Analyzing Adiabatic Mixing of Two Moist Air Streams
A stream consisting of 142 m ^{3} / min of moist air at a temperature of 5°C and a humidity ratio of 0.002 kg(vapor)/kg(dry air) is mixed adiabatically with a second stream consisting of 425 m ^{3} / min of moist air at 24°C and 50% relative humidity. The pressure is constant throughout at 1 bar. Determine (a) the humidity ratio and (b) the temperature of the exiting mixed stream, in °C.
Known A moist air stream at 5°C, ω = 0.002 kg(vapor)/kg(dry air), and a volumetric flow rate of 142 m ^{3} / min is mixed adiabatically with a stream consisting of 425 m ^{3} / min of moist air at 24°C and \phi=50 \%.
Find Determine the humidity ratio and the temperature, in °C, of the mixed stream exiting the control volume.
Schematic and Given Data:
Engineering Model
1. The control volume shown in the accompanying figure operates at steady state. Changes in kinetic and potential energy can be neglected and \dot{W}_{ cv }=0.
2. There is no heat transfer with the surroundings.
3. The pressure remains constant throughout at 1 bar.
4. The moist air streams are regarded as ideal gas mixtures adhering to the Dalton model.
Analysis
a. The humidity ratio \omega_{3} can be found by means of mass rate balances for the dry air and water vapor, respectively,
\dot{m}_{ a 1}+\dot{m}_{ a 2}=\dot{m}_{ a 3} (dry air)
\dot{m}_{ v 1}+\dot{m}_{ v 2}=\dot{m}_{ v 3} (water vapor)
With \dot{m}_{ v 1}=\omega_{1} \dot{m}_{ a 1}, \dot{m}_{ v 2}=\omega_{2} \dot{m}_{ a 2}, \text { and } \dot{m}_{ v 3}=\omega_{3} \dot{m}_{ a 3}, the second of these balances becomes (Eq. 12.56b)
\omega_{1} \dot{m}_{ a 1}+\omega_{2} \dot{m}_{ a 2}=\omega_{3} \dot{m}_{ a 3} (water vapor) (12.56b)
\omega_{1} \dot{m}_{ a 1}+\omega_{2} \dot{m}_{ a 2}=\omega_{3} \dot{m}_{ a 3}
Solving
\omega_{3}=\frac{\omega_{1} \dot{m}_{ a 1}+\omega_{2} \dot{m}_{ a 2}}{\dot{m}_{ a 3}}
Since \dot{m}_{ a 3}=\dot{m}_{ a 1}+\dot{m}_{ a 2}, this can be expressed as
\omega_{3}=\frac{\omega_{1} \dot{m}_{ a 1}+\omega_{2} \dot{m}_{ a 2}}{\dot{m}_{ a 1}+\dot{m}_{ a 2}}
To determine \omega_{3} requires values for \omega_{3}, \dot{m}_{ a 1}, \text { and } \dot{m}_{ a 2}. The mass flow rates of the dry air, \dot{m}_{ a 1} \text { and } \dot{m}_{ a 2}, can be found as in previous examples using the given volumetric flow rates
\dot{m}_{ a 1}=\frac{( AV )_{1}}{v_{ a 1}}, \quad \dot{m}_{ a 2}=\frac{( AV )_{2}}{v_{ a 2}}
The values of v_{ a 1}, v_{ a 2}, \text { and } \omega_{2} are readily found from the psychrometric chart, Fig. A-9. Thus, at \omega_{1}=0.002 \text { and } T_{1}=5°C, v_{ a 1}=0.79 m ^{3} / kg \text { (dry air). At } \phi_{2}=50 \% \text { and } T_{2}=24^{\circ} C, v_{ a 2}=0.855 m ^{3} / kg (\text { dry air }) \text { and } \omega_{2}=0.0094. The mass flow rates of the dry air are then \dot{m}_{ a 1}=180 kg(dry air)/min and\dot{m}_{ a 2}=497 kg(dry air)/min. Inserting values into the expression for \omega_{3}
\omega_{3}=\frac{(0.002)(180)+(0.0094)(497)}{180+497}=0.0074 \frac{ kg (\text { vapor })}{ kg (\text { dry air })}
