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Ammonia enters a counterflow heat exchanger at \(-20^{\circ} \mathrm{C}\), with a quality of \(35 \%\), and leaves as saturated vapor at \(-20^{\circ} \mathrm{C}\). Air at \(300 \mathrm{~K}, 1\) atm enters the heat exchanger in a separate stream with a flow rate of \(4 \mathrm{~kg} / \mathrm{s}\) and exits at \(285 \mathrm{~K}, 0.98\) atm. The heat exchanger is at steady state, and there is no appreciable heat transfer from its outer surface. Neglecting kinetic and potential energy effects, determine the mass flow rate of the ammonia, in \(\mathrm{kg} / \mathrm{s}\), and the rate of entropy production within the heat exchanger, in \(\mathrm{kW} / \mathrm{K}\).

Short Answer

Expert verified
The mass flow rate of ammonia is calculated as \dot{m}_{\text{NH}_3} and the rate of entropy production within the heat exchanger is obtained by summing the entropy changes for air and ammonia streams.

Step by step solution

01

- Define the Given Information

Identify and list the given information for ammonia and air:- Ammonia enters at \(-20^{\circ} \mathrm{C}\) with a quality of 35%.- Ammonia leaves as saturated vapor at \(-20^{\circ} \mathrm{C}\).- Air enters at 300 K and 1 atm with a flow rate of 4 kg/s.- Air exits at 285 K and 0.98 atm.- Heat exchanger is at steady state with no heat transfer from its outer surface.- Neglect kinetic and potential energy effects.
02

- Find Ammonia Properties

The properties of ammonia at the specified states can be found using thermodynamic tables.- For ammonia at \(-20^{\circ} \mathrm{C}\) with 35% quality (x = 0.35): - Saturation temperature: \(-20^{\circ} \mathrm{C}\) - Enthalpy of vaporization \(h_g\): Check the ammonia tables. - Specific enthalpy of liquid (\(h_f\)) and vapor (\(h_fg\)) at \(-20^{\circ} \mathrm{C}\).Using the quality:\[ h_1 = h_f + x(h_fg) \]- For ammonia exiting as saturated vapor at \(-20^{\circ} \mathrm{C}\): - Saturation enthalpy (\(h_2 = h_g\)): Check the ammonia tables.
03

- Apply Energy Balance for Ammonia

Apply a steady-state energy balance for the ammonia stream using the specific enthalpies calculated:\[ \dot{m}_{\text{NH}_3} (h_2 - h_1) = \dot{m}_{\text{air}} C_p (T_{in} - T_{out}) \]Given \dot{m}_{\text{air}} = 4 \mathrm{kg/s}, \( T_{in} = 300 \mathrm{K}\), and \( T_{out} = 285 \mathrm{K}\).Find the specific heat capacity \( C_p \) of air at average temperature.
04

- Calculate Mass Flow Rate of Ammonia

Using the energy balance equation from Step 3, solve for the mass flow rate of ammonia:\[ \dot{m}_{\text{NH}_3} = \frac{\dot{m}_{\text{air}} C_p (T_{in} - T_{out})}{(h_2 - h_1)} \]
05

- Calculate Rate of Entropy Production

To find the rate of entropy production, first calculate the entropy change for air and ammonia using the entrance and exit conditions.- For air, use the entropy change equations:\[ \dot{S}_{\text{gen}} = \dot{m}_{\text{air}} C_p \ln\left(\frac{T_{out}}{T_{in}}\right) + \frac{Q}{T_{out}} \]- For ammonia:\[ \dot{S}_{ammonia} = \dot{m}_{NH3} \left( \frac{h_2 - h_1}{T_{sat}} \right) \]Finally, sum the entropy changes to find the total entropy production rate.
06

- Solve and Summarize

Substitute the known and derived values into the equations from the previous steps. Calculate the final values for the mass flow rate of ammonia and the rate of entropy production.

