Question 12.4: Considering an Ideal Gas Mixture Expanding Isentropically Th...

Known A gas mixture consisting of CO _{2} \text { and } O _{2} in specified proportions expands isentropically through a nozzle from specified inlet conditions to a given exit pressure.

Find Determine the temperature at the nozzle exit, in K, the entropy changes of the CO _{2} \text { and } O _{2} from inlet to exit, in kJ/kmol ⋅ K, and the exit velocity, in m/s.

Schematic and Given Data:

Engineering Model

1. The control volume shown by the dashed line on the accompanying figure operates at steady state.

2. The mixture composition remains constant as the mixture expands isentropically through the nozzle.

3. The Dalton model applies: The overall mixture and each mixture component act as ideal gases. The state of each component is defined by the temperature and the partial pressure of the component.

4. The change in potential energy between inlet and exit can be ignored.

Analysis

a. The temperature at the exit can be determined using the fact that the expansion occurs isentropically: \bar{s}_{2}-\bar{s}_{1}=0. As there is no change in the specific entropy of the mixture between inlet and exit, Eq. 12.35 can be used to write

\Delta \bar{s}=\sum_{i=1}^{j} y_{i}\left[\bar{s}_{i}\left(T_{2}, p_{i 2}\right)-\bar{s}_{i}\left(T_{1}, p_{i 1}\right)\right] (12.35)

\bar{s}_{2}-\bar{s}_{1}=y_{ O _{2}} \Delta \bar{s}_{ O _{2}}+y_{ CO _{2}} \Delta \bar{s}_{ CO _{2}}=0 (a)

Since composition remains constant, the ratio of partial pressures equals the ratio of mixture pressures. Accordingly, the change in specific entropy of each component can be determined using Eq. 12.36. Equation (a) then becomes

\Delta \bar{s}_{i}=\bar{s}_{i}^{\circ}\left(T_{2}\right)-\bar{s}_{i}^{\circ}\left(T_{1}\right)-\bar{R} \ln \frac{p_{2}}{p_{1}} (12.36)

On rearrangement

The sum of mole fractions equals unity, so the coefficient of the last term on the right side is \left(y_{ O _{2}}+y_{ CO _{2}}\right)=1.

Introducing given data and values of \bar{s}^{\circ} \text { for } O _{2} \text { and } CO _{2} at T_{1}=700 K from Table A-23

or

To determine the temperature T_{2} requires an iterative approach with the above equation: A final temperature T_{2} is assumed, and the \bar{s}^{\circ} \text { values for } O _{2} \text { and } CO _{2} are found from Table A-23. If these two values do not satisfy the equation, another temperature is assumed. The procedure continues until the desired agreement is attained. In the present case

Linear interpolation between these values gives T_{2}=517.6 K.

Alternative Solution

Alternatively, the following IT program can be used to evaluate T_{2} without resorting to iteration with table data. In the program, yO2 denotes the mole fraction of O _{2}, p1_O2 denotes the partial pressure of O _{2} at state 1, s1_O2 denotes the entropy per mole of O _{2} at state 1, and so on.

Using the Solve button, the result is T_{2}=517.6 K, which agrees with the value obtained using table data. Note that IT provides the value of specific entropy for each component directly and does not return \bar{s}^{\circ} of the ideal gas tables.

1 b. The change in the specific entropy for each of the components can be determined using Eq. 12.36. For O _{2}

Inserting \bar{s}^{\circ} values for O _{2} from Table A-23 at T_{1}=700 K and T_{2}=517.6 K

Similarly, with CO _{2} data from Table A-23

2        = −0.92 kJ/kmol ⋅K

c. Reducing the energy rate balance for the one-inlet, one-exit control volume at steady state

where h_{1} \text { and } h_{2} are the enthalpy of the mixture, per unit mass of mixture, at the inlet and exit, respectively. Solving for V _{2}

The term \left(h_{1}-h_{2}\right) in the expression for V _{2} can be evaluated as

where M is the apparent molecular weight of mixture, and the molar specific enthalpies of O _{2} \text { and } CO _{2} are from Table A-23. With Eq. 12.9, the apparent molecular weight of the mixture is

M=\sum_{i=1}^{j} y_{i} M_{i} (12.9)

Then, with enthalpy values at T_{1}=700 K \text { and } T_{2}=517.6 K from Table A-23,

= 194.7 kJ/kg

Finally,

3 V _{2}=\sqrt{\left(3 \frac{ m }{ s }\right)^{2}+2\left(194.7 \frac{ kJ }{ kg }\right)\left|\frac{1 kg \cdot m / s ^{2}}{1 N }\right|\left|\frac{10^{3} N \cdot m }{1 kJ }\right|}=624 m / s

1 Parts (b) and (c) can be solved alternatively using IT. These parts also can be solved using a constant c_{p} together with Eqs. 12.38 and 12.39. Inspection of Table A-20 shows that the specific heats of CO _{2} \text { and } O _{2} increase only slightly with temperature over the interval from 518 to 700 K, and so suitable constant values of c_{p} for the components and the overall mixture can be readily determined. These alternative solutions are left as exercises.

\Delta \bar{h}=\bar{c}_{p}\left(T_{2}-T_{1}\right), \quad \Delta \bar{h}_{i}=\bar{c}_{p, i}\left(T_{2}-T_{1}\right) (12.38)

\Delta \bar{s}=\bar{c}_{p} \ln \frac{T_{2}}{T_{1}}-\bar{R} \ln \frac{p_{2}}{p_{1}}, \quad \Delta \overline{s_{i}}=\bar{c}_{p, i} \ln \frac{T_{2}}{T_{1}}-\bar{R} \ln \frac{p_{2}}{p_{1}} (12.39)

2 Each component experiences an entropy change as it passes from inlet to exit. The increase in entropy of the oxygen and the decrease in entropy of the carbon dioxide are due to entropy transfer accompanying heat transfer from the CO _{2} to the O _{2} as they expand through the nozzle. However, as indicated by Eq. (a), there is no change in the entropy of the mixture as it expands through the nozzle.

3 Note the use of unit conversion factors in the calculation of V _{2}.

Skills Developed

Ability to…

• analyze the isentropic expansion of an ideal gas mixture flowing through a nozzle.

• apply ideal gas mixture principles together with mass and energy balances to calculate the exit velocity of a nozzle.

• determine the exit temperature for a given inlet state and a given exit pressure using tabular data and alternatively using IT.

Quick Quiz

What would be the exit velocity, in m/s, if the isentropic nozzle efficiency were 90%? Ans. 592 m/s.