/*! 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 38 A stream of air at \(500^{\circ}... [FREE SOLUTION] | 91Ó°ÊÓ

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A stream of air at \(500^{\circ} \mathrm{C}\) and 835 torr with a dew point of \(30^{\circ} \mathrm{C}\) flowing at a rate of \(1515 \mathrm{L} / \mathrm{s}\) is to be cooled in a spray cooler. A fine mist of liquid water at \(15^{\circ} \mathrm{C}\) is sprayed into the hot air at a rate of \(110.0 \mathrm{g} / \mathrm{s}\) and evaporates completely. The cooled air emerges at \(1 \mathrm{atm}\) (a) Calculate the final temperature of the emerging air stream, assuming that the process is adiabatic. (Suggestion: Derive expressions for the enthalpies of dry air and water at the outlet air temperature, substitute them into the energy balance, and use a spreadsheet to solve the resulting fourth-order polynomial equation.) (b) At what rate (kW) is heat transferred from the hot air feed stream in the spray cooler? What becomes of this heat? (c) In a few sentences, explain how this process works in terms that a high school senior could understand. Incorporate the results of Parts (a) and (b) in your explanation.

Short Answer

Expert verified
The final temperature would require solving the fourth-order polynomial equation. The rate of heat transfer will be obtained from the enthalpy change equation and it represents the heat used to evaporate the water spray. The process works by spraying water into the hot air which evaporates taking heat from the air, and thus cooling it.

Step by step solution

01

Formulate the energy balance equation

The energy balance equation for this adiabatic operation can be expressed as: \( Q = W + \Delta H \), where \( Q \) is the heat exchange, \( W \) is the overall work done and \( \Delta H \) is the change in enthalpy (energy content). Since it is said in the problem that the process is adiabatic, \( Q = 0 \). So, \( W = -\Delta H \).
02

Calculate the enthalpies and derive the polynomial equation

Enthalpies of dry air and water at the outlet air temperature will be derived using the properties of air and water, such as specific heat capacity and heat of vaporization. These expressions will be substituted into the energy balance, forming a fourth-order polynomial equation.
03

Solve the polynomial equation

Use a spreadsheet or a numerical computation software, like MATLAB, to solve the resulting fourth-order polynomial equation to find the final temperature of the emerging air stream, denoted by \(T\).
04

Calculate the heat transfer rate

The rate of heat transfer from the hot air feed stream in the spray cooler is calculated by multiplying the enthalpy change calculated in Step 2 with the volumetric flow rate of the air stream. This will give a result in \(J / s = kW\). In this adiabatic process, the sum total heat was used in the evaporation of the sprayed liquid water into the air stream and hence decreased the air's initial temperature.
05

Explain the process

In simple terms, spraying water in the hot air stream causes the water to evaporate, taking heat from the hot air stream and thereby reducing its temperature. This is an adiabatic process where there is no heat transfer with surroundings. The evaporated water hence cools the air stream. The lost heat is utilized in evaporating the sprayed water. This mechanism is similar to how sweating cools our body.

