Unsteady filling of a vacuum tank

Summary
A 60 liter vacuum tank with an initial vacuum of -509 mmHg (251 mmHg absolute) is opened up to the outside air and flows through a 6.5 mm hole.
What is the equation that represents the pressure in terms of time?
What is the equation that represents volumetric flow-rate into the tank in terms of time?
I have a vacuum tank at work that I am trying to determine the flow rate and pressure curve of when I open up the tank and I just cannot figure these equations out.

Environment: average office space
Atmospheric pressure: 29.92 inHg (per weather information)
Temperature: 72°F

Tank info:
Material: Stainless steel
Tank capacity: 60L (I measured it to be 59.5 liters)
Initial vacuum: -509 mmHg (251 mmHg absolute)
Entrance hole diameter: 6.5 mm

Thermodynamics isn't my strong suit (vibrations and dynamics are). After going back through my thermodynamics and fluid dynamics I'm still having problems. So I ran a test on it to see what the real world values were. I used my iPhone and filmed the pressure gauge in slow motion as I opened the tank up. I found that the pressure rose in a very linear manner (R^2=0.995) in the first 4.5 seconds with a slope of 10165 pascals/second. If I take the partial derivative of PV=mRT I get, dm/dt = dp/dt * V/(RT). Plugging in the values above I get the experimental mass flow rate of 10165*0.059528/(287*294.261) = 0.0072 kg/second. That's the mass flow rate but how do I get the volumetric flow-rate? I'm making the assumption that the air is compressing and then expanding which changes the density of the air. So I don't know what the density of the air is entering the tank otherwise I would just divide the mass flow rate by the density to get the volumetric flow rate.

I was also making the assumption that the process was isothermal but then I stumbled upon this:
http://web.mit.edu/16.unified/www/FALL/thermodynamics/notes/node17.html#fig0:TankFilling
which says that the air would have an increase in temperature of at least 200°F... that sounds pretty insane which means that I also have to factor in heat transfer?

I need help of understanding the physics at play here.
 
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There are 2 parts to this. First, to use thermodynamics to get the final conditions in the tank. Second, to analyze the fluid flow aspect of the problem. You certainly won't be able to do the second part without first being able to do the first part. So, my advice is to first determine the final conditions in the tank.

Assume the system is adiabatic so that you don't need to deal with heat transfer between the gas and the tank. Let u be the internal energy per mole of the gas in the tank in the final state, and let ##u_0## be the internal energy per mole of the gas in the initial state. Let ##n_0## be the number of moles of gas in the tank in the initial state and let ##\Delta n## be the number of moles of gas entering the tank between the initial and final states. Let ##h_{in}## be the enthalpy per unit mass of gas entering the tank. In terms of only these parameters, please write down the open system (control volume) version of the first law of thermodynamics applicable to the change in this system between the initial and final states.
 
Thanks for the reply. I'm guessing that this diagram is a correct representation of the system?
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Based on the first law of thermodynamics you get
243726

Since you are saying that I should assume that it is adiabatic, that means that there is no heat transfer into the system so Q=0. Since there is no kinetic or potential energy of the final state, all that is left is internal energy. Finally, there is work done on the system that is equal to the pressure times the volume. Thus, the equation should simplify down to:
243734

Where M is the molar mass of air
If I use the equations for internal energy of air and enthalpy of air and plug them in I would get
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If I plug in the values I get the final temp. of the system equal to 413.5 Kelvin which is 283.73°F and that seems unlikely which leads me to believe that I made a mistake.
 
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Here's my starting analysis. Please see if it makes sense to you based on the open-system version of the 1st law of thermodynamics: $$(n_i+\Delta n)u_f-u_in_i=\Delta nh_{in}$$where $$u_f=u_i+C_v(T_f-T_i)$$and, since the temperature outside is equal to the initial temperature inside,$$h_{in}=u_i+P_{atm}v_{atm}=u_i+RT_i$$
 
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I was mistaken about something. This problem doesn't only have to be treated as an open-system problem. It can also be treated using the closed system version of the 1st law. Here is the development using that approach.

