Introduction to Hydraulics
4.1 Hydrostatic Pressure
4.1.2 Hydrostatic Pressure for Compressible Fluids
The density ρ of a compressible fluid (gases in general are compressible) is not constant and depends on the pressure and temperature. In order to account for this, we must consider the equation of state for real gases:
pV = z n ¯R T = z m M
R T ,¯
where z is the real gas deviation factor (see Figure 4.2) of the real gas at pressure p and temperature T . Solving for ρ = m/V results in:
ρ = M
z ¯R T p . (4.2)
Considering the expression (4.1) for the pressure differential we have:
dp = M g
z ¯R T p dD .
Since z depends on p, separating variables and integrating results in:2
p2
2And assuming that the temperature is constant in the gas column.
Figure 4.2: Real gas deviation factor.
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The pressure integral can only be calculated if we know how z depends on p. This is normally very complicated. For short columns, z cam be considered constant and we can write:
z
where z1 is the compressibility factor at pressure p1 and T . A more accurate approach is to use an average value for z given by:
¯
z = z1+ z2
2 ,
where z2 is the deviation factor for p2 calculated using the expression above.
Using this new average value of the compressibility factor, a new pressure p2 is obtained from
p2 = p1ez ¯¯M gR T(D2−D1),
and compared with the previous one. The process is repeated until conver-gence is obtained.
Values for the universal gas constant ¯R for various units are:
R = 10.732¯ psi ft3
It is important to note that pressure and temperature must be given in absolute scales, as required by the gas equation of state. The absolute temperatures are normally the Rankine and the Kelvin scales and given by:
T [◦R] = t[◦F] + 459.67 , T [K] = t[◦C] + 273.15 .
The figures above are normally approximated to 460 and 273 respectively.
The Fahrenheit and Celsius temperature are converted using the following re-lations:
t[◦F] = 9
5(t[◦C] + 40) − 40 , t[◦C] = 5
9(t[◦F] + 40) − 40 .
Example 9: What is the density of the air at 13 psia and 60◦F? (assume ideal gas)
Solution:
With sufficient accuracy, the molecular weight of the air is:
Mair = 22% × 32 + 78% × 28 = 28.88 lbm
Example 10: Consider a 10,000 ft deep borehole with a drillstring and bit to the bottom. The annular is completely filled with methane (CH4), and the drillstring is filled with a 8.4 lbm/gal mud. After closing the BOP, the pressure in the drill pipe at the surface is 640 psia. What is the expected pressure in the casing at the surface, assuming ideal gas and average temperature of 150◦F.
Solution:
The mass of one lb–mole of methane is 1x12+4x1 = 16 lbm. The drillstring–
annular system form a U–tube system. The pressure at the bottom of the bore-hole can be calculated using the fluid inside the drill pipe and the surface pres-sure:
pbottom = 640 psi + 0.0519 × 8.4 lbm/gal × 10000 ft = 5000 psi
This pressure is balanced by the pressure of the casing at the surface and the hydrostatic pressure of the gas. Using Equation (4.3) gives:
psur = 5000 e1×1545.4×(150+460)16×1 (0−10000)
psur = 4220 psi
4.2 Buoyancy
Archimedes principle of buoyancy states that the buoyant force exerted on a body fully or partially immersed in a fluid is equal in magnitude (and opposite in direction) to the weight of the volume of fluid which is displaced by that body.
For homogeneous bodies immersed in homogeneous fluids, the net or buoyed weight of the body can be calculated from
Wnet=
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where W is the weight of the body (in the air), ρf is the density of the fluid, and ρb is the density of the body. The term
1 −ρρf
b
is called buoyancy fac-tor. This expression is valid only for homogeneous bodies fully immersed in homogeneous fluids.
For homogeneous bodies, the geometric center of the body coincides with the center of mass. For non–homogeneous bodies, an equivalent density (total mass/volume) can be used, but it is important to keep in mind that the geometric center (where the buoyant force applies) may not coincide with the center of mass. In these cases, stable or instable equilibrium may exist.
