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Helium at \(400 \mathrm{~K}, 1\) bar enters an insulated mixing chamber operating at steady state, where it mixes with argon entering at \(300 \mathrm{~K}, 1\) bar. The mixture exits at a pressure of 1 bar. If the argon mass flow rate is \(x\) times that of helium, plot versus \(x\) (a) the exit temperature, in \(\mathrm{K}\). (b) the rate of exergy destruction within the chamber, in \(\mathrm{kJ}\) per \(\mathrm{kg}\) of helium entering. Kinetic and potential energy effects can be ignored. Let \(T_{0}=300 \mathrm{~K}\).

Short Answer

Expert verified
Plot the exit temperature as \( T_{ex} = \frac{5.193 \times 400 + x \times 0.5203 \times 300}{5.193 + x \times 0.5203} \). For exergy destruction, use \( e_{D} = 5.193 \times 100 - (1 + x) \times (5.193 + x \times 0.5203)(T_{ex} - 300) \).

Step by step solution

01

- Define the problem

Determine the exit temperature and the rate of exergy destruction for a mixing process of helium and argon in an insulated steady-state mixing chamber. The argon mass flow rate is given as a multiple of the helium mass flow rate, and we aim to plot the results versus this multiple.
02

- Write the energy balance equation

Since the mixing chamber is insulated, there is no heat transfer. Thus, the energy balance equation is: \[ \frac{m_{He}c_{p,He}T_{He,in} + m_{Ar}c_{p,Ar}T_{Ar,in}}{m_{He} + m_{Ar}} = T_{ex} \]where: \( T_{He,in} = 400 \text{ K} \),\( T_{Ar,in} = 300 \text{ K} \).Also, \( m_{Ar} = x \times m_{He} \).
03

- Solve for the exit temperature

Substitute \( m_{Ar} = x \times m_{He} \) into the energy balance equation:\[ \frac{m_{He}c_{p,He}T_{He,in} + x \times m_{He}c_{p,Ar}T_{Ar,in}}{m_{He} + x \times m_{He}} = T_{ex} \] Simplify to:\[ T_{ex} = \frac{c_{p,He}T_{He,in} + x \times c_{p,Ar} T_{Ar,in}}{c_{p,He} + x \times c_{p,Ar}} \].With:\( c_{p,He} = 5.193 \text{ kJ/kg.K} \), \( c_{p,Ar} = 0.5203 \text{ kJ/kg.K} \).
04

- Insert specific heats and solve

Substitute the given specific heats into the expression from Step 3:\[ T_{ex} = \frac{5.193 \times 400 + x \times 0.5203 \times 300}{5.193 + x \times 0.5203} \].This equation provides the exit temperature, \( T_{ex} \), in Kelvin as a function of \( x \). Plot \( T_{ex} \) versus \( x \).
05

- Write the exergy balance equation

The exergy balance equation for a steady-state system is:\[ \frac{m_{He} e_{He,in} + m_{Ar} e_{Ar,in} - (m_{He} + m_{Ar}) e_{ex}}{m_{He}} \]where exergy destruction is:\[ e_{D} = e_{He,in} + x \times e_{Ar,in} - (1 + x) e_{ex} \]Replace:\( e = c_p \times (T - T_0) \).
06

- Calculate exergy terms

Substitute the values of specific heat and temperatures:\[ e_{He,in} = 5.193 \times (400 - 300) \text{ kJ/kg} \],\[ e_{Ar,in} = 0.5203 \times (300 - 300) = 0 \text{ kJ/kg} \],\[ e_{ex} = (5.193 + x \times 0.5203) \times (T_{ex} - 300) \text{ kJ/kg} \].
07

- Determine the rate of exergy destruction

Calculate using:\[ e_{D} = 5.193 \times 100 + x \times 0 - (1 + x) \times (5.193 + x \times 0.5203)(T_{ex} - 300) \text{ kJ/kg} \].Substitute \( T_{ex} \) in terms of \( x \) to get the exergy destruction rate. Plot the exergy destruction versus \( x \).

