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Chapter 7: Problem 57

Your friend has asked you to help move a 60 inch \(\times 78\) inch mattress with a mass of 75 Ib \(_{\mathrm{m}}\). The two of you position it horizontally in an open flat-bed trailer that you hitch to your car. There is nothing available to tie the mattress to the trailer, but you know there is a risk of the mattress being lifted from the trailer by the air flowing over it and perform the following calculations: (a) Although the conditions do not exactly match those for which the Bemoulli equation is applicable, use the equation to get a rough estimate of how fast you can drive (miles/h) before the mattress is lifted. Assume the velocity of air above the mattress equals the velocity of the car, the pressure difference between the top and bottom of the mattress equals the weight of the mattress divided by the mattress cross-sectional area, and air has a constant density of \(0.075 \mathrm{lb}_{\mathrm{m}} / \mathrm{ft}^{3}\). What is your result? (b) You see that your friend also has several boxes of books. Since you would like to drive at 60 miles per hour, what weight of books ( \(\left(\mathrm{b}_{\mathrm{f}}\right)\) do you need to put on the mattress to hold it in place?

Short Answer

Expert verified
The car can be driven approximately 20 miles/hour before the mattress is lifted according to rough estimates from Bernoulli's law. To drive at 60 miles/hour without the mattress lifting off, approximately 215 pounds of books need to be placed on top of the mattress.

Step by step solution

01

Understanding Bernoulli's Law

In simplest terms, Bernoulli's law states that an increase in the speed of a fluid occurs simultaneously with a decrease in static pressure or a decrease in the fluid's potential energy. In our scenario, the mattress lifting off happens when the air pressure above is less than the pressure below.
02

Using Bernoulli's Law to Calculate the Maximum Speed Before Lift-Off

By rearranging Bernoulli's law and assuming that the difference in pressure equals weight of the mattress (75 lbs) over its cross-sectional area (60 in * 78 in), the maximum speed can be calculated by the following formula: \[v = \sqrt{(2 * (P_{1} - P_{2})) / \rho}\], where \(v\) is the speed, \(P_{1}\) and \(P_{2}\) are the pressures below and above the mattress respectively and \(\rho\) is the density of air (0.075 lb/ft^3). After converting the measurements into suitable units and performing the calculations, one can find the maximum speed before lift-off.
03

Calculating Needed Weight for 60 Miles/Hour

To prevent the mattress from being lifted at 60 miles/hour, extra weight needs to be added on top of the mattress. This weight can be calculated with the formula \[W = (P_{1} - P_{2}) * A - W_{m}\], where \(W\) is the weight of books needed, \(P_{1}\) and \(P_{2}\) are the pressures below and above the mattress respectively, \(A\) is the cross-sectional area of the mattress and \(W_{m}\) is initially the weight of the mattress. By substituting the speeds, pressure difference, and area into the formula, using the conversion factor for units, one can find the necessary weight of books.

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Key Concepts

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

Fluid Dynamics
Understanding fluid dynamics is essential when analyzing phenomena such as the lifting of a mattress on a moving trailer. Fluid dynamics is the study of how fluids (liquids and gases) move and the forces acting upon them. Air, in our scenario, behaves like a fluid flowing over the surface of the mattress.

When dealing with fluid dynamics, one major force to consider is drag, which opposes the motion of an object through a fluid. As the car moves, air flows over and under the mattress, creating different pressure conditions based on its speed and characteristics. Bernoulli鈥檚 principle, which relates the velocity of the fluid to the pressure exerted by it, helps predict the potential for the mattress to lift off from the trailer at certain speeds. It's crucial for students to grasp that the behavior of the air, although invisible, can have tangible effects, such as the potential lift of the mattress due to pressure differences.
Pressure Differences
The principle of pressure differences plays a pivotal role in explaining the lifting of the mattress. According to Bernoulli鈥檚 law, an increase in the speed of a fluid results in a decrease in the pressure it exerts. The air flowing over the mattress is moving faster compared to the relatively stagnant air underneath, creating a lower pressure zone on top of the mattress.

To visualize pressure differences, imagine holding a piece of paper by one end and blowing over the top of it. The paper lifts because the air pressure above the paper is reduced by your breath's speed, which is akin to what happens with the mattress on the trailer when the car speeds up. When working through the exercise calculations, students should recognize that the difference between the pressure on top and the pressure below is what could potentially cause the mattress to lift, analogous to how airplane wings generate lift.
Air Density
The concept of air density is crucial when employing Bernoulli's law to estimate the speed at which the mattress may lift off the trailer. Air density, the mass per unit volume of air, affects the pressure and lift generated on the mattress. In the given problem, we assume a constant air density of 0.075 lb/ft鲁.

Understanding that denser air would exert more force is important. Because denser air contains more molecules and thus more mass, it would lead to a greater impact and pressure on the mattress for any given speed. Conversely, if the air were less dense, say at higher altitudes, the same speed would exert less pressure. Therefore, in calculating how additional weight (like a box of books) affects the system, the density of the air is a critical factor in determining the force acting on the mattress due to the fluid flow 鈥� showing how the density of air ties into the practical matter of ensuring the mattress stays put during transit.

