# Activity: U Tube Effect

Problem 1:
Case 1:

In this example, the densities in the two sides of the U tube are different, so the well will flow in the direction where the hydrostatic pressure is the lowest and it stops when the equilibrium is reached. The equilibrium is faced when the fluids level drops by "X " at the B side, so according to the data delivered in the sketch,what is " X "?

To get to the equilibrium the hydrostatic pressures on both sides have to be the same.
Phyd_A = Phyd_B
0.052 x MWA x Za = 0.052 x MWB x Zb
Zb = (MWA x Za ) / MWB
Zb = ( 10 x 10000 ) / 10.30
Zb = 9708 ft
X = Z – Zb
X = 10000 – 9708
X = 292 ft
The level at the B side has to drop by 292 ft to get to the equilibrium.

Case 2:

In the case where the well is closed, the difference in hydrostatic pressure will be transferred to the other side and it is represented by a Surface Drill Pipe Pressure "SDPP"
The pressure exerted by the column of fluid in the A side is
Pbottom_A= 0.052 x MWA x Z
Pbottom_A = 0.052 x 10 x 10000
Pbottom_A = 5200 psi

The pressure exerted by the column of fluid in the B side is
Pbottom_B= 0.052 x MWB x Z
Pbottom_B = 0.052 x 10.30 x 10000
Pbottom_B = 5356 psi

The surface drill pipe pressure SDPP is
SDPP = Pbottom_B – Pbottom_A
SDPP = 5356 – 5200
SDPP = 156 psi

Problem 2:

In this example, the surface drill pipe pressure and the annular pressure (casing pressure) are calculated. The well is shut in and the circulation is ceased.

The surface drill pipe pressure (SDPP) is
SDPP = Ppore – Pbottom_A
Pbottom_A = 0.052 x MW1 x Z
Pbottom_A = 0.052 x 10 x 10000
Pbottom_A = 5200 psi
SDPP = 6520 – 5200
SDPP = 1320 psi

The Annular pressure (Casing Pressure CP) equals to the difference between the pore pressure and the pressures on bottom exerted by the fluids (Gas and Drilling fluids) on the B side.

CP = Ppores - Pgas - Pmud
Pgas is the pressure exerted by the column of gas
Pgas = 0.052 x Hgas x Wgas
Pgas = 0.052 x 800 x 2.50
Pgas = 104 psi

Where
Hgas is the gas column length
Wgas is the gas weight
Pmud is the pressure exerted by the column of drilling fluids on B side
Pmud = 0.052 x (Z-Hgas) x MW1
Pmud = 0.052 x (10000-800) x 10
Pmud = 4784 psi

So the annular pressure (casing pressure) is
CP = 6520 - 104 – 4784
CP = 1632 psi

The required density of the drilling fluid (Kill Mud) to balance the well can be estimated by
MWre = (Ppores) / (0.052 x Z)
MWre = 6520 / (0.052 x 10000)
MWre = 12.54 ppg

Case 2:

If we consider that the influx is completely evacuated from the annulus and the required mud weight (Kill mud) is pumped to a certain depth, what will be the surface pressures (drill pipe pressure and the casing pressure)? Taking in the consideration that the circulation is stopped.

Surface Drill pipe pressure will be:
SDPP= Ppores - Pre – Pmud
Pre is the pressure of the column of the kill mud
Pre = 0.052 x MWre x Zr
Pre = 4108 psi

Pmud is the pressure of the column of original drilling fluids
Pmud = 0.052 x MW1 x (Z-Zr)
Pmud = 0.052 x 10 x (10000-6300)
Pmud = 1924 psi

So surface drill pipe pressure is
SDPP = 6520 - 4108 – 1924
SDPP = 488 psi

Casing pressure can be get by:
CP = Ppores - Pbottom_B
CP = Ppore - 0.052 x MW1 x Z
CP = 6520 - 0.052 x 10 x 10000
CP = 6520 - 5200
CP = 1320 psi

Case 3:
In this example, we consider that the kill mud with the required density is pumped while evacuating the influx out of the hole. In this situation what will be the surface pressures and also what will be the equivalent density at certain depths in the annulus? (** the circulation is stopped)

Surface Drill pipe pressure will be:
SDPP= Ppores - Pre – Pmud
Pre is the pressure of the column of the kill mud
Pre = 0.052 x MWre x Zr

MWre is the required density and it is calculated as
MWre = Ppores / (0.052 x Z)
MWre = 6000 / (0.052 x 10000)
MWre = 11.54 ppg

So Pre will be
Pre = 0.052 x 11.54 x 8000
Pre = 4800 psi
Pmud is the pressure of the column of original drilling fluids
Pmud = 0.052 x MW1 x (Z-Zr)
Pmud = 0.052 x 10 x (10000-8000)
Pmud = 1040 psi

So surface drill pipe pressure is
SDPP = 6000 - 4800 – 1040
SDPP = 160 psi

The casing pressure is calculated as follows
CP = Ppores - Pgas – Pmud
Pgas is the pressure exerted by the column of the influx
Pgas = 0.052 x Hgas x Wgas
Pgas = 0.052 x 1650 x 1.15
Pgas = 99 psi

Pmud is the pressure generated by the column of drilling fluids
Pmud = 0.052 x MW1 x (Z-Zgas)
Pmud = 0.052 x 10 x (10000-1650)
Pmud = 4342 psi

So the casing pressure is
CP = 6000 - 99 - 4342
CP = 1559 psi

Now, we want to know what will be the equivalent density above and below the bubble of gas.
Pressure above the bubble of gas equals to the sum of pressures exerted above it (casing pressure and the mud pressure)

Pabove = CP + Pmud
Pabove = CP + 0.052 x MW1 x Z1
Pabove = 1559 + 0.052 x 10 x 3300
Pabove = 3275 psi

The equivalent density at this point is
dequi = Pabove / (0.052 x Z1)
dequi = 3275 / (0.052 x 3300)
dequi = 19.08 ppg

The density below the bubble can be estimated as follows
Pbelow = Pabove + 0.052 x Hgas x Wgas
Pbelow = 3275 + 0.052 x 1650 x 1.15
Pbelow = 3374 psi

So the equivalent density below the bubble is
dequi = Pbelow / (0.052 x (Z1+Zgas))
dequi = 3374 / (0.052 x (3300+1650))
dequi = 13.10 ppg