Question

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 \(1 \mathrm{bar}\). 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

The exit temperature is \( \frac{400 + 300x}{1+x} \text{K} \). Calculate exergy destruction by determining entropy generation due to mixing, applying \( \text{Exergy Destruction} = T_0 \times \text{ds} \). Plot results against 'x'.

Step-by-step solution

State the Given Data and Assumptions

Identify the known variables: Helium at 400 K, 1 bar; Argon at 300 K, 1 bar. The mixture exits at 1 bar. The argon mass flow rate is 'x' times that of helium. Assume steady-state operation with negligible kinetic and potential energy effects. The reference temperature, \(T_0\), is 300 K. The system is insulated, so there is no heat transfer.

Apply the Conservation of Mass

Using the mass flow rates, write the mass balance equation: \[ \text{m}_{\text{He}} + \text{m}_{\text{Ar}} = \text{m}_{\text{mix}} \] Given \( \text{m}_{\text{Ar}} = x \text{m}_{\text{He}} \) and \( \text{m}_{\text{mix}} = (1+x)\text{m}_{\text{He}} \), the mass flow rate of the mixture (exit stream) can be expressed.

Apply the Conservation of Energy (First Law of Thermodynamics)

For an insulated system, the energy balance equation is: \[ \text{m}_{\text{He}} h_{\text{He}} + \text{m}_{\text{Ar}} h_{\text{Ar}} = (1+x) \text{m}_{\text{He}} h_{\text{mix}} \] Using the specific enthalpies (typically derived from ideal gas properties) and mass flow rates, substitute \( h_{\text{He}} = c_{p,\text{He}} T_{\text{He}} \) and \( h_{\text{Ar}} = c_{p,\text{Ar}} T_{\text{Ar}} \): \[ c_{p,\text{He}} T_{\text{He}} + xc_{p,\text{Ar}} T_{\text{Ar}} = (1+x) c_{p,\text{mix}} T_{\text{mix}} \]

Solve for Exit Temperature

Substitute \( T_{\text{He}} = 400 \text{K} \) and \( T_{\text{Ar}} = 300 \text{K} \) into the energy balance equation. Rearrange to solve for \( T_{\text{mix}} \), the exit temperature: \[ 400 + 300x = (1+x) T_{\text{mix}} \] \[ T_{\text{mix}} = \frac{400 + 300x}{1+x} \]

Define Entropy Generation and Exergy Destruction

Using the concept of entropy generation for an irreversible adiabatic process, the exergy destruction rate can be expressed as: \[ \text{Exergy Destruction} = T_0 \times (\text{S}_{\text{gen}}) \] Compute the entropy generation as the difference in entropy change between mixing and individual gases.

Calculate Entropy Generation

The change in entropy for each gas can be calculated using: \[ \text{ds} = c_{p} \text{ln}(\frac{T_{\text{final}}}{T_{\text{initial}}}) - R \text{ln}(\frac{P_{\text{final}}}{P_{\text{initial}}}) \] Assume ideal gas behavior and constant pressure mixing, entropy generation can be simplified to changes in temperature.

Find Exergy Destruction Rate

Substitute the calculated entropy generation into the exergy destruction rate formula: \[ \text{Exergy Destruction} = T_0 \times \text{ds} \] where \( T_0 = 300 \text{K} \). Ensure \( ds \) is expressed per kg of Helium entering.

Plotting

Using a graphing tool, plot the values obtained for: (a) Exit temperature \( T_{\text{mix}} \) versus the mass flow rate ratio 'x'. (b) Exergy destruction rate versus the mass flow rate ratio 'x'.

Key concepts

Conservation of Mass

In thermodynamics, the conservation of mass principle is vital for analyzing systems involving mixing or separation. For a steady-state system, the mass entering and leaving must be equal. In our exercise, Helium and Argon mix in a chamber. The mass flow rates are given as \( \text{m}_{\text{He}} \) and \( \text{m}_{\text{Ar}} \), with Argon flowing at a rate ‘x’ times that of Helium. This can be expressed with the mass balance equation:
\[ \text{m}_{\text{He}} + \text{m}_{\text{Ar}} = \text{m}_{\text{mix}} = (1 + x) \text{m}_{\text{He}} \]
Here, \( \text{m}_{\text{mix}} \) represents the total mass flow rate of the exit mixture. Understanding this balance helps in accurately determining the properties of the mixture after both gases have combined. This principle ensures no mass is lost in the process, which is essential for further thermodynamic analysis.

First Law of Thermodynamics

The first law of thermodynamics, also known as the law of energy conservation, states that energy cannot be created or destroyed, only transformed. In this exercise, we deal with an insulated mixing chamber, meaning no heat transfer occurs. Therefore, we use the energy balance equation:
\[ \text{m}_{\text{He}} h_{\text{He}} + \text{m}_{\text{Ar}} h_{\text{Ar}} = (1 + x) \text{m}_{\text{He}} h_{\text{mix}} \]
Substituting the specific enthalpies with terms involving specific heat capacities \( c_p \) and temperatures, we get:
\[ c_{p,\text{He}} T_{\text{He}} + x c_{p,\text{Ar}} T_{\text{Ar}} = (1 + x) c_{p,\text{mix}} T_{\text{mix}} \]
This equation allows us to solve for the exit temperature \( T_{\text{mix}} \) of the combined gases. By applying the known values, we arrive at:
\[ T_{\text{mix}} = \frac{400 + 300x}{1 + x} \]
This formula helps in plotting the exit temperature against varying flow rates of Argon.

Entropy Generation

Entropy generation is a measure of irreversibility in a process. In an ideal scenario, processes would be reversible and generate no entropy, but real systems always have some degree of entropy generation. This mixing process is adiabatic and irreversible, leading to entropy generation given by:
\[ \text{ds} = c_{p} \text{ln}\bigg(\frac{T_{\text{mix}}}{T_{\text{initial}}}\bigg) - R \text{ln}\bigg(\frac{P_{\text{final}}}{P_{\text{initial}}}\bigg) \]
For each gas, Argon and Helium, entropy change calculations are simplified as pressure remains constant. Therefore, we focus on temperature change:
\[ \text{ds} = \text{sum of entropy changes of both gases as they mix and reach } T_{\text{mix}} \]
This measure is crucial in determining system efficiency, as higher entropy generation points to more significant losses.

Exergy Destruction

Exergy is the useful work potential of a system's energy. Exergy destruction occurs because of irreversibilities leading to a loss of useful energy. In this exercise, exergy destruction rate is tied to entropy generation through the equation:
\[ \text{Exergy destruction} = T_0 \times \text{ds} \]
Where \( T_0 = 300 \text{ K} \) is the reference temperature.
Considering the entropy generated during the mixing of Helium and Argon, we substitute \( \text{ds} \) with the calculated value per kg of Helium:

\[ \text{Exergy destruction} = 300 \times \text{ds} \]
This result helps in evaluating the system's overall efficiency and the impact of different mixing ratios (x). Plotting exergy destruction against 'x' highlights how varying flow rates affect energy losses.