$IfNot ParametricTable= P_ratio_gas. P ratio,gas = 14; Raport comprimare compresor aer  Pressure ratio for gas compressor (2) $EndIf


 Sherilyn May
 1 years ago
 Views:
Transcription
1 P Equations Thermodynamics  An Engineering Approach (5th Ed)  Cengel, Boles  McgrawHill (2006)  pg. 598 Centrala cu cicluri combinate Se considera o centrala electrica cu ciclu combinat gazeabur care are o putere neta de 450 MW. Raportul de comprimare al ciclului turbinei cu gaze este 14. Aerul intra în compresor la 300 K si în turbina la 1400 K. Gazele de ardere care ies din turbina cu gaze sunt folosite pentru a încalzi aburul la 8 MPa pâna la 400 C întrun schimbator de caldura. Gazele de ardere ies din schimbatorul de caldura la 460 K. Un preîncalzitor de apa de alimentare prin amestec (deschis) încorporat în ciclul cu abur functioneaza la o presiune de 0.6 MPa. Presiunea de condensare este de 20 kpa. Considerând toate procesele de comprimare si destindere ca fiind izoentropice, sa se determine (a) raportul dintre debitele masice de aer si abur, (b) fluxul necesar de caldura introdusa în camera de ardere, si (c) randamentul termic al ciclului combinat. Sa se studieze efectul raportului de comprimare al ciclului cu gaze daca acesta variaza de la 10 la 20 asupra raportului dintre debitele de gaze si abur si asupra randamentului termic al ciclului. Sa se reprezinte rezultatele în functie de raportul de comprimare al ciclului cu gaze, si sa se discute rezultatele. $UnitSystem K kpa Marimi de intrare: T 8 = 300 [K] ; P 8 = 14.7 [kpa] ; Gaze (aer la intrarea in compresor)  Gas compressor inlet (1) 1
2 $IfNot ParametricTable= P_ratio_gas P ratio,gas = 14; Raport comprimare compresor aer  Pressure ratio for gas compressor (2) $EndIf T 10 = 1400 [K] ; Intrare turbina cu gaze  Gas turbine inlet (3) T 12 = 460 [K] ; Gas exit temperature from Gastosteam heat exchanger (4) P 12 = P 8 ; Presiune iesire aer  Assumed air exit pressure (5) kw Ẇ net = 450 [MW] 1000 MW ; (6) $IfNot ParametricTable= eta η comp = 1.0; Eta gas,turb = 1.0; η pump = 1.0; Eta steam,turb = 1.0; (7) (8) $EndIf P 5 = 8000 [kpa] ; DELT AT 5,11 = 40 [K] Diferenta de temperatura la capatul cald al HEx (9) ; Abur la intrare in turbina  Steam turbine inlet (10) $IfNot ParametricTable= P[6] P 6 = 600 [kpa] ; Extraction pressure for steam open feedwater heater (11) $EndIf P 7 = 20 [kpa] ; Presiunea de condensare a aburului  Steam condenser pressure (12) Ciclul de forta cu gaze (aer)  GAS POWER CYCLE ANALYSIS Gas Compressor anaysis Starea [8]: h 8 = h (Air, T = T 8 ) ; s 8 = s (Air, T = T 8, P = P 8 ) ; (13) Starea [9s]: s s,9 = s 8 ; For the ideal case the entropies are constant across the compressor (14) P 9 = P ratio,gas P 8 ; (15) T s,9 = T (Air, s = s s,9, P = P 9 ) ; Ts9 is the isentropic value of T[9] at compressor exit (16) 2