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Key Concepts

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

energy balance equation
In thermodynamics, a critical concept for analyzing systems is the energy balance equation. When substances mix in an insulated chamber, there is no heat transfer with the environment. This simplifies our energy balance equation. For this scenario involving helium and argon mixing, the equation becomes:
     \[ \frac{m_{He}c_{p,He}T_{He,in} + m_{Ar}c_{p,Ar}T_{Ar,in}}{m_{He} + m_{Ar}} = T_{ex} \] 
Here:
  • \( m_{He} \) is the mass flow rate of helium
  • \( c_{p,He} \) is the specific heat capacity of helium
  • \( T_{He,in} \) is the inlet temperature of helium
  • \( m_{Ar} \) is the mass flow rate of argon
  • \( c_{p,Ar} \) is the specific heat capacity of argon
  • \( T_{Ar,in} \) is the inlet temperature of argon
Combining the mass flow rates into one expression using the factor \( x \) defined as the ratio of argon to helium flow rates simplifies notation:\( m_{Ar} = x \times m_{He} \). With this simplified form, we can calculate the exit temperature \( T_{ex} \). This temperature depends on the inlet temperatures and specific heat capacities of the gases involved.
specific heat capacity
Understanding specific heat capacity is essential for solving many thermodynamic problems. Specific heat capacity (\( c_p \)) indicates how much energy is needed to raise the temperature of one kilogram of a substance by one degree Kelvin. For helium, \( c_{p,He} \) is approximately 5.193 kJ/kg.K, and for argon, \( c_{p,Ar} \) is approximately 0.5203 kJ/kg.K. These values show how helium requires much more energy to change its temperature compared to argon. In our energy balance equation:
    \[ T_{ex} = \frac{m_{He}c_{p,He}T_{He,in} + x \times m_{He}c_{p,Ar}T_{Ar,in}}{m_{He} + x \times m_{He}} \] 
Specific heat capacities help determine the final exit temperature \( T_{ex} \) because they directly affect how energy is distributed in the system. Since helium and argon mix, we use their specific heat capacities combined with their temperatures to find a balanced exit temperature.
exergy destruction
Exergy destruction quantifies the loss of useful energy during a process. It's a measure of irreversibility in a system. For our mixing process, we use the exergy balance equation. We start by expressing the exergy of each flow using specific heat values and temperatures as follows:
    \[ e = c_p \times (T - T_0) \]
where \( T_0 \) is the reference temperature, 300 K in this case; \( T_{He,in} = 400 \text{ K} \) for helium, and \( T_{Ar,in} = 300 \text{ K} \) for argon.We substitute these values into our exergy balance equation:
    \[ e_{D} = e_{He,in} + x \times e_{Ar,in} - (1 + x) \times (5.193 + x \times 0.5203)(T_{ex} - 300) \] 
Here, \( e_{He,in} = 5.193 \times 100 \text{ kJ/kg} \) and \( e_{Ar,in} = 0 \text{ kJ/kg} \). Finally, we plot the exergy destruction rate \( e_D \) against the mass flow rate factor \( x \) to see how irreversibility changes with different proportions of helium and argon.
thermal efficiency
Thermal efficiency measures how well a system converts heat into useful work or energy. In this mixing process, while we are not directly calculating thermal efficiency, understanding it helps grasp the broader implications of exergy destruction. Higher exergy destruction implies lower efficiency because more useful energy is lost. For an ideal, fully efficient process, exergy destruction would be zero. However, real processes will always have some degree of irreversibility.Efficiency for a generic cycle can be represented as:
    \[ \eta = \frac{Work \; Output}{Heat \; Input} \] 
The closer we get to minimizing exergy destruction, the higher our thermal efficiency will be, meaning we are making better use of the energy available. Understanding this concept is crucial for improving system designs and achieving more sustainable engineering practices.

