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Sizing Sizing a goose neck vent October 18 by jbz
This is This is one of those random things that came up, through work, where I was surprised surpri sed to find a relative relative void of info on the internet: sizing a gooseneck for an aboveground storage tank. storage tank. Venting Venting for aboveground storage tanks is dealt with in standards such as API 2000 , 2000 , which gives the venting as an equivalent flowrate of air at standard state. Most off the shelf venting ven ting you would want want is sized in terms of pressure drop and SCF/hr through it. For example for an e mergency vent from Morrison from Morrison brothers: (https://downc (https://downcomer.files.wordpress.com/2014/10/morrisson‑vents.png) omer.files.wordpress.com/2014/10/morrisson‑vents.png) (https://downcomer.files.wordpress.com/2014/10/morrisson‑venting.png) (https://downcomer.files.wordpress.com/2014/10/morrisson‑venting.png) But if you want a goose neck vent, it is up to you to figure out the size you need for the required venting. venting. Which Which really isn’t super challenging, so let’s step through it.
The The gooseneck (https://downcomer.files.wordpress.com/2014/10/img_20141017_222955.jpg) This particular gooseneck is a vertical length of pipe, two 90 degree bends, and a mesh screen exit. The gooseneck is a constant diameter throughout. Some basic dimensions are needed, if only as an initial guess: 1. The diameter diameter of of the pipe, D
2. 3. 4. 5.
The length of the vertical length of pipe, L The pipe roughness, ε The bend radius of the first elbow, r1 The bend radius of the second elbow, r1
Normal conditions or Actual conditions? I’m assuming the flowrate comes from API 2000, and is a flowrate of air at normal conditions. In API 2000 the reference state is different if the calculations are done in SI versus USC, oddly, and furthermore the definition of STP is slightly different than the IUPAC definition of STP, the reference pressure is 1atm instead of 1bar. So keep that in the back of your head. In terms of real numbers, “Normal” means: 1. 2. 3. 4. 5. 6.
Molar mass, M = 0.02896 kg/mol Temperature, T = 273.15 K Pressure, P = 101.325 kPa Heat capacity ratio, k = 1.4 Density, ρ = 1.293 kg/m3 Viscosity, μ = 1.72×10‑5 Pa*s
But what about the flow through the gooseneck? Well I suppose you have two choices, model the flow through the gooseneck at normal conditions, or look at a more realistic temperature and pressure. My preference is to convert the volumetric flowrate at normal conditions into a mass flowrate of air, , and do the calculations at actual operating conditions of the tank. Though I can see the other argument of doing the calculations at normal conditions, since that is what the code correlates things to.
Flowrates, mass flux, Reynolds number If the vent has a constant cross sectional area the mass velocity, G , is constant throughout, which is convenient since the Reynold’s number is then largely only dependent on temperature. The mass velocity is given by: And the Reynolds number:
The only parameter in the Reynolds number that changes is the viscosity, μ , which is mostly dependent upon temperature and not pressure. Perry’s has a correlation of the viscosity of air as a function of absolute temperature:
Where μ is in Pa*s and T is in K.
Interlude on compressibility Compressible flow can occur in a variety of ways, the most common models are isothermal and adia batic fanno flow. Isothermal is fairly straight forward. As already mentioned the Reynold’s number depends only on temperature, which is constant, so the Reynold’s number is constant throughout. This means the frictional head loss is constant throughout, and it is a simple matter of calculating the pressure drop. Adiabatic flow can be annoying since the temperature varies throughout the pipe, and thus the Reynold’s number varies, sometimes quite significantly, through the pipe. This leads to an iterative series of calculations, starting by assuming a constant friction factor, calculating the pressure and temperature changes, adjusting the friction factor and iterating until everything converges.
Frictional head loss Books like Crane’s TP‑410 or Perry’s are great places to start in search of K factors for fittings and correlations for Darcy or Fanning friction factors of the straight sections. I use Crane’s, and so everything is in terms of Darcy friction factors, which are 4× the fanning friction factor. The K factors are given by the following: Gooseneck Vent Component
K
Entrance to gooseneck 0.5 Length of vertical pipe First 90 bend Second 90 bend Mesh Screen
Exit to atmosphere
1.0
The straight pipe friction factor, f , can be calculated using the Serghide equation:
The turbulent friction factor, f T , is given by the following: Each bend has a constant, C , that depends upon the bend radius, the values of the constant are tabulated in Crane TP‑410 pg. A‑30).
Isothermal Compressible Flow Suppose the tank is at it’s max operating temperature, , then the conditions at the entrance and exit of the gooseneck are: Pressure kPa
Temperature K
Entrance Exit
101.325
The pressure inside is given by: Which can be solved numerically, or by fixed point iteration (if you are a masochist).
Adiabatic (Fanno) Flow Adiabatic flow, or Fanno flow, adds some wrinkles. The temperature is no longer constant, which means the Reynold’s number is no longer constant. The two unknowns are the pressure inside the tank (which we assume is at max operating temperature), and the temperature outside the tank (which we assume is at atmospheric pressure). The two are related to each other through the equations for adiabatic flow, but disentangling all that can be a head ache. One solution is to two, nested, iterations. The outer iteration is to find the temperature at the exit of the vent, the inner iteration to find the pressure at the entrance to the vent.
As a first guess suppose the temperature at the exit of the gooseneck is the same as the entrance, . With that the Mach number can be calculated at exit conditions:
And the Fanno parameter at the exit:
The Fanno parameter at the entrance is given by:
Where has been calculated at the exit conditions. Now comes the main part of the inner loop: calculating the Mach number at the entrance. Using whatever numerical method you like, Newton’s method works pretty well, find the Mach number for the entrance that is the root of:
The pressure at the entrance can be found by recalling that G is constant throughout, and using the ideal gas law:
Now to re‑start the outer loop! The temperature at the exit is given by:
Given this new set of conditions start again, calculating the Mach number at the exit, Reynolds number at the exit, and the friction factor. Furthermore calculate the Reynolds number and friction factor at the entrance. The overall frictional head loss is then the average of the entrance and exit. With this in hand, start again calculating the fanno parameter at the exit, the mach number at the entrance, and the pressure at the entrance. Continue iterating until it converges. aboveground storage tanks
compressible gas flow
fanno flow
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