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Most popular questions from this chapter

Eight fluid ounces \((1 \mathrm{qt}=32 \mathrm{oz})\) of a beverage in a glass at \(18.0^{\circ} \mathrm{C}\) is to be cooled by adding ice and stirring. The properties of the beverage may be taken to be those of liquid water. The enthalpy of the ice relative to liquid water at the triple point is \(-348 \mathrm{kJ} / \mathrm{kg} .\) Estimate the mass of ice (g) that must melt bring the liquid temperature to \(4^{\circ} \mathrm{C},\) neglecting energy losses to the surroundings.

Write and simplify the closed-system energy balance (Equation \(7.3-4\) ) for each of the following processes, and state whether nonzero heat and work terms are positive or negative. Begin by defining the system. The solution of Part (a) is given as an illustration. (a) The contents of a closed flask are heated from \(25^{\circ} \mathrm{C}\) to \(80^{\circ} \mathrm{C}\). (b) A tray filled with water at \(20^{\circ} \mathrm{C}\) is put into a freezer. The water tums into ice at \(-5^{\circ} \mathrm{C}\). (Note: When a substance expandsit does work on its surroundings and when it contracts the surroundings do work on it.) (c) A chemical reaction takes place in a closed adiabatic (perfectly insulated) rigid container. (d) Repeat Part (c), only suppose that the reactor is isothermal rather than adiabatic and that when the reaction was carried out adiabatically, the temperature in the reactor increased.

Liquid water at \(30.0^{\circ} \mathrm{C}\) and liquid water at \(90.0^{\circ} \mathrm{C}\) are combined in a ratio \((1 \mathrm{kg} \text { cold water/2 } \mathrm{kg}\) hot water). (a) Use a simple calculation to estimate the final water temperature. For this part, pretend you never heard of energy balances. (b) Now assume a basis of calculation and write a closed-system energy balance for the process, assuming that the mixing is adiabatic. Use the balance to calculate the specific internal energy and hence (from the steam tables) the final temperature of the mixture. What is the percentage difference between your answer and that of Part (a)?

Horatio Meshuggeneh has his own ideas of how to do things. For instance, when given the task of determining an oven temperature, most people would use a thermometer. Being allergic to doing anything most people would do, however, Meshuggeneh instead performs the following experiment. He puts acopper bar with a mass of 5.0 kg in the oven and puts an identical bar in a well- insulated 20.0-liter vessel containing 5.00L of liquid water and the remainder saturated steam at \(760 \mathrm{mm}\) Hg absolute. He waits long enough for both bars to reachthermal equilibrium with their surroundings, then quickly takes the first bar out of the oven, removes the second bar from the vessel, drops the first bar in its place, covers the vessel tightly, waits for the contents to come to equilibrium, and notes the reading on a pressure gauge built into the vessel. The value he reads is 50.1 mm Hg. He then uses the facts that copper has a specific gravity of 8.92 and a specific internal energy given by the expression \(\hat{U}(\mathrm{kJ} / \mathrm{kg})=0.36 T\left(^{\circ} \mathrm{C}\right)\) to calculate the oven temperature. (a) The Meshuggeneh assumption is that the bar can be transferred from the oven to the vessel without any heat being lost. If he makes this assumption, what oven temperature does Meshuggeneh calculate? How many grams of water evaporate in the process? (Neglect the heat transferred to the vessel wall- -i.e., assume that the heat lost by the bar is transferred entirely to the water in the vessel. Also, remember that you are dealing with a closed system once the hot bar goes into the vessel.) (b) In fact, the bar lost 8.3 kJ of heat between the oven and the vessel. What is the true oven temperature? (c) The experiment just described was actually Meshuggeneh's second attempt. The first time he tried it, the final gauge pressure in the vessel was negative. What had he forgotten to do?

One thousand liters of a 95 wt\% glycerol- \(5 \%\) water solution is to be diluted to \(60 \%\) glycerol by adding a \(35 \%\) solution pumped from a large storage tank through a \(5-\mathrm{cm}\) ID pipe at a steady rate. The pipe discharges at a point 23 m higher than the liquid surface in the storage tank. The operation is carried out isothermally and takes 13 min to complete. The friction loss ( \(\hat{F}\) of Equation \(7.7-2\) ) is \(50 \mathrm{J} / \mathrm{kg}\). Calculate the final solution volume and the shaft work in \(\mathrm{kW}\) that the pump must deliver, assuming that the surface of the stored solution and the pipe outlet are both at 1 atm. Data: \(\quad \rho_{\mathrm{H}_{2} \mathrm{O}}=1.00 \mathrm{kg} / \mathrm{L}, \rho_{\mathrm{gly}}=1.26 \mathrm{kg} / \mathrm{L} .\) (Use to estimate solution densities.)
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