math book 8

8 • 1 Essential Math

Section 8

Fluid Calculations

Table of Contents Fluid Calculations...................................................................................................................................... 8-3

Introduction............................................................................................................................................ 8-3 Learning Objectives ............................................................................................................................... 8-3

Unit A: Hydrostatic and Differential Pressure Calculations...................................................................... 8-3 Hydrostatic Pressure Calculations.......................................................................................................... 8-4 Dual Densities Pressure Calculations..................................................................................................... 8-5 Differential Pressure Calculations.......................................................................................................... 8-6 Unit A Quiz............................................................................................................................................ 8-7

Unit B: Force and Buoyancy Calculations ................................................................................................ 8-8 Force, Pressure and Area Calculation .................................................................................................... 8-8 Buoyancy Calculations ........................................................................................................................ 8-10 Interpolation......................................................................................................................................... 8-13 Unit B Quiz .......................................................................................................................................... 8-14

Self-Check Test for Section 8.................................................................................................................. 8-15 Answers Keys .......................................................................................................................................... 8-17

Fluid Calculations


Use for Section notes…

Fluid Calculations


Fluid Calculations

Introduction

Pressure is the exertion of force upon a body by another body when in contact with it. The standard measure of pressure is pounds per square inch (abbreviated lb/in.2 or psi). This expression means that the pressure being exerted in the area or space concerned has the indicated number of pounds of pressure on each square inch. For instance, our atmosphere exerts a pressure on each square inch of the earth’s surface of 14.7 lb. It is common to have pressures both higher and lower than this atmospheric pressure. A vacuum pump may be used to withdraw some of the air from a closed container, leaving pressure well below atmospheric pressure. Pumping more air into a tire will result in pressure above atmospheric pressure.

A regular pressure gauge reads zero when the pressure on it is normal atmospheric pressure. When the pressure reading goes above zero, it is actually showing the pressure beyond normal atmospheric pressure, not the absolute total pressure. As a result, such readings are said to be in pounds per square inch gauge (abbreviated

psig). A pressure reading that combines atmospheric pressure and gauge pressure is in pounds per square inch (psi absolute).

In your work in the oilfield, you will deal with several forms of pressure. This section breaks down this topic into the various types of pressure. An understanding of pressure calculations is necessary for cementing, stimulation, special tools, drill stem testing and nearly all other aspects of oilfield service. Therefore, accuracy in pressure calculations is critical.

Learning Objectives

Upon completion of this section, you will be able to:

• calculate hydrostatic and differential pressures

• understand the relationships among force, pressure, and area

• find buoyancy factors

Unit A: Hydrostatic and Differential Pressure Calculations

Hydrostatic pressure is the pressure exerted by a column of fluid. The size or the shape of the container in which the fluid is stored makes no difference. The vertical height of the fluid column and the density of the fluid (lb/gal) are the only factors involved in calculating hydrostatic pressure.

Differential pressure occurs when two fluid columns with different densities at the same vertical height are present in a hole at the same

time. Differential pressure can only occur when the two unequal density columns are connected or related to each other so that the pressures can work across a barrier.

Fluid Calculations


Hydrostatic Pressure Calculations

All the figures in Figure 8.1 have the same pressure at their base, 0.433 psi. They are all one foot deep and filled with fresh water weighing 8.33 lb/gal at 68°F.

1 ft

Figure 8.1

Figure 8.2 illustrates that the amount of fluid in a container has nothing to do with hydrostatic pressure. Even though all the figures have different pressures at their bases, they have the same volume.

