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cricks11611 (Petroleum)
16 Oct 09 13:48
So we all know that the tubing pressure decreases during liquid loading of a gas well, but why?
Here's the equation I have:
Bottom hole pressure = Tubing pressure + Hydrostatic Head + Frictional Losses.
So if you begin to accumulate fluids and you get a column of fluid in your tubing, your bottom hole pressure increases, which exerts a backpressure on the formation, reducing production.
No matter how i look at it, i keep going in circles and creating loops in my equations.
zdas04 (Mechanical)
16 Oct 09 15:18
Your reasoning is fine up to a point, but lets look at if from the other end.
First, think of a well that is not liquid loading. The flowing tubing pressure is:
FTP = P(reservoir) - Losses in near-Wellbore - P(hydrostatic) - Friction in Tubing 字串5
So if you start loading, P(reservoir) and near well-bore losses don't change, P(hydrostatic) increases due to the liquid accumulation, and Tubing Friction decreases due to reduced flow. The question is "which change is dominant?" Well, if the hydrostatic pressure increases more than the friction decreases then FTP goes down. That is the normal condition.
Now, in cases where you are flowing a bunch of gas up the tubing, I have seen wells start to load up and decrease friction slightly more than the hydrostatic increase and FTP actually go up a small amount during loading events. This is rare and requires a pretty big well, but it happens.
David
cricks11611 (Petroleum)
16 Oct 09 15:38
Right. That makes sense, and when you say the reservoir pressure and near well-bore pressure drop doesn't change, that makes sense too. But then doesn't that mean your bottom hole pressure isn't changing? If your bottom hole pressure isn't changing, what is causing the drop in production? 字串1
Am i confusing this by assuming that the bottom hole pressure is the same as the sandface pressure just outside the perfs? (neglecting minor pressure drop through perfs)
zdas04 (Mechanical)
16 Oct 09 17:11
Yep, you are. Reservoir pressure approximately is constant in the short term (i.e., nothing you do to a well is going to change something the size of a reservoir in the short term).
The losses in the near wellbore are dominated on one end by reservoir pressure and on the other end by the wellbore. As wellbore pressure goes up (due to accumulated liquid), the pressure immediately on the other side of the perfs also goes up, but assuming nearly radial diffusion that effect is a pretty short distance. The number changes, but it is definitely a second order effect and can safely assumed to be zero change in the short term.
The wellbore side of the perfs is a different issue. Flowing bottomhole pressure changes from millisecond to millisecond. I think of the perfs, skin, etc. much like I think of a choke--pressure changes downstream of the choke are reluctantly transmitted to the upstream side of the choke, they are transmitted but it is sluggish and there is a significant time lag. 字串4
David
cricks11611 (Petroleum)
16 Oct 09 17:28
So your bottom hole pressure is increasing due to the hydrostatic head of the liquid. But then if you start your nodal analysis from the bottom hole pressure point, and you want to calculate the tubing pressure, it would be
Tubing pressure=bottom hole pressure-hydrostatic head-friction.
The increase in bottom hole pressure is the same as the increase in the hydrostatic head, and your tubing pressure would remain the same.
I understand what you're trying to say, but i can't seem to make it make mathematical sense.
zdas04 (Mechanical)
16 Oct 09 18:47
One clarification, if the well logs-off then the friction in the tubing goes to zero. So
BHP=Hydrostatic + applied tubing pressure
If BHP < Hydrostatic + Tubing then the liquid would flow back into the reservoir until everything is back in equilibrum. 字串4
Maybe someone else can help you, I'm obviously not explaining it in a way that works for you.
David
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