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Most popular questions from this chapter

At steady state, moist air at \(42^{\circ} \mathrm{C}, 1\) atm, \(30 \%\) relative humidity is mixed adiabatically with a second moist air stream entering at \(1 \mathrm{~atm}\). The mass flow rates of the two streams are the same. A single mixed stream exits at \(29^{\circ} \mathrm{C}\), \(1 \mathrm{~atm}, 40 \%\) relative humidity with a mass flow rate of \(2 \mathrm{~kg} / \mathrm{s}\). Kinetic and potential energy effects are negligible. For the second entering moist air stream, determine, using data from the psychrometric chart, (a) the relative humidity. (b) the temperature, in \({ }^{\circ} \mathrm{C}\).

An equimolar mixture of helium (He) and carbon dioxide \(\left(\mathrm{CO}_{2}\right)\) enters an insulated nozzle at \(260^{\circ} \mathrm{F}, 5 \mathrm{~atm}\), \(100 \mathrm{ft} / \mathrm{s}\) and expands isentropically to a velocity of \(1110 \mathrm{ft} / \mathrm{s}\). Determine the temperature, in \({ }^{\circ} \mathrm{F}\), and the pressure, in atm, at the nozzle exit. Neglect potential energy effects.

Outside air at \(50^{\circ} \mathrm{F}, 1\) atm, and \(40 \%\) relative humidity enters an air-conditioning device operating at steady state. Liquid water is injected at \(45^{\circ} \mathrm{F}\) and a moist air stream exits with a volumetric flow rate of \(1000 \mathrm{ft}^{3} / \mathrm{min}\) at \(90^{\circ} \mathrm{F}, 1 \mathrm{~atm}\) and a relative humidity of \(40 \%\). Neglecting kinetic and potential energy effects, determine (a) the rate water is injected, in \(\mathrm{lb} / \mathrm{min}\). (b) the rate of heat transfer to the moist air, in Btu/h.

Figure P12.82 shows a compressor followed by an aftercooler. Atmospheric air at \(14.7 \mathrm{lbf} / \mathrm{in}^{2}, 900^{\circ} \mathrm{F}\), and \(75 \%\) relative humidity enters the compressor with a volumetric flow rate of \(100 \mathrm{ft}^{3} / \mathrm{min}\). The compressor power input is \(15 \mathrm{hp}\). The moist air exiting the compressor at \(100 \mathrm{lbf} / \mathrm{in}^{2}, 400^{\circ} \mathrm{F}\) flows through the aftercooler, where it is cooled at constant pressure, exiting saturated at \(100^{\circ} \mathrm{F}\). Condensate also exits the aftercooler at \(100^{\circ} \mathrm{F}\). For steady-state operation and negligible kinetic and potential energy effects, determine (a) the rate of heat transfer from the compressor to its surroundings, in Btu/min. (b) the mass flow rate of the condensate, in \(\mathrm{lb} / \mathrm{min}\). (c) the rate of heat transfer from the moist air to the refrigerant circulating in the cooling coil, in tons of refrigeration.

Using the ideal gas model with constant specific heats, determine the mixture temperature, in \(\mathrm{K}\), for each of two cases: (a) Initially, \(0.6 \mathrm{kmol}\) of \(\mathrm{O}_{2}\) at \(500 \mathrm{~K}\) is separated by a partition from \(0.4 \mathrm{kmol}\) of \(\mathrm{H}_{2}\) at \(300 \mathrm{~K}\) in a rigid insulated vessel. The partition is removed and the gases mix to obtain a final equilibrium state. (b) Oxygen \(\left(\mathrm{O}_{2}\right)\) at \(500 \mathrm{~K}\) and a molar flow rate of \(0.6 \mathrm{kmol} / \mathrm{s}\) enters an insulated control volume operating at steady state and mixes with \(\mathrm{H}_{2}\) entering as a separate stream at \(300 \mathrm{~K}\) and a molar flow rate of \(0.4 \mathrm{kmol} / \mathrm{s}\). A single mixed stream exits. Kinetic and potential energy effects can be ignored.

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