1 gallon

0.433 psi

0.433 psi

0.866 psi

1.299 psi

1 ft

1 ft

2 ft

3 ft

Figure 8.2

Hydrostatic pressure can be calculated at any depth in a hole or a container. The best method for calculating hydrostatic pressure is to use a Hydrostatic Pressure and Fluid Weight

Conversion Table as seen in Figure 8.3. The extreme left-hand column of the table gives fluid densities in lb/gal and the fourth column provides pressure in psi for one foot of depth. The formula for hydrostatic pressure is:

Hydrostatic pressure = psi/ft × depth

Sample Problem

You have a 500 ft column of water. What is the hydrostatic pressure exerted at the bottom of the column? (Water weighs 8.33 lb/gal)

Solution

Look up the psi/ft for 8.33 lb/gal water in Figure 8.3. Then multiply by the depth of the fluid:

0.4330 psi/ft × 500 ft = 216.5 psi hydrostatic

Fluid Calculations


Figure 8.3

Dual Densities Pressure Calculations

What happens if you have two fluid densities in the same hole? To determine hydrostatic pressure, you must calculate each pressure separately and then sum the pressures.

100 ft

500 ft

100 ft

400 ft

FluidA

FluidB

Total Hydrostatic Pressure (HP) = Fluid A HP + Fluid B HP

Figure 8.4

Do not add the fluid heights together because of the differing densities. Do not add the psi/ft together since this will give you a greater density than you have in the hole.

Sample Problem

What is the hydrostatic pressure at 500 feet under the conditions as shown in Figure 8.4?

• The first 100 ft of the annulus is filled with 8.33 lb/gal water

• From 100 to 500 ft there is 16 lb/gal weighted fluid in the annulus.

Solution

Determine the psi exerted by each fluid independently, and then sum the pressures.

Hydrostatic pressure = psi/ft × depth

Psi from 8.33 lb/gal water: 0.4330 psi/ft × 100 ft = 43.3 psi

Psi from 16 lb/gal fluid: 0.8312 psi/ft × 400 ft = 332.48 psi

Total hydrostatic pressure at 500 ft = 43.3 psi + 332.48 psi = 375.78 psi hydrostatic

Fluid Calculations


Differential Pressure Calculations

A U-tube as compared to a hole condition can be used to illustrate fluid flow from the effects of differential pressure. This is illustrated in Figure 8.5. A fluid with a higher density will always try to push one with a lighter density to an equalization point.

WeightedFluid

Water

WeightedFluid

Water

Figure 8.5

Figure 8.6 shows an example where no differential pressure exists. There are equal densities and equal vertical heights. There is hydrostatic pressure at the base of this object, but no differential pressure; therefore, the system is balanced

Figure 8.6

Sample Problem

The fluid columns in the cased hole in Figure 8.7 have unequal densities (8.33 lb/gal water in the casing, 15.6 lb/gal fluid in annulus) and equal vertical heights (3 ft). The pressure exerted by a higher density fluid pushing against a lower density fluid is recorded at the surface on a gauge. What is the differential pressure in this case?

Figure 8.7

Solution

Weighted fluid at 15.6 lb/gal 0.8104 psi/ft

Water at 8.33 lb/gal – 0.4330 psi/ft

0.3744 psi/ft

0.3744 psi/ft × 3 ft = 1.1322 psi differential

The hydrostatic pressure at the base of the hole is 2.4312 psi (0.8104 psi/ft × 3 ft). The differential pressure (1.1322 psi) plus the hydrostatic pressure of the water column (0.4330 psi/ft × 3 ft = 1.299 psi) is equal to the hydra static pressure (1.1322 psi + 1.299 psi = 2.4312 psi) at the base of the hole.

Sample Problem

Figure 8.8 illustrates a typical situation. There is weighted fluid in the annulus and in the bottom part of the casing. The upper part of the casing is filled with water. Note that the weighted fluid column for one foot on each side of the barrier is balanced. Therefore, this part of the fluid column will not contribute to the differential pressure. Determine the differential pressure in the unbalanced part of the fluid column.

Solution

Weighted fluid at 15.6 lb/gal 0.8104 psi/ft

Water at 8.33 lb/gal - 0.4330 psi/ft

0.3774 psi/ft

Fluid Calculations


WeightedFluid

Water

1 ft

2 ft

2.4312 psi = HydrostaticPressure (0.8104 psi/ft x 3 ft

0.7548 psi = Differential Pressure

Figure 8.8

Because a two-foot portion of the column is unbalanced:

2 ft × 0.3774 psi/ft = 0.7548 psi differential

This hole is not balanced because of the difference in fluid densities and the difference in

the vertical heights of the two fluid columns. To keep the weighted fluid from pushing the water out of the hole, it must be, closed off at the top of the casing. This pressure (0.7548 psi) would be recorded at the surface on a gauge.

