/*! 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 64 A sheet of cellulose acetate fil... [FREE SOLUTION] | 91Ó°ÊÓ

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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?

Short Answer

Expert verified
To determine the feed ratio, the air inlet temperature as a function of the film temperature change, the film temperature change for a given inlet air temperature, and the required value of the air inlet temperature for given film temperatures, a detailed heat balance is necessary with the necessitated assumptions. Explanation on how the film temperature can drop even with an air temperature consistently higher, lies in the latent heat consumed in the evaporation of acetone, thus cooling the film.

Step by step solution

01

Calculate the mass

As 90% of the acetone evaporates, only 10% which is equivalent to 0.5 kg remains in the 100 kg film. The mass of the dry film can therefore be calculated as \(100 - 0.5 = 99.5\) kg .
02

Calculate the mass of evaporated acetone

The 90% of the 5 wt% acetone in 100 kg film becomes 4.5 kg.
03

Formulate heat balance for the film

The heat gained by the film is equal to the heat supplied by the air and the heat of vaporization of acetone. Mathematically, this can be written as: \[99.5 \times 1.33 \times (T_{f2}-35) = 4.5 \times \Delta H_v + 99.5 \times 1.33 \times (T_{f2}-35) \] where \( \Delta H_v \) is the heat of vaporization for acetone.
04

Derive the expression for \(T_{al}\)

Using the fact that the heat lost by the air is the heat gained by the film, we can state that:\[V_{air} \times 1.01 \times 10^3 \times C_{p_{air}} \times (T_{al}-49) = 99.5 \times 1.33 \times (T_{f2} - 35)\]Solving for \(T_{al}\) in terms of \(T_{f2}\), we get: \[T_{al} = \frac{99.5 \times 1.33 \times (T_{f2} - 35)}{V_{air} \times 1.01 \times 10^3 \times C_{p_{air}}} + 49\]
05

Calculate the film temperature change

For an inlet air temperature of \(120^{\circ} C\), we will plug into the above equation to solve for \(T_{f2}\) which will give us the change in temperature of the film.
06

Derive the value for \(T_{al}\) for different film temperatures

We will replace \(T_{f2}\) by the given \(34^{\circ} C\) or \(36^{\circ} C\) in the equation from step 4 and calculate the necessary \(T_{al}\)
07

Understand the heat transfer phenomenon

The film temperature can drop despite the air temperature being consistently higher because of the latent heat consumed in the evaporation of acetone which effectively cools the film.

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Key Concepts

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

Adiabatic Dryer
An adiabatic dryer is a device where materials are dried using a stream of air or gas, and no external heat supply is provided during the process. The drying involves the removal of moisture through evaporation, which is accompanied by a drop in temperature due to the latent heat of vaporization. As this process occurs, the air becomes increasingly humid and leaves the dryer at a higher temperature and humidity level than when it entered.
This process is significant in chemical process engineering because it allows for drying without an external heat source, which can be both energy-efficient and cost-effective. In the case of our exercise, the acetone in the cellulose acetate film evaporates, cooling the film as it does so, even though the air itself might be warmer than the film.
Heat Capacity
Heat capacity is a measure of the amount of heat energy required to raise the temperature of a substance by a given amount; this property is crucial for understanding energy changes in chemical processes.
In the context of the adiabatic drying exercise, the heat capacity of the film, air, and acetone play a significant role in calculating the energy balance. The specific heat capacities provided allow us to estimate the amount of heat necessary to bring about the observed temperature changes, whether it be the drying film or the air stream. The heat capacity of dry air, which we assumed in the calculation, typically depends on temperature and pressure but is often approximated for engineering calculations like this one.
Heat of Vaporization
The heat of vaporization is the amount of heat energy required to convert a certain amount of a substance from its liquid state to its vapor state at a constant temperature. It's also known as enthalpy of vaporization and is a form of latent heat.
In the drying process, heat of vaporization is a critical factor because it represents the energy needed to evaporate the solvent—in this case, acetone—from the film. Although it is usually temperature-dependent, in this textbook problem, it is assumed to be constant for the sake of simplification. This assumption greatly streamlines the calculations but is often a close approximation in practical scenarios.
Vapor-Liquid Equilibrium
Vapor-liquid equilibrium (VLE) is the state where a liquid and its vapor are in equilibrium with each other at a given temperature and pressure, meaning that the rate of evaporation equals the rate of condensation.
Understanding VLE helps us to predict how the solvent will behave within the adiabatic dryer and at what temperature and pressure it will evaporate, which is essential for optimizing the drying process. In industrial settings, VLE data are foundational for designing distillation columns, evaporators, and other separation process units.
Mass and Energy Balances
Mass and energy balances are the cornerstone of chemical process calculations. They ensure that all material and energy entering a process equal all material and energy leaving, accounting for changes within the process.
In the example problem, we used mass balances to determine the amount of acetone evaporating and the mass of the dry film. For energy balances, we calculated the energy change of the film due to temperature change and the energy absorbed during acetone evaporation.
Understanding these balances is fundamental for engineers to design, analyze, and optimize chemical processes, ensuring sustainability and cost-efficiency of operations.

