5_8esm_09_053.pdf - 9.53 The cycle of Problem 9.42 is...

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Unformatted text preview: 9.53 The cycle of Problem 9.42 is modified to include the effects of irreversibilities in the adiabatic expansion and compression processes. If the states at the compressor and turbine inlets remain unchanged, the cycle produces 10 MW of power, and the compressor and turbine isentropic efficiencies are both 80%, determine (a) the pressure (kPa), temperature (K), and enthalpy (kJ/kg) at each principal state of the cycle and sketch the T-s diagram. (b) the mass flow rate of air, in kg/s. (c) the rate of heat transfer, in kW, to the working fluid passing through the heat exchanger. (d) the thermal efficiency. KNOWN: An air-standard Brayton cycle operates with known states at the turbine and compressor inlets and lmown compressor and turbine isentropic efficiencies. The net power output of the cycle is given ' FIND: Determine the mass flow rate of air, the rate of heat transfer to the working fluid passing through the heat exchanger, and the thermal efficiency. SCHEMATIC AND GIVEN DATA: WW=IOMW r._..a.s_....-._......-..~._...._.H.... ENGINEERING MODEL: 1. Each component is analyzed as a control volume at steady state. The control volumes are shown on the accompanying sketch by dashed lines. 2. The turbine and compressor operate adiabatical 1y. 3. There are no pressure drops for flow through the heat exchangers. 4. Kinetic and potential energy effects are negligible. 5. The working fluid is air modeled as an ideal gas. a? Problem 9.53 (Continued) — Page 2 ANALYSIS: (a) The T-s diagram for the cycle is shown below. 300K States 1 and 3 are the same as the corresponding states in Problem 9.42. Thus, h1 = 300.19 kJ/kg and k3 = 1575.57 kJ/kg. Furthermore, States 2 and 4 in Prob. 9.42 correspond to States 23 and 45 in the current problem, so h), = 610.65 kJ/kg and I14, = 800.78 kJ/kg. State 2 can be determined using the isentropic compressor efficiency h2: — h] ’12-’11 7,9: Solving for h; and inserting values h; = h; + (11;, — Ira/r]c = 300.19 kJ/kg + (610.65 kJ/kg — 300.19 kJ/kg)/(0. 80) = 688.27 kJ/kg Interpolating in Table A-22, T 2 = 676.7 K. Similarly, State 4 can be determined using the isentropic turbine efficiency h3_h4 hi—hh h. = 1:3 — 77.013 — h“) = 1575.57 kJ/kg — (0.80)(1575.57 kJ/kg — 800.78 kJ/kg) = 955.74 kJ/kg 7].: From Table A-22, H 2 920.3 K. 48 Problem 9.53 (Continued) — Page 3 Insumm unr- mas—mm- _—_I_-E__ ——E_—— -_—I_—_ -—-I___ (b) The mass flow rate of air is found as follows. Mass and energy rate balances for control volumes enclosing the turbine and compressor give Wt = p510:3 — h,,) and We = mo:2 — h!) The net power of the cycle is Wm = W. —W" = rh[(h3 -h4)-(h2 —h1)] Solving for )5: W cycle ”’=——‘[(h,-h4)-(h2—h,)1 Inserting values L19. m = ~————————U 123000 kw U U 113w = 43.15 15213 1575.57——955.74— — 688.27——300.19— kg kg kg kg (c) The rate of heat transfer to the working fluid passing through the heat exchanger can be determined by applying mass and energy balances to a control volume around the heat exchanger to give kg 1:] k5 lkW I. = ' — = 43.15— 1575.57——688.27— —— = 8 ”(W Qum(lhh2)( 5 kg REL s (d) The thermal efficiency is W 1] = Q?” = (10,000 kW)/(38,287.0 kw = 0.2612 (26.12%! m ...
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