The differential pressure (0.7548 psi) plus the hydrostatic pressure of the water column (0.4330 psi/ft × 2 It = 0.8660 psi) plus the fluid column across the barrier (0.8104 psi/ft × 1 ft = 0.8104 psi) is equal to the hydrostatic pressure at the base of the hole.

0.7848 psi + 0.866 psi + 0.8104 psi = 2.4312 psi

Differential pressure occurs in one form or another in many service jobs and will govern many things that can or cannot be done in oilfield operations.

Unit A Quiz

For items 1 through 4, fill in the blanks. The remaining items should be calculated, check the answer to check your progress in Unit A.

1. Hydrostatic pressure is the_____________ exerted by a column of________________.

2. Different-shaped figures may have the same volume, but a_____________ hydrostatic pressure at the base.

3. For differential pressure to exist, it must work across a _______________.

4. Fluid with a heavier density will always try to push a lighter fluid to an _____________________________ _______________.

5. What is the hydrostatic pressure at the bottom of a 10,000 ft column of 7.3 lb/gal fluid? Use Figure 8.3 as a reference. You have pipe in a hole that is 600 ft deep. If there is 15.6 lb/gal cement in the annulus and 9 lb/gal mud in the pipe, what is the differential pressure?

6. Inside 2000 ft of casing there is 30 ft of 16 lb/gal fluid and 1970 ft of 9.5 lb/gal mud. Outside the casing, there is 2000 ft of 16 lb/gal fluid. What is the differential pressure? Now, look up the suggested answers in the Answer Key.

Fluid Calculations


Unit B: Force and Buoyancy Calculations

Force is the effect of pressure (psi) applied to an area (sq in.). Force can also be defined as power that tends to cause motion. Force is expressed in pounds and has a direction (indicated by arrows).

Buoyancy is the upward force exerted by a fluid. The buoyant force is equal to the weight of fluid displaced when the pipe is run in the hole.

This unit will define and discuss the relationships among force, pressure, and area and give examples to help you understand how they are used. In the last part of this unit, buoyancy calculations will be shown.

Force, Pressure and Area Calculation

Force is equal to the pressure multiplied by the area:

Force (lb) = Pressure (psi) × Area (sq in.)

or

F = P × A

Note that if any two of these quantities are known, the third can be calculated by transposing this formula. Therefore, the formula for pressure is:

P = F ÷ A

and the formula for area is:

A = F ÷ P

Figure 8.9 illustrates the relationships among force, pressure, and area.

3 in.

1 in.

3 in.

1 in.

100 lb

100 lb

100 lb

100 lb

100 lb

100 lb

100 lb

100 lb

100 lb

900 lb ofForce

Pressure = 100 lb/sq in.

Area = 9 sq in.

Force = 900 lb

Each square inch has 100 lb exertedagainst it; therefore, the total force is

900 lb working over 9 sq in.

Figure 8.9

Sample Problem

What is the upward force created by the hydraulic cylinder’s piston in Figure 8.10?

Fluid Calculations


3 in.

W eight

3000 psi

Figure 8.10

Solution

P = 3000 psi

A = 3 in. × 3 in. × 0.7854 = 7.0686 sq in.

F = P × A

F = 3000 psi × 7.0686 sq in. = 21,205.8 1b⇑

If this were a hydraulic jack, the piston would be capable of lifting 21,205.8 lb of physical weight when 3000 psi was applied to the area of this piston.

Sample Problem

What is the force created on the cap of the 5-1/2 in. plug container head illustrated in Figure 8.11?

5000 psi

5.12 in.

Figure 8.11

Solution

Pressure = 5000 psi

Area = 5.12 in. × 5.12 in. × 0.7854 = 20.588789 sq in.