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Most popular questions from this chapter

A natural gas containing 95 mole \(\%\) methane and the balance ethane is burned with \(20.0 \%\) excess air. The stack gas, which contains no unburned hydrocarbons or carbon monoxide, leaves the furnace at \(900^{\circ} \mathrm{C}\) and \(1.2 \mathrm{atm}\) and passes through a heat exchanger. The air on its way to the furnace also passes through the heat exchanger, entering it at \(20^{\circ} \mathrm{C}\) and leaving it at \(245^{\circ} \mathrm{C}\). (a) Taking as a basis \(100 \mathrm{mol} / \mathrm{s}\) of the natural gas fed to the furnace, calculate the required molar flow rate of air, the molar flow rate and composition of the stack gas, the required rate of heat transfer in the preheater, \(\dot{Q}\) (write an energy balance on the air), and the temperature at which the stack gas leaves the preheater (write an energy balance on the stack gas). Note: The problem statement does not give you the fuel feed temperature. Make a reasonable assumption, and state why your final results should be nearly independent of what you assume. (b) What would \(\dot{Q}\) be if the actual feed rate of the natural gas were 350 SCMH [standard cubic meters per hour, \(\left.\mathrm{m}^{3}(\mathrm{STP}) / \mathrm{h}\right] ?\) Scale up the flowchart of Part (a) rather than repeating the entire calculation.

Saturated propane vapor at \(2.00 \times 10^{2}\) psia is fed to a well- insulated heat exchanger at a rate of \(3.00 \times 10^{3} \mathrm{SCFH}\) (standard cubic feet per hour). The propane leaves the exchanger as a saturated liquid (i.e., a liquid at its boiling point) at the same pressure. Cooling water enters the exchanger at \(70^{\circ} \mathrm{F},\) flowing cocurrently (in the same direction) with the propane. The temperature difference between the outlet streams (liquid propane and water) is \(15^{\circ} \mathrm{F}\). (a) What is the outlet temperature of the water stream? (Use the Antoine equation.) Is the outlet water temperature less than or greater than the outlet propane temperature? Briefly explain. (b) Estimate the rate (Btu/h) at which heat must be transferred from the propane to the water in the heat exchanger and the required flow rate \(\left(1 \mathrm{b}_{\mathrm{m}} / \mathrm{h}\right)\) of the water. (You will need to write two separate energy balances.) Assume the heat capacity of liquid water is constant at \(1.00 \mathrm{Btu} /\left(\mathrm{lb}_{\mathrm{m}} \cdot^{\circ} \mathrm{F}\right)\) and neglect heat losses to the outside and the effects of pressure on the heat of vaporization of propane.

Estimate the specific enthalpy of steam (kJ/kg) at \(100^{\circ} \mathrm{C}\) and 1 atm relative to steam at \(350^{\circ} \mathrm{C}\) and 100 bar using: (a) The steam tables. (b) Table B.2 or APEx and assuming ideal-gas behavior. What is the physical significance of the difference between the values of \(\hat{H}\) calculated by the two methods?

On a cold winter day the temperature is \(2^{\circ} \mathrm{C}\) and the relative humidity is \(15 \% .\) You inhale air at an average rate of \(5500 \mathrm{mL} / \mathrm{min}\) and exhale a gas saturated with water at body temperature, roughly \(37^{\circ} \mathrm{C} .\) If the mass flow rates of the inhaled and exhaled air (excluding water) are the same, the heat capacities \(\left(C_{p}\right)\) of the water-free gases are each \(1.05 \mathrm{J} /\left(\mathrm{g} \cdot^{\circ} \mathrm{C}\right),\) and water is ingested into the body as a liquid at \(22^{\circ} \mathrm{C},\) at what rate in \(\mathrm{J} /\) day do you lose energy by breathing? Treat breathing as a continuous process (inhaled air and liquid water enter, exhaled breath exits) and neglect work done by the lungs.

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?

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