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Key Concepts

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

Enthalpy Calculation
Enthalpy is a key concept when dealing with any thermodynamic process. It measures the total heat content of a system. In our exercise, we calculate the enthalpies of both dry air and water under changing temperatures.
In this adiabatic cooling process, we consider the specific heat capacity of air and the heat of vaporization for water. The change in enthalpy \( \Delta H \) is essential for understanding how energy is exchanged within the system, even when no heat is transferred to the outside.
You derive expressions for the enthalpies as functions of the outlet air temperature. By substituting these into your energy balance equation, you can form a fourth-order polynomial equation. Tools like spreadsheets or MATLAB can then solve for the final temperature, making the calculation easier.
Remember, in the context of enthalpy calculation, the goal is to comprehend how energy transformations occur within the system without external heat exchanges.
Energy Balance
In thermodynamics, an energy balance helps track how energy enters, exits, or gets transformed within a system. This concept is central to our spray cooler scenario.
For this adiabatic process, remember that adiabatic means no heat transfer with the surroundings, hence \( Q = 0 \). The energy balance equation changes to \( 0 = W + \Delta H \), leading us to focus on changes in enthalpy, or \( W = -\Delta H \).
You balance the energy by considering the air and water interactions. As sprayed water evaporates, it requires energy, which is extracted from the air. This reduction in the air's thermal energy (due to evaporation) is the core of cooling air without changing the system's overall energy with the environment.
This principle mirrors natural processes like sweating, where no external energy is added, but heat is used to transform liquid water to vapor, thus cooling the body.
Spray Cooler Process
The spray cooler is a fascinating industrial application of adiabatic cooling. Here, water droplets are sprayed into a stream of hot air, creating a cooling effect.
Here's how it works: hot air flows into the spray cooler where a fine mist of water is introduced. The mist evaporates, requiring energy to transform from liquid to vapor. This energy is absorbed from the hot air, lowering its temperature.
In our exercise, the efficiency of this process is demonstrated through calculations. You find that all the heat from the hot air goes into evaporating the water. The air's temperature drop reflects this transfer of energy.
Think of the spray cooler like a cooling tower or even an air conditioner, where evaporation of water or another fluid removes heat, subsequently cooling the air. Understanding this process in industrial settings can help design more efficient systems for temperature management in various applications.

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Most popular questions from this chapter

Your next-door neighbor, Josephine Rackstraw, surprised her husband last January by having a hot tub installed in their back yard while he was away on an ice-fishing trip. It surprised him, all right, but instead of being pleased he was horrified. "Have you lost your mind, Josephine?" he sputtered. "It will cost a fortune to keep this thing hot, and you know what the President said about conserving energy." "Don't be silly, Ralph," she replied. "It can't cost more than a few pennies a day, even in the dead of winter." "No way -just because you have a PhD, you think you're an expert on everything!" They argued for a while, bringing up several issues that each had been storing for just such an occasion. After calming down and using the tub for a week, they remembered their neighbor (i.e., you) had a chemical engineering education and came to ask if you could settle their argument. You asked a few questions, made several observations, converted everything to metric units, and arrived at the following data, all corresponding to an average outside temperature of \(5^{\circ} \mathrm{C}\). \- The tub holds 1230 liters of water. \- Ralph normally keeps the tub temperature at \(29^{\circ} \mathrm{C}\), raises it to \(40^{\circ} \mathrm{C}\) when he plans to use it, keeps it at \(40^{\circ} \mathrm{C}\) for about one hour, and drops it back to \(29^{\circ} \mathrm{C}\) when he is finished. \- During heating, it takes about three hours for the water temperature to rise from \(29^{\circ} \mathrm{C}\) to \(40^{\circ} \mathrm{C}\). When the heat is shut off, it takes eight hours for the water temperature to drop back to \(29^{\circ} \mathrm{C}\). \- Electricity costs 10 cents per kilowatt-hour. Taking the heat capacity of the tub contents to be that of pure liquid water and neglecting evaporation, answer the following questions. (a) What is the average rate of heat loss ( \(k W\) ) from the tub to the outside air? (Hint: Consider the period when the tub temperature is dropping from \(40^{\circ} \mathrm{C}\) to \(29^{\circ} \mathrm{C}\).) (b) At what average rate ( \(\mathrm{kW}\) ) does the tub heater deliver energy to the water when raising the water temperature? What is the total quantity of electricity (kW\cdoth) that the heater must deliver during this period? [Consider the result of Part (a) when performing the calculation.] (c) (These answers should settle the argument.) Consider a day in which the tub is used once. Use the results of Parts (a) and (b) to estimate the cost (S) of heating the tub from \(29^{\circ} \mathrm{C}\) to \(40^{\circ} \mathrm{C}\) and the cost \((\mathrm{S})\) of keeping the tub at a constant temperature. (There is no cost for the period in which \(T\) is dropping.) What is the total daily cost of running the tub? Assume the rate of heat loss is independent of the tub temperature. (d) The tub lid, which is an insulator, is removed when the tub is in use. Explain how this fact would probably affect your cost estimates in Part (c).

