/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 16 In April \(2010,\) the worst oil... [FREE SOLUTION] | 91Ó°ÊÓ

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In April \(2010,\) the worst oil spill ever recorded occurred when an explosion and fire on the Deepwater Horizon offshore oil-drilling rig left 11 workers dead and began releasing oil into the Gulf of Mexico. One of the attempts to contain the spill involved pumping drilling mud into the well to balance the pressure of escaping oil against a column of fluid (the mud) having a density significantly higher than those of seawater and oil. In the following problems, you may assume that seawater has a specific gravity of 1.03 and that the subsea wellhead was 5053 ft below the surface of the Gulf. (a) Estimate the gauge pressure (psig) in the Gulf at a depth of \(5053 \mathrm{ft}\). (b) Measurements indicate that the pressure inside the wellhead is 4400 psig. Suppose a pipe between the surface of the Gulf and the wellhead is filled with drilling mud and balances that pressure. Estimate the specific gravity of the drilling mud. (c) The drilling mud is a stable slurry of seawater and barite (SG \(=4.37\) ). What is the mass fraction of barite in the slurry? (d) What would you expect to happen if the barite weight fraction were significantly less than that estimated in Part (c)? Explain your reasoning.

Short Answer

Expert verified
The gauge pressure in the Gulf at a depth of 5053 ft is approximately 1522.8 psig. The specific gravity of the drilling mud that would balance this pressure at this depth is estimated to be around 2.18. The mass fraction of barite in the slurry is estimated to be around 0.31 or 31\%. If the barite mass fraction decreases significantly, it will fail to counter the oil pressure leading to more spillages.

Step by step solution

01

Gauge Pressure Calculation

The problem asks for the gauge pressure (psig) at a depth of 5053 ft. By definition, gauge pressure in a fluid is given by \( P = \rho g h \), where \( \rho \) is the density, \( g \) is the acceleration due to gravity, and \( h \) is the height (or depth in this case). The specific gravity of sea water is 1.03, hence density of sea water \( \rho = 1.03 \times 62.4 \, lb/ft^{3} = 64.272 \, lb/ft^{3} \). (Note: 62.4 lb/ft³ is the density of water).We insert these findings into the pressure equation together with the gravitational acceleration \( g = 32.2 \, ft/s^{2} \), to find out the pressure: \( P = \rho gh = 64.272 \times 32.2 \times 5053 \, lb/ft^{2} \). Since \( 1 \, psig = 144 \, lb/ft^{2} \), we divide by 144 to convert to psig.
02

Estimate Specific Gravity

Next, we need to estimate the specific gravity of the drilling mud. Given that the pressure inside the wellhead is 4400 psig, the mud using its specific gravity and the same depth needs to exert the same amount of pressure to balance this. Therefore, the specific gravity of the mud \( SG_{mud} = P_{mud} / (g \times h \times SG_{water}) \). Here, \( P_{mud} = 4400 \times 144 \, lb/ft^{2} \), \( SG_{water} = 1.03 \), and all other variables are the same as in step 1.
03

Mass Fraction Calculation

The third part of the exercise asks for the mass fraction of barite in the slurry. The specific gravity of barite is stated as 4.37. The specific gravity of a mix can be represented by the equation \( SG_{mix} = x \times SG_{barite} + (1-x) \times SG_{seawater} \), where \( x \) is the mass fraction of barite. Using the values from step 2 and solving for \( x \) will give us the mass fraction of barite.
04

Observations on Changes in Mass Fraction

A significant decrease in the mass fraction of barite would mean less of heavy material counteracting the oil pressure. The mud would have a lower specific gravity. It would therefore not be able to sufficiently counterbalance the high oil pressure which would result in the oil continuing to spew into the Gulf.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Gauge Pressure Calculation
Understanding gauge pressure is crucial for diverse fields such as engineering, meteorology, and even cooking. Gauge pressure is the pressure of a fluid relative to the ambient atmospheric pressure. As you saw in the example regarding the oil spill in the Gulf of Mexico, calculating gauge pressure involves considering the fluid's density, the depth at which the pressure is being measured, and the acceleration due to gravity.

