Adrian M. Fuxman B.Sc. M.Sc. Ph.D. Candidate Edmonton, Alberta, Canada email@example.com
In this report, we show how to compute the volumetric ﬂow rate of fugitive gas that is emitted to the atmosphere during the closing operation of an ONIS Line Blind . During the closingprocess, three events occur, see Figure 1: 1. Phase #1: Rotation of the level to separate the two sections of the pipe. 2. Phase #2: Movement of the slide gate (see Figure 2) . 3. Phase #3: Rotation of the level to join together the two sections of the pipe.
Upper Δ Section
Figure 1: Drawing of the closing process
Figure 2: Drawing of the slide gate
Duringeach of the three events, gas is emitted to the atmosphere (fugitive gas). In this report, we show how to compute the ﬂow rate of fugitive gas for each event. As a ﬁrst approximation to the solution of the problem, we consider that the dynamic of the system is fast and thus mass and mechanical energy balances can be written without dynamic term. To calculate the gas that escape from the pipe atany time, we develop a mass balance around the line blind, see Figure 3 : ρ1 < v1 > A1 = ρ2 < v2 > A2 + ρ3 < v3 > A3 (1)
where v1 is the ﬂow velocity upstream and A1 is the total cross-sectional area of the pipe. All process variables upstream are considered to be known.
V1 A1 upstream
V A2 fugitive gas
V A2 fugitive gas
V A3 upstream
Figure 3: Control volume around the ONIS lineblind
To compute the velocity of fugitive gas, v2 , and the velocity of the gas right after the moving gate, v3 , we develop a mechanical energy balance. The mechanical energy balance that relates 1→3, see Figure 4 is given by : < v3 >2 < v1 >2 − + 2 2
1 dP = 0 ρ
where friction is neglected and the ﬂuid is assumed to be turbulent. To solve equation (2) we assume thatthe gas will undergo an adiabatic process without friction. For ideal gases, ˆ ˆ adiabatic behavior without friction results in P ρ−γ = cte, where γ = Cp /Cv . In general, γ can be considered to be 1.67 for monoatomic gases, 1,4 for diatomic gases and 1.3 for polyatomic gases such as CO2 , SO2 , NH3 and CH4 .
2 1 1’ 3
Figure 4: Reference points for the mechanical energy balance Equation(2) becomes: < v3 >2 < v1 >2 − − 2 2 P1 ρ1 γ γ−1 1− P3 P1
From the consideration of an adiabatic process, we know that P ρ−γ = cte, and thus equation (1) becomes < v1 > A1 = P2 P1
< v2 > A2 +
< v3 > A3
The mechanical energy balance that relates 1’ → 2, see Figure 4, can be developed in a similar way as we did for 1→ 3, to give: < v2 >2
Assuming that the velocity in the radial direction is zero at the centerline, v1 = 0 (which is a reasonable assumption for large diameter pipes with small area for the gas to escape), 3
and assuming that the pressure at 1’ is an average between P1 and P3 , we obtain: < v2 >= 2 (P1 + P3 )/2 ρ1 +
P2 (P1 + P3 )/2
The simultaneous solution of equations (3), (4) and (6) gives velocity of the fugitive gas, < v2 >, and the pressure of the gas after the sliding gate for prescribed conditions upstream and atmospheric pressure (P2 = Patmospheric ). By solving for < v3 > in equation (3), and replacing < v3 > and < v2 > (from equation (6)) in equation (4) we obtain thefollowing nonlinear equation γ−1 γ P2 1/γ (P1 + P3 )/2 γ P2 1− 2 A2 + γ−1 1 P1 (P1 + P3 )/2 P3 γ ρ1 + P1 /2 P3 P1
< v1 >2 +2 P1 ρ1 γ γ−1 1− P3 P1
A3 =< v1 > A1 (7)
that needs to be solved to obtain the solution to the problem. From equations (3) and (6), we ﬁnd that the minimum requirement for the simultaneous solution of the...