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A gas stream containing \(n\) -hexane in nitrogen with a relative saturation of \(90 \%\) is fed to a condenser at \(75^{\circ} \mathrm{C}\) and 3.0 atm absolute. The product gas emerges at \(0^{\circ} \mathrm{C}\) and 3.0 atm at a rate of \(746.7 \mathrm{m}^{3} / \mathrm{h}\). (a) Calculate the percentage condensation of hexane (moles condensed/mole fed) and the rate \((\mathrm{kW})\) at which heat must be transferred from the condenser. (b) Suppose the feed stream flow rate and composition and the heat transfer from the condenser are the same as in Part (a), but the condenser and outlet stream pressure is only 2.5 atm instead of 3.0 atm. How would the outlet stream temperatures and flow rates and the percentage condensations of hexane calculated in Parts (a) and (b) change (increase, decrease, no change, no way to tell)? Don't do any calculations, but explain your reasoning.

Short Answer

Expert verified
The percentage condensation of hexane and the rate of heat transfer from the condenser can be calculated using the given information and relevant formulae. Decreasing the pressure will increase the percentage of hexane that condenses out of the gas phase.

Step by step solution

01

Calculation of moles of hexane

Using the relative saturation, we can write a formula. For 90% saturation, the moles of hexane is equal to 0.9 times the saturation limit moles of hexane at 75 C and 3 atm. The saturation limit can be found from vapor pressure data for n-hexane.
02

Determination of moles condensed

After passing through the condenser, the temperature of the gas stream drops to 0 degrees Celsius. At this reduced temperature and a pressure of 3 atm, the saturation limit for n-hexane will be lower than the number of moles we determined in Step 1. Therefore, some hexane will condense out of the gas phase.
03

Percentage condensation calculation

The percentage condensation of hexane is calculated by subtracting the number of moles of hexane at 0 degrees Celsius from the number of moles of hexane at 75 degrees Celsius, and then dividing by the number of moles at 75 degrees Celsius. Multiply by 100 to convert to a percentage.
04

Heat transfer rate calculation

The rate of heat transfer from the condenser can be calculated using the formula Q = mcp(T1-T2), where m is the mass flow rate of the gas, cp is the specific heat of the gas, and T1 and T2 are the initial and final temperatures of the gas in Kelvin, respectively.
05

Conceptual understanding on effect of pressure change

Lowering the pressure will decrease the saturation limit for n-hexane at 0 degrees Celsius, leading to an increase in the percentage of hexane that condenses out of the gas phase.

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Key Concepts

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

Percentage Condensation Calculation
Understanding the percentage condensation calculation is crucial for students working on chemical process problems. This concept involves determining the proportion of a vapor that becomes liquid during condensation.

In the context of the exercise, this calculation is performed by comparing the moles of n-hexane before and after passing through a condenser. Initially, the gas stream is saturated with n-hexane vapor at a given temperature and pressure. As the gas stream cools down in the condenser, the amount of n-hexane that can remain vapor decreases, and excess n-hexane condenses out of the vapor phase.

The formula used for this calculation is as follows: \[ \text{Percentage condensation of hexane} = \frac{\text{moles of hexane before condensation} - \text{moles of hexane after condensation}}{\text{moles of hexane before condensation}} \times 100\% \]
It's important for students to note that precise percentage calculations require accurate vapor pressure data for the specific compound, in this case, n-hexane. Furthermore, incorporating the concept of relative saturation into the calculation allows for a more realistic and applicable result when dealing with industrial processes.
Heat Transfer Rate Calculation
The heat transfer rate reflects the amount of heat energy transferred per unit time and is a fundamental concept in chemical engineering thermodynamics. The calculation of this rate is pivotal when assessing the efficiency of condensers and other heat-exchanging devices within a chemical process.

The formula to calculate the rate at which heat must be transferred from the condenser, denoted as \( Q \) in kilowatts (kW), can be represented as: \[ Q = \frac{m \cdot c_p \cdot (T_1 - T_2)}{1000} \] where \( m \) is the mass flow rate, \( c_p \) is the specific heat capacity of the gas at constant pressure, and \( T_1 \) and \( T_2 \) are the initial and final temperatures of the gas stream in Celsius, respectively. The division by 1000 is required to convert the energy from Joules (J) to kilowatts (kW).

When performing this calculation, it is vital for students to remember to convert all temperature measurements to Kelvin by adding 273.15 to the Celsius values. This step ensures that all thermodynamic equations utilized in the process are consistent with the Kelvin scale, which is the standard scale used in scientific heat calculations.
Vapor Pressure and Temperature Relationship
The relationship between vapor pressure and temperature is fundamental to understanding phase changes such as condensation and boiling. This relationship is especially important in the context of chemical processes that involve temperature changes under constant pressure conditions.

For a given substance, its vapor pressure increases with temperature. This behavior is described by the Clausius-Clapeyron equation, which relates the change in vapor pressure to temperature. As part of the improved exercise advice, it's critical to highlight this dependence: at higher temperatures, the amount of vapor that pressure can support grows, while at lower temperatures, the vapor pressure decreases, and more of the substance condenses into liquid.

