/*! 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 88 A stream of oxygen enters a comp... [FREE SOLUTION] | 91影视

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A stream of oxygen enters a compressor at \(298 \mathrm{K}\) and 1.00 atm at a rate of \(127 \mathrm{m}^{3} / \mathrm{h}\) and is compressed to \(358 \mathrm{K}\) and 1000 atm. Estimate the volumetric flow rate of compressed \(\mathrm{O}_{2},\) using the compressibility-factor equation of state.

Short Answer

Expert verified
The volumetric flow rate of the compressed oxygen gas is 0.153 m鲁/h.

Step by step solution

01

Understand the Compressibility factor

The compressibility factor or \(Z\) is a factor used in equations of state calculations for gases, which helps us to adjust the behavior of gases away from the ideal gas law. It is defined as \(Z = \frac{pV}{nRT}\), where p is pressure, V is volume, n is number of moles, R is gas constant, and T is temperature. In this exercise, the volume isn't constant so we'll adjust the formula to \(Z = \frac{p鈧乂鈧亇{T鈧亇 = \frac{p鈧俈鈧倉{T鈧倉\) where subscripts 1 and 2 refer to initial and final states.
02

Inserting Given Values into Equation

Now, insert the values from the exercise into the formula. Given: \(p鈧 = 1 \text{ atm}, V鈧 = 127 \text{ m鲁/h}, T鈧 = 298 \text{ K}, p鈧 = 1000 \text{ atm}, T鈧 = 358 \text{ K}\). The unknown in this equation is \(V鈧俓), which represents the final volume after compression of the gas, so solve the equation for \(V鈧俓). \nWith this, our equation looks like this: \(V鈧 = \frac{p鈧乂鈧乀鈧倉{p鈧俆鈧亇\).
03

Calculate V鈧

Now we have all variables to compute the result: \(V鈧 = \frac{1 \text{ atm} 脳 127 \text{ m鲁/h} 脳 358 \text{ K}}{1000 \text{ atm} 脳 298 \text{ K}} = 0.153 \text{ m鲁/h}\)
04

Interpretation of the Result

The final volume, after compression of gas is 0.153 m鲁/h. As expected, volume has decreased due to compression as more gas molecules are packed in a smaller volume when the pressure is increased.

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Key Concepts

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

compressor process
A compressor process involves the mechanical action of compressing a gas to increase its pressure and decrease its volume. Compressors are used in various applications, from industrial machinery to everyday appliances like refrigerators.
When a gas, such as oxygen, enters a compressor, it undergoes a change in state due to the work done on it.
As the gas is compressed, its pressure rises, leading to a reduction in volume. The principle behind compression is to apply external force to the gas particles, bringing them closer together. This results in:
  • An increase in the kinetic energy of the gas particles.
  • A rise in pressure, since the particles collide more frequently.
  • A decrease in gas volume, according to principles of gas behavior as outlined by basic gas laws.
During compression, maintaining a balance of energy and system stability is crucial. The compressibility factor, discussed later, aids in understanding deviations from ideal behavior that gases might exhibit under high-pressure conditions.
oxygen compression
Compressing oxygen involves increasing its pressure to pack the molecules into a smaller space, which drastically reduces its volume. This is particularly useful in various industrial processes, medical applications, and scientific research. Oxygen's properties mean it needs careful handling during compression.
Under such conditions, oxygen may behave differently from ideal gas behavior. Therefore, adjustments using factors like the compressibility factor are necessary.
Several considerations are essential during oxygen compression:
  • The potential for high-temperature increases, which might pose safety risks if not managed.
  • The need for specialized materials and equipment to withstand high-pressure conditions.
  • Safety protocols to prevent leaks because oxygen can accelerate combustion of materials.
Understanding oxygen's behavior under varying temperatures and pressures is fundamental. This knowledge ensures processes are efficient and safe for both operators and the equipment used.
equation of state
An equation of state is a mathematical model describing the state properties of a gas. The most widely known is the Ideal Gas Law, represented as \(PV = nRT\).
However, real gases often deviate from this behavior due to interactions between gas molecules.
The compressibility factor \(Z\) is introduced to these equations when real gas behavior is evident. For example, in our scenario with oxygen, which is subjected to extreme pressures and temperatures, \(Z\) modifies the Ideal Gas Law:\[ Z = \frac{pV}{nRT} \]This equation becomes crucial in adjusting calculations to account for:
  • Intermolecular forces that may be significant at high pressures.
  • The volume occupied by gas molecules, which becomes considerable.
The ability to predict gas behavior accurately with equations of state is vital in fields such as chemical engineering and thermodynamics. This ensures efficient and safe process design, especially when dealing with high-pressure systems like compressors.
volumetric flow rate
Volumetric flow rate is a measure of the volume of fluid flowing through a given surface per unit time. It is crucial in analyzing systems involving fluid movement, such as gases in compressors.
Expressed typically in units like cubic meters per hour (m鲁/h), it determines the efficiency and capacity of systems in processing the fluid.
In the context of gas compression, knowing the volumetric flow rate helps engineers:
  • Design systems suitable for the desired output and efficiency.
  • Determine the size and capability of compressors and other equipment components.
  • Predict system performance under varying conditions of temperature and pressure.
Accurate computation of volumetric flow rate becomes especially significant when gases deviate from ideal behavior, as they often do in high-pressure applications. Such calculations allow for effective process management and optimization, ensuring that systems meet operational demands efficiently.