To simplify, we can think of it like this: deep under the ocean's surface, water exerts pressure on any object because of its weight. The deeper you go, the greater the pressure. This scale doesn't start at zero, though; it's 'zeroed' against atmospheric pressure - meaning it represents the excess pressure over and above atmospheric pressure. For scientists and engineers working in the field, these calculations are critical for designing equipment that can withstand underwater pressure, such as subsea oil wells or pipelines.
Specific Gravity Estimation
Specific gravity (SG) is a measure of the density of a substance compared to the density of water. This concept becomes handy when differentiating between materials or substances based on their density without getting into complex units. For example, estimating the specific gravity of drilling mud, as needed for the oil spill scenario, helps engineers determine if the mud will be effective at counteracting the high pressure from the oil wellhead.

Practically, if a substance has a specific gravity less than 1, it floats on water, and if it's greater than 1, it sinks. The drilling mud in the case of the Deepwater Horizon spill needed a specific gravity higher than that of seawater to neutralize the upthrust of the escaping oil. Understanding specific gravity is hence pivotal in environmental management and industrial processes—like choosing the right materials for flotation devices or anticipating how pollutants might spread in water.
Mass Fraction in Slurry
The mass fraction is a dimensionless number representing the ratio of a substance's mass to the total mass of a mixture. In chemical engineering, particularly in preparing mixtures like slurry, mass fraction is essential to ensure the desired properties of the final product. For the Deepwater Horizon oil spill, engineers had to calculate the mass fraction of barite in the drilling mud slurry to ensure it was dense enough to counter the oil's escape pressure.

By balancing the specific gravity equation with known values of seawater and barite, you can determine the exact composition required for the drilling mud. This careful calculation exemplifies how the industry uses mass fraction to control processes and achieve specific outcomes, essential for tasks ranging from manufacturing to environmental remediation.
Oil Spill Environmental Impact
The environmental impact of an oil spill extends far beyond the immediate vicinity of the spill. It can devastate entire ecosystems, from the seabed to the shorelines, and affect the flora and fauna for years to come. Containment and clean-up efforts are challenging and costly, with an emphasis on preventing the oil from reaching coastal areas and harming wildlife or disrupting local economies that rely on fishing and tourism.

In the context of the Deepwater Horizon spill, the extended consequences included harm to marine life, widespread economic loss, and health issues among cleanup workers and local residents. By relating this example to the theoretical and practical aspects of chemical engineering education, we highlight the importance of responsible engineering practices and the mitigation of environmental risks. Understanding the mechanics behind gauge pressure, specific gravity, and mass fractions is not just academic; it is instrumental in anticipating and preventing disasters, measuring their potential impact, and orchestrating effective responses.

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Most popular questions from this chapter

At \(25^{\circ} \mathrm{C},\) an aqueous solution containing \(35.0 \mathrm{wt} \% \mathrm{H}_{2} \mathrm{SO}_{4}\) has a specific gravity of \(1.2563 .\) A quantity of the \(35 \%\) solution is needed that contains 195.5 kg of \(\mathrm{H}_{2} \mathrm{SO}_{4}\). (a) Calculate the required volume (L) of the solution using the given specific gravity. (b) Estimate the percentage error that would have resulted if pure-component specific gravities of \(\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{SG}=1.8255)\) and water had been used for the calculation instead of the given specific gravity of the mixture.

