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Thermodynamics
Entropy & Second Law
GATE 2021
Hard

An adiabatic vortex tube 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).

2.24 kW/K(± 0.15 accepted)

Explanation

Relative to inlet state 1: s2 = cp·ln(340/300) − R·ln(100/500) ≈ 125.8+461.9 = 587.7 J/kg.K. s3 = cp·ln(240/300) − R·ln(100/500) ≈ −224.3+461.9 = 237.7 J/kg.K. Ṡgen = m2·s2 + m3·s3 − m1·s1 = 3×587.7 + 2×237.7 − 0 ≈ 2238 W/K ≈ 2.24 kW/K.

Source: GATE ME 2021 Session 2 (ME-2) (2021)

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