PETE 325
Petroleum Production Systems
Week 4
Nodal Analysis, or
Well Deliverability, or
“NODAL Analysis” ™
Production System Analysis
Divide system at a point (node) into upstream
and downstream components
•Upstream components are inflow
•Downstream components are outflow
•Endpoint pressures are fixed: Pres; Psep
It is all one system
• Preservoir – Pseparator = ΔPsystem = Σ ΔPcomponent
• A restriction at any component can hamper production
 M ak es com pletion design im portant
 Also im portant for flow assurance
• ( ΔP, ΔT can cause solids deposition)
• Whole system must be analyzed for proper initial design
and/or “debottlenecking”
• Production Engineers Often Focus on the region from well
inflow to wellhead outflow—is the well performing as it
should?
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Well Inflow Performance
Wellbore/Tubing Flow Performance
Preservoir  Pwf  Pwh  surface flow system
Basically, we superimpose IPR and VLP
Each component has a corresponding q and P relationship
for inflow and outflow
(for a given flow rate, the pressure at a component is fixed)
Pnode
Inflow performance governed by Pres - Pwf
VLP governed by Pwf - P*wh
Flow through the system is determined from
the facts that…
…only one pressure is present at a node
…inflow must equal outflow at a node—mass
balance requirement
q
*PPS uses
Ptf
We can compute IPR Curves and VLP Curves for given sets
of conditions. Their combination determines well deliverability.
Pnode
Inflow performance impacted by Pres - Pwf
Plus - ?
VLP impacted by Pwf - Pwh
plus
-?
q
We can compute IPR Curves and VLP Curves for given sets
of conditions. Their combination determines well deliverability.
Pnode
Inflow performance influenced by Pres - Pwf
Plus - fluid properties
- k, h
- ln(re/rw)
- skin
VLP influenced by Pwf - Pwh
plus
- ΔPpotential energy
- ΔPkinetic energy
- ΔPfriction
For a given IPR and well configuration,
We can determine flow rate as a
function of required wellhead pressure, for example.
q
Example
μ = 1.7
Reservoir Data:
B = 1.1 bbl/bbl; µ= 1.7 cp; kh = 8.3 md; h = 53 ft
re = 2980 ft; rw = 0.328 ft (7 7/8 in)
What can we say about ΔP for the VLP
in this well and for this oil?
- ΔPKE? ΔPPE? ΔPF?
Radial inflow relationship for steady-state black oil:
q=
Rearrange:
For our example:
k H h( pe − pwf )
 r  
141.2 Bo µ o  ln  e  + s 
  rw  
 r  
141.2 Bo µ o  ln  e  + s 
  rw   q
pwf = pe −
kh
pwf = 5651− 5.54q
IPR Curve for this Well
Now for the VLP
Single-Phase Liquid Flow: Incompressible
or
Fanning friction factor or Moody friction factor
Example, Cont’d
ΔPKE = 0 (incompressible fluid, no diameter change)
(Independent of flow rate.)
But:
ΔPF is a function of flow rate (friction factor)
ff =16μ/ρvd = 16/NRE
N RE
Dvρ
qρ
=
=
Dµ
µ
or
qρ
1.48
in oilfield units
Dμ
Example, Continued
Must assume turbulent flow.
Instead of implicit eqn. shown last week, we can use
Chen’s explicit method (shudder) from PPS, Ch 7:
Continued
Likewise:
300 BPD ΔPF = 8 psi
500 BPD ΔPF = 20 psi
700 BPD ΔPF = 35 psi
Therefore total pressure
drop, if surface pressure
is zero?
Thus Well Deliverability:
The VLP Curve crosses
the IPR Curve at
about 3060 psi (+/-)
So the flow rate at
zero wellhead pressure
is ~470 bpd or so.