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Compressible Fluid Flow

  1. Jun 23, 2015 #1
    I have a quick question regarding compressible fluid flow, specifically: for a given amount of time, how much pressure is lost when attaching a balloon valve to a small air tank. I decided to model my solution after the following problem (from an old fluid mechanics text book of mine),

    Air is extracted from a large tank in which the temperature and pressure are 70C and 101 kPa (abs), respectively, through a nozzle. At one location in the nozzle the static pressure is 25 kPa and the diameter is 15 cm. What is the mass flow rate? Assume isentropic flow.

    and have solved up to the mass flow rate for my case (with the exception of rho). From here I would like to convert from my current mass flow rate to mass loss and eventually to pressure loss. I have assumed that: my case is a compressible fluid flow problem, the air contained within the tank can be considered an ideal gas, flow is isentropic (my analysis needs only to be a general estimate of the pressure lost), and standard temperature and pressure.

    My original plan was to calculate the volumetric flow rate of the tank Q=V*A=(velocity*area)=dv/dt=(change in volume/change in time), integrate with respect to time and solve for the change in volume, then use p=nRT/v to find the pressure change. However, after some post-solving analysis, I came to the conclusion that, for my case, the volume is constant (seeing as how a tank will not expand under the pressures I am considering) and therefore my solution is not correct. I have so far solved for the velocity and area of the mass flow rate m_dot=rho*V*A, but unsure of how to proceed. Am I safe in using the density of air within the tank (standard temperature inside and outside the tank, standard atmospheric pressure outside the tank and 200 psi inside the tank) for my calculation of the mass flow rate. Moreover, how should I continue from here (total mass of air lost) to the total pressure loss during attachment? Thank you for your time.
  2. jcsd
  3. Jun 28, 2015 #2
    Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
  4. Jun 28, 2015 #3


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    It's not clear what problem you are trying to solve. All we know is that the pressure in the tank is 200 psi. Anything else? Size of the tank? How the tank is vented?

    BTW, the ratio of pressure inside the tank, 200 psi, to ambient pressure, 14.7 psi, suggests that some sort of choked flow might be created, depending on how the tank is vented. When flow is choked, there is a maximum amount of mass flow which occurs, and this rate remains constant until the pressure inside the tank is reduced sufficiently.
  5. Jun 29, 2015 #4


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    Umm... No. When the flow is choked, the mass flow remains constant no matter how much lower you drop the back pressure. As the total pressure drops in the tank, the mass flow will also drop directly proportionally to ##p_0## and inversely proportionally to ##\sqrt{T_0}##. Eventually, once the tank pressure gets low enough, this relationship changes.
  6. Jun 29, 2015 #5


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    That's what I said. In the choked condition, the mass flow rate cannot increase further, and remains constant until choking no longer occurs. Lowering the back pressure only means that choking will continue to occur. Choking disappears when the ratio of reservoir pressure to ambient pressure drops below a certain amount.

    Last edited: Jun 29, 2015
  7. Jun 29, 2015 #6
    Thank you for the replies! As far as the size of the tank goes, it is somewhat small (the actual size depends on the results of this analysis), but for now we can treat it as 36000 inches3 (49 inch diameter and 19 inch height). The tank is vented similar to this http://imgur.com/L8uvRDq [Broken], SteamKing, so the diameter for A* is .150 inches. I unfortunately do not have any other details for this project as it is intentionally ambiguous; I would ultimately like to be able to apply this analysis to a number of other tank cases, and so instead of solving for values, I am trying to create a series of equations to solve for the pressure loss as a result from the attaching of the balloon valve.

    Hopefully this rewording of the problem clarifies a bit. I have a 36000 inch3, 200 psi tank sitting at STP (standard temperature and pressure), and I wish to attach a balloon valve to the tank. The design of the balloon valve or vent for the tank are arbitrary so if the schematic I provided above over complicates the analysis, we can simplify the schematic. While the balloon valve is being attached, the pressure valve on the tank (similar in design to a Schrader valve on a bike) is briefly opened and a certain amount of air is allowed to escape. What I need to calculate is how much pressure (air) is lost when attaching this balloon valve. Assumptions I have made so far are that this is a compressible flow problem, the exit valve is a simple cylinder (in actuality the exit valve will have a small obstruction, so the air lost will be less than what is calculated), flow is isentropic, and standard temperature and pressure.

    Using the webpage that SteamKing provided, I have a general equation for the mass flow of my problem. The time taken for full attachment is unknown at this point in time, however I have estimated it to be about two seconds. From the mass flow and the time for full attachment I can solve for the total mass lost, however this is where I am confused on how to proceed. I know the mass of the air inside the tank prior to attachment, m = (density of air at standard temperature and 200 psi) * (volume of the tank), and the mass lost during attachment, however I do not know how to convert my mass change into a pressure change.
    Last edited by a moderator: May 7, 2017
  8. Jun 29, 2015 #7
    Alright, I have decided to solve this as follows: after solving for the mass flow rate during attachment, I used p = rho * R * T, with rho = m / v, to get delta_p = delta_rho * R * T = (delta_m / v) * R * T (equation 1). Using the mass flow rate from above, I multiplied this by the estimated time taken to attach the valve (two seconds) and used this new value, total mass lost during attachment, in my delta_p equation (equation 1) along with previous values of R and T to solve for the change in pressure. Am I correct in how I ended my analysis?
  9. Jun 29, 2015 #8