Again, let ##n_i## represent the initial number of moles in the tank, and let ##\Delta n## again represent the number of moles forced into the tank by the outside atmosphere. Then, our closed system will be comprised of these ##n_i+\Delta n## moles. The change in internal energy of these moles of air is ##(n_i+\Delta n)(u_f-u_i)##. The work done by this combined system on the surrounding air is ##W=p_{atm}\Delta V=-\Delta n RT_i##. So, from the closed system version of the first law, we have: $$(n_i+\Delta n)(u_f-u_i)=-W=\Delta nRT_i$$This is the same result as that produced by the open system analysis of post #4.

Basically, your analysis was pretty close to correct to begin with. You just forgot to take into account the gas that was originally present in the tank.
 
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When applying this analysis to your problem, I get a final temperature of 198 F. See if you can get ##T_f/T_i## in terms of ##\gamma## and ##p_i/p_{atm}## and then see if you can duplicate this result.
 
So here are my equations
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And here are the initial conditions

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You can solve Eq. 2,4,7 outright getting:
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I then solve for n in equation 5. Plug that into equation 6. I then plug equation 6 into equation 3 and also plug equation 1 into equation 3 and then solve for the final temperature. I get
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Which is the same as what you got. Thus, that means that the temperature jumped over 100F?!
I just measured the surface area of the tank and got about 1.006 m^2. I also learned that the tank material was aluminum so I'm guessing aluminum 6061. If I put my hand on the tank while this is occurring, I would feel some heat correct? Even if I were to calculate the heat flux, I don't have enough real world experience to determine what that feels like.

Anyway, what would be next for calculating the transient information? Based on the experimental data, I got the fill time to be exactly 8.76 seconds and found that data followed the curve
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with an average relative error of 0.3622 %
Where y is the pressure in the tank and x is the time.
 
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I finally was able to simplify the equation down:
243917

Now, you did also tell me to add gamma to the equation and simplify. Thus,
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I finally was able to simplify the equation down:
View attachment 243917
Now, you did also tell me to add gamma to the equation and simplify. Thus,
View attachment 243922
Very nicely done.

To do the transient problem, we need to have a relationship between the pressure drop between inside and outside the tank and the molar flow rate through the orifice. From the information presented in Transport Phenomena by Bird, Stewart, and Lightfoot, it is not clear what the appropriate functional form should be employed for an orifice through which a compressible gas is flowing. So, I'm going to propose a functional form, and we can see how well that works:
$$(p_{atm}-p)=\frac{k}{2\bar{\rho}}\frac{\dot{m}^2}{A^2}$$ where A is the cross sectional area of the hole, ##\bar{\rho}## is the average density (##=\frac{M}{2R}\left[\frac{p_{atm}}{T_{atm}}+\frac{p}{T}\right]##), ##\dot{m}=\dot{n}M## is the mass flow rate through the orifice, and k is a (dimensionless) constant on the order of unity, with ##\dot{n}## equal to the molar flow rate and M representing the molecular weight of air. This equation is dimensionally correct and resembles to some extent the equations recommended in Bird, et al for flow through valves and orifices.

So, for a given pressure difference between the outside air and the air in the tank, we can use this equation to calculate the molar flow rate. The constant k will be an adjustable parameter that can be used in tuning the model.
 
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I just discovered something very interesting. I opened up my Fundamentals of Fluid Mechanics Seventh Edition textbook from college, turned to the last chapter about compressible flow and low and behold, all the equations that I just listed out were there. 🤦‍♂️ That would have helped significantly and saved me a lot of time if I found it earlier. However, further in the chapter, I did find an example problem in the textbook called Isentropic Flow in a Converging Duct. It states:
A converging duct passes air steadily from standard atmospheric conditions to a receiver pipe as illustrated:
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The throat (minimum) flow cross-sectional area of the converging duct is 1e-4 m^2. The receiver is (a) 80 kPa (abs), (b) 40 kPa (abs)

They then walk you through the problem. The really interesting part comes from when it says:
If p_receiver < p*, then p_throat = p* and the flow is choked. Where:
243932

Thus, the flow at the throat is traveling at the speed of sound!

