Unformatted text preview: + m1h1 + m2 h2 = m3h3
kJ h1 ≅ h f @ T1 = 104.83 P = 0.1 MPa kg 1 kJ T1 = 25o C s1 ≅ s f @ T1 = 0.3672 kg ⋅ K kJ h2 = 2675.8 P2 = 0.1 MPa kg kJ T2 = 100o C s2 = 7.3611 kg ⋅ K kJ h3 = 417.51 P3 = 0.1 MPa kg kJ Sat. liquid s3 = 1.3028 kg ⋅ K Chapter 7  78 Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles 779 Qnet = m3 h3 − m1h1 − m2 h2 kg kJ kg kJ kg kJ 417.51 − 10 104.83 − 0.5 2675.8 s kg s kg s kg kJ = +1997.7 s = 10.5
So, 1996.33 kJ/s of heat energy must be transferred from the surroundings to this mixing process, or Qnet , surr = −Qnet , CV . For the process to be possible, the second law must be satisfied. Write the second law for the isolated system, Qk ∑ T + ∑ mi si − ∑ me se + S gen = ∆SCV k
For steadyflow
∆SCV = 0 . Solving for entropy generation, we have S gen = ∑ me se − ∑ mi si − ∑ = m3 s3 − m1s1 − m2 s2 − = 10.5 Qk Tk Qcv Tsurr kg kJ kg kJ − 10 0.3672 1.3028 s kg ⋅ K s kg ⋅ K kg kJ 1997.7 kJ / s − 0.5 7.3611 − s kg ⋅ K (20 + 273) K kJ = −0.491 K ⋅s
Chapter 7  79 Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles 780 Since S gen must be ≥ 0 to satisfy the second law, this process is impossible, and the inventor's claim is false. To find the minimum value of the surrounding temperature to make this mixing process possible, set S gen = 0 and solve for Tsurr. S gen = ∑ me se − ∑ mi si − ∑ Tsurr = = Qcv m3 s3 − m1s1 − m2 s2 Qk =0 Tk 1997.7 kJ / s kg kJ kg kJ kg kJ − 10 0.3672 − 0.5 7.3611 10.5 1.3028 s kg ⋅ K s kg ⋅ K s kg ⋅ K = 315.75K
One way to think about this process is as follows: Heat is transferred from the surroundings at 315.75 K (42.75oC) in the amount of 1997.7 kJ/s to increase the water temperature to approximately 42.75oC before the water is mixed with the superheated steam. Recall that the surroundings must be at a temperature greater than the water for the heat transfer to take place from the surroundings to the water. Chapter 7  80 Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles 781 Answer to Example 74 Find the entropy and/or temperature of steam at the following states: P 5 MPa 1 MPa T 120oC 50oC Region Compressed Liquid and in the table Compressed liquid but not in the table Superheated Quality, x = 0.9 Saturated mixture sf<s<sg at P Saturated mixture X = (ssf)/sfg = 0.9262 s kJ/kg⋅K 1.5233 s = sf at 50oC = 0.7038 7.1794 s = sf + x sfg = 7.0056 7.1794 1.8 400oC MPa 40 kPa T=Tsat =75.87o C 40 kPa T=Tsat =75.87o C Chapter 7  81...
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This note was uploaded on 09/17/2009 for the course MAE 301 taught by Professor Hassan during the Fall '08 term at N.C. State.
 Fall '08
 Hassan

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