Question

Consider a one-inlet, one-exit control volume at steady state through which the flow is internally reversible and isothermal. Show that the work per unit of mass flowing can be expressed in terms of the fugacity \(f\) as $$ \left(\frac{\dot{W}_{\mathrm{cv}}}{\dot{m}}\right)_{\operatorname{int}}=-R T \ln \left(\frac{f_{2}}{f_{1}}\right)+\frac{\mathrm{V}_{1}^{2}-\mathrm{V}_{2}^{2}}{2}+g\left(z_{1}-z_{2}\right) $$

Short answer

The work per unit mass flow is expressed as \( \frac{\text{W}_{\text{cv}}}{\text{m}} = -R T \text{ln}(\frac{f_2}{f_1}) + \frac{V_1^2 - V_2^2}{2} + g(z_1 - z_2) \)

Step-by-step solution

Apply the First Law of Thermodynamics

For a control volume at steady state, the First Law of Thermodynamics can be written as \[ \frac{\text{d}}{\text{d}t} \bigg( E_{\text{in}} - E_{\text{out}}\bigg) + \frac{\text{d}}{\text{d}t} \bigg( W_{\text{cv}} \bigg) = \frac{\text{d}}{\text{d}t} \bigg( \frac{\text{d} U_{\text{cv}}}{\text{d}t} \bigg) = h_{\text{in}} + \frac{V_{\text{in}}^2}{2} + gz_{\text{in}} - h_{\text{out}} - \frac{V_{\text{out}}^2}{2} - gz_{\text{out}} - w_{\text{cv/in}} \] Where: - \(h\): Enthalpy- \(V\): Velocity- \(z\): Elevation- \(w_{\text{cv}}\): Specific work done

Simplify Enthalpy Using Isothermal and Internally Reversible Assumptions

Since the process is internally reversible and isothermal, the change in internal energy (\(dU\)) is zero, and thus the change in enthalpy is also zero. Therefore, we can simplify \[ \frac{d h_{\text{in}}}{\text{dt}} = \frac{d h_{\text{out}}}{\text{dt}}. \]

Express Work in Terms of Flow Properties

By integrating the first point, work (\(w_{\text{cv, in}}\)) can be expressed as:\[ \frac{\text{d}W_{\text{cv}}}{\text{d}m} = h_{\text{in}} + \frac{V_{\text{in}}^2}{2} + g z_{\text{in}} - h_{\text{out}} - \frac{V_{\text{out}}^2}{2} - g(z_{\text{out}})\] However due to isothermal and reversible flow, \(h_{\text{in}} = h_{\text{out}}\). Hence: \[ \frac{d W_{\text{cv}}}{d m} = \frac{V_1^2 - V_2^2}{2} + g(z_1 - z_2)\]

Link Work with Fugacity

By integrating the Second Law of Thermodynamics for reversible isothermal processes for a one-inlet and one-outlet system, you get an expression for work with fugacity terms: \[d W_{\text{cv, isothermal/reversible}} = -R T \text{d} \text{ln}(\frac{f_2}{f_1})\] The above can be written as \[ \frac{W_{\text{cv}}}{\text{d}m} = -R T \text{ln}(\frac{f_2}{f_1}) + \frac{V_1^2 - V_2^2}{2} + g(z_1 - z_2)\]

Key concepts

First Law of Thermodynamics

The First Law of Thermodynamics is a crucial concept in understanding the energy balance within a system. Essentially, it states that energy cannot be created or destroyed; it can only be transferred or converted from one form to another.
For a control volume at steady state with one inlet and one exit, the First Law can be written as: \[ \frac{\text{d}}{\text{d}t} \bigg( E_{\text{in}} - E_{\text{out}}\bigg) + \frac{\text{d}}{\text{d}t} \bigg( W_{\text{cv}} \bigg) = \frac{\text{d}}{\text{d}t} \bigg( \frac{\text{d} U_{\text{cv}}}{\text{d}t} \bigg) \]

• \( h \): Enthalpy
• \( V \): Velocity
• \( z \): Elevation
• \( w_{\text{cv}} \): Specific work done

This equation helps us understand how energy changes within a system by considering the different forms of energy at the inlet and outlet, as well as the work done by the control volume.

Internally Reversible Process

An internally reversible process refers to a theoretical process in which the system goes through a series of equilibrium states. In this context, if the direction of the process were reversed, the system would return through the same states.
For our purpose, assuming a process is internally reversible simplifies many calculations. For instance, if a flow is internally reversible and isothermal, then the change in internal energy (\(dU\)) is zero. Consequently, the change in enthalpy (\(dH\)) is also zero, simplifying our energy equations significantly.

Isothermal Process

An isothermal process is one in which the temperature remains constant. When a process is isothermal, the internal energy of an ideal gas doesn't change (since internal energy is a function of temperature).
In the given problem, the term 'isothermal' aids in simplifying our calculations as follows:\[ \frac{d h_{\text{in}}}{\text{dt}} = \frac{d h_{\text{out}}}{\text{dt}}. \] This means that the enthalpy (\(h\)) remains constant during the process. This assumption is vital for expressing specific work with flow properties and helps in integrating other thermodynamic equations.

Fugacity

Fugacity is an adjusted pressure that accounts for deviations from ideal gas behavior in real gases. It's a measure of a substance's tendency to escape or expand and can be considered a 'realistic' pressure that would exert the same amount of escaping tendency as the actual substance at that condition.
In the problem, we're given that work per unit of mass can be expressed in terms of fugacity as shown: \[ \frac{W_{\text{cv}}}{\text{d}m} = -R T \text{ln} \bigg( \frac{f_2}{f_1} \bigg) + \frac{V_1^2 - V_2^2}{2} + g(z_1 - z_2). \] This equation links the work required with the ratio of fugacities at the inlet and exit, combining both pressure-related terms and mechanical energy changes.

Specific Work

Specific work refers to the work done per unit mass of the working substance. In thermodynamic processes, this is crucial as it allows us to analyze work without depending on the total volume or mass.
For the isothermal, internally reversible process described in the problem, specific work (\( w_{\text{cv}} \)) can be simplified as:\[ \frac{W_{\text{cv}}}{\text{d}m} = \frac{V_1^2 - V_2^2}{2} + g(z_1 - z_2), \] where the specific work is related to velocity changes and gravitational potential energy changes (due to height differences). This expression helps isolate the effects of kinetic and potential energy variations, ignoring thermal energy changes due to the isothermal condition.

Enthalpy

Enthalpy (\(h\)) is a measure of the total energy of a thermodynamic system. It includes internal energy as well as the energy required to make space for the system by displacing its environment.
Mathematically, enthalpy is defined as: \[ h = U + pV, \] where:

• \( U \): Internal Energy
• \( p \): Pressure
• \( V \): Volume

In the given problem, the enthalpy change between inlet and outlet is zero due to the isothermal assumption. This helps in simplifying the calculation of specific work as it removes the need to consider heat transfer into or out of the system, allowing us to focus on mechanical energy forms.