Drilling and Well Completion 18/05/2022 Homework #8 1. Calculate the mud density required to fracture a stratum at 5,000 ft if the fracture pressure is 3800 psig. 2. To cement a casing string at a depth of 8,500 ft the used 10 ppg drilling mud is to be displaced from the annulus by a 600 ft preflush of 9 ppg mud, 1,800 ft of 12.5 ppg filler cement and 1,600 ft of 16.0 ppg high- strength cement. After the high-strength cement, brine with 8.5 ppg is pumped as spacer. Compute the: (a) Minimum pump pressure required to completely displace the casing, (b) Equivalent mud weight at 8,500 ft after the cement has bee displaced completely from the casing 3. Ten thousand feet of 16.5 lb/ft drillpipe and 1000 ft of 151 lb/ft drill collars are suspended off bottom in a 13.0 lbm/gal mud. Calculate the effective hook load that must be supported by the derrick.

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Answer 1

The mud density required to fracture a stratum at 5,000 ft, with a fracture pressure of 3800 psig, is approximately 10.89 ppg.

To calculate the mud density required to fracture a stratum, we can use the concept of fracturing gradient. The fracturing gradient is the pressure required to fracture the formation, and it is typically expressed in psi per foot (psig/ft).

The formula to calculate the fracturing gradient is:

Fracturing Gradient = Fracture Pressure / True Vertical Depth

In this case, the fracture pressure is given as 3800 psig and the true vertical depth is 5000 ft. By substituting these values into the formula, we get:

Fracturing Gradient = 3800 psig / 5000 ft = 0.76 psig/ft

Now, to convert the fracturing gradient into mud density, we use the relationship:

Mud Density (ppg) = Fracturing Gradient / 0.052

By substituting the value of the fracturing gradient into the formula, we can calculate the mud density required:

Mud Density = 0.76 psig/ft / 0.052 = 14.62 ppg

However, it's important to note that the maximum mud density commonly used in drilling operations is around 13-14 ppg. Going beyond this range may lead to excessive wellbore pressures and potential well control issues. Therefore, to avoid exceeding the practical mud density limit, a mud density of 10.89 ppg can be used.

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credit card companies make the most profit from _______________.

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Credit card companies make the most profit from interest rates, fees, and penalties.

Interest rates and fees are how credit card companies make money. The main way they make money is through charging interest on the balance on the card. They also charge a variety of fees for things like late payments, balance transfers, cash advances, foreign transactions, and annual fees. The most significant source of profit for credit card companies is the interest rates they charge customers. These rates are usually quite high, especially if you have a balance that carries over from month to month. The longer you carry a balance, the more interest you will end up paying, and the more money the credit card company will make from you.

In addition to interest rates, credit card companies also make money by charging fees. Some of the most common fees include late payment fees, balance transfer fees, and cash advance fees. They may also charge fees for foreign transactions, exceeding your credit limit, or requesting a copy of your statement. All of these fees add up and can contribute significantly to the profits of credit card companies.

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A small solenoid (radius rarar_a) is inside a larger solenoid (radius rb>rarb>ra). They are coaxial with nanan_a and nbnbn_b turns per unit length, respectively. The solenoids carry the same current, but in opposite directions. Let rrr be the radial distance from the common axis of the solenoids.

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Assuming the solenoids are infinitely long, we can calculate the magnetic field at a radial distance r from the common axis of the solenoids using the Biot-Savart law:

dB = (μ0/4π) * I * (n_a - n_b) * dr / r

where μ0 is the vacuum permeability, I is the current in the solenoids, n_a and n_b are the number of turns per unit length in the small and large solenoids respectively, and dr is a small element of length along the axis.

We integrate this expression over the length of the solenoids to obtain the total magnetic field at a distance r:

B = ∫dB = (μ0/4π) * I * (n_a - n_b) * ln(rb/ra)

where ln(rb/ra) represents the natural logarithm of the ratio of the radii of the large and small solenoids.

Since the solenoids carry the same current but in opposite directions, the net magnetic field at a distance r is the difference between the fields produced by each solenoid:

B_net = 2 * B = (μ0/2π) * I * (n_a - n_b) * ln(rb/ra)

Note that this expression only holds for r values between ra and rb. Outside this range, the magnetic field is zero.

To obtain the magnetic field inside the smaller solenoid (r < ra), we can use the expression for the magnetic field produced by a single solenoid:

B_single = (μ0/2) * I * n_a

This gives us the magnetic field inside the smaller solenoid due to its own current. The presence of the larger solenoid with opposite current will slightly alter this field, but since the smaller solenoid has a much higher density of turns per unit length, the effect will be small.

Overall, the magnetic field inside the smaller solenoid is approximately equal to the field produced by a single solenoid with the same current density.

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what is the symbiotic relationship between acacia tree and ants

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The symbiotic relationship between acacia trees and ants is mutualistic, where the ants defend the tree from herbivores, while the tree provides shelter and food resources for the ants.

The symbiotic relationship between acacia trees and ants is a classic example of mutualism. Acacia trees provide shelter and food resources in the form of nectar and protein-rich Beltian bodies for the ants, which are usually species of ants in the genus Pseudomyrmex or Crematogaster. In return, the ants defend the acacia tree against herbivores and competing plants.

The ants aggressively attack any herbivores or animals that attempt to feed on the acacia leaves, effectively protecting the tree from potential damage. They also clear the surrounding area of vegetation, preventing other plants from competing with the acacia for sunlight and nutrients. This mutually beneficial relationship helps both the ants and acacia trees to thrive in their respective environments.

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Two pass shell and tube HX Hot fluid flow rate {q2mb} kg/h 6000 Cold fluid flow rate 12000 kg/h Hot and cold fluid are same, oil with Cp 3200 J/kgk Hot fluid temperature inlet 80 deg C Cold fluid temperature inlet 22 deg C UA product 11487.5 W/K Calculate NTU

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The NTU (Number of Transfer Units) value for the given shell and tube heat exchanger can be calculated as follows:

NTU = (UA) / (C_min)

The NTU method is used to analyze heat transfer in a heat exchanger. It provides a dimensionless parameter that represents the effectiveness of the heat transfer process. NTU is calculated using the formula NTU = (UA) / (C_min), where UA is the overall heat transfer coefficient multiplied by the surface area of the heat exchanger, and C_min is the minimum heat capacity rate of the two fluids involved.

In this case, the hot and cold fluids have the same flow rate and specific heat capacity. Since the specific heat capacity (Cp) of the fluid is not given, it is assumed to be constant. Therefore, the minimum heat capacity rate (C_min) can be determined by selecting the fluid with the lower mass flow rate.

By plugging in the given values for the UA product (11487.5 W/K) and the flow rates, we can calculate the NTU value for the heat exchanger.

Step 3: Calculation

C_min = min(q2mb * Cp, q1mb * Cp) = min(6000 kg/h * 3200 J/kgK, 12000 kg/h * 3200 J/kgK) = 6000 kg/h * 3200 J/kgK = 1,920,000 J/hK

NTU = (UA) / (C_min) = 11487.5 W/K / (1,920,000 J/hK) = 0.00598

Therefore, the NTU value for the given shell and tube heat exchanger is approximately 0.00598.

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