An adiabatic vortex tube, shown in the figure given below is supplied with 5 kg/s of air (inlet 1) at 500 kPa and 300 K. Two separate streams of air are leaving the device from outlets 2 and 3. Hot air leaves the device at a rate of 3 kg/s from outlet 2 at 100 kPa and 340 K, and 2 kg/s of cold air stream is leaving the device from outlet 3 at 100 kPa and 240 K.

Assume constant specific heat of air is 1005 J/kg.K and gas constant is 287 J/kg.K. There is no work transfer across the boundary of this device. The rate of entropy generation is ______________kW/K (round off to one decimal place).

The correct answer is 2.24.

1. Analysis of the Given System:
From the provided schematic images of the adiabatic vortex tube:
• Inlet 1 (High pressure air inlet): mass flow rate $m_1=5\text{kg/s}$, pressure $P_1=500\text{kPa}=5\text{bar}$, temperature $T_1=300\text{K}$.
• Outlet 2 (Low pressure hot air outlet): mass flow rate $m_2=3\text{kg/s}$, pressure $P_2=100\text{kPa}=1\text{bar}$, temperature $T_2=340\text{K}$.
• Outlet 3 (Low pressure cold air outlet): mass flow rate $m_3=2\text{kg/s}$, pressure $P_3=100\text{kPa}=1\text{bar}$, temperature $T_3=240\text{K}$.
• Specific heat at constant pressure: $C_p=1005\text{J/(kg}\cdot \text{K)}=1.005\text{kJ/(kg}\cdot \text{K)}$.
• Gas constant: $R=287\text{J/(kg}\cdot \text{K)}=0.287\text{kJ/(kg}\cdot \text{K)}$.

2. Governing Thermodynamic Equations:
For a steady-state, open, and adiabatic system with no work interactions:

$\frac{\partial S_{\text{cv}}}{\partial t}=\sum m_i{s}_i-\sum m_e{s}_e+\frac{\dot{Q}}{T}+{\dot{S}}_{\text{gen}}$

Since the flow is steady ($\frac{\partial S_{\text{cv}}}{\partial t}=0$) and the device is adiabatic ($\dot{Q}=0$), the entropy generation rate simplifies to:

${\dot{S}}_{\text{gen}}=m_2{s}_2+m_3{s}_3-m_1{s}_1$

By conservation of mass, we have $m_1=m_2+m_3$. Substituting this yields:

${\dot{S}}_{\text{gen}}=m_2(s_2-s_1)+m_3(s_3-s_1)$

The change in specific entropy for an ideal gas with constant specific heats is calculated using:

$\Delta s=C_p\ln \left(\frac{T_f}{T_i}\right)-R\ln \left(\frac{P_f}{P_i}\right)$

3. Step-by-Step Calculations:
Let us evaluate the specific entropy changes for both exit streams relative to the inlet state:
• For the hot stream (outlet 2):

$s_2-s_1=1.005\ln \left(\frac{340}{300}\right)-0.287\ln \left(\frac{100}{500}\right)$

$s_2-s_1=1.005\left(0.12516\right)-0.287(-1.60944)$

$s_2-s_1=0.12579+0.46191=0.5877\text{kJ/(kg}\cdot \text{K)}$

• For the cold stream (outlet 3):

$s_3-s_1=1.005\ln \left(\frac{240}{300}\right)-0.287\ln \left(\frac{100}{500}\right)$

$s_3-s_1=1.005(-0.22314)-0.287(-1.60944)$

$s_3-s_1=-0.22426+0.46191=0.23765\text{kJ/(kg}\cdot \text{K)}$

4. Rate of Entropy Generation:
Multiply by the respective mass flow rates:

${\dot{S}}_{\text{gen}}=3\left(0.5877\right)+2\left(0.23765\right)$

${\dot{S}}_{\text{gen}}=1.7631+0.4753=2.2384\text{kW/K}$

Rounding to two decimal places gives 2.24 kW/K.