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Chapter 8: Problem 5

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?

Short Answer

Expert verified
The specific enthalpy of steam differs when calculated using steam tables versus the ideal gas equation due to the assumptions inherent in the ideal gas law which becomes less accurate at higher temperatures and pressures. The discrepancy between the values indicates the extent to which these assumptions are invalid under the given conditions.

Step by step solution

01

Using steam tables

First, refer to the steam tables to find the specific enthalpy values directly at the given conditions. To do this,\n\nFor steam at \(100^{\circ} \mathrm{C}\) and 1 atm, read the specific enthalpy, \( \hat{H}_{100C,1atm} \), directly from the steam tables. \n\nNext, for steam at \(350^{\circ} \mathrm{C}\) and 100 bar, read the specific enthalpy, \( \hat{H}_{350C,100bar} \), directly from the steam tables.\n\nThe difference between the two can then be calculated as follows: \( \Delta\hat{H}_{1} = \hat{H}_{100C,1atm} - \hat{H}_{350C,100bar} \).
02

Using Table B.2 or APEx

Next, calculate the specific enthalpy assuming ideal gas behavior using an appropriate equation of state. This typically involves calculating specific enthalpy as \( \hat{H} = C_{p}T \) where \(C_{p}\) is the heat capacity at constant pressure and T is temperature in Kelvin. The heat capacities can be found in Table B.2 or through APEx software.\n\nFind the specific enthalpy at the two conditions and then calculate the difference as follows: \( \Delta\hat{H}_{2} = \hat{H}_{100C,1atm} - \hat{H}_{350C,100bar} \).
03

Analyzing the results

Once the specific enthalpies are calculated using both methods, discuss the difference between the values and the physical significance of this difference.\n\nWhy might there be a discrepancy? Remember, steam tables are more accurate because they account for the specific behavior of water. By contrast, the ideal gas assumption, while easier to calculate, is less accurate because it assumes that the gas behaves ideally, which is not often the case in real life, especially for steam at high temperatures and pressures. Remember also to consider if the calculated values are reasonable, given known values for the specific enthalpy of steam.

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Key Concepts

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

Understanding Specific Enthalpy
Specific enthalpy is a crucial concept in thermodynamics, especially when dealing with energy transfer in processes involving gases and vapors. It represents the total energy content of a substance for a given mass and is expressed in units like kJ/kg.
Specific enthalpy combines two forms of energy:
  • Internal energy (the energy stored within the molecules)
  • Flow work (the energy required to push the fluid into or out of a control volume)
The formula for specific enthalpy is often expressed as:\[ \hat{H} = U + PV \]Where:
  • \(U\) is the specific internal energy
  • \(P\) is pressure
  • \(V\) is specific volume
Knowing specific enthalpy helps engineers and scientists calculate the efficient transfer of energy in open systems like turbines and engines.
Reading Steam Tables
Steam tables are essential tools for thermodynamic calculations involving water and steam. They provide the specific enthalpy values at various temperatures and pressures, which are not always straightforward to calculate from first principles.
  • Steam tables give data for both the saturated and superheated states of water and steam.
  • They help avoid complexities in calculating thermodynamic properties and ensure precision.
To use a steam table effectively:
  1. Identify the state of steam (saturated or superheated)
  2. Locate the corresponding temperature and pressure in the table
  3. Read off the specific enthalpy values provided
Steam tables are more reliable for calculating properties in applications because they are derived from experimental data specific to water and steam under various states.
Exploring Ideal Gas Behavior
The ideal gas law is a simplified model used to describe the behavior of gases. An ideal gas is an imaginary gas that perfectly follows the gas law equation:\[ PV = nRT \]Though this model is often idealized, it can provide reasonable approximations for real gases under certain conditions.
  • It assumes no interactions between gas particles (no attractions or repulsions).
  • It presumes that the volume of individual gas molecules is negligible compared to the volume of the container.
While this approach simplifies calculations, it's important to note:
  • Real gases deviating significantly from ideal behavior at high pressures and low temperatures.
  • For steam, especially at high pressures and temperatures, such deviations can lead to inaccuracies in enthalpy calculations.
Thus, care must be taken when applying ideal gas principles, particularly for substances like steam.
The Process of Enthalpy Calculation
Calculating enthalpy, whether through steam tables or ideal gas equations, involves an understanding of both empirical data and mathematical approximations.
  • Using steam tables gives immediate results based on conducted experiments.
  • Ideal gas calculations involve integrating specific heat capacity over a temperature range:
The formula for enthalpy at constant pressure is:\[ \hat{H} = C_p(T_2 - T_1) \]Where:
  • \(C_p\) is the heat capacity at constant pressure
  • \(T_2\) and \(T_1\) are the final and initial temperatures, respectively
By comparing these methods, one gains insight into the practical applications of thermodynamics and the trade-offs between accuracy and simplicity.
Explaining Heat Capacity
Heat capacity is a material-specific property that signifies the amount of heat required to change the temperature of an object by a given amount. It is an essential factor in calculating specific enthalpy for thermodynamic systems.
  • At constant pressure, it is denoted as \(C_p\) and at constant volume as \(C_v\).
  • The heat capacity at constant pressure is pertinent in many enthalpy calculations as it appears frequently in the ideal gas formula for enthalpy.
Knowing the heat capacity helps determine how much energy is needed to affect the temperature, and assists in comprehending the thermal properties of gases and vapors during processes.Understanding how heat capacity changes with temperature and pressure helps refine thermodynamic calculations and develop better models of energy exchange.

