Reduce the pressure in an air reservoir

In summary: You substitute using the ideal gas law to remove density from the RHS.I wrote ##\pm## to let you decide on the sign. What can you substitute for ##\rho_{\textrm{air}}R'T##?Well, p is pressure, T is temperature, V is volume, so I'm not sure. I'm sorry, I don't understand your question. You should be able to express ##dp/dt## in terms of just ##p## and some constants.Sorry for the confusion. Let me explain it in more detail.You have the equation ##\dfrac{dp}{dt}=\pm\dfrac{\dot{V}}{V}\left(\rho_{\textrm
  • #1
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Homework Statement
An air reservoir at ##T=20\, \textrm{ºC}## and ##p_0=150\, \textrm{kPa}## is emptied by a pump. The volume of the reservoir is ##V=1,5\, \textrm{m}^3##, and the pump evacuates ##\dot{V}=90\, \textrm{l}/\textrm{min}## of air, independent of pressure. Assuming that air is an ideal gas and the process isothermal, determine the time it takes to reduce the pressure to ##p=50\, \textrm{kPa}##.

Solution: ##t=18,3\, \textrm{min}##
Relevant Equations
##\dfrac{d}{dt}\int_{VC}\rho\, dV=\dot{m}##
TRANSITIONAL REGIME: ##\dot{V}=90\, \textrm{L}/\textrm{min}=0,0015\, \textrm{m}^3/\textrm{s}##
$$\dfrac{d}{dt}\int_{VC}\rho\, dV=\dot{m}\rightarrow \dfrac{d}{dt}\rho V=\dot{m}\rightarrow $$
$$\boxed{\rho_{\textrm{air}}}\rightarrow pV=R'T\rightarrow \rho =\dfrac{p}{R'T}\rightarrow 287$$
$$\rightarrow \dfrac{dp}{dt}=\dfrac{\dot{m}R'T}{V}=\dfrac{\rho_{\textrm{air}}\dot{V}R'T}{V}=103,011\, \textrm{Pa}/\textrm{s}$$
$$50000=150000-\dfrac{dp}{dt}t\rightarrow \boxed{t=16,67\, \textrm{min}}$$
$$\rho_{\textrm{average}}\rightarrow p_{\textrm{avg}}=100000\rightarrow \rho_{\textrm{avg}}=1,189\, \textrm{kg}/\textrm{m}^3$$
Would it be well done like this?
 
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  • #2
Guillem_dlc said:
##\dot{V}=90\, \textrm{L}/\textrm{min}=0,0015\, \textrm{m}^3/\textrm{s}##
Ok

Guillem_dlc said:
$$\boxed{\rho_{\textrm{air}}}\rightarrow pV=R'T\rightarrow \rho =\dfrac{p}{R'T}\rightarrow 287$$
This doesn't look right. Did you mean to write ##pV = mR'T##, where ##R'## is a constant?

Then ##\rho = \dfrac{p}{R'T}## looks OK. However, ##\rho## clearly changes with time as the amount of air in the tank decreases. So it's not clear how you got 287. What are the units and for what value of time did you calculate this? Did you ever make use of this?

Guillem_dlc said:
$$\rightarrow \dfrac{dp}{dt}=\dfrac{\dot{m}R'T}{V}=\dfrac{\rho_{\textrm{air}}\dot{V}R'T}{V}=103,011\, \textrm{Pa}/\textrm{s}$$
The equation ## \dfrac{dp}{dt}=\dfrac{\dot{m}R'T}{V}=\dfrac{\rho_{\textrm{air}}\dot{V}R'T}{V}## looks OK except for a sign issue. ##\frac{dp}{dt}## and ##\dot m## are negative since the pressure and the mass of the air in the tank decrease with time. But your expression after the second equal sign is positive since ##\dot V## is specified as a positive number.

Note that ## \dfrac{dp}{dt}## is not a constant. You can see this by examining ##\dfrac{\rho_{\textrm{air}}\dot{V}R'T}{V}## which is clearly time-dependent due to the time dependence of ##\rho_{\textrm{air}}##. So, the numerical value that you calculated for ## \dfrac{dp}{dt}## will not be useful.

Guillem_dlc said:
$$50000=150000-\dfrac{dp}{dt}t\rightarrow \boxed{t=16,67\, \textrm{min}}$$
This equation assumes that ##\frac{dp}{dt}## is constant. But, as mentioned above, it is time-dependent.My hint would be to work with your equation ## \dfrac{dp}{dt}= \pm \dfrac{\rho_{\textrm{air}}\dot{V}R'T}{V}## and write it as ## \dfrac{dp}{dt}= \pm \dfrac{\dot{V}}{V}\left(\rho_{\textrm{air}}R'T\right)##.

I wrote ##\pm## to let you decide on the sign. What can you substitute for ##\rho_{\textrm{air}}R'T##?
 
