Compressor driven by an electromagnet

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SUMMARY

The forum discussion centers on the calculations involved in determining the pressure increase of a compressor driven by an electromagnet. Users present various calculations, including the total volume of the system and the density of nitrogen at different states. A significant point of contention is the reported pressure increase of 2583 times, which participants agree is unrealistic. The discussion highlights the need for verification of the provided data to ensure accurate results.

PREREQUISITES
  • Understanding of thermodynamic principles, specifically isothermal processes.
  • Familiarity with fluid dynamics and gas laws.
  • Basic knowledge of electromagnetism as it relates to compressor design.
  • Proficiency in mathematical calculations involving volume and density.
NEXT STEPS
  • Research the ideal gas law and its application in isothermal processes.
  • Learn about the design and operation of electromagnet-driven compressors.
  • Explore methods for verifying thermodynamic data in engineering problems.
  • Investigate common errors in pressure calculations for gas compression systems.
USEFUL FOR

Engineers, physicists, and students involved in mechanical design, thermodynamics, and fluid mechanics will benefit from this discussion, particularly those focused on compressor technology and pressure calculations.

Guillem_dlc
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Homework Statement
The device shown in figure 1 is a compressor driven by an electromagnet. When the piston is in position 1 it sucks in nitrogen at ##p_1=101300\, \textrm{Pa}## and ##T_1=20\, \textrm{ºC}##.

Determine what is the pressure, at ##\textrm{kPa}##, when the piston reaches position 2 and the gas exits through the exhaust valve, assuming the process is isothermal (the valves are not represented in the figure). Other data: ##\bar{M}_N=28\, \textrm{kg}/\textrm{kmol}##; ##R=8314\, \textrm{J}/(\textrm{kmolK})##; ##D_1=35\, \textrm{mm}##; ##D_2=20\, \textrm{mm}##; ##L=40\, \textrm{mm}##; ##l=15\, \textrm{mm}##. See also the other figures. The second figure represents the stroke volume ##V_A## and the exhaust volume ##V_B##. And the third figure represents the volume of a truncated cone.

Solution: ##p_2=261,692\, \textrm{kPa}##
Relevant Equations
Volumes formulas, ##\rho =\dfrac{p}{R^2T}##
Figures:
1.png

2.png

3.png


Attempt at a Solution:
$$L=0,015\, \textrm{m}\qquad R_2=0,01\, \textrm{m}$$
$$R_1-R_2=R_D=0,0075\, \textrm{m}$$
$$L=0,04$$
$$\rho_1=\dfrac{p_1}{R^2T_1}=\dfrac{p_1\bar{M}}{RT_1}=1,164\, \textrm{kg}/\textrm{m}^3$$
Calculate total volume ##V_A+V_B##:
$$V_B=\pi \cdot R_2^2\cdot L=4,712\cdot 10^{-6}\, \textrm{m}^3$$
$$V_A=\dfrac{\pi L(R_1^2+R_1R_2+R_2^2)}{3}=24,347\cdot 10^{-6}\, \textrm{m}^3$$
$$\rightarrow m=\rho_1 (V_A+V_B)$$
The piston then reaches position 2.
$$\rho_2=\dfrac{m}{V_B}=\dfrac{\rho_1 (V_A+V_B)}{V_B}=7,178\, \textrm{kg}/\textrm{m}^3$$I have done this and I don't know how to continue.
 
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If the process is isothermal, then ##P_fV_f=P_iV_i##, where ##i## and f indicate initial and final. The answer given doesn't seem correct. I get 625 kPa.
 
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According to the solution value, the compressor increases the pressure of nitrogen 2583 times, which is evidently impossible for this problem.
Could you verify the given data, @Guillem_dlc ?
 
Lnewqban said:
According to the solution value, the compressor increases the pressure of nitrogen 2583 times, which is evidently impossible for this problem.
Could you verify the given data, @Guillem_dlc ?
What do you mean if they are correct? What the statement gives?
 
Yes, it seems to be an error in the provided information.
 
Lnewqban said:
Yes, it seems to be an error in the provided information.
Well, maybe, I don't know
 
Guillem_dlc said:
Well, maybe, I don't know
I got similar value to the one calculated by @Chestermiller above, which seems much more reasonable than the supposed solution of 261 692 kPa.
That is the number we would like you to verify, if possible.
 
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