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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___ (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
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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