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Most popular questions from this chapter

Methanol is produced by reacting carbon monoxide and hydrogen at \(644 \mathrm{K}\) over a \(\mathrm{ZnO}-\mathrm{Cr}_{2} \mathrm{O}_{3}\) catalyst. A mixture of \(\mathrm{CO}\) and \(\mathrm{H}_{2}\) in a ratio \(2 \mathrm{mol} \mathrm{H}_{2} / \mathrm{mol}\) CO is compressed and fed to the catalyst bed at \(644 \mathrm{K}\) and 34.5 MPa absolute. A single-pass conversion of 25\% is obtained. The space velocity, or ratio of the volumetric flow rate of the feed gas to the volume of the catalyst bed, is The product gases are passed through a condenser, in which the methanol is liquefied. (a) You are designing a reactor to produce \(54.5 \mathrm{kmol} \mathrm{CH}_{3} \mathrm{OH} / \mathrm{h}\). Estimate (i) the volumetric flow rate that the compressor must be capable of delivering if no gases are recycled, and (ii) the required volume of the catalyst bed. (Use Kay's rule for pressure-volume calculations.) (b) If (as is done in practice) the gases from the condenser are recycled to the reactor, the compressor is then required to deliver only the fresh feed. What volumetric flow rate must it deliver assuming that the methanol produced is completely recovered in the condenser? (In practice it is not; moreover, a purge stream must be taken off to prevent the buildup of impurities in the system.)

Lewis \(^{12}\) describes the hazards of breathing air containing appreciable amounts of an asphyxiant (a gas that has no specific toxicity but, when inhaled, excludes oxygen from the lungs). When the mole percent of the asphyxiant in the air reaches \(50 \%,\) marked symptoms of distress appear, and at \(75 \%\) death occurs in a matter of minutes. A small storage room whose dimensions are \(2 \mathrm{m} \times 1.5 \mathrm{m} \times 3 \mathrm{m}\) contains a number of expensive and dangerous chemicals. To prevent unauthorized entry, the door to the room is always locked and can be opened with a key from either side. A cylinder of liquid carbon dioxide is stored in the room. The valve on the cylinder is faulty and some of the contents have escaped over the weekend. The room temperature is \(25^{\circ} \mathrm{C}\). (a) If the concentration of \(\mathrm{CO}_{2}\) reaches the lethal 75 mole \(\%\) level, what would be the mole percent of \(\mathrm{O}_{2} ?\) (b) How much \(\mathrm{CO}_{2}(\mathrm{kg})\) is present in the room when the lethal concentration is reached? Why would more than that amount have to escape from the cylinder for this concentration to be reached? (c) Describe a set of events that could result in a fatality in the given situation. Suggest at least two measures that would reduce the hazards associated with storage of this scemingly harmless substance.

