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Chapter 3: Problem 12

An object of density \(\rho_{\mathrm{a}},\) volume \(V_{\mathrm{a}},\) and weight \(W_{\mathrm{a}}\) is thrown from a rowboat floating on the surface of a small pond and sinks to the bottom. The weight of the rowboat without the jettisoned object is \(W_{\mathrm{b}}\). Beforethe object was thrown out, the depth of the pond was \(h_{\mathrm{pl}}\), and the bottom of the boat was a distance \(h_{\mathrm{b} 1}\) above the pond bottom. After the object sinks, the values of these quantities are \(h_{\mathrm{p} 2}\) and \(h_{\mathrm{b} 2}\). The area of the pond is \(A_{\mathrm{p}}\); that of the boat is \(A_{b} . A_{b}\) may be assumed constant, so that the volume of water displaced by the boat is \(A_{\mathrm{b}}\left(h_{\mathrm{p}}-h_{\mathrm{b}}\right)\). (a) Derive an expression for the change in the pond depth \(\left(h_{\mathrm{p} 2}-h_{\mathrm{p} 1}\right) .\) Does the liquid level of the pond rise or fall, or is it indeterminate? (b) Derive an expression for the change in the height of the bottom of the boat above the bottom of the pond \(\left(h_{b 2}-h_{b 1}\right) .\) Does the boat rise or fall relative to the pond bottom, or is it indeterminate?

Short Answer

Expert verified
The change in the pond's depth after the object is thrown out of the boat is \(\frac{V_{a}}{A_{p}}\), hence the liquid level in the pond rises. The change in the height of the boat relative to the bottom of the pond is \(\frac{h_{p1}}{1 + W_{b} / W_{a}} - h_{b1}\), which could be positive or negative depending on the ratio of the weights of the boat and the object.

Step by step solution

01

Consider the change in volume in the pond

To calculate the change in the pond's depth, consider the change in total volume in the pond once the object sinks. The total volume of the pond without the object can be represented as \(A_{p}h_{p1}\). After the object is thrown into the pond, its volume \(V_{a}\) will contribute to the total volume. Therefore the volume of the pond with the object would be \(A_{p}h_{p2} = A_{p}h_{p1} + V_{a}\).
02

Express change in pond depth

Rearrange the equation from step 1 to express the change in pond depth \(h_{p2} - h_{p1}\) in terms of the given parameters. Solving for \(h_{p2} - h_{p1}\), we get \(h_{p2} - h_{p1} = \frac{V_{a}}{A_{p}}\). This shows that the liquid level of the pond rose since the change in pond depth is positive.
03

Analyze the change in height of the boat

The weight of the boat must equal the weight of the displaced water, both before and after the object is thrown out. Before the object sinks, the weight of the system (boat + object) displaces a volume of water \(A_{b}(h_{p1} - h_{b1})\). After the object sinks, the boat alone displaces a volume of water \(A_{b}(h_{p2} - h_{b2})\). Setting these equal gives the equation \(W_{b} + W_{a} = W_{b}\left(1 + \frac{h_{p1} - h_{b1}}{h_{p2} - h_{b2}}\right)\).
04

Express change in height of the boat

Rearrange the equation from step 3 to express the change in the height of the boat \(h_{b2} - h_{b1}\) in terms of the given parameters. Solving for \(h_{b2} - h_{b1}\), we get \(h_{b2} - h_{b1} = \frac{h_{p1}}{1 + W_{b}/W_{a}} - h_{b1}\). This shows that the height difference can be positive or negative depending on the ratio of the weights of the boat and the object, hence it is indeterminate.

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Key Concepts

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

Density and Buoyancy

Understanding density and buoyancy is fundamental to grasping why objects float or sink. Density, symbolized as \( \rho \), is a property that measures how much mass is contained within a unit volume of a material. Mathematically, it's defined by the formula \( \rho = \frac{m}{V} \), where \( m \) is mass and \( V \) is volume. Buoyancy, on the other hand, refers to the force exerted by a fluid that supports the weight of an object placed in it.

The principle of buoyancy can be explained by Archimedes' Principle, which states that the buoyant force on an object is equal to the weight of the fluid displaced by the object. If an object is less dense than the fluid, it will float, otherwise it will sink. In chemical engineering, this concept is crucial for separating components by density, designing ships, and understanding fluid flow around submerged objects.

For the textbook exercise, the object sinks because its density is greater than that of the water, indicating a stronger gravitational force overpowering the buoyant force exerted by the water. After the object is thrown in, it displaces an additional volume of water, causing the pond level to rise, a direct illustration of the principles of buoyancy.

Conservation of Volume

The conservation of volume as it relates to fluids says that the volume of a fluid is conserved within a closed system. If you add or remove a specified volume of substance to a system, the total volume of the system must adjust to accommodate this change while maintaining an unchanging total volume.

In the exercise's context, when the object with volume \( V_a \) is thrown into the pond, the volume of water displaced by the object will be equal to the volume of the object itself, due to the conservation of volume. This additional displaced water results in a rise in the pond's depth, as highlighted in the exercise solution, in alignment with both the principles of buoyancy and conservation of volume.