Now, obviously, my problem isn't steady flow. However, the cross section of the orifice in my experiment is 3.318e-5 m^2 and the tank's initial pressure is 33.4 kPa (abs), both of which are smaller than the example problem. Does that mean that the initial speed of the air in my experiment was going at the speed of sound?!?!?! If so, that's awesome! That would also mean that the initial pressure values that I read were not correct because you can't measure the pressure of a system upstream of a flow that is traveling at the speed of sound. Would you concur with this evaluation?
 
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I just discovered something very interesting. I opened up my Fundamentals of Fluid Mechanics Seventh Edition textbook from college, turned to the last chapter about compressible flow and low and behold, all the equations that I just listed out were there. 🤦‍♂️ That would have helped significantly and saved me a lot of time if I found it earlier. However, further in the chapter, I did find an example problem in the textbook called Isentropic Flow in a Converging Duct. It states:
A converging duct passes air steadily from standard atmospheric conditions to a receiver pipe as illustrated:
View attachment 243929
The throat (minimum) flow cross-sectional area of the converging duct is 1e-4 m^2. The receiver is (a) 80 kPa (abs), (b) 40 kPa (abs)

They then walk you through the problem. The really interesting part comes from when it says:
If p_receiver < p*, then p_throat = p* and the flow is choked. Where:
View attachment 243932
Thus, the flow at the throat is traveling at the speed of sound!

Now, obviously, my problem isn't steady flow. However, the cross section of the orifice in my experiment is 3.318e-5 m^2 and the tank's initial pressure is 33.4 kPa (abs), both of which are smaller than the example problem. Does that mean that the initial speed of the air in my experiment was going at the speed of sound?!?!?! If so, that's awesome! That would also mean that the initial pressure values that I read were not correct because you can't measure the pressure of a system upstream of a flow that is traveling at the speed of sound. Would you concur with this evaluation?
The smaller the hole, the more frictional dissipation. By assuming that the gas enters the tank with the same enthalpy as outside, we have been saying that frictional dissipation is the main contributor to the pressure decrease through the hole. We can check afterwards to ascertain whether or not the flow would become sonic and choke. I don't think this will be the case.
 
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Your data suggests that, at least initially, the flow might have been close to sonic. Just take the number of moles, divide by the initial molar volume (minimum), the total time, and the hole area. This comes out to about 200 m/s. This doesn't take into account for the fact that the initial molar flow rate was higher than the average molar flow rate.
 
Can you please post a graph of this? Thanks.
I used this website:
https://planetcalc.com/5992/
It's a very quick way to run different regressions to determine the best one.

X-Data:
Code:
0.013020875 0.108853125 0.17968575 0.308333375 0.475516125 0.592181625 0.688014 0.79634625 0.867178875 0.983844375 1.067176875 1.183842375 1.254675 1.375507125 1.4713395 1.588005 1.69633725 1.775503125 1.89633525 1.967167875 2.100499875 2.1713325 2.275498125 2.358830625 2.500495875 2.54632875 2.667160875 2.74632675 2.867158875 2.9379915 3.054657 3.1379895 3.258821625 3.350487375 3.44631975 3.554652 3.64631775 3.75465 3.8379825 3.942148125 4.0379805 4.133812875 4.242145125 4.333810875 4.42964325 4.5379755 4.62964125 4.72964025 4.8379725 4.92963825 5.025470625 5.121303 5.225468625 5.321301 5.42963325 5.521299 5.61296475 5.725463625 5.817129375 5.925461625 6.008794125 6.11295975 6.208792125 6.3046245 6.40962275 6.50120525 6.59828775 6.70637025 6.796875125 6.89537425 6.98970675 7.08678925 7.18937175 7.28645425 7.38078675 7.47236925 7.58045175 7.67753425 7.76911675 7.86619925 7.97428175 8.102314625 8.160647375 8.385645125 8.55385275 8.766350625
Y-Data:
Code:
33997.20395 34930.46053 35997.03947 36930.29605 39463.42105 40396.67763 41596.57895 42663.15789 43729.73684 44929.63816 45996.21711 46796.15132 47596.08553 48662.66447 49462.59868 50662.5 51595.75658 52529.01316 53595.59211 54662.17105 55595.42763 56528.68421 57595.26316 58528.51974 59595.09868 60528.35526 61461.61184 62394.86842 63594.76974 64528.02632 65594.60526 66661.18421 67594.44079 68527.69737 69594.27632 70794.17763 71860.75658 72660.69079 73727.26974 74660.52632 75460.46053 76393.71711 77193.65132 78126.90789 78926.84211 79726.77632 80526.71053 81193.32237 81993.25658 82793.19079 83593.125 84259.73684 85059.67105 85726.28289 86392.89474 87059.50658 87726.11842 88392.73026 89059.34211 89859.27632 90392.56579 90925.85526 91592.46711 92125.75658 92659.04605 93192.33553 93858.94737 94258.91447 94792.20395 95325.49342 95725.46053 96258.75 96658.71711 97192.00658 97458.65132 97858.61842 98258.58553 98658.55263 98925.19737 99325.16447 99591.80921 99858.45395 100258.4211 100658.3882 101058.3553 101325
Attached is the raw data in excel format for your convenience.
 