A slurry contains crystals of copper sulfate pentahydrate \(\left[\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}(\mathrm{s}), \text { specific gravity }=2.3\right]\) suspended in an aqueous copper sulfate solution (liquid SG \(=1.2\) ). A sensitive transducer is used to measure the pressure difference, \(\Delta P(\mathrm{Pa}),\) between two points in the sample container separated by a vertical distance of \(h\) meters. The reading is in turn used to determine the mass fraction of crystals in the slurry, \(x_{\mathrm{c}}(\mathrm{kg}\) crystals/kg slurry). (a) Derive an expression for the transducer reading, \(\Delta P(\mathrm{Pa}),\) in terms of the overall slurry density, \(\rho_{\mathrm{s}}\left(\mathrm{kg} / \mathrm{m}^{3}\right),\) assuming that the equation used to calculate the pressure head in Chapter 3 \(\left(P=P_{0}+\rho g h\right)\) is valid for this two-phase system. (b) Validate the following expression relating the overall slurry density to the liquid and solid crystal densities \(\left(\rho_{1} \text { and } \rho_{c}\right)\) and the mass fraction of crystals in the slurry: $$\frac{1}{\rho_{\mathrm{si}}}=\frac{x_{\mathrm{c}}}{\rho_{\mathrm{c}}}+\frac{\left(1-x_{\mathrm{c}}\right)}{\rho_{1}}$$ (c) Suppose \(175 \mathrm{kg}\) of the slurry is placed in the sample container with \(h=0.200 \mathrm{m}\) and a transducer reading \(\Delta P=2775\) Pa is obtained. Calculate \((\mathrm{i}) \rho_{\mathrm{s}_{\mathrm{s}},(\mathrm{ii})} x_{\mathrm{c}},\) (iii) the total slurry volume, (iv) the mass of crystals in the slurry, (v) the mass of anhydrous copper sulfate (CuSO \(_{4}\) without the water of hydration) in the crystals, (vi) the mass of liquid solution, and (vii) the volume of liquid solution. (d) Prepare a spreadsheet to generate a calibration curve of \(x_{c}\) versus \(\Delta P\) for this device. Take as inputs \(\rho_{\mathrm{c}}\left(\mathrm{kg} / \mathrm{m}^{3}\right), \rho_{1}\left(\mathrm{kg} / \mathrm{m}^{3}\right),\) and \(h(\mathrm{m}),\) and calculate \(\Delta P(\mathrm{Pa})\) for \(x_{\mathrm{c}}=0.0,0.05,0.10, \ldots, 0.60\) Run the program for the parameter values in this problem \(\left(\rho_{\mathrm{c}}=2300, \rho_{1}=1200, \text { and } h=0.200\right)\) Then plot \(x_{c}\) versus \(\Delta P\) (have the spreadsheet program do it, if possible), and verify that the value of \(x_{c}\) corresponding to \(\Delta P=2775\) Pa on the calibration curve corresponds to the value calculated in Part (c). (e) Derive the expression in Part (b). Take a basis of \(1 \mathrm{kg}\) of slurry \(\left[x_{\mathrm{c}}(\mathrm{kg}), V_{c}\left(\mathrm{m}^{3}\right)\right.\) crystals, \(\left.\left(1-x_{\mathrm{c}}\right)(\mathrm{kg}), V_{l}\left(\mathrm{m}^{3}\right) \text { liquid }\right],\) and use the fact that the volumes of the crystals and liquid are additive.

The label has come off a cylinder of gas in your laboratory. You know only that one species of gas is contained in the cylinder, but you do not know whether it is hydrogen, oxygen, or nitrogen. To find out, you evacuate a 5 -liter flask, seal it and weigh it, then let gas from the cylinder flow into it until the gauge pressure equals 1.00 atm. The flask is reweighed, and the mass of the added gas is found to be 13.0g. Room temperature is \(27^{\circ} \mathrm{C}\), and barometric pressure is 1.00 atm. What is the gas?

A distillation column is being used to separate methanol and water at atmospheric pressure. The column temperature varies from approximately \(65^{\circ} \mathrm{C}\) at the top to \(100^{\circ} \mathrm{C}\) at the bottom. Liquid enters the top of the column and flows down to the bottom; vapor is generated in a reboiler at the bottom of the column, flows upward, and leaves at the top. The molar flow rate of vapor up the column may be assumed to be constant from top to bottom. The vapor velocity is kept below \(5.0 \mathrm{ft} / \mathrm{s}\) to keep the vapor from entraining liquid (suspending and carrying away liquid droplets). (a) Where in the column is the greatest risk of liquid entrainment? Explain your answer. (b) Assuming that the liquid flowing down the column and the column internals (equipment inside the column) occupy a negligible fraction of the column cross-sectional area, estimate the minimum column diameter if the vapor flow rate is 25.0 lb-mole/min. (c) Suppose the column is constructed with a diameter \(10 \%\) greater than that determined in Part (b). What are the vapor velocities at the top and bottom of the column if the vapor molar flow rate in both locations is 25.0 ib-mole/min? How much can the vapor molar flow rate be increased without causing liquid entrainment? (d) There is a need to increase process throughput, which would require the vapor molar flow rate to be doubled. It has been suggested that increasing the pressure in the column would allow that to be done without risking excessive liquid entrainment. Again applying a vapor velocity limit of \(5 \mathrm{ft} / \mathrm{s}\) what would the new pressure be?

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