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.

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

# 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:

\[\begin{split} \dot{m}_1 u_1 = \dot{m}_3 u_3 + \dot{m}_2 u_2 \\ \dot{m} c_p T_1 = f \dot{m} c_p T_3 + (1-f) \dot{m} c_p T_2 \\ T_3 = \frac{T_1 - (1-f) T_2}{f} \end{split}\]

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:

\[\begin{split} \dot{m} s_1 + \dot{S}_{\text{gen}} = f \dot{m} s_3 + (1-f) \dot{m} s_2 \\ \frac{\dot{S}_{\text{gen}}}{\dot{m}} = f s_3 + (1-f) s_2 = f(s_3 - s_2) + (s_2 - s_1) \;. \end{split}\]

We can obtain the \(\Delta s\) values by using the relationship for an ideal gas with constant specific heat:

(19)\[\begin{equation} \Delta s_{1-2} = c_p \ln \left(\frac{T_2}{T_1}\right) - R \ln \left( \frac{p_2}{p_1} \right) \end{equation}\]

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.