In the exercise provided, as the temperature of the n-hexane and nitrogen gas stream decreases in the condenser, the vapor pressure of n-hexane falls, leading to a higher percentage of n-hexane condensing out of the gas phase. Students should also grasp that changes in system pressure can affect the vapor pressure of a substance, which subsequently impacts condensation behavior and the temperature at which the phase change occurs.

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Most popular questions from this chapter

Ever wonder why espresso costs much more per cup than regular drip coffee? Part of the reason is the expensive equipment needed to brew a proper espresso. A high-powered burr grinder first shears the coffee beans to a fine powder without producing too much heat. (Heating the coffee in the grinding stage prematurely releases the volatile oils that give espresso its rich flavor and aroma.) The ground coffee is put into a cylindrical container called a gruppa and tamped down firmly to provide an even flow of water through it. An electrically heated boiler inside the espresso machine maintains water in a reservoir at 1.4 bar and \(109^{\circ} \mathrm{C}\). An electric pump takes cold water at \(15^{\circ} \mathrm{C}\) and 1 bar, raises its pressure to slightly above 9 bar, and feeds it into a heating coil that passes through the reservoir. Heat transferred from the reservoir through the coil wall raises the water temperature to \(96^{\circ} \mathrm{C}\). The heated water flows into the top of the gruppa at \(96^{\circ} \mathrm{C}\) and 9 bar, passes slowly through the tightly packed ground beans, and dissolves the oils and some of the solids in the beans to become espresso, which decompresses to 1 atm as it exits the machine. The water temperature and uniform flow through the bed of packed coffee in the gruppa lead to the more intense flavor of espresso relative to normal drip coffee. Water drawn directly from the reservoir is expanded to atmospheric pressure where it forms steam, which is used to heat and froth milk for lattes and cappuccinos. (a) Sketch this process, using blocks to represent the pump, reservoir, and gruppa. Label all heat and work flows in the process, including electrical energy. (b) To make a 14-oz latte, you would steam 12 ounces of cold milk (3^'C) until it reaches 71 ^ C and pour it over 2 ounces of espresso. Assume that the steam cools but none of it condenses as it bubbles through the milk. For each latte made, the heating element that maintains the reservoir temperature must supply enough energy to heat the espresso water plus enough to heat the milk, plus additional energy. Assuming $$ \left(C_{p}\right)_{\operatorname{milk}}=3.93 \frac{\mathrm{J}}{\mathrm{g} \cdot^{\circ} \mathrm{C}}, \quad \mathrm{SG}_{\mathrm{milk}}=1.03 $$ calculate the quantity of electrical energy that must be provided to the heating element to accomplish those two functions. Why would more energy than what you calculate be required? (There are several reasons.) (c) Coffee beans contain a considerable amount of trapped carbon dioxide, not all of which is released when the beans are ground. When the hot pressurized water percolates through the ground beans, some of the carbon dioxide is absorbed in the liquid. When the liquid is then dispensed at atmospheric pressure, fine \(\mathrm{CO}_{2}\) bubbles come out of solution. In addition, one of the chemical compounds formed when the coffee beans are roasted and extracted into the espresso is melanoidin, a surfactant. Surfactant molecules are asymmetrical, with one end being hydrophilic (drawn to water) and the other end hydrophobic (repelled by water). When the bubbles (thin water films containing \(\mathrm{CO}_{2}\) ) pass through the espresso liquid, the hydrophilic ends of the melanoidin molecules attach to the bubbles and the dissolved bean oils in turn attach to the hydrophobic ends. The result is that the bubbles emerge coated with the oils to form the crema , the familiar reddish brown stable foam at the surface of good espresso. Speculate on why you don't see crema in normal drip coffee. (Hint: Henry's law should show up in your explanation.) Note: All soaps and shampoos contain at least one surfactant species. (A common one is sodium lauryl sulfate.) Its presence explains why if you have greasy hands, washing with plain water may leave the grease untouched but washing with soap removes the grease. (d) Explain in your own words (i) how espresso is made, (ii) why espresso has a more intense flavor than regular drip coffee, (iii) what the crema in espresso is, how it forms, and why it doesn't appear in regular drip coffee, and (iv) why washing with plain water does not remove grease but washing with soap does. (Note: Many people automatically assume that all chemical engineers are extraordinarily intelligent. If you can explain those four things, you can help perpetuate that belief.)

