COURSE: Detailed Explanation on Entropy Questions Lesson : Questions -(4-6
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Question 4 In a Carnot cycle, heat is supplied at 350 C and rejected at 27 C. The working fluid is water which, while receiving heat, evaporates from liquid at 350 C to steam at 350 C. The associated entropy change is 1.44 kJ/kg K. (a) If the cycle operates on a stationary mass of 1 kg of water, how much is the work done per cycle, and how much is the heat supplied? (b) If the cycle operates in steady flow with a power output of 20 kW, what is the steam flow rate?
Solution 4 If heat required for evaporation is Q kJ/kg then Q-1.44 (350 273) or Q = 897.12 kJ/kg It is a Carnot eycle so 1 - 1-301 272 (273 + 27) (350 + 273) :. W n.Q 465.12 kJ 20 b) P mW or m w W 465.12 kg/s -0.043 kg/s
Question 5 A heat engine receives reversibly 420 kJ/cycle of heat from a source at 327 C, and rejects heat reversibly to a sink at 27 C. There are no other heat transfers. For each of the three hypothetical amounts of heat rejected, in (a), (b), and (c) below, compute the cyclic integral of dQIT from these results show which case is irreversible, which reversible, and which impossible: (a) 210 kJ/cycle rejected (b) 105 kJ/cycle rejected (c) 315 kJ/cycle rejected
Solution 5 +420 210 T (327 +273) (27+273) .-. Cycle is Reversible, Possible dQ_420 105_ 600 300 . Cycle is Impossible 420 315 -0.35 600 300 .. Cycle is irreversible but possible.
Question 6 Water is heated at a constant pressure of 0.7 MPa. The boiling point is 164.97 C. The initial temperature of water is 0 C. The latent heat of evaporation is 2066.3 kJ/kg. Find the increase of entropy of water, if the final state is steam
Solution 6 (AS)Water 437.97 dT = [ 1 4187 T 273 4.187 In (437.97) = 1.979 kJ/K (AS)Eva pour p 700 kPa T 437.97 K 273 273 1 x 2066.3 kJ/ K 437.97 4.7179 kJ/K (As) system 6.697 kJ/kg - K
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