5_8esm_09_053.pdf

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

• Homework Help
• 3

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

This is the end of the preview. Sign up to access the rest of the document.

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 ...
View Full Document

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern