Showing That the Final Result Should Not Depend on the Choice of System A compressor is operating in a continuous, steady-state manner to produce a gas at temperature T2 and pressure P2 from one at T1 and P1. Show that for the time interval Δt Q + Ws = (Hˆ2 − Hˆ1) ΔM where ΔM is the mass of gas tha

Question

Showing That the Final Result Should Not Depend on the Choice of System

A compressor is operating in a continuous, steady-state manner to produce a gas at temperature [latex]T_{2} ext { and pressure } P_{2} ext { from one at } T_{1} ext { and } P_{1}[/latex]. Show that for the time interval Δt

[latex]Q+W_{s}=\left(\hat{H}_{2}-\hat{H}_{1} ight) \Delta M[/latex]

where ΔM is the mass of gas that has flowed into or out of the system in the time Δt. Establish this result by (a) first writing the balance equations for a closed system consisting of some convenient element of mass, and then (b) by writing the balance equations for the compressor and its contents, which is an open system.

Accepted Answer

a. The closed-system analysis
Here we take as the system the gas in the compressor and the mass of gas ΔM, that will enter the compressor in the time interval Δt. The system is enclosed by dotted lines in the figure.

At the later time t + Δt, the mass of gas we have chosen as the system is as shown below.

We use the subscript c to denote the characteristics of the fluid in the compressor, the subscript 1 for the gas contained in the system that is in the inlet pipe at time t, and the subscript 2 for the gas in the system that is in the exit pipe at time t + Δt. With this notation the mass balance for the closed system is

[latex]M_{2}(t+\Delta t)+M_{c}(t+\Delta t)=M_{1}(t)+M_{c}(t)[/latex]

The energy balance for this system, neglecting the potential and kinetic energy terms (which, if retained, would largely cancel), is

[latex]\left.M_{2} \hat{U}_{2} ight|_{t+\Delta t}+\left.M_{c} \hat{U}_{c} ight|_{t+\Delta t}-\left.M_{1} \hat{U}_{1} ight|_{t}-\left.M_{c} \hat{U}_{c} ight|_{t}=W_{s}+Q+P_{1} \hat{V}_{1} M_{1}-P_{2} \hat{V}_{2} M_{2}[/latex] (a)

In writing this equation we have recognized that the flow terms vanish for the closed system and that there are two contributions of the [latex]\int P d V[/latex] type, one due to the deformation of the system boundary at the compressor inlet and another at the compressor outlet. Since the inlet and exit pressures are constant at [latex]P_{1} ext { and } P_{2}[/latex], these terms are

[latex]\begin{aligned}-\int P d V &=-\left.P_{1} \int d V ight|_{ ext {inlet }} \quad-\left.P_{2} \int d V ight|_{ ext {outlet }} \&=-P_{1}\left\{V_{1}(t+\Delta t)-V_{1}(t) ight\}-P_{2}\left\{V_{2}(t+\Delta t)-V_{2}(t) ight\}\end{aligned}[/latex]

However, [latex]V_{1}(t+\Delta t)=0 ext { and } V_{2}(t)=0[/latex] , so that

[latex]-\int P d V=+P_{1} V_{1}-P_{2} V_{2}=P_{1} \hat{V}_{1} M_{1}-P_{2} \hat{V}_{2} M_{2}[/latex]

Now, using the energy balance and Eq. a above, and recognizing that since the compressor is in steady-state operation,

[latex]\left.M_{c} \hat{U}_{c} ight|_{t+\Delta t}=\left.M_{c} \hat{U}_{c} ight|_{t}[/latex]

we obtain

[latex]\Delta M\left(\hat{U}_{2}-\hat{U}_{1} ight)=W_{s}+Q+P_{1} \hat{V}_{1} \Delta M-P_{2} \hat{V}_{2} \Delta M[/latex]

or

[latex]\Delta M\left(\hat{U}_{2}+P_{2} \hat{V}_{2}-\hat{U}_{1}-P_{1} \hat{V}_{1} ight)=\Delta M\left(\hat{H}_{2}-\hat{H}_{1} ight)=W_{s}+Q[/latex]

b. The open-system analysis Here we take the contents of the compressor at any time to be the system. The mass balance for this system over the time interval Δt is

