Analyzing Adiabatic Mixing of Two Streams
At steady state, 100 m ^{3} / min of dry air at 32°C and 1 bar is mixed adiabatically with a stream of oxygen \left( O _{2}\right) at 127°C and 1 bar to form a mixed stream at 47°C and 1 bar. Kinetic and potential energy effects can be ignored. Determine (a) the mass flow rates of the dry air and oxygen, in kg/min, (b) the mole fractions of the dry air and oxygen in the exiting mixture, and (c) the time rate of entropy production, in kJ/K ⋅ min.
Known At steady state, 100 m ^{3} / min of dry air at 32°C and 1 bar is mixed adiabatically with an oxygen stream at 127°C and 1 bar to form a mixed stream at 47°C and 1 bar.
Find Determine the mass flow rates of the dry air and oxygen, in kg/min, the mole fractions of the dry air and oxygen in the exiting mixture, and the time rate of entropy production, in kJ/K ⋅ min.
Schematic and Given Data:
Engineering Model
1. The control volume identified by the dashed line on the accompanying figure operates at steady state.
2. No heat transfer occurs with the surroundings.
3. Kinetic and potential energy effects can be ignored, and \dot{W}_{ cv }=0.
4. The entering gases can be regarded as ideal gases. The exiting mixture can be regarded as an ideal gas mixture adhering to the Dalton model.
5. The dry air is treated as a pure component.
Analysis
a. The mass flow rate of the dry air entering the control volume can be determined from the given volumetric flow rate ( AV )_{1}
\dot{m}_{ a 1}=\frac{( AV )_{1}}{v_{ a 1}}
where v_{ a 1} is the specific volume of the air at 1. Using the ideal gas equation of state
v_{ a 1}=\frac{\left(\bar{R} / M_{ a }\right) T_{1}}{p_{1}}=\frac{\left(\frac{8314 N \cdot m }{28.97 kg \cdot K }\right)(305 K )}{10^{5} N / m ^{2}}=0.875 \frac{ m ^{3}}{ kg }
The mass flow rate of the dry air is then
\dot{m}_{ a 1}=\frac{100 m ^{3} / min }{0.875 m ^{3} / kg }=114.29 \frac{ kg }{ min }
The mass flow rate of the oxygen can be determined using mass and energy rate balances. At steady state, the amounts of dry air and oxygen contained within the control volume do not vary. Thus, for each component individually it is necessary for the incoming and outgoing mass flow rates to be equal. That is,
\dot{m}_{ a 1}=\dot{m}_{ a 3} (dry air)
\dot{m}_{o2}=\dot{m}_{o3} (oxygen)
Using assumptions 1–3 together with the foregoing mass flow rate relations, the energy rate balance reduces to
0=\dot{m}_{ a } h_{ a }\left(T_{1}\right)+\dot{m}_{o} h_{o}\left(T_{2}\right)-\left[\dot{m}_{ a } h_{ a }\left(T_{3}\right)+\dot{m}_{o} h_{o}\left(T_{3}\right)\right]
where \dot{m}_{ a } \text { and } \dot{m}_{ o } denote the mass flow rates of the dry air and oxygen, respectively. The enthalpy of the mixture at the exit is evaluated by summing the contributions of the air and oxygen, each at the mixture temperature. Solving for \dot{m}_{ o }
\dot{m}_{o}=\dot{m}_{ a }\left[\frac{h_{ a }\left(T_{3}\right)-h_{ a }\left(T_{1}\right)}{h_{ o }\left(T_{2}\right)-h_{ o }\left(T_{3}\right)}\right]
The specific enthalpies can be obtained from Tables A-22 and A-23. Since Table A-23 gives enthalpy values on a molar basis, the molecular weight of oxygen is introduced into the denominator to convert the molar enthalpy values to a mass basis
\dot{m}_{o}=\frac{(114.29 kg / min )(320.29 kJ / kg -305.22 kJ / kg )}{\left(\frac{1}{32 kg / kmol }\right)(11,711 kJ / kmol -9,325 kJ / kmol )}
=23.1 \frac{ kg }{\min }
b. To obtain the mole fractions of the dry air and oxygen in the exiting mixture, first convert the mass flow rates to molar flow rates using the respective molecular weights
\dot{n}_{ a }=\frac{\dot{m}_{ a }}{M_{ a }}=\frac{114.29 kg / min }{28.97 kg / kmol }=3.95 kmol / min
\dot{n}_{o}=\frac{\dot{m}_{o}}{M_{o}}=\frac{23.1 kg / min }{32 kg / kmol }=0.72 kmol / min