F = P × A

F = 5000 psi × 20.588789 sq in. = 102,943.94 lb⇑

If there were no threads to hold the cap on the head, it would take the physical weight of two pump trucks to hold the cap down since the force created is over 102,000 lb.

Sample Problem

What is the force created on the cap in the 10-3/4 in. plug container head pictured in Figure 8.12?

5000 psi

10.31 in.

Figure 8.12

Solution

Pressure = 5000 psi

Area = 10.31 in. × 10.31 in. × 0.7854 = 83.484956 sq in.

F = 5000 psi × 83.484956 sq in. = 417,424.78 lb⇑

Compare the force in this sample problem and the problem immediately preceding it. The force on the 10-3/4 in. cap (417,424.78 lb⇑) is

Fluid Calculations


approximately four times greater than the force on the 5-1/2 in. cap (102,943.94 lb⇑). Double the diameter and the force will be four times greater. This change in force is due to the change in area, even the applied pressure (5000 psi) remained the same.

Sample Problem

What is the pressure created with 2000 lb of force on the hydraulic cylinder’s piston rod in Figure 8.13?

2 in.

2000 lb

636.618 psi

Figure 8.13

Solution

Force = 2000 lb⇓

Area = 2 in. × 2 in. × 0.7854 = 3.1416 sq in.

P = F + A

P = 2000 lb⇓ ÷ 3.1416 sq in. = 636.618 psi

Sample Problem

An HT-400 pump can withstand a maximum force of 180,000 lb against its power end. The force that is transmitted to the power end is created by the plungers working against the fluid pressure that the pump is pumping. Find the maximum pressure that can be applied with the following plunger sizes:

• 6 in. plunger • 5 in. plunger • 4-1/2 in. plunger • 4 in. plunger • 3-3/8 in. plunger

Solution

6-in. plunger: 180,000 lb ÷ 28.2744 sq in.= 6366.18 psi

5-in. plunger: 180,000 lb ÷ 19.635 sq in. = 9173.3 psi

4-1/2 in. plunger: 180,000 lb ÷ 15.90435 sq in.= 11,317.658 psi

4-in. plunger: 180,000 lb ÷ 12.5664 sq in. = 14,323.91 psi

3-3/8 in. plunger: 180,000 lb ÷ 8.9462 sq in. = 20,120.27 psi

The pressures calculated above are the maximum pressures to reach the maximum force limitations on an HT-400 pump power end. The recommended working pressure for each plunger size as listed in the HT-400 pump manual is:

• 6 in. - 6250 psi • 5 in. - 9000 psi • 4-1/2 in. - 11,200 psi • 4 in. - 14,000 psi • 3-3/8 in. - 20,000 psi

The concepts of force, pressure and area cover most downhole situations as well as surface conditions. Short strings of pipe or large diameter pipes can be pumped out of the hole easily. Therefore, it is important to know where the forces are created. Other factors to be considered with the pipe suspended in a hole full of fluid are the direction of the force created and the buoyancy of the pipe.

Buoyancy Calculations

As stated earlier in this unit, buoyancy is the upward force exerted by a fluid. In other words, the fluid is trying to float whatever is placed in it. In oilfield applications, this means that even

Fluid Calculations


very heavy items such as drill pipe and drill collars will be lighter in fluid than in air.

Calculating the weight of steel tubular goods in a liquid is easy when you use your Red Book. Figure 8.14 shows the appropriate table used in identifying the buoyancy factor of a given fluid. It supplies you with the first critical piece of information you need to know in finding the weight of tubular goods in a liquid.

Buoyancy factor can be calculated if the density and weight per unit volume of the fluid and pipe are known.

We use pounds per gallon (ppg or lb/gal) as the weight per unit volume to arrive at the buoyancy factor of a liquid because we are using a volume standard that says 1 gal of steel will weigh 65.447 lb/gal and the weight of the steel will be affected by the buoyant tendencies of the liquid.

