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Course: ME 3322, Summer 2011
School: Georgia Tech
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Word Count: 480

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W. George Woodruff School of Mechanical Engineering Quiz #3 Summer 2011 ME3322 NAME ____________Key_______________ _______________________________________________________________________________________ 1) Fill-in the blanks in the following statements regarding the Clausius inequality: Q a. The cyclic integral is always negative. (1) T b b. If the boundary temperature, Tb, is constant over the entire...

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W. George Woodruff School of Mechanical Engineering Quiz #3 Summer 2011 ME3322 NAME ____________Key_______________ _______________________________________________________________________________________ 1) Fill-in the blanks in the following statements regarding the Clausius inequality: Q a. The cyclic integral is always negative. (1) T b b. If the boundary temperature, Tb, is constant over the entire cycle then the Clausius inequality simply states that Q0. (1) 2) The temperature vs. entropy (T-s) diagram of an internally reversible power cycle is shown below. a. Find the net heat input during the cycle. (3) s3 s3 Q= Tds Tds =Area of Triangle 123 s2 2 700 s1 Q=4002/2=400 KJ. Thus 400 KJ of Heat is added. b. Find the net work developed during the cycle. (2) U=Q-W=0W=Q= 400 KJ 300 1 1 3 S (KJ/oK) 3 3) Consider a closed thermal system going from the initial state 1 to final state 2. a. Suppose that the process (12) is adiabatic (no heat transfer). What can you say about the entropy change S2-S1? Why? (2) By the 2nd law of thermodynamics for closed systems, S2-S1=(Q/T)b+=0 b. Is it possible to have S2<S1 if heat transfer were allowed? If no explain why and if yes explain how. (2) Yes. If heat is removed, the entropy could go down. c. Suppose that the process (12) is isothermal expansion of an ideal gas. What can you say about the entropy change S2-S1? Why? (2) S2-S1=-m ln(p2/p1)=-mln(v1/v2)>0 4) Consider a single channel open (control volume) thermal system operating at steady-state: h2,s2 h1,s1 system The specific enthalpies and entropies of the inlet and outlet given are by h1=2500 KJ/Kg, h2=1500 KJ/Kg, s1=2 KJ/Kg oK and s2=1 KJ/ Kg oK. The mass flowrate of the fluid through the control volume is dm/dt=10 Kg/sec. a. Is it possible for this system to be adiabatic (perfectly insulated)? Why? (2) No because it implies entropy destruction. To show this more precisely, consider the entropy flow Q equation: j + m( si se ) + cv = 0 . If Qj=0, then m(2 1) + cv = m + cv = 0 , which is j Tj b impossible since m > 0 and cv 0 . ME3322 Quiz #3 Summer 2011 NAME ____________________________ _______________________________________________________________________________________ b. Find the maximum amount of power (dWCV/dt) that this system can develop if it has a uniform boundary temperature of Tb=500 oK. (5) Wcv Qcv Qcv Wcv + (hi he ) = 0 = + 1000 m m (1) Energy Balance: m Entropy Balance: Q Qcv + ( si se ) + cv = 0 cv = 500(1 + cv ) (2) m mTb Wcv = 500(1 cv ) . Since cv 0 , the maximum work output is m W achieved when cv = 0 . Thus cv = 500 KJ/Kg and dWcv/dt=5000 KJ/sec. m max Combining eqs. (1) and (2) gives ME3322 Quiz #3 Summer 2011 NAME ____________________________ _______________________________________________________________________________________ Useful Formulas Ideal Gas Entropy Change: s2-s1=so(T2)-so(T1)-Rln(p2/p1). 2 Heat Transfer to an Internally Reversible Process: Qint = Tds rev 1 Q Closed System Entropy Balance: S = + T b 1 Steady-state Control Volume Energy and Entropy Balance (Single Channel, K.E.=P.E.=0): Qcv Wcv + (hi he ) = 0 m Q T j + m(si se ) + cv = 0 j j b 2
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