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    I'm attaching some quick calculations for your scenario:
    Code (Text):

    Air in Tank

    P = 200 psig

    V = 36000 in^3  (35,829)

    g = 32.2 ft/s^2

    60 F = 520 R
    14.7 psia

    dia.:  0.15 in
    length = 3 x 0.35 in = 1.05 in

    A = 0.0177 in^2 = 1.227 E-4 ft^2 for fitting

    f = 0.039 (approx)

    K = 1.5 (std. entr. & exit)
    fL/d = 0.039 * 1.05 / 0.15
    fL/d = 0.27

    Ktot = 1.77

    M = 29 lbm/lb-mol for air
    k = 1.4 for air

    R = 1545 ft-lbf/R-lb-mol

    R' = R / M = 53.28 ft-lbf / R-lbm

    rho = (144 P) / (R'T)

    rho = 144 (14.7) / (53.28 * 520) = 0.0764 lbm/ft^3 for air at P = 14.7 psia, T = 60 F

    rho = 144 (214.7) / (53.28 * 520) = 1.116 lbm/ ft^3 for air at P = 200 psig, T = 60 F

    Tank contains 36000 / 1728 = 20.833 ft^3 of air at 200 psig, 60 F

    mass of air in tank initially = 20.833 ft^3 * 1.116 lbm / ft^3 = 23.25 lbm air

    c^2 = k*g*R'T  c = speed of sound in a gas

    c^2 = 1.4* 32.2 * 53.28 * 520; c = 1117.6 ft/sec

    @ delta t = 2 sec

    delta m = 2 * 1117.6 ft/s * 1.227E-4 ft^2 * 1.116 lbm/ft^3

    delta m = 0.306 lbm of air

    At 14.7 psia and 60 F

    PV = mR'T

    V = mR'T/P = 0.306 * 53.28 * 520 / (144*14.7) = 4.01 ft^3

    for K = 1.77, the pressure ratio for choked flow is approx.

    Delta P / P1 = 0.595  From Crane TP-410, p. A-22

    Delta P = 0.595 * 214.7 = 127.8 psi

    choked flow occurs until P = 86.6 psia = 72.2 psig in the tank
    This is a rough calculation of the amount of air released in a 2-sec burst when the internal pressure of the tank is 200 psig.

    After each release, the total mass of air in the tank is reduced incrementally, and the pressure remaining in the tank can be calculated using PV = mR'T, where m is the amount of air still in the tank.

    As you can see, the flow from the tank remains choked until the pressure in the tank drops to about 70 psig.
  10. Jul 1, 2015 #9


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    But it doesn't remain constant. That's my point. Since the total pressure in the reservoir is getting lower, the mass flow rate is also getting lower. Even while it remains choked, the mass flow rate can change when either ##p_0## or ##T_0## change. That is the case here. The fact that it remains choked just means the mass flow rate continues to obey the relatively simple relationship with these variables that choked flow implies.

    Also, you can solve this analytically. You can easily find the relationship between total temperature, pressure, and mass flow rate, so you can get ##\dot{m} = f(p_0,T_0)## and you can convert that into ##\dot{m} = g(m)## if you assume ##T_0## is constant, where ##g## is linear in ##m##. That's a pretty simple differential equation. It's easily solvable to get a complete time trace up until the flow unchokes. You could make it more complicated by making different assumptions or models of ##T_0## to get a more accurate answer. Then it's ##\dot{m} = g(m,T_0)##.
    Last edited: Jul 1, 2015
  11. Sep 22, 2016 #10
  12. Sep 22, 2016 #11


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    As bonh3ad has pointed out, the flow rate will immediately start declining because in nozzle critical flow the flow rate is proportional to the nozzle (orifice) inlet pressure and at some point as the balloon nozzle is installed the pressure between the tank valve and the balloon valve the orifice will rise to the point that subcritical flow will result and finally the flow rate will continue to decline and totally cease at the point that the pressure between the tank valve and the balloon fill valve equalizes with the tank. Ultimately, there will two sources of tank pressure loss, the first will be the vented air as the balloon fill valve is attached the other will be the amount of air to fill the volume between the two valves.
    Considering the above, it appears to me that it makes sense (from a practical engineering standpoint) to determine if the loss is significant enough to warrant a rigorous calculation of this magnitude; and, this can be done by calculating the worst case amount of loss assuming the .150 diameter tank valve orifice is flowing at the maximum tank 200 psig pressure for the entire 2 second flow time.

    Taking that assumption, I have utilized a program with a standard critical nozzle flow formula to establish the results of this type of venting 200 psig inlet pressure air through a .150 diameter orifice and the volumetric conversions to determine the tank pressure loss simplified by assuming that all results will ultimately stabilize at the same temperature as before the flow started.

    By this process the results are:
    Nozzle flow rate (scfm) = 67.23
    Total volume lost in 2 sec (acf) = 0.153
    Tank Volume (cu ft) = 20.833
    Tank Air Volume @ 200 psig (acf) = 304.28
    Percent of Tank Volume/Pressure Loss = 0.050%
    Residual Tank Pressure (psig) = 199.90

    So based upon that result, is a rigorous calculation really required?
    Note: it would have been more appropriate to use helium for the calculation but the program does not include helium in its gas selections.
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