/*! 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 16 A vessel contains carbon dioxide... [FREE SOLUTION] | 91影视

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Chapter 7: Problem 16

A vessel contains carbon dioxide. Using the ideal gas model (a) determine the specific exergy of the gas, in Btu/lb, at \(p=80 \mathrm{lbf}^{2} \mathrm{in}^{2}\) and \(T=180^{\circ} \mathrm{F}\). (b) plot the specific exergy of the gas, in Btu/b, versus pressure ranging from 15 to \(80 \mathrm{lbf} / \mathrm{in}^{2}\), for \(T=80^{\circ} \mathrm{F}\). (c) plot the specific exergy of the gas, in Btu/lb, versus temperature ranging from 80 to \(180^{\circ} \mathrm{F}\), for \(p=15 \mathrm{lbf} / \mathrm{in}^{2}{ }^{2}\) The gas is at rest and zero elevation relative to an exergy reference environment for which \(T_{0}=80^{\circ} \mathrm{F}, p_{0}=15 \mathrm{lbf} / \mathrm{in} .^{2}\)

Short Answer

Expert verified
Specific exergy at the given conditions is 165.36 Btu/lb. Graphs can be plotted after repetitive calculations for other conditions.

Step by step solution

01

- Convert Temperatures

Convert the given temperatures from Fahrenheit to Rankine using the formula: \( T_R = T_F + 459.67 \). For part (a), convert 180掳F to Rankine: \( T = 180 + 459.67 = 639.67 \, \text{R} \). Similarly, convert the reference temperature 80掳F to Rankine: \( T_0 = 80 + 459.67 = 539.67 \, \text{R} \).
02

- Calculate the Specific Enthalpy

For part (a), calculate the specific enthalpy using the formula: \( h = c_p(T - T_0) \), where the specific heat at constant pressure \( c_p \) for CO2 can be taken as 0.204 Btu/(lb路R). So, \( h = 0.204(639.67 - 539.67) = 0.204 \times 100 = 20.4 \, \text{Btu/lb} \).
03

- Calculate the Specific Entropy

Calculate the specific entropy change using the formula: \( \Delta s = c_p \ln\left( \frac{T}{T_0} \right) - R \ln\left( \frac{p}{p_0} \right) \), where the specific gas constant for CO2, \( R \), is 0.1889 Btu/(lb路R). So, \( \Delta s = 0.204 \ln\left( \frac{639.67}{539.67} \right) - 0.1889 \ln\left( \frac{80}{15} \right) \). Calculate these terms: \( 0.204 \ln\left( \frac{639.67}{539.67} \right) \approx 0.204 \times 0.1729 = 0.0353 \). Similarly, \( 0.1889 \ln\left( \frac{80}{15} \right) \approx 0.1889 \times 1.6094 = 0.3041 \). Hence, \( \Delta s = 0.0353 - 0.3041 = -0.2688 \, \text{Btu/(lb路R)} \).
04

- Calculate Specific Exergy

The specific exergy is given by: \( e = h - T_0 \Delta s \). Substituting the values, we get: \( e = 20.4 - 539.67 \times (-0.2688) \). Therefore, \( e = 20.4 + 144.96 = 165.36 \, \text{Btu/lb} \).
05

- Plot Specific Exergy vs Pressure

For part (b), the temperature is given as 80掳F. Convert this temperature to Rankine: \( T = 80 + 459.67 = 539.67 \, \text{R} \). Follow Steps 2, 3, and 4 to calculate the specific exergy for different pressures (15, 30, 45, 60, and 80 lbf/in虏) and plot the graph accordingly.
06

- Plot Specific Exergy vs Temperature

For part (c), the pressure is given as 15 lbf/in虏. Follow Steps 2, 3, and 4 to calculate the specific exergy for different temperatures (80掳F, 110掳F, 140掳F, 170掳F, and 180掳F) and plot the graph accordingly.

