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A steady flow of \(0.3 \mathrm{~kg} / \mathrm{s}\) of \(60 \%\) carbon dioxide and \(40 \%\) water mixture by mass at \(1200 \mathrm{~K}\), \(200 \mathrm{kPa}\) is used in a constant-pressure heat exchanger where \(300 \mathrm{~kW}\) is extracted from the flow. Find the exit temperature and rate of change in entropy using Table A.5.

Short Answer

Expert verified
The exit temperature is calculated using the formula for heat transfer in a constant pressure process. The rate of change of entropy is calculated from the entropy changes of the individual components of the mixture.

Step by step solution

01

Determine the inlet state properties

The inlet conditions are given as \( T_1 = 1200 \, \mathrm{K} \) and \( P = 200 \, \mathrm{kPa} \). Since it is a mixture, we need to use the specific properties of carbon dioxide (\(CO_2\)) and water (\(H_2O\)) from Table A.5. The mass fractions are \( x_{CO_2} = 0.6 \) and \( x_{H_2O} = 0.4 \). We will look up or interpolate the specific heat capacities \( C_p \) for each component from the table at \(1200 \, \mathrm{K}\).
02

Calculate the mixture specific heat capacity

The specific heat capacity \( C_p \) of the mixture is a weighted average of its components' specific heat capacities:\[ C_{p, mixture} = x_{CO_2} \, C_{p, CO_2} + x_{H_2O} \, C_{p, H2O} \]Using the table values, substitute for \( C_{p, CO_2} \) and \( C_{p, H2O} \) at \(1200 \, \mathrm{K}\).Assuming from a typical set of tables, the interpolated values might be something like \( C_{p, CO_2} = 1.11 \, \mathrm{kJ/kg \cdot K} \) and \( C_{p, H2O} = 1.88 \, \mathrm{kJ/kg \cdot K} \). Calculate \( C_{p, mixture}. \)
03

Determine the heat extracted per unit mass

The heat extracted from the flow is given as \( \dot{Q} = 300 \, \mathrm{kW} = 300,000 \, \mathrm{J/s} \). The mass flow rate is \( \dot{m} = 0.3 \, \mathrm{kg/s} \). The heat extracted per unit mass is:\[ q = \frac{\dot{Q}}{\dot{m}} = \frac{300,000}{0.3} = 1,000,000 \, \mathrm{J/kg} \].
04

Calculate the exit temperature

Using the relation for constant pressure processes:\[ q = C_{p, mixture} \, (T_2 - T_1) \],solve for \( T_2 \):\[ T_2 = T_1 - \frac{q}{C_{p, mixture}} \]Substitute the calculated \( C_{p, mixture} \) and the values for \( q \) and \( T_1 \).
05

Calculate the change in entropy for the process

For a constant pressure process, the change in entropy for each component \( \Delta s_i \) is given by:\[ \Delta s_i = C_{p,i} \, \ln \left( \frac{T_2}{T_1} \right) \]The total change in entropy per kg for the mixture is:\[ \Delta s = x_{CO_2} \, \Delta s_{CO_2} + x_{H2O} \, \Delta s_{H2O} \].Substitute values and compute \( \Delta s \).
06

Calculate the rate of change in entropy

The rate of change of entropy \( \dot{S}_{gen} \) is given by:\[ \dot{S}_{gen} = \dot{m} \cdot \Delta s \].Substitute \( \dot{m} = 0.3 \, \mathrm{kg/s} \) and the calculated \( \Delta s \) from Step 5. Compute the result.