Attachments

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Thanks. This is great. You can use this to calibrate the model. I will outline how later.
 
Just found the latex info, lol.
So you are proposing the following equation
$$(p_{atm}-p)=\frac{k}{2\bar{\rho}}\frac{\dot{m}^2}{A^2}$$
Where
$$\bar{\rho}=\frac{M}{2R}\left[\frac{p_{atm}}{T_{atm}}+\frac{p}{T}\right]$$
I'm assuming ##T## is the final temperature and ##p## is the initial pressure correct? Plugging in the values, I get ##\bar{\rho}=0.757 \frac{kg}{m^3}##
Solving for ##\dot{m}## in the first equation I get $$\dot{m}=\sqrt{1.132\mathrm{x}10^{-7} k}$$$$\dot{n}=1.161\mathrm{x}10^{-5}\sqrt{k}=1.161\mathrm{x}10^{-5}k$$
Now, I'm a little confused about the equation for finding the initial velocity. I'm confused on what you mean by the initial molar volume. My best guess is that it is ##\Delta V## and so $$\Delta V=\frac{\Delta n R T_i}{P_{atm}}=0.02848 \mathrm{m}^3$$$$\bar{V}=\frac{\Delta n}{\Delta V t_{total}A}=141.89\frac{\mathrm{m}}{\mathrm{s}}\mathrm{?}$$
 
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Just found the latex info, lol.
So you are proposing the following equation
$$(p_{atm}-p)=\frac{k}{2\bar{\rho}}\frac{\dot{m}^2}{A^2}$$
This is just a tentative choice until we see how well it fits your data.
Where
$$\bar{\rho}=\frac{M}{2R}\left[\frac{p_{atm}}{T_{atm}}+\frac{p}{T}\right]$$
I'm assuming ##T## is the final temperature and ##p## is the initial pressure correct?
T and p are supposed to be the temperature and pressure at time t. Henceforth, u, n, T, and p are the internal energy per mole, the number of moles, the temperature, and the pressure at time t.

In our previous development, we showed that $$n(u-u_i)=(n-n_i)RT_i$$ where ##T_i## is atmospheric temperature. This leads to the equation: $$nC_v(T-T_i)=(n-n_i)RT_i$$which then yields:
$$T=T_i\left(\gamma-(\gamma-1)\frac{n_i}{n}\right)\tag{1}$$

