1. Consider a closed system or control mass (CM), an open system (OS or CV, cont
ID: 1766551 • Letter: 1
Question
1. Consider a closed system or control mass (CM), an open system (OS or CV, control volume), or an isolated system (IS). Note that any system may be operated in a steady state (SS), a cycle, a process, or a transient condition (transient). For the conditions or interactions given, indicate under "Finding" if the system and operation is always possible (OK), impossible (IMP), or possible only under some restricted condition such as reversibility (make a note); and identify the restriction. The systems may only exchange heat with one high temperature HT-TER at constant and uniform TH or one low temperature LT-TER at constant and uniform TL gnore KE and PE. Stay aware of both energy and entropy considerations Heat Work Work Sen in Usystem Finding em B Isolated System C CM in a process positive none none Dositive ositive ositive ositive sitive increasin zero ECM for a cycle none positive positive none ositive zero ositive none zero none ositive zeroExplanation / Answer
A - Finding : impossible (IMP)
Reason : for isolated neither mass nor energy (heat) crosses system boundary
B - Finding :
ds (system)= cyclic integral of (del Q/ T)(Entropy tranfer) + S generation
if entropy transfer is negative and equal to entropy generation then condition valid. It means change in entropy of system is zero
Entropy change increases (irreversibility)
Case Finding Reason A impossible (IMP) for an isolated neither mass nor energy (heat) crosses system boundary B possible (OK) for an isolated system (reversible process) (ds = 0) and internal energy of an isolated system is constant so change in internal energy (du = 0) C possible (OK) (E in - E out ) net energy transfer by work and heat = (delta E) system D impossible (IMP) cyclic integral of (del Q) = cyclic integral of (del W) for contolled mass E impossible (IMP) controlled mass(system) for a cycle exchanges heat with only one thermal reservior which violates kelvin-plancks statement(second law of thermodynamics) F impossilbe (IMP) ds (system)= cyclic integral of (del Q/ T) + S generation G Possible (OK)ds (system)= cyclic integral of (del Q/ T)(Entropy tranfer) + S generation
if entropy transfer is negative and equal to entropy generation then condition valid. It means change in entropy of system is zero
H impossilbe (IMP) del Q = du + del W and S generation can never be negative I Impossible (IMP)Entropy change increases (irreversibility)
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