[latex]M(t+\Delta t)-M(t)=\int_{t}^{t+\Delta t} \dot{M}_{1} d t+\int_{t}^{t+\Delta t} \dot{M}_{2} d t=(M)_{1}+(M)_{2}[/latex]

and the energy balance is

[latex]\begin{aligned}\left\{U+M\left(\frac{v^{2}}{2} ight. ight.&+\psi)\}_{t+\Delta t}-\left\{U+M\left(\frac{v^{2}}{2}+\psi ight) ight\}_{t} \&=\int_{t}^{t+\Delta t} \dot{M}_{1}\left(\hat{H}_{1}+v_{1}^{2} / 2+\psi_{1} ight) d t+\int_{t}^{t+\Delta t} \dot{M}_{2}\left(\hat{H}_{2}+v_{2}^{2} / 2+\psi_{2} ight) d t+Q+W\end{aligned}[/latex]

These equations may be simplified as follows:
1. Since the compressor is operating continuously in a steady-state manner, its contents must, by definition, have the same mass and thermodynamic properties at all times. Therefore,

[latex]M(t+\Delta t)=M(t)[/latex]

and

[latex]\left\{U+M\left(\frac{v^{2}}{2}+\psi ight) ight\}_{t+\Delta t}=\left\{U+M\left(\frac{v^{2}}{2}+\psi ight) ight\}_{t}[/latex]

2. Since the thermodynamic properties of the fluids entering and leaving the turbine do not change in time, we can write

[latex]\begin{aligned}\int_{t}^{t+\Delta t} \dot{M}_{1}\left(\hat{H}_{1}+v_{1}^{2} / 2+\psi_{1} ight) d t &=\left(\hat{H}_{1}+v_{1}^{2} / 2+\psi_{1} ight) \int_{t}^{t+\Delta t} \dot{M}_{1} d t \&=\left(\hat{H}_{1}+v_{1}^{2} / 2+\psi_{1} ight) \Delta M_{1}\end{aligned}[/latex]

with a similar expression for the compressor exit stream.
3. Since the volume of the system here, the contents of the compressor, is constant

[latex]\int_{V_{1}}^{V_{2}} P d V=0[/latex]

so that

[latex]W=W_{s}[/latex]

4. Finally, we will neglect the potential and kinetic energy changes of the entering and exiting fluids.
With these simplifications, we have

[latex]0=\Delta M_{1}+\Delta M_{2} \quad ext { or } \quad \Delta M_{1}=-\Delta M_{2}=\Delta M[/latex]

and

[latex]0=\Delta M_{1} \hat{H}_{1}+\Delta M_{2} \hat{H}_{2}+Q+W_{s}[/latex]

Combining these two equations, we obtain

[latex]Q+W_{s}=\left(\hat{H}_{2}-\hat{H}_{1} ight) \Delta M[/latex]

This is the same result as in part (a).

Comment
Notice that in the closed-system analysis the surroundings are doing work on the system (the mass element) at the inlet to the compressor, while the system is doing work on its surroundings at the outlet pipe. Each of these terms is a [latex]\int P d V[/latex] –type work term. For the open system this work term has been included in the energy balance as a [latex]P \hat{V} \Delta M[/latex] term, so that it is the enthalpy, rather than the internal energy, of the flow streams that appears in the equation. The explicit [latex]\int P d V[/latex] term that does appear in the open-system energy balance represents only the work done if the system boundaries deform; for the choice of the compressor and its contents as the system here this term is zero unless the compressor (the boundary of our system) explodes.

This illustration demonstrates that the sum [latex]Q+W_{s}[/latex] is the same for a fluid undergoing some change in a continuous process regardless of whether we choose to compute this sum from the closed-system analysis on a mass of gas or from an open-system analysis on a given volume in space. In Illustration 3.2-2 we consider another problem, the compression of a gas by two different processes, the first being a closed-system pistonand-cylinder process and the second being a flow compressor process. Here we will find that the sum Q + W is different in the two processes, but the origin of this difference is easily understood.

Question 3.2-1: Showing That the Final Result Should Not Depend on the Choic...

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