Example 11: What is the weight of 0.4 ft3of carbon steel? What is its buoyed weight when submerged in a 9.3 lbm/gal fluid? What is the equivalent density of the buoyed body in lbm/gal?
Solution:
The average density of carbon steel is 490 lb/ft3 = 65.5 lb/gal, so that the weight of the body is
W = ρs g Vb = 490 lbm
The equivalent density is the density that would result in the same buoyed weight:
Example 12: A 12,000 ft long drillstring is composed of the following elements (bottom up): 420 ft of 8 in OD–3 in ID DC, 840 ft of 7 in OD – 3 in ID DC, and 5 in E-75 19.5lb/ft drill pipe (22.28 lb/ft). The fluid density is 8.9 lb/gal. Calculate the expected hook load when the drillstring is hanging on the elevator (out of the bottom).
Solution:
The linear weight of the DCs are 8 in DC: w = π
4 × (82− 32) ×
1 ft2 144 in2
× 490 lbm/ft3 = 147.0 lbf/ft
Figure 4.3: Drillstring schematics for Example 12.
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Master of Petroleum Well Engineering Drilling Engineering Fundamentals The weight in the air of the drillstring is
Wair = 420 × 147.0 + 840 × 106.9 + (12000 − 420 − 840) × 22.28 = 390823 lbf The buoyed weight of the drillstring is:
Whook = Wbuoyed =
1 − 8.9 65.5
× 390823 = 337719 lbf
If the body is either not totally submerged, or submerged in a inhomoge-neous fluid, the expression above cannot be used. This may be complicated for complex shape bodies. A more general way to calculate the buoyed weight, even for partially immersed bodies and for complex fluid column is to calculate the effect of the hydrostatic pressure on the body. This is demonstrated in the next example.
Example 13: Recalculate the hook load of Example 12 using the effect of the hydrostatic pressure on the drillstring.
Solution:
Here we face a problem: the tool joints of the drill pipes causes a considerable increase in the average linear weight os the drill pipe body. Although we can consider each tool joint of the drillstring, a most appropriate way is to consider an equivalent cross section of the drill pipe. Since the purpose is to determine the effect of the fluid in the weight, it is not important if we choose to change the inside or the outside diameter (or both). In the present example, we chose to change both, such that the average diameter remains the same. In this case, the average diameter is Dave = 5+4.2762 = 4.638 in, and the area is given by
As = π Dave t ,
where t is the thickness of the equivalent drill pipe. The equivalent outside diameter is Dave + t and the inside diameter is Dave − t. Therefore, for a drill pipe with 22.28 lb/ft, we have:
π × 4.638 in × t × 490 lb
Note that these figures are for buoyancy calculation only, and should never be used for strength calculations (torque, tensile, burst, collapse, etc).
Now, the fluid pressure acts in the cross section areas at 10,740 ft, 11,580 ft, and 12,000 ft. In the first and second areas, the forces are downwards, and in the third case the force is upward. These forces add and subtract to the weight of the drillstring. To calculate the force we need the pressures at each depth and the area exposed to the fluid. The values are:
p10740= 0.0519 × 8.9 × 10740 = 4961 psi p11580= 0.0519 × 8.9 × 11580 = 5349 psi p12000= 0.0519 × 8.9 × 12000 = 5543 psi
A10740= π
4 [72− 5.0872] + [4.1892− 32] = 24.87 in2 A11580 = π
4 82 − 72 = 11.78 in2 A12000 = π
4 82 − 32 = 43.20 in2 The hook load is:
Whook = 390823 + 4961 × 24.87 + 5349 × 11.78 − 5543 × 43.20 = 337756 lbf This result should be compared with the previous example. The discrepancy is due rounding errors.
This method should always be used when either the annular or the drillstring is filled with non–homogeneous fluids, partially filled, etc.
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