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Key Concepts

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

Ideal gas model
The ideal gas model is a simplification used in thermodynamics to describe the behavior of gases. An ideal gas is one in which the molecules do not interact with each other and occupy no volume. The behavior of an ideal gas can be described using the ideal gas equation of state:
pv = RT
where:
- **p** is the pressure
- **v** is the specific volume
- **R** is the gas constant
- **T** is the temperature
For this exercise, we use the ideal gas model to simplify calculations of properties like specific enthalpy and specific entropy.
Specific Entropy
Specific entropy is a measure of the disorder or randomness of a system. In simpler terms, it quantifies the energy dispersion in a thermal process. The formula to calculate the specific entropy change ( \Delta s ) for an ideal gas is:
\[ \Delta s = c_p \ln \left(\frac{T}{T_0}\right) - R \ln \left(\frac{p}{p_0}\right) \]
Here,
- **c_p** is the specific heat at constant pressure
- **T** is the temperature
- **T_0** is the reference temperature
- **R** is the gas constant
- **p** is the pressure
- **p_0** is the reference pressure
For the specific entropy change in carbon dioxide (CO2), you will use coefficients specific to CO2. The calculated example from the exercise shows how small changes in temperature and pressure can affect entropy.
Temperature conversion (Fahrenheit to Rankine)
When dealing with thermodynamic calculations for ideal gases, it is essential to work in absolute temperatures. The Rankine scale is an absolute temperature scale used in American engineering systems.
The conversion from Fahrenheit to Rankine is straightforward: \[ T_R = T_F + 459.67 \]
where:
- **T_R** is the temperature in Rankine
- **T_F** is the temperature in Fahrenheit
For example, to convert 180掳F to Rankine:
\[ T = 180 + 459.67 = 639.67 \mathrm{R} \]
This conversion is necessary to simplify calculations and avoid negative temperatures in thermodynamic equations.
Specific Enthalpy
Specific enthalpy is the amount of heat content used or released in a system at constant pressure. It is an important property in thermodynamics. For an ideal gas, it can be calculated using: \[ h = c_p (T - T_0) \]
where:
- **h** is the specific enthalpy
- **c_p** is the specific heat at constant pressure
- **T** is the absolute temperature
- **T_0** is the reference temperature
For CO2 with a specific heat at constant pressure ( \ c_p ) of 0.204 Btu/(lb路R), the exercise demonstrates how the value of specific enthalpy can be derived to be 20.4 Btu/lb under the conditions provided.
Exergy
Exergy represents the usable or available energy within a system. It measures a system's potential to perform work in reference to the environment. The specific exergy formula for an ideal gas is:
\[ e = h - T_0 \Delta s \]
where:
- **e** is the specific exergy
- **h** is the specific enthalpy
- **T_0** is the reference absolute temperature
- **\Delta s** is the specific entropy change
The exercise demonstrates how to compute the specific exergy of CO2, resulting in 165.36 Btu/lb. This computation requires converting temperatures to Rankine and then finding specific enthalpy and entropy, as illustrated in the provided steps.

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Most popular questions from this chapter

Steam enters a turbine operating at steady state at \(4 \mathrm{MPa}\), \(500^{\circ} \mathrm{C}\) with a mass flow rate of \(50 \mathrm{~kg} / \mathrm{s}\). Saturated vapor exits at \(10 \mathrm{kPa}\) and the corresponding power developed is \(42 \mathrm{MW}\). The effects of motion and gravity are negligible. (a) For a control volume enclosing the turbine, determine the rate of heat transfer, in MW, from the turbine to its surroundings Assuming an average turbine outer surface temperature of \(50^{\circ} \mathrm{C}\), determine the rate of exergy destruction, in MW. (b) If the turbine is located in a facility where the ambient temperature is \(27^{\circ} \mathrm{C}\), determine the rate of exergy destruction for an enlarged control volume including the turbine and its immediate surroundings so heat transfer takes place at the ambient temperature. Explain why the exergy destruction values of parts (a) and (b) differ. Let \(T_{0}=300 \mathrm{~K}, p_{0}=100 \mathrm{kPa}\).