The following example illustrates how you would use your Red Book to identify the buoyancy factor of a liquid on 1 gal of steel (we say 1 gallon of steel instead of 1 pound because we are dealing with liquid and our answer must reflect this). The steel pipe we are using will be submerged in 2 % KCL water, which weighs 8.43 lb/gal.

By looking up 8.43 lb/gal on Table 132 (See right) of your Red Book, you will find that the pipe is expected to be “buoyed,” resulting in a factor of 0.8716 lb/gal (the number to the right of the lb/gal fluid).

Thus, 1 lb steel, submerged in an 8.43 lb/gal fluid, will weigh 0.8716 lb/gal, i.e., 1 lb of steel in liquid will weigh 0.8716 lb/gal.

You could then use this information to calculate the weight of an entire string of pipe submerged in a fluid.

But what if your Red Book is not available? You would then be expected to calculate the buoyancy factor of a liquid on your own.

Here is a formula for calculating the weight of open-ended steel pipe suspended in a fluid of a known density:

(Ds - Df) ÷ Ds = BF

Where: Ds = Density of steel

Df = Density of the fluid

BF = Buoyancy factor of the liquid

Note: This formula will not work for material other than normal steel.

Figure 8.14 - Section 130, Table 132, Page 3 of the Halliburton Cementing Tables (Red Book) shows you the Buoyancy Factor connected with a fluid’s density expressed in lb/gal.

To use this formula, let’s say that we are running into a hole using 1 ¾ in. O.D. Coiled Tubing

Fluid Calculations


that weighs 2.169 lb/ft., The hole is full of 8.6 lb/gal fluid and the tubing will be run to a depth of 9,500 ft. How much will the tubing string weigh, top to bottom, when it is run to this depth?

First you need to calculate what the weight of the string would be in air:

2.169 lb/ft × 9,500 ft = 20,605 lb

Now, using the previous formula, you would calculate the buoyancy factor of the 8.6 lb/gal fluid:

(Density of Steel – Fluid Weight) � Density of Steel = Buoyancy Factor

(65.447 lb/gal – 8.6 lb/gal) � 65.447 lb/gal = 0.8686 (rounded)

You can double check this answer by looking up buoyancy factor of the 8.6 lb/gal fluid in your on the chart in Figure 8.14 or in your Red Book.

So, with a tubing weight of 2.169 lb/ft, times 9,500 ft, the pipe would weigh 20,605 lb in air. To calculate the weight of the string in the 8.6 lb/gal fluid, you would calculate the additional factor of the effect of the liquid’s buoyancy would have on the string:

20,605 lb × 0.8686 (BF) = 17,897 lb

Sample Problem

Find the weight of 5-1/2 in., 17 lb/ft casing suspended in 10.3 lb/gal fluid. Use Figure 8.14 as a reference.

Solution

17 lb/ft × 0.8426 BF = 14.3242 lb/ft in fluid

Buoyancy is the force acting upward on the pipe. It is equal to the hydrostatic pressure at the end of the pipe acting on the area of the wall of the pipe (pipe OD area - pipe ID area). As you can see, buoyancy can be calculated in several ways. For simplicity, use the table in the Red Book.

Sample Problem

How much does 1400 ft of 5-1/2 in., 17 lb/ft casing weigh if it is suspended in 10.3 lb/gal fluid?

Solution

17 lb/ft × 0.8426 BF = 14.3242 lb/ft

To find the string's total weight, multiply by its length:

14.3242 lb/ft × 1400 ft = 20,053.88 lb

The total weight is 20,053.88 lb. This weight represents the Weight Indicator Reading. Since this is the weight of the casing in fluid, and this weight would have to be overcome by pressure to pump the pipe up or out of the hole, some distance, this weight becomes a downward force (lb ⇓).

Sample Problem

Calculate the pressure to overcome the 20,053.88 lb downward force created in the above sample problem.

Solution

P = F ÷ A

The formula for area is based on the OD of the casing.

Area = 5.5 in. × 5.5 in. × 0.7854 = 23.75835 sq in.