Remembering that in a scenario where no water is added or removed from the system, aside from what is displaced by the object, the pond's water level change serves a perfect example of this conservation in practice.

Hydrostatics

Hydrostatics, which is the study of fluids at rest, deals with the forces exerted by a fluid at equilibrium. It is within this area of fluid mechanics that we explore concepts such as pressure within fluids, buoyant forces, and the behavior of submerged objects in fluid media.

One of the key principles in hydrostatics is that the pressure at a point within a fluid at rest is the same in all directions. This pressure increases with depth and is given by \( P = \rho g h \), where \( P \) is the pressure, \( \rho \) is the fluid density, \( g \) is acceleration due to gravity, and \( h \) is the height of the fluid column above the point.

By analyzing the textbook exercise, we can connect the rise in the boat to the hydrostatic conditions of the pond. The boat rises once the object is removed because of the redistribution of buoyant force due to the differential in water displaced at the boat鈥檚 location. This is part of hydrostatic equilibrium, which helps us understand the equilibrium positioning of boats on water and offers broader insights into the design of aquatic structures and vessels.

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Most popular questions from this chapter

The reaction \(A \rightarrow B\) is carried out in a laboratory reactor. According to a published article the concentration of A should vary with time as follows: \(C_{\mathrm{A}}=C_{\mathrm{A} 0} \exp (-k t)\) where \(C_{\mathrm{A} 0}\) is the initial concentration of \(\mathrm{A}\) in the reactor and \(k\) is a constant. (a) If \(C_{\mathrm{A}}\) and \(C_{\mathrm{A} 0}\) are in \(\mathrm{Ib}-\) moles \(/ \mathrm{ft}^{3}\) and \(t\) is in minutes, what are the units of \(k ?\) (b) The following data are taken for \(C_{\mathrm{A}}(t):\) $$\begin{array}{cc}\hline t(\min ) & C_{\mathrm{A}}\left(\mathrm{lb}-\mathrm{mole} / \mathrm{ft}^{3}\right) \\\\\hline 0.5 & 1.02 \\\1.0 & 0.84 \\\1.5 & 0.69 \\\2.0 & 0.56 \\\3.0 & 0.38 \\\ 5.0 & 0.17 \\\10.0 & 0.02 \\\\\hline\end{array}$$ Verify the proposed rate law graphically (first determine what plot should yield a straight line), and calculate \(C_{\mathrm{A} 0}\) and \(k\) (c) Convert the formula with the calculated constants included to an expression for the molarity of A in the reaction mixture in terms of \(t\) (seconds). Calculate the molarity at \(t=265 \mathrm{s}\).

A mixture of methanol and propyl acetate contains 25.0 wt\% methanol. (a) Using a single dimensional equation, determine the g-moles of methanol in \(200.0 \mathrm{kg}\) of the mixture. (b) The flow rate of propyl acetate in the mixture is to be 100.0 ib-mole/h. What must the mixture flow rate be in \(\mathrm{Ib}_{\mathrm{m}} / \mathrm{h} ?\)

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.

A mixture of methane and air is capable of being ignited only if the mole percent of methane is between 5\% and 15\%. A mixture containing 9.0 mole\% methane in air flowing at a rate of 7.00 \(\times 10^{2} \mathrm{kg} / \mathrm{h}\) is to be diluted with pure air to reduce the methane concentration to the lower flammability limit. Calculate the required flow rate of air in mol/h and the percent by mass of oxygen in the product gas. (Note: Air may be taken to consist of \(\left.21 \text { mole } \% \mathrm{O}_{2} \text { and } 79 \% \mathrm{N}_{2} \text { and to have an average molecular weight of } 29.0 .\right)\)

In the manufacture of pharmaceuticals, most active pharmaceutical ingredients (APIs) are made in solution and then recovered by separation. Acetaminophen, a pain-killing drug commercially marketed as Tylenol", is synthesized in an aqueous solution and subsequently crystallized. The slurry of crystals is sent to a centrifuge from which two effluent streams emerge: ( 1 ) a wet cake containing 90.0 wt\% solid acetaminophen \((\mathrm{MW}=\) 151 g/mol) and 10.0 wt\% water (plus some acetaminophen and other dissolved substances, which we will neglect), and (2) a highly dilute aqueous solution of acetaminophen that is discharged from the process. The wet cake is fed to a dryer where the water is completely evaporated, leaving the residual acetaminophen solids bone dry. If the evaporated water were condensed, its volumetric flow rate would be \(50.0 \mathrm{Lh}\). Following is a flowchart of the process, which runs 24 h/day, 320 days/yr. A denotes acetaminophen. (a) Calculate the yearly production rate of solid acetaminophen (tonne/yr), using as few dimensional equations as possible. (b) A proposal has been made to subject the liquid solution leaving the centrifuge to further processing to recover more of the dissolved acetaminophen instead of disposing of the solution. On what would the decision depend?
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