Question

A vortex tube receives 0.3 m^3 /min of air at 600 kPa and 300 K. The discharge from the cold end of the tube is 0.6 kg/min at 245 K and 100 kPa. The discharge from the hot end is at 325 K and 100 kPa. Determine the irreversibility.

Answer 1

Answer:

Irreversibility = 5.361 kW

Explanation:

From the given information:

By applying ideal gas equation at entry:

PV =  mRT

600 × 0.3 = m × 0.287 × 300      (where R = 0.287 kJ/kg)

180 = m × 86.1

m = 180/86.1

m = 2.0905 kg/min

At the hot end, using the same ideal gas equation:

PV = mRT

100 × V = 1.4905 × 0.287 × 325

V = 139.026/100

V = 1.3903 m³/ min

This implies that: The total entropy change = Entropy of the universe

So,

$m\bigg [ c_p \ In \dfrac{T_2}{T_o}-R \ In \dfrac{P_2}{P_o} \bigg] + m_2 \bigg [ c_p \ In \dfrac{T_2}{T_o}- R \ In\dfrac{P_2}{P_o} \bigg]$

$= 0.6\bigg [ 1.004 \ In \dfrac{245}{300}-0.287 \ In \dfrac{100}{600} \bigg] +1.4905\bigg [1.004 \ In \dfrac{325}{300}- 0.287\ In\dfrac{100}{600} \bigg]$

= 0.6[-0.2033 + 0.5142] + 1.4905 [0.08036 + 0.5142]

= 1.0727 kJ/min.K

= 0.01787 kw/K

Irreversibility = $T_o [ \Delta S]$

Irreversibility = 300 × 0.01787

Irreversibility = 5.361 kW