I Equalizing Pressure in Pipe for Safe Container Design

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The discussion focuses on the design of a pressure vessel with a vent to equalize pressure and prevent wall failure during rapid pressurization. The initial conditions include a pipe at atmospheric pressure and a container vented to allow air flow, with concerns about maintaining a pressure difference below 200 psi to avoid collapse. Participants analyze the flow dynamics, suggesting calculations based on the vent size and pressure changes, while emphasizing the need for experimental validation of flow rates. The conversation also touches on the complexities of differential equations related to pressure changes and flow rates, with a call for further assistance in solving these equations. Overall, the design's safety hinges on effective pressure equalization and accurate flow rate calculations.
  • #31
I've restructured the equation for this problem in a little different form so that we can solve directly for the pressure p (psi) in the inner chamber as a function of time: $$\frac{dp}{dt}=\left(\frac{(14.7+95t)-p^2}{G}\right)^{0.8333}$$where $$G=(0.0791)\frac{64LV}{\pi D^5}\left(\frac{MV}{RT}\right)^{0.2}\left(\frac{\pi D\mu}{4}\right)^{0.8}$$
The first step in getting a solution is to evaluate the constant G (once and for all) in units of ##(psi)^{0.8}(sec)^{1.2}##. Please provide a calculation of G in these units.

The initial condition on p is p = 14.7 psi.

Chet
 
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  • #32
I've spent some more time on this problem, and figured out a way of simplifying things even more. Rather than using the Blasius friction factor equation, since we are looking for an upper bound to the difference between the two pressures, we can use an upper bound of 0.01 for the friction factor f in the turbulent region. In this case, our pressure equation reduces to:
$$\frac{64LV}{\pi D^5}\left(\frac{MV}{RT}\right)(0.01)\left(\frac{dp}{dt}\right)^2=p^2(L,t)-p^2$$
From the data you provided, the absolute temperature is 403K. The universal gas constant is ##R=8.314\times 10^7## erg/(mole K), and the volume of the inner chamber is 106 in^3. So,
$$\left(\frac{MV}{RT}\right)=\frac{(29)(106)(2.54)^3}{(8.314\times 10^7)(403)}=5.92\times 10^{-7}\ sec^2\ cm$$
So, $$\frac{64LV}{\pi D^5}\left(\frac{MV}{RT}\right)(0.01)=\frac{(64)(100)(12)(2.54)(106)(2.54)^3}{(\pi)(0.625)^5(2.54)^5}(5.92\times 10^{-7})(0.01)=0.63\ sec^2$$So, the differential equation becomes:
$$\frac{dp}{dt}=\frac{\sqrt{(14.7+15t)^2-p^2}}{\sqrt{0.63}}$$where t is in seconds, p is in psi, and the initial value of p is 14.7 psi.
 
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  • #33
I integrated this equation, and found that the pressure in the inner chamber never differs from that outside by more than 3 psi, and this maximum difference occurs at about 0.8 seconds.
 
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  • #34
Are there some errors in the numbers you put in the above equations? If so, I think you may be human.
 
  • #35
anolan23 said:
Are there some errors in the numbers you put in the above equations? If so, I think you may be human.
Is there something specific that troubles you?
 
  • #36
The MV/RT equation doesn't compute to that value, unless it's (2.54)^2 instead of 2.54^3 for the conversion of the volume
 
  • #37
anolan23 said:
The MV/RT equation doesn't compute to that value, unless it's (2.54)^2 instead of 2.54^3 for the conversion of the volume
I'll try to dig out my notes to check on the algebra. I don't have it to go through this from scratch again.
 
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  • #38
anolan23 said:
Are there some errors in the numbers you put in the above equations? If so, I think you may be human.
I confirm the arithmetic error you identified. Are there any other errors that you have identified? Otherwise the 0.63 seconds should be 2.54 times as large. So the 0.8 sec. would be a little larger, and the 3 psi would be a little larger.
 
  • #39
I tried to solve the integral through SMath, but when calculating the Constant at the initial condition p(0), it says it cannot solve for C because there are no real roots.
http://imgur.com/a/mxSeB
 
  • #40
anolan23 said:
I tried to solve the integral through SMath, but when calculating the Constant at the initial condition p(0), it says it cannot solve for C because there are no real roots.
http://imgur.com/a/mxSeB
I had solved the equation numerically on a spread sheet. I no longer have the spread sheet file.
 
  • #41
what value should I use for μ, air viscosity?
 
  • #42
anolan23 said:
what value should I use for μ, air viscosity?
It is not necessary to specify a value. See post #31.
 
  • #43
Oh that was for the older solution.

How do you solve the equation numerically?
 
  • #44
Foward Euler with a time step of 0 01 sec
 
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  • #45
Okay same thing. Thx
 
  • #46
http://imgur.com/a/IfYvT

Chestermiller,
I'm doing the Forward Euler to estimate the solution, but when calculating the first solution it's making me take the square root of a negative number. Did you have this problem?
 
  • #47
anolan23 said:
http://imgur.com/a/IfYvT

Chestermiller,
I'm doing the Forward Euler to estimate the solution, but when calculating the first solution it's making me take the square root of a negative number. Did you have this problem?
Do the first two time steps by hand, and show me the work please.
 
  • #48
Nvm I made an error. It calculates out.

I adjusted the SQRT(0.63) that you calculated and it comes out to be SQRT(0.1608).

Attached is my numerical solution to the dp/dt equation. IT doesn't make sense that the pressure inside the container is already at 100psi at 1 sec since the rate of increase of pressure is 95 psi/sec. The last sheet is where I did it.
 

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  • #49
anolan23 said:
Nvm I made an error. It calculates out.

I adjusted the SQRT(0.63) that you calculated and it comes out to be SQRT(0.1608).
Isn't 0.63 times 2.54 equal to 1.6?
Attached is my numerical solution to the dp/dt equation. IT doesn't make sense that the pressure inside the container is already at 100psi at 1 sec since the rate of increase of pressure is 95 psi/sec. The last sheet is where I did it.
I thought you said that the rate of pressure rise was 15 psi/sec. Anyhow, if it were 95 psi/sec, wouldn't that mean that the pressure outside was 110 psi after 1 sec?

As I said earlier, I want to see your hand calculations for the first few time intervals. Until I see that, I'm not even going to look at the spreadsheet.[/QUOTE]
 
  • #50
0.63 X 2.54 is equal to 1.6, but there was another error in the denominator I believe.

Yes, 95psi/sec. You're right I didn't add 95+14.7 when comparing.

Let me know if you can't read the had calculations
 

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  • #51
##p_2=p_1+0.01f_1=14.7+(0.01)(6.765)=14.768##
 
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  • #52
yeah that's a mistake thx
 
  • #53
How can I account for gases other than air being used? Just change the R value?
 
  • #54
Instead of 29, you use the molecular weight of your other gas.
 
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  • #55
It's saying a lighter gas allows for a higher rate of pressurization. I would have thought the opposite
 

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