5 pv-85 kJ/kg? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about 5. A gas of mass 1.5 kg undergoes a quasi-static expansion which follows a relationship p-a-hV. where a and b are constants. The initial and final pressures are 1000 kPa and 200 kPa respectively and the corresponding volumes are 0.20 m and 1.20 m. The specific internal energy of the gas is given by the relation u=1.5 pv-85 kJ/kg? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 5. A gas of mass 1.5 kg undergoes a quasi-static expansion which follows a relationship p-a-hV. where a and b are constants. The initial and final pressures are 1000 kPa and 200 kPa respectively and the corresponding volumes are 0.20 m and 1.20 m. The specific internal energy of the gas is given by the relation u=1.5 pv-85 kJ/kg?.
Solutions for 5. A gas of mass 1.5 kg undergoes a quasi-static expansion which follows a relationship p-a-hV. where a and b are constants. The initial and final pressures are 1000 kPa and 200 kPa respectively and the corresponding volumes are 0.20 m and 1.20 m. The specific internal energy of the gas is given by the relation u=1.5 pv-85 kJ/kg? in English & in Hindi are available as part of our courses for Mechanical Engineering.
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