A sheet of cellulose acetate film containing 5.00 wt\% liquid acetone enters an adiabatic dryer where \(90 \%\) of the acetone evaporates into a stream of dry air flowing over the film. The film enters the dryer at \(T_{\mathrm{f} 1}=35^{\circ} \mathrm{C}\) and leaves at \(T_{\mathrm{f} 2}\left(^{\circ} \mathrm{C}\right) .\) The air enters the dryer at \(T_{\mathrm{al}}\left(^{\circ} \mathrm{C}\right)\) and 1.01 atm and exits the dryer at \(T_{\mathrm{a} 2}=49^{\circ} \mathrm{C}\) and 1 atm with a relative saturation of \(40 \% . C_{p}\) may be taken to be \(1.33 \mathrm{kJ} /\left(\mathrm{kg} \cdot^{\circ} \mathrm{C}\right)\) for dry film and \(0.129 \mathrm{kJ} /\left(\mathrm{mol} \cdot^{\circ} \mathrm{C}\right)\) for liquid acetone. Make a reasonable assumption regarding the heat capacity of dry air. The heat of vaporization of acetone may be considered independent of temperature. Take a basis of \(100 \mathrm{kg}\) film fed to the dryer for the requested calculations. (a) Estimate the feed ratio [liters dry air (STP)/kg dry film]. (b) Derive an expression for \(T_{\mathrm{al}}\) in terms of the film temperature change, \(\left(T_{\mathrm{f} 2}-35\right),\) and use it to answer Parts (c) and (d). (c) Calculate the film temperature change if the inlet air temperature is \(120^{\circ} \mathrm{C}\). (d) Calculate the required value of \(T_{\mathrm{al}}\) if the film temperature falls to \(34^{\circ} \mathrm{C},\) and the value if it rises to \(36^{\circ} \mathrm{C}.\) (e) If you solved Parts (c) and (d) correctly, you found that even though the air temperature is consistently higher than the film temperature in the dryer, so that heat is always transferred from the air to the film, the film temperature can drop from the inlet to the outlet. How is this possible?

Estimate the heat of vaporization of diethyl ether at its normal boiling point using Trouton's rule and Chen's rule and compare the results with a tabulated value of this quantity. Calculate the percentage error that results from using each estimation. Then estimate \(\Delta \hat{H}_{\mathrm{v}}\) at \(100^{\circ} \mathrm{C}\) using Watson's correlation.

The off-gas from a reactor in a process plant in the heart of Freedonia has been condensing and plugging up the vent line, causing a dangerous pressure buildup in the reactor. Plans have been made to send the gas directly from the reactor into a cooling condenser in which the gas and liquid condensate will be brought to \(25^{\circ} \mathrm{C}.\) (a) You have been called in as a consultant to aid in the design of this unit. Unfortunately, the chief (and only) plant engineer has disappeared and nobody else in the plant can tell you what the offgas is (or what anything else is, for that matter). However, a job is a job, and you set out to do what you can. You find an elemental analysis in the engineer's notebook indicating that the gas formula is \(\mathrm{C}_{5} \mathrm{H}_{12} \mathrm{O}\). On another page of the notebook, the off-gas flow rate is given as \(235 \mathrm{m}^{3} / \mathrm{h}\) at \(116^{\circ} \mathrm{C}\) and 1 atm. You take a sample of the gas and cool it to \(25^{\circ} \mathrm{C}\), where it proves to be a solid. You then heat the solidified sample at 1 atm and note that it melts at \(52^{\circ} \mathrm{C}\) and boils at \(113^{\circ} \mathrm{C}\). Finally, you make several assumptions and estimate the heat removal rate in \(\mathrm{kW}\) required to bring the off-gas from \(116^{\circ} \mathrm{C}\) to \(25^{\circ} \mathrm{C}\). What is your result? (b) If you had the right equipment, what might you have done to get a better estimate of the cooling rate?

Fish and wildlife managers have determined that a sudden temperature increase greater than \(5^{\circ} \mathrm{C}\) would be harmful to the marine ecosystem of a river. Warmer waters contain less dissolved oxygen and cause organisms in a river to increase their metabolism; if the temperature increase is sudden, the organisms do not have time to adapt to the new environment and likely will die. (Changes in river temperatures of five degrees and more due to seasonal temperature variations are common, but those temperature changes are gradual.) A proposed chemical plant plans to use river water for process cooling. The river flows at a rate of \(15.0 \mathrm{m}^{3} / \mathrm{s}\) at a temperature of \(15^{\circ} \mathrm{C}\), and a fraction of it will be diverted to the plant. Preliminary calculations reveal that the cooling water will remove \(5.00 \times 10^{5} \mathrm{kJ} / \mathrm{s}\) of heat from the plant. A portion of the extracted water will evaporate from the plant into the atmosphere, and the remainder will be returned to the river at a temperature of \(35^{\circ} \mathrm{C}\). (a) Draw and completely label a flowchart of the process and prove that there is enough information available to calculate all of the unknown stream flow rates on the chart. (b) Estimate the fraction of the river flow that must be diverted to the plant and the percentage of the cooling water that evaporates. Assume that water has a constant heat capacity of \(4.19 \mathrm{kJ} /\left(\mathrm{kg} \cdot^{\circ} \mathrm{C}\right)\) and a heat of vaporization roughly that of water at the normal boiling point, and also assume that the specific enthalpy of the water vapor relative to liquid water at \(15^{\circ} \mathrm{C}\) equals the heat of vaporization. (c) Write (but don't evaluate) an expression for the enthalpy change neglected by the assumption about the specific enthalpy of the steam.

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