Refrigerant \(134 a\) enters a counterflow heat exchanger operating at steady state at \(-32^{\circ} \mathrm{C}\) and a quality of \(40 \%\) and exits as saturated vapor at \(-32^{\circ} \mathrm{C}\). Air enters as a separate sticam with a unass fluw 1 ate of \(5 \mathrm{~kg} / \mathrm{s}\) and is couleal al a constant pressure of 1 bar from 300 to \(250 \mathrm{~K}\). Heat transfer between the heat exchanger and its surroundings can be ignored, as can the effects of motion and gravity. (a) As in Fig. E7.6, sketch the variation with position of the temperature of each stream. Locate \(T_{0}\) on the sketch. (b) Determine the rate of exergy destruction within the heat exchanger, in \(\mathrm{kW}\). (c) Devise and evaluate an exergetic efficiency for the heat exchanger. Let \(T_{0}=300 \mathrm{~K}, p_{0}=1\) bar.

Steam at 30 bar and \(700^{\circ} \mathrm{C}\) is available at one location in an industrial plant. At another location, steam at 20 bar and \(400^{\circ} \mathrm{C}\) is required for use in a certain process. An engineer suggests that steam at this condition can be provided by allowing the higher-pressure steam to expand through a valve to 20 bar and then cool to \(400^{\circ} \mathrm{C}\) through a heat exchanger with heat transfer to the surroundings, which are at \(20^{\circ} \mathrm{C}\). (a) Evaluate this suggestion by determining the associated exergy destruction rate per mass flow rate of steam \((\mathrm{kJ} / \mathrm{kg})\) for the valve and heat exchanger. Discuss. (b) Evaluating exergy at 8 cents per \(\mathrm{kW} \cdot \mathrm{h}\) and assuming continuous operation at steady state, determine the total annual cost, in \(\$$, of the exergy destruction for a mass flow rate of \)1 \mathrm{~kg} / \mathrm{s}\(. (c) Suggest an alternative method for obtaining steam at the desired condition that would by preferable thermodynamically, and determine the total annual cost, in \)\$$, of the exergy destruction for a mass flow rate of \(1 \mathrm{~kg} / \mathrm{s}\). Let \(T_{0}=20^{\circ} \mathrm{C}\), \(p_{0}=1 \mathrm{~atm}\).

Air enters a compressor operating at steady state at \(T_{1}=\) \(320 \mathrm{~K}, p_{1}=2\) bar with a velocity of \(80 \mathrm{~m} / \mathrm{s}\). At the exit, \(T_{2}=550 \mathrm{~K}, p_{2}=6\) bar and the velocity is \(180 \mathrm{~m} / \mathrm{s}\). The air can be modeled as an ideal gas with \(c_{p}=1.01 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\). Stray heat transfer can be ignored. Determine, in kJ per \(\mathrm{kg}\) of air flowing, (a) the power required by the compressor and (b) the rate of exergy destruction within the compressor. Let \(T_{0}=\) \(300 \mathrm{~K}, p_{0}=1\) bar. Ignore the effects of motion and gravity.

Argon enters a nozzle operating at steady state at \(1300 \mathrm{~K}\), \(360 \mathrm{kPa}\) with a velocity of \(10 \mathrm{~m} / \mathrm{s}\) and exits the nozzle at \(900 \mathrm{~K}, 130 \mathrm{kPa}\). Stray heat transfer can be ignored. Modeling argon as an ideal gas with \(k=1.67\), determine (a) the velocity at the exit, in \(\mathrm{m} / \mathrm{s}\), and (b) the rate of exergy destruction, in \(\mathrm{kJ}\) per \(\mathrm{kg}\) of argon flowing. Let \(T_{0}=293 \mathrm{~K}, p_{0}=1\) bar.
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