Question

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 \(\underline{\hspace{2cm}}\) kW/K (round off to one decimal place). 
 

Hint:

Entropy generation in adiabatic systems can be calculated using the mass flow rate and specific entropy changes for each stream.

Answer 1

Correct Answer: 2.1 - 2.3

Step 1: Verify Mass Balance

$$\dot{m}_1 = \dot{m}_2 + \dot{m}_3$$

$$5 = 3 + 2 = 5 \text{ kg/s}$$

Step 2: Verify Energy Balance

For an adiabatic device with no work transfer:

$$\dot{m}_1 h_1 = \dot{m}_2 h_2 + \dot{m}_3 h_3$$

For an ideal gas: $h = c_p T$

$$\dot{m}_1 c_p T_1 = \dot{m}_2 c_p T_2 + \dot{m}_3 c_p T_3$$

$$5 \times 300 = 3 \times 340 + 2 \times 240$$

$$1500 = 1020 + 480 = 1500 \text{ kJ/s} $$

Step 3: Calculate Entropy Change for Each Stream

For an ideal gas, the entropy change is:

$$\Delta s = c_p \ln\left(\frac{T_2}{T_1}\right) - R \ln\left(\frac{P_2}{P_1}\right)$$

For stream 2 (hot outlet):

$$s_2 - s_1 = 1005 \ln\left(\frac{340}{300}\right) - 287 \ln\left(\frac{100}{500}\right)$$

$$s_2 - s_1 = 1005 \ln(1.1333) - 287 \ln(0.2)$$

$$s_2 - s_1 = 1005 \times 0.1252 - 287 \times (-1.6094)$$

$$s_2 - s_1 = 125.83 + 461.90 = 587.73 \text{ J/kgK}$$

For stream 3 (cold outlet):

$$s_3 - s_1 = 1005 \ln\left(\frac{240}{300}\right) - 287 \ln\left(\frac{100}{500}\right)$$

$$s_3 - s_1 = 1005 \ln(0.8) - 287 \ln(0.2)$$

$$s_3 - s_1 = 1005 \times (-0.2231) - 287 \times (-1.6094)$$

$$s_3 - s_1 = -224.22 + 461.90 = 237.68 \text{ J/kgK}$$

Step 4: Calculate Total Rate of Entropy Generation

Using the entropy balance for the control volume:

$$\dot{S}_{gen} = \dot{m}_2(s_2 - s_1) + \dot{m}_3(s_3 - s_1)$$

$$\dot{S}_{gen} = 3 \times 587.73 + 2 \times 237.68$$

$$\dot{S}_{gen} = 1763.19 + 475.36 = 2238.55 \text{ W/K}$$

$$\dot{S}_{gen} = 2.239 \text{ kW/K}$$

Answer: The rate of entropy generation is 2.2 kW/K (rounded to one decimal place).