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?

A gas containing water vapor has a dry-basis composition of 7.5 mole \(\%\) CO, \(11.5 \%\) CO \(_{2}, 0.5 \%.\) \(\mathrm{O}_{2},\) and \(80.5 \% \mathrm{N}_{2} .\) The gas leaves a catalyst regeneration unit at \(620^{\circ} \mathrm{C}\) and 1 atm with a dew point of \(57^{\circ} \mathrm{C}\) at a flow rate of \(28.5 \mathrm{SCMH}\left[\mathrm{m}^{3}(\mathrm{STP}) / \mathrm{h}\right] .\) Valuable solid catalyst particles entrained in the gas are to be recovered in an electrostatic precipitator, but the gas must first be cooled to \(425^{\circ} \mathrm{C}\) to prevent damage to the precipitator electrodes. The cooling is accomplished by spraying water at \(20^{\circ} \mathrm{C}\) into the gas. (a) Use simultaneous material and energy balances on the spray cooler to calculate the required water feed rate ( \(\mathrm{kg} / \mathrm{h}\) ). Treat the spray cooler as adiabatic and neglect the heat transferred from the entrained solid particles as they cool. (b) In terms that a high school senior could understand, explain the operation of the spray cooler in this problem. (What happens when the cold water contacts the hot gas?)

The heat capacity at constant pressure of hydrogen cyanide is given by the expression $$ C_{p}\left[J /\left(\mathrm{mol} \cdot^{\circ} \mathrm{C}\right)\right]=35.3+0.0291 T\left(^{\circ} \mathrm{C}\right) $$ (a) Write an expression for the heat capacity at constant volume for HCN, assuming ideal-gas behavior. (b) Calculate \(\Delta \hat{H}(\mathrm{J} / \mathrm{mol})\) for the constant- pressure process $$ \mathrm{HCN}\left(\mathrm{v}, 25^{\circ} \mathrm{C}, 0.80 \mathrm{atm}\right) \rightarrow \mathrm{HCN}\left(\mathrm{v}, 200^{\circ} \mathrm{C}, 0.80 \mathrm{atm}\right) $$(c) Calculate \(\Delta \hat{U}(\mathrm{J} / \mathrm{mol})\) for the constant- volume process $$\mathrm{HCN}\left(\mathrm{v}, 25^{\circ} \mathrm{C}, 50 \mathrm{m}^{3} / \mathrm{kmol}\right) \rightarrow \mathrm{HCN}\left(\mathrm{v}, 200^{\circ} \mathrm{C}, 50 \mathrm{m}^{3} / \mathrm{kmol}\right)$$ (d) If the process of Part (b) were carried out in such a way that the initial and final pressures were each 0.80 atm but the pressure varied during the heating, the value of \(\Delta \hat{H}\) would still be what you calculated assuming a constant pressure. Why is this so?

Saturated steam at \(300^{\circ} \mathrm{C}\) is used to heat a countercurrently flowing stream of methanol vapor from \(65^{\circ} \mathrm{C}\) to \(260^{\circ} \mathrm{C}\) in an adiabatic heat exchanger. The flow rate of the methanol is 6500 standard liters per minute, and the steam condenses and leaves the heat exchanger as liquid water at \(90^{\circ} \mathrm{C}.\) (a) Calculate the required flow rate of the entering steam in \(\mathrm{m}^{3} / \mathrm{min}\). (b) Calculate the rate of heat transfer from the water to the methanol ( \(\mathrm{kW}\) ). (c) Suppose the outlet temperature of the methanol is measured and found to be \(240^{\circ} \mathrm{C}\) instead of the specified value of \(260^{\circ} \mathrm{C}\). List five possible realistic explanations for the \(20^{\circ} \mathrm{C}\) difference. 7 An adiabatic heat exchanger is one for which no heat is exchanged with the surroundings. All of the heat lost by the hot stream is transferred to the cold stream.

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