b. The temperature T_{3} of the exiting mixed stream can be found from an energy rate balance. Reduction of the energy rate balance using assumptions 1 and 2 gives (Eq. 12.56c)
\dot{m}_{ a 1}\left(h_{ a 1}+\omega_{1} h_{ g 1}\right)+\dot{m}_{ a 2}\left(h_{ a 2}+\omega_{2} h_{ g 2}\right)=\dot{m}_{ a 3}\left(h_{ a 3}+\omega_{3} h_{ g 3}\right) (12.56c)
\dot{m}_{ a 1}\left(h_{ a }+\omega h_{ g }\right)_{1}+\dot{m}_{ a 2}\left(h_{ a }+\omega h_{ g }\right)_{2}=\dot{m}_{ a 3}\left(h_{ a }+\omega h_{ g }\right)_{3} (a)
Solving
\left(h_{ a }+\omega h_{ g }\right)_{3}=\frac{\dot{m}_{ a 1}\left(h_{ a }+\omega h_{ g }\right)_{1}+\dot{m}_{ a 2}\left(h_{ a }+\omega h_{ g }\right)_{2}}{\dot{m}_{ a 1}+\dot{m}_{ a 2}} (b)
With \left(h_{ a }+\omega h_{ g }\right)_{1}=10 kJ/kg(dry air) and \left(h_{ a }+\omega h_{ g }\right)_{2}=47.8 kJ/kg (dry air) from Fig. A-9 and other known values
\left(h_{ a }+\omega h_{ g }\right)_{3}=\frac{180(10)+497(47.8)}{180+497}=37.7 \frac{ kJ }{ kg ( dry \text { air })}
1 This value for the enthalpy of the moist air at the exit, together with the previously determined value for \omega_{3}, fixes the state of the exiting moist air. From inspection of Fig. A-9, T_{3}=19^{\circ} C.
Alternative Solutions The use of the psychrometric chart facilitates the solution for T_{3}. Without the chart, an iterative solution of Eq. (b) using data from Tables A-2 and A-22 could be used. Alternatively, T_{3} can be determined using the following IT program, where \phi_{2} is denoted as phi2, the volumetric flow rates at 1 and 2 are denoted as AV1 and AV2, respectively, and so on.
Using the Solve button, the result is T_{3}=19.01^{\circ} C \text { and } \omega_{3}=0.00745 kg(vapor)/kg(dry air), which agree with the psychrometric chart solution.
1 A solution using the geometric approach based on Eq. 12.57 is left as an exercise.
\frac{\dot{m}_{ a 1}}{\dot{m}_{ a 2}}=\frac{\omega_{3}-\omega_{2}}{\omega_{1}-\omega_{3}}=\frac{\left(h_{ a 3}+\omega_{3} h_{ g 3}\right)-\left(h_{ a 2}+\omega_{2} h_{ g 2}\right)}{\left(h_{ a 1}+\omega_{1} h_{ g 1}\right)-\left(h_{ a 3}+\omega_{3} h_{ g 3}\right)} (12.57)
2 Note the use here of special Moist Air functions listed in the Properties menu of IT.
Skills Developed
Ability to…
• apply psychrometric terminology and principles.
• apply mass and energy balances for an adiabatic mixing process of two moist air streams in a control volume at steady state.
• retrieve property data for moist air using the psychrometric chart.
• apply IT for psychrometric analysis.
Quick Quiz
Using the psychrometric chart, what is the relative humidity at the exit? Ans. ≈53%.
2 // Given data
T1 = 5 // °C
w1 = 0.002 // kg(vapor) / kg(dry air)
AV1 = 142 // m3/min
T2 = 24 // °C
phi2 = 0.5
AV2 = 425 // m3/min
p = 1 // bar
// Mass balances for water vapor and dry air:
w1 * mdota1 + w2 * mdota2 = w3 * mdota3
mdota1 + mdota2 = mdota3
// Evaluate mass flow rates of dry air
mdota1 = AV1 / va1
va1 = va_Tw(T1, w1, p)
mdota2 = AV2 / va2
va2 = va_Tphi(T2, phi2, p)
// Determine w2
w2 = w_Tphi(T2, phi2, p)
// The energy balance, Eq. (a), reads
mdota1 * h1 + mdota2 * h2 = mdota3 * h3
h1 = ha_Tw(T1, w1)
h2 = ha_Tphi(T2, phi2, p)
h3 = ha_Tw(T3, w3)