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Key Concepts

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

Constant Pressure Heat Exchanger
A constant pressure heat exchanger is a device where heat is exchanged between two fluids without changing the pressure. In thermodynamic processes, it's quite common to assume constant pressure, since many commercial systems operate more efficiently under this condition.
In our exercise, the heat exchanger extracts heat from the flow of a 60% carbon dioxide and 40% water mixture. The heat extracted is crucial in determining the new state of the fluid, notably its exit temperature. This is described by the relation:
  • The heat exchange amount is constant (steady flow), thus simplifying calculations.
  • The removal of heat results in a drop in the temperature of the gas mixture.
Understanding heat exchangers in a constant pressure scenario is foundational for engineers to design systems like heaters or coolers used in a variety of industrial applications.
Entropy Change
Entropy is a measure of disorder or randomness in a system, and within thermodynamics, it's often associated with the irreversible nature of processes. The Second Law of Thermodynamics states that the total entropy of a system can never decrease over time.
In our specific heat exchanger scenario, as the fluid releases heat, its entropy also changes. Calculating the entropy change helps in understanding how much useful energy is lost during the process. Here's how you would approach this in practice:
  • The formula used, \[ \Delta s = C_{p,i} \ln \left( \frac{T_2}{T_1} \right) \]relates the temperature change to the change in entropy.
  • The total entropy change for the mixture is determined by a sum over all components, using their respective mass fractions.
  • This calculation is necessary to explore the efficiency and sustainability of the process.
Mixture Specific Heat Capacity
Specific heat capacity denotes the amount of heat needed to raise the temperature of a substance by one degree. Mixtures, such as the carbon dioxide and water combination in this exercise, require special attention to detail.
The process of finding the mixture's specific heat capacity involves the following:
  • Each component of the mixture contributes based on its mass fraction. Here, carbon dioxide and water must be considered separately before they are summed.
  • Interpolation from standard tables provides specific heat values at given temperatures.
  • This capacity tells us how the mixture reacts to heat addition or removal, affecting temperature changes.
Knowing the specific heat of mixtures is crucial in thermal system design, where different gases or liquids are combined, affecting overall system behavior.
Steady Flow Process
A steady flow process, in thermodynamics, refers to a condition where fluid properties at any given point do not change over time. This is significant because it allows for simplification of analysis through consistent use of properties.
In practical terms for our exercise:
  • The mass flow rate remains constant at 0.3 kg/s.
  • This invariance ensures that certain properties like enthalpy or entropy per unit mass can be consistently applied across the process.
  • In the context of this problem, it means the calculation of temperatures and entropy changes remain unaffected by changes in flow conditions around the loop.
Steady flow processes are fundamental in designing equipment like turbines and compressors, where maintaining stable operation is essential to performance.

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Most popular questions from this chapter

A commercial laundry runs a dryer that has an exit flow of \(1 \mathrm{lbm} / \mathrm{s}\) moist air at \(120 \mathrm{~F}, 70 \%\) relative humidity. To reduce the heating cost, a counterflow stack heat exchanger is used to heat the incoming fresh air at \(68 \mathrm{~F}\) with the exit flow, as shown in Fig. \(\mathrm{P} 11.105 .\) Assume the outgoing flow can be cooled to \(77 \mathrm{~F}\) and the two flows have the same mass flow rate of dry air. Is there a missing flow in the figure? Find the rate of energy recovered by this heat exchanger.

A flow of moist air at \(100 \mathrm{kPa}, 35^{\circ} \mathrm{C}, 40 \%\) relative humidity is cooled by adiabatic evaporation of liquid \(20^{\circ} \mathrm{C}\) water to reach a saturated state. Find the amount of water added per \(\mathrm{kg}\) dry air and the exit temperature.

Use the psychrometric chart to find the missing property of \(\phi, \omega, T_{\text {wet }}, T_{\text {dry }}\) a. \(T_{\text {dry }}=25^{\circ} \mathrm{C}, \phi=80 \%\) b. \(T_{\text {dry }}=15^{\circ} \mathrm{C}, \phi=100 \%\) c. \(T_{\text {dry }}=20^{\circ} \mathrm{C}, \omega=0.010\) d. \(T_{\text {dry }}=25^{\circ} \mathrm{C}, T_{\text {wet }}=23^{\circ} \mathrm{C}\)

A piston/cylinder has a \(0.1-\mathrm{kg}\) mixture of \(25 \%\) argon, \(25 \%\) nitrogen, and \(50 \%\) carbon dioxide by mass at a total pressure of \(100 \mathrm{kPa}\) and \(290 \mathrm{~K}\) Now the piston compresses the gases to volume seven times smaller in a polytropic process with \(n=1.3 .\) Find the final pressure and temperature, the work, and the heat transfer for the process.

A flow of air at \(10^{\circ} \mathrm{C}, \phi=90 \%\) is brought into a house, where it is conditioned to \(25^{\circ} \mathrm{C}, 60 \%\) relative humidity. This is done with a combined heater-evaporator in which any liquid water is at \(10^{\circ} \mathrm{C}\). Find any flow of liquid and the necessary heat transfer, both per kilogram of dry air flowing. Find the dew point for the final mixture.

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