Calculating Force Needed to Compress Air From 1 atm

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Discussion Overview

The discussion revolves around calculating the force required to compress air from a low atmospheric pressure to a higher pressure using a bellows mechanism. Participants explore the implications of different compression processes, particularly adiabatic conditions, and the energy needed as the pressure in the reservoir increases over time.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant inquires about a formula for calculating the force needed to compress air in a chamber from 1 atm to a higher pressure using a bellows system.
  • Another participant provides a detailed explanation involving adiabatic processes, presenting equations for work done during compression, including variables such as mass, specific heat, and temperature.
  • The second participant emphasizes that the efficiency of the device must be considered and notes that lower initial temperatures can reduce the work required for compression.
  • A later reply clarifies that while the process is adiabatic due to a fast cycle rate, the air drawn into the chamber comes from outside, which may affect the temperature dynamics as air is compressed into the reservoir.
  • One participant expresses uncertainty about agreeing with the provided formulation, indicating a lack of understanding and a desire to work through the concepts presented.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the formulation provided, with one expressing uncertainty about its validity and another clarifying the conditions of the process. The discussion remains unresolved regarding the agreement on the mathematical approach and its implications.

Contextual Notes

Participants mention the need for additional information about the process, such as whether it is adiabatic or involves heat relaxation, which could influence the calculations. There is also a discussion about the impact of temperature on the work required for compression.

swabbie58
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Is there a formula to calculate the amount of force needed to compress air from a low atmosphere to a higher one? In my situation, I have a chamber wherein the pressure is one (1) atm and I ned to push the air from that area into a reservoir where the pressure will increase with each cycle. Am not using a fan, but rather a sort of bellows (one side of the rectangular chamber moves toward an opposite side with hoses to the reservoir). As the reservoir pressure mounts, how can I calculate for the growing amount of energy needed to push more air in and raise the atm pressure?
 
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swabbie58 said:
Is there a formula to calculate the amount of force needed to compress air from a low atmosphere to a higher one? In my situation, I have a chamber wherein the pressure is one (1) atm and I ned to push the air from that area into a reservoir where the pressure will increase with each cycle. Am not using a fan, but rather a sort of bellows (one side of the rectangular chamber moves toward an opposite side with hoses to the reservoir). As the reservoir pressure mounts, how can I calculate for the growing amount of energy needed to push more air in and raise the atm pressure?

It is necessary to have more information about the process on your chamber. Is is adiabatic (i.e fast) or it has times of heat relaxing?. If it is adiabatic, you can obtain the energy needed (work):

[tex]W=\Delta U=m c_v \Delta T[/tex] where m is the mass contained into the chamber in one cycle.

Adiabatic process: [tex]W= m c_v T_o \Big(\frac{T_f}{T_o}-1\Big)= m c_v T_o \Big(\Big(\frac{P_f}{P_o}\Big)^\frac{\gamma-1}{\gamma}-1\Big)[/tex]

where T_o is the temperature at which you start the process of compression,
c_v=714 J/kg
gamma=1.4
and if you define your compression ratio as: [tex]r=\frac{P_f}{P_o}[/tex] then:

[tex]W= m c_v T_o (r^{\frac{\gamma-1}{\gamma}}-1)[/tex] is the work per cycle (Joules). Surely you will have to multiply this by some kind mechanic efficiency of your device.

HINT: the less [tex]T_o[/tex] the less [tex]W[/tex]. Do you understand the meaning of cooling before compressing air?
 
While it is adiabatic in that it uses a very fast cycle rate, the air being drawn into the chamber for the subsequent cycle comes from outside the system. So, while the temperature of the first-cycle air will rise, that heat will be carried with it down the line to the reservoir. There it will build up as each cycle of air is forced into the reservoir.
 
So you agree with the formulation posted or not?
 
Sorry--my computer had to be taken down to install a new graphics card. Frankly, I don't know enough to know whether I "agree" with it or not! That's why I am posting the question on the forum. I'll have to work through it and see what happens.
 

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