where \dot{n} denotes molar flow rate. The molar flow rate of the mixture \dot{n} is the sum
\dot{n}=\dot{n}_{ a }+\dot{n}_{ o }=3.95+0.72=4.67 kmol / min
The mole fractions of the air and oxygen in the exiting mixture are, respectively,
1 y_{ a }=\frac{\dot{n}_{ a }}{\dot{n}}=\frac{3.95}{4.67}=0.846 \quad \text { and } \quad y_{o}=\frac{\dot{n}_{ o }}{\dot{n}}=\frac{0.72}{4.67}=0.154
c. For the control volume at steady state, the entropy rate balance reduces to
\begin{aligned}0=& \dot{m}_{ a } s_{ a }\left(T_{1}, p_{1}\right)+\dot{m}_{ o } s_{o}\left(T_{2}, p_{2}\right)-\left[\dot{m}_{ a } s_{ a }\left(T_{3}, y_{ a } p_{3}\right)\right.\\&\left.+\dot{m}_{o} s_{o}\left(T_{3}, y_{o} p_{3}\right)\right]+\dot{\sigma}\end{aligned}
The specific entropy of each component in the exiting ideal gas mixture is evaluated at its partial pressure in the mixture and at the mixture temperature. Solving for \dot{\sigma}
\dot{\sigma}=\dot{m}_{ a }\left[s_{ a }\left(T_{3}, y_{ a } p_{3}\right)-s_{ a }\left(T_{1}, p_{1}\right)\right]+\dot{m}_{ o }\left[s_{o}\left(T_{3}, y_{ o } p_{3}\right)-s_{o}\left(T_{2}, p_{2}\right)\right]
Since p_{1}=p_{3}, the specific entropy change of the dry air is
s_{ a }\left(T_{3}, y_{ a } p_{3}\right)-s_{ a }\left(T_{1}, p_{1}\right)=s_{ a }^{\circ}\left(T_{3}\right)-s_{ a }^{\circ}\left(T_{1}\right)-\frac{\bar{R}}{M_{ a }} \ln \frac{y_{ a } p_{3}}{p_{1}}
=s_{ a }^{\circ}\left(T_{3}\right)-s_{ a }^{\circ}\left(T_{1}\right)-\frac{\bar{R}}{M_{ a }} \ln y_{ a }
The s _{ a }^{\circ} terms are evaluated from Table A-22. Similarly, since p_{2}=p_{3}, the specific entropy change of the oxygen is
s_{o}\left(T_{3}, y_{o} p_{3}\right)-s_{o}\left(T_{2}, p_{2}\right)=\frac{1}{M_{o}}\left[\bar{s}_{o}^{\circ}\left(T_{3}\right)-\bar{s}_{o}^{\circ}\left(T_{2}\right)-\bar{R} \ln y_{o}\right]
The \bar{s}_{ o }^{\circ} terms are evaluated from Table A-23. Note the use of the molecular weights M_{ a } \text { and } M_{ o } in the last two equations to obtain the respective entropy changes on a mass basis.
The expression for the rate of entropy production becomes
\dot{\sigma}=\dot{m}_{ a }\left[s_{ a }^{\circ}\left(T_{3}\right)-s_{ a }^{\circ}\left(T_{1}\right)-\frac{\bar{R}}{M_{ a }} \ln y_{ a }\right]
+\frac{\dot{m}_{ o }}{M_{ o }}\left[\bar{s}_{ o }^{\circ}\left(T_{3}\right)-\bar{s}_{o}^{\circ}\left(T_{2}\right)-\bar{R} \ln y_{ o }\right]
Substituting values
\dot{\sigma}=\left(114.29 \frac{ kg }{\min }\right)\left[1.7669 \frac{ kJ }{ kg \cdot K }-1.71865 \frac{ k J }{ kg \cdot K }\right.
\left.-\left(\frac{8.314}{28.97} \frac{ kJ }{ kg \cdot K }\right) \ln 0.846\right]
+\left(\frac{23.1 kg / min }{32 kg / kmol }\right)\left[207.112 \frac{ kJ }{ kmol \cdot K }-213.765 \frac{ kJ }{ kmol \cdot K }\right.
\left.-\left(8.314 \frac{ kJ }{ kmol \cdot K }\right) \ln 0.154\right]
2 =17.42 \frac{ kJ }{ K \cdot \min }
1 This calculation is based on dry air modeled as a pure component (assumption 5). However, since O _{2} is a component of dry air (Table 12.1), the actual mole fraction of O _{2} in the exiting mixture is greater than given here.
2 Entropy is produced when different gases, initially at different temperatures, are allowed to mix.
Skills Developed
Ability to…
• analyze the adiabatic mixing of two ideal gas streams at steady state.
• apply ideal mixture principles together with mass, energy, and entropy rate balances.
Quick Quiz
What are the mass fractions of air and oxygen in the exiting mixture? Ans. m f_{\text {air }}=0.832, m f_{O_{2}}=0.168.
| TABLE 12.1 Approximate Composition of Dry Air | |
| Component | Mole Fraction (%) |
| Nitrogen | 78.08 |
| Oxygen | 20.95 |
| Argon | 0.93 |
| Carbon dioxide | 0.03 |
| Neon, helium, methane, and others | 0.01 |
