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Chapter 8: Problem 38

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

A stream of air at \(77^{\circ} \mathrm{F}\) and 1.2 atm absolute flowing at a rate of \(225 \mathrm{ft}^{3} / \mathrm{h}\) is blown through ducts that pass through the interior of a large industrial motor. The air emerges at \(500^{\circ} \mathrm{F}\). Calculate the rate at which the air is removing heat generated by the motor. What assumption have you made about the pressure dependence of the specific enthalpy of air?

The specific internal energy of formaldehyde (HCHO) vapor at 1 atm and moderate temperatures is given by the formula $$\hat{U}(\mathrm{J} / \mathrm{mol})=25.96 T+0.02134 T^{2}$$ where \(T\) is in \(^{\circ} \mathrm{C}\) (a) Calculate the specific internal energies of formaldehyde vapor at \(0^{\circ} \mathrm{C}\) and \(200^{\circ} \mathrm{C}\). What reference temperature was used to generate the given expression for \(\hat{U} ?\) (b) The value of \(\hat{U}\) calculated for \(200^{\circ} \mathrm{C}\) is not the true value of the specific internal energy of formaldehyde vapor at this condition. Why not? (Hint: Refer back to Section 7.5a.) Briefly state the physical significance of the calculated quantity. (c) Use the closed system energy balance to calculate the heat (J) required to raise the temperature of 3.0 mol HCHO at constant volume from 0^0 C to 200^'C. List all of your assumptions. (d) From the definition of heat capacity at constant volume, derive a formula for \(C_{v}(T)\left[\mathrm{J} /\left(\mathrm{mol} \cdot^{\circ} \mathrm{C}\right)\right]\) Then use this formula and Equation \(8.3-6\) to calculate the heat \((\) J) required to raise the temperature of 3.0 mol of HCHO(v) at constant volume from 0^ C to 200^'C. [You should get the same result you got in Part (c).]

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.

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.

Polyvinylpyrrolidone (PVP) is a polymer product used as a binding agent in pharmaceutical applications as well as in personal-care items such as hairspray. In the manufacture of \(\mathrm{PVP}\), a spray-drying process is used to collect solid PVP from an aqueous suspension, as shown in the flowchart on the next page. A liquid solution containing 65 wt\% \(\mathrm{PVP}\) and the balance water at \(25^{\circ} \mathrm{C}\) is pumped through an atomizing nozzle at a rate of \(1500 \mathrm{kg} / \mathrm{h}\) into a stream of preheated air flowing at a rate of \(1.57 \times 10^{4}\) SCMH. The water evaporates into the stream of hot air and the solid PVP particles are suspended in the humidified air. Downstream, the particles are separated from the air with a filter and collected. The process is designed so that the exiting solid product and humid air are in thermal equilibrium with each other at \(110^{\circ} \mathrm{C}\). For convenience, the spray-drying and solid- separation processes are shown as one unit that may be considered adiabatic. (a) Draw and completely label the process flow diagram and perform a degree- of-freedom analysis. (b) Calculate the required temperature of the inlet air, \(T_{0}\), and the volumetric flow rate \(\left(\mathrm{m}^{3} / \mathrm{h}\right)\) and relative humidity of the exiting air. Assume that the polymer has a heat capacity per unit mass one third that of liquid water, and only use the first two terms of the polynomial heat-capacity formula for air in Table B.2. (c) Why do you think the polymer solution is put through an atomizing nozzle, which converts it to a mist of tiny droplets, rather than being sprayed through a much less costly nozzle of the type commonly found in showers? (d) Due to a design flaw, the polymer solution does not remain in the dryer long enough for all the water to evaporate, so the solid product emerging from the separator is a wet powder. How will this change the values of the outlet temperatures of the emerging gas and powder and the volumetric flow rate and relative humidity of the emerging gas (increase, decrease, can't tell without doing the calculations)? Explain your answers.
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