/*! 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

Magnesium sulfate has a number of uses, some of which are related to the ability of the anhydrate form to remove water from air and others based on the high solubility of the heptahydrate \(\left(\mathrm{MgSO}_{4} \cdot 7 \mathrm{H}_{2} \mathrm{O}\right)\) form, also known as Epsom salt. The densities of the anhydrate and heptahydrate crystalline forms are 2.66 and \(1.68 \mathrm{g} / \mathrm{mL},\) respectively. Suppose you wish to form a 20.0 wt\% \(\mathrm{MgSO}_{4}\) aqueous solution by simply pouring crystals of one of the forms into a tank of water while the temperature is held constant at \(30^{\circ} \mathrm{C}\). The specific gravity of the 20.0 wt\% solution at \(30^{\circ} \mathrm{C}\) is \(1.22 .\) Answer the following questions for both forms of the \(\mathrm{MgSO}_{4}\) crystals: (a) What volume of water should be in the tank before crystals are added if the final product is to be 1000 kg of the 20 wt\% solution? (b) Suppose the tank diameter is \(0.30 \mathrm{m}\). What is the height of liquid in the tank before the crystals are added? (c) What is the height of the water in the tank after addition of the crystals but before they begin to dissolve? (d) What is the height of liquid in the tank after all the MgSO \(_{4}\) has dissolved?

Air in industrial plants is subject to contamination by many different chemicals, and companies must monitor ambient levels of hazardous species to be sure they are below limits specified by the National Institute for Occupational Safety and Health (NIOSH). In personal breathing-zone sampling (as opposed to area sampling), workers wear devices that periodically collect air samples less than 10 inches away from their noses. Breathing-zone sampling and analysis methods for hundreds of species are set forth in the NIOSH Manual of Analytical Methods. \(^{13}\) For benzene, NIOSH specifies a recommended exposure limit (REL) of 0.1 ppm time-weighted average exposure (TWA), and the Occupational Safety and Health Administration (OSHA) permissible exposure limit (PEL) is 1.0ppm TWA. A worker in a petrolcum refinery has a personal breathing-zone sampler for benzenc clipped to her shirt collar. Following the NIOSH prescription, air is pumped through the sampler at a rate of \(0.200 \mathrm{L} / \mathrm{min}\) by a small battery-operated pump attached to the worker's belt. The sampler contains an adsorbent that removes essentially all of the benzene from the air passing through it. After several hours, the sampler is removed and sent to a lab for analysis, and the worker puts on a fresh sampler. On a particular day when the temperature is \(21^{\circ} \mathrm{C}\) and barometric pressure is \(730 \mathrm{mm}\) Hg, samples are collected during a 4-h period before lunch and a 3.5-h period after lunch. The analytical laboratory reports \(0.17 \mathrm{mg}\) of benzene in the first sample and \(0.23 \mathrm{mg}\) in the second. (a) Calculate the average benzene concentration, \(C_{\mathrm{B}}(\mathrm{ppm}),\) in the worker's breathing zone during each sampling period, where 1 ppm = 1 mol C \(_{6} \mathrm{H}_{6} / 10^{6}\) mol air. (b) The worker's TWA is the average concentration of benzene in her breathing zone during the eight hours of her shift. It is calculated by multiplying \(C_{\mathrm{B}}\) in each sampling period by the time of that period, summing the products over all periods during the shift, and dividing by the total time of the shift. Assume that the worker's exposure during the unsampled 30 minutes was zero, and calculate her TWA. (c) If the worker's exposure is above the recommended limits, what actions might the company take?

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?

The absolute pressure within a 35.0 -liter gas cylinder should not exceed 51.0 atm. Suppose the cylinder contains \(50.0 \mathrm{mol}\) of a gas. Use the SRK equation of state to calculate the maximum permissible cylinder temperature if the gas is (a) carbon dioxide and (b) argon. Finally, calculate the values that would be predicted by the ideal-gas equation of state.

The flow rate required to yield a specified reading on an orifice meter varies inversely as the square root of the fluid density; that is, if a fluid with density \(\rho_{1}\left(g / \mathrm{cm}^{3}\right)\) flowing at a rate \(\dot{V}_{1}\left(\mathrm{cm}^{3} / \mathrm{s}\right)\) yields a meter reading \(\phi\), then the flow rate of a fluid with density \(\rho_{2}\) required to yield the same reading is $$\dot{V}_{2}=\dot{V}_{1}\left(\rho_{1} / \rho_{2}\right)^{1 / 2}$$ (a) An orifice meter has been calibrated with nitrogen at \(25^{\circ} \mathrm{C}\) and \(758 \mathrm{mm}\) Hg, but it now has methane flowing through it at \(50^{\circ} \mathrm{C}\) and \(1800 \mathrm{mm}\) Hg. Applying the nitrogen calibration to the reading indicates that the flow rate is \(21 \mathrm{L} / \mathrm{min}\). Estimate the true volumetric flow rate of the methane. (b) Repeat Part (a) but suppose the stream contains 10.0 mole \(\% \mathrm{CO}_{2}\) and 5.0 mole \(\%\) cthane in addition to methane.

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