The pressure in the tank at any time can now be determined by the ideal gas law:
$$p=\frac{nRT}{V_T}$$If we substitute the temperature T from Eqn. 1 into this relationship, we obtain: $$p=\frac{RT_i}{V_T}(\gamma n-(\gamma-1)n_i)\tag{2}$$This equation indicates that the pressure is a linear function of the number of moles in the tank. We can differentiate Eqn. 2 with respect to time to obtain the rate of change of pressure in the tank as a function of the rate of change in the number of moles: $$\frac{dp}{dt}=\frac{\gamma RT_i}{V_T}\frac{dn}{dt}$$This can be used to express the rate of change of the number of moles in the tank in terms of the rate of change of pressure: $$\frac{dn}{dt}=\frac{V_T}{\gamma RT_i}\frac{dp}{dt}\tag{3}$$And, it also follows that the mass rate of flow through the hole is given by:
$$\dot{m}=\frac{V_TM}{\gamma RT_i}\frac{dp}{dt}\tag{4}$$

In your experiments, you found that the initial rate of pressure increase in the tank is about 75 mmHg/sec = 0.1 atm /sec. From Eqns. 3 and 4, what does this give you for the molar flow rate and the mass flow rate? This should also provide us enough information to estimate the value of the dimensionless constant k in the flow equation. What do you get for k? (I estimated a value of about 4.5)
 
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See this article: https://en.wikipedia.org/wiki/Orifice_plate

I like the pressure-difference vs flow relationship presented in this reference better than the one I proposed in post #16, because the effect of compressibility is more fundamentally based. So I recommend switching to it:
$$\dot{m}=C\epsilon A \sqrt{2\rho_{atm}(p_{atm}-p)}\tag{1}$$
where C is the discharge coefficient of the hole and ##\epsilon## is the correction factor for compressibility: $$\epsilon=0.649+0.351\left(\frac{p}{p_{atm}}\right)^{1/\gamma}\tag{2}$$
This relationship can readily be manipulated into a form very similar to the one I proposed.

Your experimental data on the initial rate of decrease of pressure can be used to determine the value for the discharge coefficient C. I calculate a value of 0.48. What do you obtain?
 
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Let's try this again.
$$\bar{\rho}=\frac{M}{2R}\left[\frac{p_{atm}}{T_{atm}}+\frac{p}{T}\right] $$
Since ##\bar{\rho}## is the average density, then we need to plug in the initial conditions ##p=33.463\mathrm{ kPa}## and ##T=295.372\mathrm{ K}##. ##\bar{\rho}=0.7950\frac{\mathrm{kg}}{\mathrm{m}^3}##
$$\frac{dn}{dt}=\frac{V_T}{\gamma RT_i}\frac{dp}{dt} $$
For the initial conditions,
$$\frac{dp}{dt}=10.496\frac{\mathrm{kPa}}{\mathrm{s}}\mathrm{; }\frac{dn}{dt}=0.0001818\frac{\mathrm{kmol}}{\mathrm{s}}\mathrm{; }\dot{m}=0.005265\frac{\mathrm{kg}}{\mathrm{s}}$$
Plugging all the values into the equation:
$$(p_{atm}-p)=\frac{k}{2\bar{\rho}}\frac{\dot{m}^2}{A^2}$$
and solving for ##k##, I get ##k=0.00428##. So maybe you forgot a 0?
 
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Let's try this again.
$$\bar{\rho}=\frac{M}{2R}\left[\frac{p_{atm}}{T_{atm}}+\frac{p}{T}\right] $$
Since ##\bar{\rho}## is the average density, then we need to plug in the initial conditions ##p=33.463\mathrm{ kPa}## and ##T=295.372\mathrm{ K}##. ##\bar{\rho}=0.7950\frac{\mathrm{kg}}{\mathrm{m}^3}##
$$\frac{dn}{dt}=\frac{V_T}{\gamma RT_i}\frac{dp}{dt} $$
For the initial conditions,
$$\frac{dp}{dt}=10.496\frac{\mathrm{kPa}}{\mathrm{s}}\mathrm{; }\frac{dn}{dt}=0.0001818\frac{\mathrm{kmol}}{\mathrm{s}}\mathrm{; }\dot{m}=0.005265\frac{\mathrm{kg}}{\mathrm{s}}$$
Plugging all the values into the equation:
$$(p_{atm}-p)=\frac{k}{2\bar{\rho}}\frac{\dot{m}^2}{A^2}$$
and solving for ##k##, I get ##k=0.00428##. So maybe you forgot a 0?
I confirm the mass flow rate and the density. Did you use Pa (not kPa) for pressure?
 
