Compressor driven by an electromagnet

AI Thread Summary
The discussion revolves around the calculations for a compressor driven by an electromagnet, focusing on the volumes and densities involved in the process. The initial calculations yield a density of 1.164 kg/m³ and a final density of 7.178 kg/m³ after the piston reaches position 2. Participants express skepticism about the provided solution, which suggests an unrealistic pressure increase of 2583 times for nitrogen, leading to a calculated pressure of 261,692 kPa. There is a consensus that the provided data may contain errors, as alternative calculations yield a more reasonable pressure of 625 kPa. Verification of the data is requested to clarify these discrepancies.
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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