# Vortex tube

A vortex tube takes in high-pressure air at 650 kPa and 305 K, and splits it into two streams at a lower pressure, 100 kPa: one at a higher temperature of 325 K and one at a lower temperature. The fraction of mass entering that leaves at the cold outlet is $f = 0.25$. The vortex tube operates continuously at steady state, is adiabatic, and performs/experiences no work. Air should be modeled as an ideal gas with constant specific heat: $R = 287$ J/kg⋅K and $c_p = 1004$ J/kg⋅K.

<figure>
  <center>
  
  <figcaption>Figure: Vortex tube</figcaption>
  </center>
</figure>

**Problem:** Determine the temperature at the cold end. Then, determine whether this device is physically possible.


```python
# Enter the known quantities
import numpy as np
from pint import UnitRegistry
ureg = UnitRegistry()
Q_ = ureg.Quantity

gas_constant = Q_(287, 'J/(kg K)')
cp = Q_(1004, 'J/(kg K)')

temp_1 = Q_(305, 'K')
pres_1 = Q_(650, 'kPa')

temp_2 = Q_(325, 'K')
pres_2 = Q_(100, 'kPa')

f = 0.25
pres_3 = Q_(100, 'kPa')
```

First, we can find the temperature at the cold outlet by performing an energy balance on the device:
\begin{align}
\dot{m}_1 u_1 &= \dot{m}_3 u_3 + \dot{m}_2 u_2 \\
\dot{m} c_p T_1 &= f \dot{m} c_p T_3 + (1-f) \dot{m} c_p T_2 \\
T_3 &= \frac{T_1 - (1-f) T_2}{f}
\end{align}


```python
temp_3 = (temp_1 - (1-f)*temp_2) / f
print(f'Temperature at cold outlet: {temp_3: .2f}')
```

    Temperature at cold outlet: 245.00 kelvin


Now, examine whether the device is physically possible by performing an entropy balance:
\begin{align}
\dot{m} s_1 + \dot{S}_{\text{gen}} &= f \dot{m} s_3 + (1-f) \dot{m} s_2 \\
\frac{\dot{S}_{\text{gen}}}{\dot{m}} &= f s_3 + (1-f) s_2 = f(s_3 - s_2) + (s_2 - s_1)
\end{align}

We can obtain the $\Delta s$ values by using the relationship for an ideal gas with constant specific heat:
\begin{equation}
\Delta s_{1-2} = c_p \ln \left(\frac{T_2}{T_1}\right) - R \ln \left( \frac{p_2}{p_1} \right)
\end{equation}


```python
delta_s_12 = (
    cp * np.log(temp_2/temp_1) - 
    gas_constant * np.log(pres_2/pres_1)
    )
delta_s_23 = (
    cp * np.log(temp_3/temp_2) - 
    gas_constant * np.log(pres_3/pres_2)
    )

entropy_gen = f * delta_s_23 + delta_s_12
print(f'Entropy generation rate: {entropy_gen: .2f}')
```

    Entropy generation rate: 530.05 joule / kelvin / kilogram


Since the rate of entropy generation is positive, this device can operate as described.

bb0d695e79ae18faf009b4ca92ea1be32656b501
ipynb
Jupyter Notebook
content/second-law/vortex-tube.ipynb
msb002/computational-thermo
9302288217a36e0ce29e320688a3f574921909a5
CC-BY-4.0 BSD-3-Clause
content/second-law/vortex-tube.ipynb
msb002/computational-thermo
9302288217a36e0ce29e320688a3f574921909a5
CC-BY-4.0 BSD-3-Clause
content/second-law/vortex-tube.ipynb
msb002/computational-thermo
9302288217a36e0ce29e320688a3f574921909a5
CC-BY-4.0 BSD-3-Clause
Qwen/Qwen-72B
1. YES
2. YES