P = 20,053.88 lb⇓ ÷ 23.75835 sq in. = 844.08 psi

Therefore, the pressure required to overcome the downward force is 844.08 psi. This pressure will bring the downward force to zero pounds. Any pressure above this 844 psi will lift the casing.

Sample Problem

Calculate the upward and downward forces at the completion of the job illustrated in Figure 8.15. Give the amount of force and direction.

Fluid Calculations


Water8.33lb/gal

Pump TruckGauge

20-in., 94 lb/ft Casing

26-in. Hole

15.6 lb/gal Cement

400-ft. Pipe Depth

Figure 8.15

Given information:

• The hole was full of fresh water when the casing was run (844.08 psi).

• Displace the cementing plug with fresh water.

• Cement back to ground level.

Solution

BF for 8.33 lb/gal water = 0.8727

94 1b × 0.8727 BE = 82.0338 lb/ft × 400 ft = 32,813.52 lb⇓

The downward force (weight indicator reading) is 32,813.52 lb. This force has to be overcome with pressure acting over the area of the OD of the casing to move the pipe up the hole.

Pressure acting over the area of the OD of the casing must now be considered. The hydrostatic pressure of the water working over the area of the casing was taken into account through the buoyancy factor. However, pressure has been added to the system by placing cement in the annulus. The added pressure is the difference between the pressure exerted by the cement and fresh water.

Differential pressure:

15.6 lb/gal cement: 0.8104 psi/ft 8.33 lb/gal water: – 0.4330 psi/ft 0.3774 psi/ft

0.3774 psi/ft × 400 ft = 150.96 psi

The pressure acting over the casing end is 150.96 psi. The area to be used is based on the

OD of the casing because the pressure works over the area of the casing end (wall thickness). The difference in pressure between the fresh water and cement is transmitted to the head or swage in the top of the casing.

Area: 20 in. × 20 in. × 0.7854 = 314.16 sq in.

Upward force: 150.96 psi × 314.16 sq in. = 47,425.593 1b⇑

For the resulting force, compare the forces and their directions. The upward force is 47,425.593 lb. The downward force is 32,813.52 lb. Therefore, the resulting force is:

47,425.593 lb⇑ - 32,813.52 lb⇓ = 14,612.073 lb⇑

The force against the chain is 14,612.073 lb⇑. The chain must be of sufficient strength to hold this upward force. The chain anchor point must also have enough physical strength and weight to hold this force.

Interpolation

Let's say you wanted to look up the buoyancy factor for a fluid that weighs 8.36 lb/gal. You'll notice in the table for buoyancy factors, the densities are listed in tenths (e.g., 8.3, 8.4, etc.). Do you say that the BF for 8.3 or 8.4 pounds per gallon fluids is "close enough"? For accurate calculations, you need to interpolate.

Interpolation between two numbers or two of any quantitative values is possible when you know two end points from which to calculate. In making an interpolation calculation, assume that the relationship is straight between the end points, and that any value can be found between the points.

Sample Problem

If an 8.3 lb/gal fluid has a BF of 0.8732, and an 8.4 lb/gal fluid has a BF of 0.8716, what is the BF for an 8.36 lb/gal fluid?

Fluid Calculations


Solution

0016.08732.0

8716.0

8.36 is 6/l0ths of the distance between 8.3 and 8.4. Therefore, BF should be 6/l0ths of the way

from 0.8732 to 0.8716, or 6/10 of 0.0016 subtracted from 0.8732:

0.8732 – (0.6 × 0.0016) = 0.87224

The BF for 8.36 lb/gal fluid is 0.87224.

Unit B Quiz

Fill in the Blanks to check your progress in Unit B

1. Force can be defined as _______________ which tends to cause ________________.

2. Buoyancy is an_________________ force exerted by a fluid.

3. To calculate the pressure to overcome a downward force, use the formula: P = ______ divided by _________________.

4. Interpolation between two numbers is possible when you know the two _________ _________ from which to calculate.

5. If you run 7 in., 32 lb/ft casing to a total depth of 5100 ft in a 9 lb/gal fluid, what does the casing weigh?

6. What is the pressure required to bring the downward force to zero if the casing in item #5 is run in 8.95 lb/gal fluid?

