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ENGR5003 / Week 1
L3 slide 18: Adiabatic steady flow, do the entropy results contradict?From lectureCurrent spec2:14 marks5 min

ENGR5003 Lecture 3, slide 18 (Entropy rate balance for open system)

For a steady-flow device with one inlet (1) and one exit (2), the entropy balance is ∑δmisi−∑δmjsj+∑δQTr+δSg=dS\sum \delta m_i s_i - \sum \delta m_j s_j + \sum \dfrac{\delta Q}{T_r} + \delta S_g = dS. For an ADIABATIC steady-flow system the slide shows two results side by side: s2−s1=sg≥0s_2 - s_1 = s_g \ge 0, and ΔS=0\Delta S = 0.

The lecturer asks: are they contradictory? Explain, and say what s2−s1≥0s_2 - s_1 \ge 0 means for a real adiabatic turbine or compressor.

Formulas you may need
  • Open-system entropy balance: ∑δmisi−∑δmjsj+∑δQ/Tr+δSg=dS\sum \delta m_i s_i - \sum \delta m_j s_j + \sum \delta Q/T_r + \delta S_g = dS (learn this; concepts only)
  • Steady flow, one inlet and one exit: s2−s1=sf+sgs_2 - s_1 = s_f + s_g (learn this)