How do I calculate specific work for cooled air with constant pressure?

In summary, the conversation discusses finding the specific work in a process where air is cooled at a constant pressure of 115 kPa. The initial specific volume of the air is 0.06 m^3/kg and the final specific volume is 0.03 m^3/kg. The specific work can be calculated using the equation Specific Work= integral (limit goes from 2 1) PxdV, but the user is unsure how to compute it. They also clarify that specific work is the amount of work done per unit mass of gas.
  • #1
t3rom
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Homework Statement



Air is cooled in a process with constant pressure of 115 kPa. Before the process begins, air has a specific volume of 0.06 m^3/kg. The final specific volume is 0.03 m^3/kg. Find the specific work in the process.


Homework Equations



none.


The Attempt at a Solution



I think this equation can be used to solve this:

Specific Work= integral (limit goes from 2 1) PxdV, but I don't understand how do I compute this. There aren't many Specific work related problems online, I've tried searching. Any help or pointers will be greatly appreciated! Thanks!
 
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  • #2
Specific work is the amount of work done per unit mass of gas.
 
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  • #3
Thanks, I got it :)
 

What is specific work and why is it important?

Specific work is a measure of the amount of work done per unit mass in a thermodynamic process. It is important because it helps us understand the efficiency of a process and can be used to compare different processes.

How is specific work calculated?

Specific work is calculated by dividing the work done by the change in mass. In thermodynamics, work is defined as the force applied over a distance, so it can be calculated by multiplying the force by the distance. The change in mass can be determined by subtracting the initial mass from the final mass.

What is the unit of specific work?

The unit of specific work is joules per kilogram (J/kg). This unit represents the amount of work done per unit mass and is commonly used in thermodynamics and fluid mechanics.

What factors can affect the specific work in a process?

The specific work in a process can be affected by factors such as the type of process (isentropic, isothermal, etc.), the type of fluid being used, the efficiency of the process, and any external forces or work being applied.

How is specific work used in real-world applications?

Specific work is used in many real-world applications, such as in the design and analysis of engines, turbines, and compressors. It is also used in the study of fluid flow and the efficiency of different processes, such as power plants and refrigeration systems.

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