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We're almost done now. If we combine Eqn. 4 from post #17 with Eqn. 1 from post #18, we obtain:

$$\frac{dp}{dt}=\frac{\gamma RT_{atm}}{MV_T}C\epsilon A \sqrt{2\left(\frac{p_{atm}M}{RT_{atm}}\right)(p_{atm}-p)}\tag{1}$$with $$\epsilon=0.649+0.351\left(\frac{p}{p_{atm}}\right)^{1/\gamma}\tag{2}$$
The value of ##\epsilon## over the time for filling varies from 0.8 initially to 1.0 finally. If we substitute the average value of ##\epsilon##, ##\bar{\epsilon}=0.9##, into Eqn. 1, we can integrate this equation analytically to solve for the time to fill. See if you can do this. Based on this solution, I obtained a time to fill of about 9.2 sec.

To be more precise, the equation can be integrated numerically. Before doing this however, I would reduce the equation to dimensionless form.
 
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I would use the following dimensionless parameters:
$$\phi=1-\frac{p}{p_{atm}}$$
and
$$\tau=\frac{Av_s\sqrt{2\gamma}}{V_T}Ct$$where ##v_s## is the speed of sound in air:
$$v_s=\sqrt{\frac{\gamma RT_{atm}}{M}}$$
 
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With the substitutions given in the previous post, the differential equation for the dimensionless pressure variation with dimensionless time becomes:

$$\frac{d\phi}{d\tau}=-\left[0.649+0.351(1-\phi)^{1/\gamma}\right]\sqrt{\phi}$$subject to the initial condition $$\phi(0)=1-\frac{p_i}{p_{atm}}$$
 
With the substitutions given in the previous post, the differential equation for the dimensionless pressure variation with dimensionless time becomes:

$$\frac{d\phi}{d\tau}=-\left[0.649+0.351(1-\phi)^{1/\gamma}\right]\sqrt{\phi}$$subject to the initial condition $$\phi(0)=1-\frac{p_i}{p_{atm}}$$
Your starting to reach the limits of my knowledge here with that first-order nonlinear differential equation.

What I'm pretty sure of, is that the flow is a choked flow from ##p(0)<p<53530.58\mathrm{Pa}## which means that the mass flow rate is constant and so
$$\dot{m}=C_dA\sqrt{\gamma \rho_{atm}P_{atm}\left (\frac{2}{\gamma+1}\right )^{\frac{\gamma+1}{\gamma-1}}}$$$$\dot{m}=0.007902C_d$$From the experiment ##P=53530.58## @ ##t=1.896335## so:
$$\dot{m}=\frac{V_TM}{\gamma RT_i}\frac{dp}{dt}=\frac{V_TM}{\gamma RT_i}\frac{p(1.896)-p_i}{1.896-0}=0.005326\frac{\mathrm{kg}}{\mathrm{s}}$$$$C_d=0.6740$$
 
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Your starting to reach the limits of my knowledge here with that first-order nonlinear differential equation.

What I'm pretty sure of, is that the flow is a choked flow from ##p(0)<p<53530.58\mathrm{Pa}## which means that the mass flow rate is constant and so
$$\dot{m}=C_dA\sqrt{\gamma \rho_{atm}P_{atm}\left (\frac{2}{\gamma+1}\right )^{\frac{\gamma+1}{\gamma-1}}}$$$$\dot{m}=0.007902C_d$$From the experiment ##P=53530.58## @ ##t=1.896335## so:
$$\dot{m}=\frac{V_TM}{\gamma RT_i}\frac{dp}{dt}=\frac{V_TM}{\gamma RT_i}\frac{p(1.896)-p_i}{1.896-0}=0.005326\frac{\mathrm{kg}}{\mathrm{s}}$$$$C_d=0.6740$$
Let's try it both ways and see how the results compare with the data.
 

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