Adiabatic Expansion of Pressurized Air in a Piston-Cylinder Setup

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SUMMARY

The discussion focuses on the adiabatic expansion of pressurized air in a piston-cylinder setup, specifically analyzing a system with a pressurized air cylinder B containing 10 liters of air at 150 °C and 20 bar, placed inside a larger cylinder A with a piston. The key equations used include the internal energy equations and the ideal gas law, with specific heat capacity at constant volume (cV = 0.718 kJ/(kg·K)) and specific gas constant (Rs = 0.287 kJ/(kg·K)). Participants identified errors in the initial energy balance calculations and emphasized the importance of correctly applying the principles of adiabatic processes and conservation of energy.

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Peter_parker
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Homework Statement
Uncertain how to take into account the temperature of the hot cylinder?
Relevant Equations
U=m*cv*dt
p*V=m*Rs*T
Task:
A thermally insulated pressurized air cylinder, B, was initially placed inside a closed-bottom, circular hollow cylinder A with an inner diameter of 50 cm. Then a tightly fitting, frictionless sliding piston with a mass of 20 kg was installed. Using the outlet valve, the height of the piston was set to h1 = 100 cm. The pressurized air cylinder B contains 10 liters of hot air with a temperature of 150 °C (TB) and an absolute pressure of 20 bar. The air in container A, that is, outside the pressurized air cylinder B, initially has the ambient temperature.

For your calculations, consider air as an ideal gas with a specific heat capacity at constant volume, cV = 0.718 kJ/(kg·K), and specific gas constant, Rs = 0.287 kJ/(kg·K).

The intrinsic volume of the bottle material should be neglected. The following environmental condition exists: ambient pressure pamb = 1 bar, ambient temperature Tamb = 20 °C.

Question of the homework:
After opening the pressurized air cylinder B, the entire system quickly reaches an equilibrium state. The process is adiabatic. At what height is the piston now located? What temperature does the system reach?

What i did:
Equation1 / Part1:
Condition 1: U1=m_a*cv*T1+m_a*cv*TB= m_both*cv*(T1+TB)
Condition 2: U2=m_both*cv*(T2)
Connection via:U12=Q+Wv with Q =0 and Wv=p*(V1-V2) to m_both*cv*(T2-(T1+TB))=p*(V1-V2)
Equation2 / Part2:
p(V2+VB)=m_both*Rs*T2

By combining Equation 1 and Equation 2, I get T2 / V2 --> then used the area of the cylinder to calculate h2.

However, I'm getting a very unrealistic value for h2. I'm wondering if I made a mistake in the way I included TB in Condition 1. Can anyone spot an mistake in my approach?

Thanks!
 
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Let’s see some numbers.

What is the initial pressure in A?
What is the mass of air in A?
What is the mass of air in B?
 
Last edited:
Peter_parker said:
Homework Statement: Uncertain how to take into account the temperature of the hot cylinder?
Relevant Equations: U=m*cv*dt
p*V=m*Rs*T

Task:
A thermally insulated pressurized air cylinder, B, was initially placed inside a closed-bottom, circular hollow cylinder A with an inner diameter of 50 cm. Then a tightly fitting, frictionless sliding piston with a mass of 20 kg was installed. Using the outlet valve, the height of the piston was set to h1 = 100 cm. The pressurized air cylinder B contains 10 liters of hot air with a temperature of 150 °C (TB) and an absolute pressure of 20 bar. The air in container A, that is, outside the pressurized air cylinder B, initially has the ambient temperature.

For your calculations, consider air as an ideal gas with a specific heat capacity at constant volume, cV = 0.718 kJ/(kg·K), and specific gas constant, Rs = 0.287 kJ/(kg·K).

The intrinsic volume of the bottle material should be neglected. The following environmental condition exists: ambient pressure pamb = 1 bar, ambient temperature Tamb = 20 °C.

Question of the homework:
After opening the pressurized air cylinder B, the entire system quickly reaches an equilibrium state. The process is adiabatic. At what height is the piston now located? What temperature does the system reach?

What i did:
Equation1 / Part1:
Condition 1: U1=m_a*cv*T1+m_a*cv*TB= m_both*cv*(T1+TB)
This algebra is incorrect.
Peter_parker said:
Condition 2: U2=m_both*cv*(T2)
Connection via:U12=Q+Wv with Q =0 and Wv=p*(V1-V2) to m_both*cv*(T2-(T1+TB))=p*(V1-V2)
Equation2 / Part2:
p(V2+VB)=m_both*Rs*T2

By combining Equation 1 and Equation 2, I get T2 / V2 --> then used the area of the cylinder to calculate h2.

However, I'm getting a very unrealistic value for h2. I'm wondering if I made a mistake in the way I included TB in Condition 1. Can anyone spot an mistake in my approach?

Thanks!
 
Chestermiller said:
Let’s see some numbers.

What is the initial pressure in A?
What is the mass of air in A?
What is the mass of air in B?
Pressure in A: pamb + F/A, here Mass of the piston * g, together = F, and the area of cylinder A
Mass of A: mA=(pA * V A ) / (RS / *TA) , same for B with pB etc.
together A and B are mboth
 
Chestermiller said:
This algebra is incorrect.
Can you go in detail please?
 
Peter_parker said:
Pressure in A: pamb + F/A, here Mass of the piston * g, together = F, and the area of cylinder A
Mass of A: mA=(pA * V A ) / (RS / *TA) , same for B with pB etc.
together A and B are mboth
Let's see some numbers for these in Pa and kg.
 
Peter_parker said:
Can you go in detail please?
$$m_ac_vT_1+m_bc_vT_B \neq m_{both}c_v(T_1+T_B)$$
 
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Peter_parker said:
Homework Statement: Uncertain how to take into account the temperature of the hot cylinder?
Relevant Equations: U=m*cv*dt
p*V=m*Rs*T

Condition 1: U1=m_a*cv*T1+m_a*cv*TB= m_both*cv*(T1+TB)
Condition 2: U2=m_both*cv*(T2)
Connection via:U12=Q+Wv with Q =0 and Wv=p*(V1-V2) to m_both*cv*(T2-(T1+TB))=p*(V1-V2)
Equation2 / Part2:
p(V2+VB)=m_both*Rs*T2
Learn how to use LaTeX so that we can read your equations. What is the principle that makes Conditions 1 and 2 true? Is it conservation of energy? Because energy is not conserved in this process.

Also, I don't see where you are using the fact that the process is adiabatic.
 
Mister T said:
Learn how to use LaTeX so that we can read your equations. What is the principle that makes Conditions 1 and 2 true? Is it conservation of energy? Because energy is not conserved in this process.

Also, I don't see where you are using the fact that the process is adiabatic.
Adiabatic: ##\Delta U=-W##
 

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