Compressor driven by an electromagnet

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Homework Help Overview

The discussion revolves around a problem involving a compressor driven by an electromagnet, focusing on the calculations related to pressure, volume, and density of gases in a thermodynamic context.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss calculations related to the volumes and densities of gases, questioning the validity of provided data and the feasibility of pressure increases suggested in the original solution.

Discussion Status

Some participants express skepticism about the accuracy of the initial data and calculations, with multiple interpretations of the problem being explored. There is a call for verification of the provided information, indicating a productive direction in the discussion.

Contextual Notes

Participants note potential errors in the provided information and question the assumptions underlying the calculations, particularly regarding the feasibility of the pressure values obtained.

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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