7. If you cement 200 ft of 16 in., 84 lb/ft casing with 16.4 lb/gal cement and displace the entire casing with water, what is the amount and direction of the resulting force.

Now, look up the answers in the Answer Key.

Fluid Calculations


Self-Check Test for Section 8

Find the solutions to these problems.

1. What is the hydrostatic pressure of a 300 ft column of 9.0 lb/gal brine water?

2. What is the differential pressure of a 3000 ft column of 9.0 lb/gal mud and a 3000 ft column of 16.4 lb/gal cement?

3. What will the recorded pressure be at the pump truck when the fresh water is in the tubing in this situation? Tubing is 2-3/8 in., 4.7 lb/ft, EUE J-55 to 8000 ft Casing is 5-1/2 in., 17 lb/ft J-55 to 8500 ft Well fluid is 10 lb/gal Tubing will be filled with fresh water (8.33 lb/gal)

4. What is the pressure required to lift a 70,000 lb block on a hydraulic cylinder if the cylinder diameter is 1 ft?

5. What size hydraulic cylinder would you need to lift 50,000 lb with a pressure of 250 psi?

6. With open-ended tubing suspended in well fluid, what is the required pressure to bring the tubing weight to zero? Tubing is 3-1/2 in., 9.3 lb/ft EUE J-55 to 3500 ft Casing is 7 in., 20 lb/ft J-55 to 4000 ft Perforations are 3500 ft to 3510 ft Well fluid is 2% KC1 water

Fluid Calculations


7. Interpolate the buoyancy factor for 15.55 lb/gal mud.

8. What is the hydrostatic pressure in the annulus at 1000 ft under these circumstances? • The top 300 ft of the annulus is filled with 8.33 lb/gal water. • From 300 ft to 1000 ft, there is 15.2 lb/gal well fluid in the annulus.

9. What is the differential pressure under these conditions?

WellFluid8.33lb/gal 15.4 lb/gal Cement

800 ft

10. Calculate the upward and downward forces at the completion of this job (see figure above, right) and the resulting force. Give the amount of force and direction. The hole was full of 9 lb/gal fluid when casing was run. Displace the cementing plug with 8.33 lb/gal fluid. Cement back to ground level.

Water8.33lb/gal

Mud 9 lbs/gal

15.2-lb/gal Cement

7-in., 20 lb/ft Casing

1,000 ft

11. What is the accurate buoyancy factor for a fluid that weighs 12.46 lb/gal? Now, look up the suggested answers in the Answer Key.

Fluid Calculations


Answers Keys

Refer to the pages provided as references if you answered any of these items incorrectly, or if you were unsure of your answers.

Refer to Items from Unit A Quiz Page

1. pressure/fluid 8-2

2. different 8-2

1. barrier 8-5

2. equalization point 8-4

3. 3.792 psi 8-3

4. 205.74 differential 8-4, 5

5. 771.649 psi differential 8-4, 5

Refer to Items from Unit B Quiz Page

1. power/motion 8-7

2. upward pressure 8-7

3. F/A 8-7

1. end points 8-13

2. 140,760 lbs 8-11

3. 3550.7474 psi 8-9

4. 2183.647 lbs⇑ 8-12, 13

Refer to Self-Check Test Page

1. 258.75 psi 8-11

2. 1153.2 psi 8-6

3. 692 psi 8-6

4. 618.93444 psi 8-10

5. 200 sq in. 8-10

6. 2949.1247 psi 8-10

7. 0.76235 BF 8-11

8. 6892.62 psi

9. 293.6 psi

10. 3750.4 lb⇓

11. 0.8096 BF for 12.46 lb/gal fluid

Refer to the page numbers provided as references if you answered any of these items incorrectly, or if you were unsure of your answers. Now review to prepare for the final test.

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