HW9 - Department of Mechanical Science and Engineering ME 300 Thermodynamics Section D Fall 2009 Homework#9 Solution 1 Problem 9.1 Recalling the

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Unformatted text preview: Department of Mechanical Science and Engineering ME 300 Thermodynamics Section D Fall 2009 Homework #9 - Solution 1 Problem 9.1 Recalling the schematic for Problem 8.17 and the results: States h (kJ/kg) 1 3321.4 ˙ Q in ˙ m 3155.5kJ/kg 2 2272.1 ˙ Q out ˙ m 2120.6kJ/kg 3 151.53 ˙ W net ˙ m 1034.9kJ/kg 4 165.9 η 0.328 (a) Determine the mass flow rate in kg/s based on the required design net power output. Since the net power output is given as ˙ W cycle =100MW we can determine the mass flow rate: ˙ W cycle = ˙ m (( h 1- h 2 )- ( h 4- h 3 )) → ˙ m = 96 . 63 kg/s (b) Determine ˙ Q in , ˙ Q out , ˙ W p in MW and the back work ratio bwr. We can use equations (8.2), (8.3), (8.4) and (8.6) straight out of the textbook: ˙ Q out = ˙ m ( h 2- h 3 ) = 204 . 917 MW ˙ W p = ˙ m ( h 4- h 3 ) = 1 . 328 MW ˙ Q in = ˙ m ( h 1- h 4 ) = 304 . 917 MW or ↔ ˙ W cycle = ˙ Q in- ˙ Q out bwr = ( h 4- h 3 ) ( h 1- h 2 ) = 0 . 01363 2 Problem 9.2 For the first modification we consider the reheat option: The reheat takes place at a pressure of p 2 = p 3 =1.5MPa and the temperature at the inlet to the second stage of the turbine is T 3 =440 ◦ C. (a) Determine the enthalpy at each state (1 through 6). We work our way systematically through all states starting with state 1. With the given pressure of p 1 =10MPa and temperature T 1 =480 ◦ C we find from Table A-4 that h 1 =3321.4kJ/kg. Since the water expands through the first stage of the turbine we need to use the turbine efficiency to calculate the enthalpy at state 2. η t, 1 = h 1- h 2 h 1- h 2 s To determine h 2 s we first need the entropy at state 1 since s 2 s = s 1 . From Table A-4 we determine s 1 =6.5282kJ/kg · K. Checking Table A-3 at p 3 =1.5MPa shows that state 2s is in the superheated region. Therefore, interpolating at p 3 =1.5MPa from Table A-4: h 2 s = 2899 . 3- 2796 . 8 6 . 6628- 6 . 4546 · (6 . 5282- 6 . 4546) + 2796 . 8 = 2833 . 03 kJ/kg When then can solve for h 2 =2930.71kJ/kg. Since the reheat process takes place at constant pressure p 2 = p 3 =1.5MPa and the tempera- ture is T 3 =440 ◦ C we get from Table A-4 that h 3 =3342.5kJ/kg. Since at state 3 the steams enters the second stage of the turbine we note that the entropy is s 3 =7.394kJ/kg · K. For state 4 we use again the efficiency equation for the second stage of the turbine: η t, 1 = h 3- h 4 h 3- h 4 s 3 To determine h 4 s we first need to check what the phase description is at state 4. Since s 4 s = s 3 we determine from Table A-3 at p 4 =6kPa the phase description is two-phase since s f < s 4 s < s g . Therefore, we need first to determine the quality for the isentropic process from 3 to 4s: x 4 s = s 4 s- s f 4 s g 4- s f 4 = 0 . 8801 Now we can determine h 4 s : h 4 s = h f 4 + x 4 s h fg 4 = 2277 . 75 When then can solve for h 4 =2490.70kJ/kg....
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This note was uploaded on 09/30/2011 for the course ME 300 taught by Professor Staff during the Fall '08 term at University of Illinois, Urbana Champaign.

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HW9 - Department of Mechanical Science and Engineering ME 300 Thermodynamics Section D Fall 2009 Homework#9 Solution 1 Problem 9.1 Recalling the

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