**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

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