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Most popular questions from this chapter

The brakes on an automobile act by forcing brake pads, which have a metal support and a lining, to press against a disk (rotor) attached to the wheel. Friction between the pads and the disk causes the car to slow or stop. Each wheel has an iron brake disk with a mass of \(15 \mathrm{lb}_{\mathrm{m}}\) and two brake pads, each having a mass of \(11 \mathrm{b}_{\mathrm{m}}\). (a) Suppose an automobile is moving at 55 miles per hour when the driver suddenly applies the brakes and brings the car to a rapid halt. Take the heat capacity of the disk and brake pads to be \(0.12 \mathrm{Btu} /\left(\mathrm{lb}_{\mathrm{m}} \cdot^{\circ} \mathrm{F}\right)\) and assume that the car stops so rapidly that heat transfer from the disk and pads has been insignificant. Estimate the final temperature of the disk and pads if the car is (i) a Toyota Camry, which has a mass of about \(3200 \mathrm{Ib}_{\mathrm{m}},\) or (ii) a Cadillac Escalade, which has a mass of about \(5.900 \mathrm{lb}_{\mathrm{m}}.\) (b) Why are the linings on brake pads no longer made of asbestos? Your answer should provide information on specific issues or concerns caused by the use of asbestos.

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.

A fuel gas containing 95 mole\% methane and the balance ethane is burned completely with 25\% excess air. The stack gas leaves the furnace at \(900^{\circ} \mathrm{C}\) and is cooled to \(450^{\circ} \mathrm{C}\) in a waste- heat boiler, a heat exchanger in which heat lost by cooling gases is used to produce steam from liquid water for heating, power generation, or process applications. (a) Taking as a basis of calculation 100 mol of the fuel gas fed to the fumace, calculate the amount of heat (kJ) that must be transferred from the gas in the waste heat boiler to accomplish the indicated cooling. (b) How much saturated steam at 50 bar can be produced from boiler feedwater at \(40^{\circ} \mathrm{C}\) for the same basis of calculation? (Assume all the heat transferred from the gas goes into the steam production.) (c) At what rate ( \(k\) mol/s) must fuel gas be burned to produce 1280 kg steam per hour (an amount required elsewhere in the plant) in the waste heat boiler? What is the volumetric flow rate \(\left(\mathrm{m}^{3} / \mathrm{s}\right)\) of the gas leaving the boiler? (d) Briefly explain how the waste-heat boiler contributes to the plant profitability. (Think about what would be required in its absence.)

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).

As part of a design calculation, you must evaluate an enthalpy change for an obscure organic vapor that is to be cooled from \(1800^{\circ} \mathrm{C}\) to \(150^{\circ} \mathrm{C}\) in a heat exchanger. You search through all the standard references for tabulated enthalpy or heat capacity data for the vapor but have no luck at all, until you finally stumble on an article in the May 1922 Antarctican Journal of Obscure Organic Vapors that contains a plot of \(C_{p}\left[\operatorname{cal} /\left(\mathrm{g} \cdot^{\cdot} \mathrm{C}\right)\right]\) on a logarithmic scale versus \(\left[T\left(^{\circ} \mathrm{C}\right)\right]^{1 / 2}\) on a linear scale. The plot is a straight line through the points \(\left(C_{p}=0.329, T^{1 / 2}=7.1\right)\) and \(\left(C_{p}=0.533, T^{1 / 2}=17.3\right)\) (a) Derive an equation for \(C_{p}\) as a function of \(T.\) (b) Suppose the relationship of Part (a) turns out to be $$ C_{p}=0.235 \exp \left[0.0473 T^{1 / 2}\right] $$ and that you wish to evaluate $$ \Delta \hat{H}(\mathrm{cal} / \mathrm{g})=\int_{1800 \mathrm{c}}^{150^{\circ} \mathrm{C}} C_{p} d T $$ First perform the integration analytically, using a table of integrals if necessary; then write a spreadsheet or computer program to do it using Simpson's rule (Appendix A.3). Have the program evaluate \(C_{p}\) at 11 equally spaced points from \(150^{\circ} \mathrm{C}\) to \(1800^{\circ} \mathrm{C}\), estimate and print the value of \(\Delta H,\) and repeat the calculation using 101 points. What can you conclude about the accuracy of the numerical calculation?
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