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  • #3
TSny said:
Then ##\rho = \dfrac{p}{R'T}## looks OK. However, ##\rho## clearly changes with time as the amount of air in the tank decreases. So it's not clear how you got 287. What are the units and for what value of time did you calculate this? Did you ever make use of this?
It looks like they might have used the average density of the air in the tank ## \rho_{air} \approx \rho_{avg}##. I see it in the last line of the OP. If I did the calculation correct, it's not a terrible approximation.

The “287” looks suspiciously like the air specific mass based ##R## value.
 
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  • #4
erobz said:
It looks like they might have used the average density of the air in the tank ## \rho_{air} \approx \rho_{avg}##. I see it in the last line of the OP. If I did the calculation correct, it's not a terrible approximation.

The “287” looks suspiciously like the air specific mass based ##R## value.
Yes, that looks like what they did. It approximates the answer with less than 10% error.
 
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  • #5
TSny said:
Yes, that looks like what they did. It approximates the answer with less than 10% error.
I did that yes
 
  • #6
Guillem_dlc said:
I did that yes
OK. The reason your answer is off somewhat is that the value of ##\rho_{\rm avg}## that you used is not the true time average of ##\rho##. This is because ##\rho## does not decrease linearly with time.

If you want to get the correct answer of 18.3 minutes for the time, then try following the hint at the end of post #2. You should be able to express ##dp/dt## in terms of just ##p## and some constants.
 
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  • #7
TSny said:
OK. The reason your answer is off somewhat is that the value of ##\rho_{\rm avg}## that you used is not the true time average of ##\rho##. This is because ##\rho## does not decrease linearly with time.

If you want to get the correct answer of 18.3 minutes for the time, then try following the hint at the end of post #2. You should be able to express ##dp/dt## in terms of just ##p## and some constants.
OK, I'll try again now, it would be instead of the average density to replace it with ideal gases, wouldn't it?
 
  • #8
Guillem_dlc said:
OK, I'll try again now, it would be instead of the average density to replace it with ideal gases, wouldn't it?
You substitute using the ideal gas law to remove density from the RHS.
TSny said:
I wrote ##\pm## to let you decide on the sign. What can you substitute for ##\rho_{\textrm{air}}R'T##?
 
  • #9
erobz said:
You substitute using the ideal gas law to remove density from the RHS.
OK yes, thank you! And the ##p## as a function of ##t## is the same? I mean the ##p## that I then put in the formula.
 
  • #10
Guillem_dlc said:
OK yes, thank you! And the ##p## as a function of ##t## is the same? I mean the ##p## that I then put in the formula.
You mean is ##p## the same as the pressure inside the tank? Yeah.
 
  • #11
erobz said:
You mean is ##p## the same as the pressure inside the tank? Yeah.
I mean here the pressure varies from ##150## to ##50##, then how should I put the pressure when I substitute the density for the ideal gases?
 
  • #12
Guillem_dlc said:
I mean here the pressure varies from ##150## to ##50##, then how should I put the pressure when I substitute the density for the ideal gases?
Quite literally ##p##. The resulting equation is first order separable ODE.
 
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  • #13
The number of moles per unit volume in the tank at any time is ##\frac{p}{RT}##, so the rate of change of the number of moles in the tank is: $$\frac{dn}{dt}=-\frac{pV'}{RT}$$The rate of change of pressure is $$\frac{dp}{dt}=\frac{RT}{V}\frac{dn}{dt}$$If we combine these two equations, we obtain:$$\frac{dp}{dt}=-\frac{pV'}{V}$$
 
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  • #14
One thing which is not related: if they give me the relative density of a gas, should I multiply by the density of air or water?
 
  • #15
Guillem_dlc said:
One thing which is not related: if they give me the relative density of a gas, should I multiply by the density of air or water?
What is your definition of "relative density." In this problem, you are working with molar density, not mass density.
 
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  • #16
Chestermiller said:
What is your definition of "relative density." In this problem, you are working with molar density, not mass density.
Density divided by density of the medium, right?
Yes yes, in the problem we worked with molar density, it was a question that is not related to the exercise.
 
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  • #17
$$\rho_{relative}=\dfrac{\rho_{substance}}{\rho_{reference}}$$
Not like this?
 
  • #18
Guillem_dlc said:
$$\rho_{relative}=\dfrac{\rho_{substance}}{\rho_{reference}}$$
Not like this?
I was about to pontificate on how this is just like specific gravity. Then I visited Wikipedia. Someone else had already pontificated.
https://en.wikipedia.org/wiki/Relative_density said:
Relative density, or specific gravity,[1][2] is the ratio of the density (mass of a unit volume) of a substance to the density of a given reference material. Specific gravity for liquids is nearly always measured with respect to water at its densest (at 4 °C or 39.2 °F); for gases, the reference is air at room temperature (20 °C or 68 °F). The term "relative density" (often abbreviated r.d. or RD) is often preferred in scientific usage, whereas the term "specific gravity" is deprecated.
[...]
Relative density (##{\displaystyle RD}##) or specific gravity (##{\displaystyle SG}## is a dimensionless quantity, as it is the ratio of either densities or weights

$${\displaystyle {\mathit {RD}}={\frac {\rho _{\mathrm {substance} }}{\rho _{\mathrm {reference} }}},}$$
 
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