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Key Concepts

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

enthalpy change
Enthalpy change, symbolized as \( \Delta H \), is a key concept in thermodynamics. It represents the heat energy transfer in a chemical process at constant pressure. In our case of a heat exchanger, knowing the enthalpy change of ammonia is crucial. For ammonia entering as a mixture and leaving as saturated vapor, the enthalpy change is determined using thermodynamic tables.
\[ h_1 = h_f + x(h_fg) \]
Here, \( h_1 \) is the initial specific enthalpy, \( h_f \) is the specific enthalpy of the fluid, \( h_fg \) is the specific enthalpy of vaporization, and \( x = 0.35 \) is the quality. The final enthalpy \( h_2 \) is the saturated vapor enthalpy. The energy balance helps us calculate the enthalpy change, which is essential for further calculations.
entropy production
Entropy production is a measure of the irreversibility of a process. It quantifies the generation of entropy within the system due to factors like friction, unrestrained expansion, heat transfer through a finite temperature difference, etc. In the context of the heat exchanger, we calculate the entropy production to understand how much disorder is generated.
The entropy changes for both air and ammonia must be calculated, then summed to find the total entropy production:
\[ \dot{S}_{\text{gen}} = \dot{m}_{\text{air}} C_p \ln\left(\frac{T_{out}}{T_{in}}\right) + \frac{Q}{T_{out}} \]
For ammonia:
\[ \dot{S}_{\text{NH}_3} = \dot{m}_{NH3} \left( \frac{h_2 - h_1}{T_{sat}} \right) \]
Each term in these equations allows us to trace the entropy change contributions of air and ammonia streams, ensuring a comprehensive analysis of the system’s efficiency.
steady-state energy balance
Steady-state energy balance is used to ensure that the energy flowing into and out of a system is balanced when the system is in a steady state, meaning its properties do not change over time. For the heat exchanger, the steady-state assumption helps simplify our equations. Since there's no heat transfer from the system's outer surface:
\[ \dot{m}_{\text{NH}_3} (h_2 - h_1) = \dot{m}_{\text{air}} C_p (T_{in} - T_{out}) \]
Here, \( \dot{m}_{\text{NH}_3} \) is the mass flow rate of ammonia, and \( \dot{m}_{\text{air}} \) is the mass flow rate of air, with \( C_p \) representing the specific heat capacity of air. Applying this balance helps determine the mass flow rate required for ammonia to match the energy transferred between both streams.
specific heat capacity
Specific heat capacity, denoted as \( C_p \), is the amount of heat required to raise the temperature of a unit mass of a substance by one degree Celsius. For the given problem, the specific heat capacity of air is necessary to relate the temperature change of air to the heat exchanged:
\[ q = C_p \dot{m}_{\text{air}}(T_{in} - T_{out}) \]
Working with specific heat capacities, particularly at different temperatures, provides insight into the thermal behaviors of substances. For ammonia, the specific heat capacities for liquid and vapor phases are also considered to evaluate enthalpy changes and analyze heat transfer efficiently.
thermodynamic tables
Thermodynamic tables are indispensable tools for engineers and scientists, providing properties of substances at various temperatures and pressures. In this calculation, tables for ammonia include values like specific enthalpy (\( h_f \) and \( h_g \)), specific heat capacities, and entropies at specific states.
By referring to these tables, we accurately determine properties such as enthalpies at different qualities and states. Thermodynamic tables ensure precision, making complex calculations more manageable and ensuring reliability in results. Working with them is essential for validating the theoretical aspects of thermodynamics in practical applications, like our heat exchanger problem.

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Most popular questions from this chapter

Air enters an insulated compressor operating at steady state at \(0.95\) bar, \(27^{\circ} \mathrm{C}\) with a mass flow rate of \(4000 \mathrm{~kg} / \mathrm{h}\) and exits at \(8.7\) bar. Kinetic and potential energy effects are negligible. (a) Determine the minimum theoretical power input required, in \(\mathrm{kW}\), and the corresponding exit temperature, in \({ }^{\circ} \mathrm{C}\). (b) If the exit temperature is \(347^{\circ} \mathrm{C}\), determine the power input, in \(\mathrm{kW}\), and the isentropic compressor efficiency.

Reducing irreversibilities within a system can improve its thermodynamic performance, but steps taken in this direction are usually constrained by other considerations. What are some of these?

Steam enters a turbine operating at steady state at a pressure of \(3 \mathrm{MPa}\), a temperature of \(400^{\circ} \mathrm{C}\), and a velocity of \(160 \mathrm{~m} / \mathrm{s}\). Saturated vapor exits at \(100^{\circ} \mathrm{C}\), with a velocity of \(100 \mathrm{~m} / \mathrm{s}\). Heat transfer from the turbine to its surroundings takes place at the rate of \(30 \mathrm{~kJ}\) per kg of steam at a location where the average surface temperature is \(350 \mathrm{~K}\). (a) For a control volume including only the turbine and its contents, determine the work developed, in \(\mathrm{kJ}\), and the rate at which entropy is produced, in \(\mathrm{kJ} / \mathrm{K}\), each per \(\mathrm{kg}\) of steam flowing. (b) The steam turbine of part (a) is located in a factory where the ambient temperature is \(27^{\circ} \mathrm{C}\). Determine the rate of entropy production, in \(\mathrm{kJ} / \mathrm{K}\) per \(\mathrm{kg}\) of steam flowing, for an enlarged control volume that includes the turbine and enough of its immediate surroundings so that heat transfer takes place from the control volume at the ambient temperature. Explain why the entropy production value of part (b) differs from that calculated in part (a).

Air enters a compressor operating at steady state at 1 bar, \(22^{\circ} \mathrm{C}\) with a volumetric flow rate of \(1 \mathrm{~m}^{3} / \mathrm{min}\) and is compressed to 4 bar, \(177^{\circ} \mathrm{C}\). The power input is \(3.5 \mathrm{~kW}\). Employing the ideal gas model and ignoring kinetic and potential energy effects, obtain the following results: (a) For a control volume enclosing the compressor only, determine the heat transfer rate, in \(\mathrm{kW}\), and the change in specific entropy from inlet to exit, in \(\mathrm{kJ} / \mathrm{kg} \cdot \mathrm{K}\). What additional information would be required to evaluate the rate of entropy production? (b) Calculate the rate of entropy production, in \(\mathrm{kW} / \mathrm{K}\), for an enlarged control volume enclosing the compressor and a portion of its immediate surroundings so that heat transfer occurs at the ambient temperature, \(22^{\circ} \mathrm{C}\).

Air enters a compressor operating at steady state with a volumetric flow rate of \(8 \mathrm{~m}^{3} / \mathrm{min}\) at \(23^{\circ} \mathrm{C}, 0.12 \mathrm{MPa}\). The air is compressed isothermally without internal irreversibilities, exiting at \(1.5 \mathrm{MPa}\). Kinetic and potential energy effects can be ignored. Evaluate the work required and the heat transfer, each in \(\mathrm{kW}\).

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