Drop-on-demand (DoD) technology is an emerging form of drug delivery in which a reservoir is filled with a solution of an active pharmaceutical ingredient (API) dissolved in a volatile liquid, and a device sprays nanometer-scale drops of the solution onto an edible substrate, such as a small strip the size of a stick of chewing gum. The liquid evaporates very rapidly, causing the API to crystallize on the substrate. The exact dose required by a patient can be administered based on the known concentration of the API in the reservoir and the volume of solution deposited on the sa1) For a man of \(245 \mathrm{lb}_{m}\) dosage is \(2.2 \mathrm{mL}=5.3 \cdot 10^{20}\) drops a2) For a child of \(65 ~ l b_{m}\) dosage is \(0.60 \mathrm{mL}=1.4 \cdot 10^{20}\) drops b) \(V_{\text {dose}}=\frac{D \cdot W_{p} \cdot 0.4536 \mathrm{kg} / \mathrm{lb}_{f}}{M W_{A P I} \cdot m_{s}} \cdot \frac{1000 \mathrm{mL}}{1 \mathrm{L}}\) c) \(\frac{A}{V}=\frac{3}{\pi},\) the nanometric drops have larger area, through which the water can evaporate faster.ubstrate, enabling greater dosage accuracy than can be provided by administering fractions of tablets. (a) A DoD device is charged with a 1.20 molar solution of ibuprofen (the API) in \(n\) -hexane. The molecular weight of ibuprofen is \(206.3 \mathrm{g} / \mathrm{mol} .\) If a prescribed dosage is \(5.0 \mathrm{mg}\) ibuprofen/kg patient weight, how many milliliters of solution should be sprayed for a 245 -pound man and a 65 -pound child? How many drops are in each dose, assuming that each drop is a sphere with a radius of \(1 \mathrm{nm} ?\) (b) The DoD device is to be automated, so that the operator enters a patient's body weight into a computer that determines the required solution volume and causes that volume to be sprayed on the substrate. Derive a formula for the volume, \(V_{\text {dare }}(\mathrm{mL}),\) in terms of the following variables: \(M_{\mathrm{s}}(\operatorname{mol} \mathrm{API} / \mathrm{L})=\) molarity of reservoir solution \(\mathrm{SG}_{\mathrm{s}}=\) specific gravity of reservoir solution \(\mathrm{MW}_{\mathrm{API}}(\mathrm{g} / \mathrm{mol})=\) molecular weight of \(\mathrm{API}\) \(D(\mathrm{mg} \mathrm{APl} / \mathrm{kg} \text { body weight })=\) prescribed dosage \(W_{\mathrm{p}}\left(\mathrm{b}_{\mathrm{f}}\right)=\) patient's weight Check your formula by verifying your solution to Part (a). (c) Calculate the surface-to-volume ratio of a sphere of radius \(r .\) Then calculate the total drop surface area of \(1 \mathrm{mL}\left(=1 \mathrm{cm}^{3}\right)\) of the solution if it were sprayed as drops of (i) radius \(1 \mathrm{nm}\) and (ii) \(1 \mathrm{mm}\). Speculate on the likely reason for spraying nanoscale drops instead of much larger drops.

The chemical reactor shown below has a cover that is held in place by a series of bolts. The cover is made of stainless steel ( \(\mathrm{SG}=8.0\) ), is 3 inches thick, has a diameter of 24 inches, and covers and seals an opening 20 inches in diameter. During turnaround, when the reactor is taken out of service for cleaning and repair, the cover was removed by an operator who thought the reactor had been depressurized using a standard venting procedure. However, the pressure gauge had been damaged in an earlier process upset (the reactor pressure had exceeded the upper limit of the gauge), and instead of being depressurized completely, the vessel was under a gauge pressure of 30 psi. (a) What force ( \(\left(\mathrm{b}_{\mathrm{f}}\right)\) were the bolts exerting on the cover before they were removed? (Hint: Don't forget that a pressure is exerted on the top of the cover by the atmosphere.) What happened when the last bolt was removed by the operator? Justify your prediction by estimating the initial acceleration of the cover upon removal of the last bolt. (b) Propose an alteration in the turnaround procedure to prevent recurrence of an incident of this kind.

Perform the following estimations without using a calculator. (a) Estimate the mass of water (kg) in an Olympic-size swimming pool. (b) A drinking glass is being filled from a pitcher. Estimate the mass flow rate of the water (g/s). (c) Twelve male heavyweight boxers coincidentally get on the same elevator in Great Britain. Posted on the elevator wall is a sign that gives the maximum safe combined weight of the passengers, \(W_{\mathrm{max}},\) in stones. (A stone is a unit of mass equal to \(14 \mathrm{lb}_{\mathrm{m}}\). It is commonly used in England as a measure of body weight, which, like the numerical equivalence between the \(1 \mathrm{b}_{\mathrm{m}}\) and \(\mathrm{Ib}_{\mathrm{f}},\) is only valid at or near sea level.) If you were one of the boxers, estimate the lowest value of \(W_{\max }\) for which you would feel comfortable remaining on the elevator. (d) The Trans-Alaska Pipeline has an outside diameter of 4 ft and extends 800 miles from the North Slope of Alaska to the northernmost ice-free port in Valdez, Alaska. How many barrels of oil are required to fill the pipeline? (e) Estimate the volume of your body \(\left(\mathrm{cm}^{3}\right)\) in two different ways. (Show your work.) (f) A solid block is dropped into water and very slowly sinks to the bottom. Estimate its specific gravity.

You purchase six oranges that weigh a total of \(2 \mathrm{Ib}_{\mathrm{f}}\) and 13 ounces. After cutting them open and squeezing all the juice your strength allows into a large measuring cup, you weigh the remaining pulp and orange peels. They weigh 1 Ib \(_{\mathrm{f}}\) and 12 ounces and the total volume of the juice is 1.75 cups. What is the specific gravity of orange juice? State any assumptions you make.

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