3 η comp = w gas,comp,isen /w gas,comp compressor adiabatic efficiency, w comp > w comp,isen (17) h 8 +w gas,comp,isen = h s,9 h s,9 = h (Air, T = T s,9 ) ; SSSF conservation of energy for the isentropic compressor, assuming: adiabatic, ke=pe=0 (19) Starea [9]: h 8 +w gas,comp = h 9 ; SSSF conservation of energy for the actual compressor, assuming: adiabatic, ke=pe=0(20) T 9 = T (Air, h = h 9 ) ; s 9 = s (Air, T = T 9, P = P 9 ) ; (21) Gas Cycle External heat exchanger analysis Starea [10]: P 10 = P 9 ; Assume process 910 is SSSF constant pressure (22) h 10 = h (Air, T = T 10 ) ; s 10 = s (Air, T = T 10, P = P 10 ) ; (23) h 9 +q in = h 10 ; Q in = ṁ gas q in ; SSSF conservation of energy for the external heat exchanger, assuming W=0, ke=pe=0(24) (25) Gas Turbine analysis Starea [11s]: P 11 = P 10 P ratio,gas ; s s,11 = s 10 ; For the ideal case the entropies are constant across the turbine (26) h s,11 = h (Air, T = T s,11 ) ; T s,11 = T (Air, s = s s,11, P = P 11 ) ; Ts11 is the isentropic value of T[11] at ga Starea [11]: h 10 = w gas,turb,isen +h s,11 ; SSSF conservation of energy for the isentropic gas turbine, assuming: adiabatic, ke= η gas,turb = w gas,turb /w gas,turb,isen ; gas turbine adiabatic efficiency, w gas,turb,isen > w gas,turb (29) h 10 = w gas,turb +h 11 ; SSSF conservation of energy for the actual gas turbine, assuming: adiabatic, ke=pe=0(30) T 11 = T (Air, h = h 11 ) ; s 11 = s (Air, T = T 11, P = P 11 ) ; (31) GastoSteam Heat Exchanger Starea [12]: h 12 = h (Air, T = T 12 ) ; s 12 = s (Air, T = T 12, P = P 12 ) ; (32) 3
4 SSSF conservation of energy for the gastosteam heat exchanger, assuming: adiabatic, W=0, ke=pe=0 ṁ gas h 11 + ṁ steam h 4 = ṁ gas h 12 + ṁ steam h 5 ; ec 1/2 ec > debitele (33) Ciclul de forta cu abur  STEAM CYCLE ANALYSIS Starea [1]: Steam Condenser exit pump or Pump 1 analysis P 1 = P 7 ; (34) h 1 = h (ST EAM, P = P 1, x = 0) ; v1 = v (ST EAM, P = P 1, x = 0) ; (35) s 1 = s (ST EAM, P = P 1, x = 0) ; T 1 = T (ST EAM, P = P 1, x = 0) ; (36) Starea [2]: P 2 = P 6 ; (37) w pump1,s = v1 (P 2 P 1 ) ; SSSF isentropic pump work assuming constant specific volume (38) w pump1 = w pump1,s /η pump ; Definition of pump efficiency (39) h 1 + w pump1 = h 2 ; Steadyflow conservation of energy (40) s 2 = s (ST EAM, P = P 2, h = h 2 ) ; T 2 = T (ST EAM, P = P 2, h = h 2 ) ; (41) Starea [3]: P 3 = P 6 ; Condensate leaves heater as sat. liquid at P[3] (42) h 3 = h (steam, P = P 3, x = 0) ; T 3 = T (steam, P = P 3, x = 0) ; (43) s 3 = s (ST EAM, P = P 3, x = 0) ; v3 = v (ST EAM, P = P 3, x = 0) ; (44) Boiler condensate pump or Pump 2 analysis P 4 = P 5 ; (45) w pump2,s = v3 (P 4 P 3 ) ; SSSF isentropic pump work assuming constant specific volume (46) w pump2 = w pump2,s /η pump ; Definition of pump efficiency (47) h 3 + w pump2 = h 4 ; Steadyflow conservation of energy (48) s 4 = s (ST EAM, P = P 4, h = h 4 ) ; T 4 = T (ST EAM, P = P 4, h = h 4 ) ; (49) 4
5 Steam Turbine analysis T 5 = T 11 T 5,11 ; (50) h 5 = h (steam, T = T 5, P = P 5 ) ; s 5 = s (ST EAM, P = P 5, T = T 5 ) ; (51) Starea [6s]: s s,6 = s 5 ; (52) h s,6 = h (ST EAM, s = s s,6, P = P 6 ) ; T s,6 = T (ST EAM, s = s s,6, P = P 6 ) ; (53) Starea [6]: h 6 = h 5 η steam,turb (h 5 h s,6 ) Definition of steam turbine efficiency (54) T 6 = T (ST EAM, P = P 6, h = h 6 ) ; s 6 = s (ST EAM, P = P 6, h = h 6 ) ; (55) y h 6 + (1 y) h 2 = 1 h 3 ; Steadyflow conservation of energy (56) w steam,pumps = (1 y) w pump1 + w pump2 ; Total steam pump work input/ mass steam (57) Starea [7s]: s s,7 = s 5 ; (58) h s,7 = h (ST EAM, s = s s,7, P = P 7 ) ; T s,7 = T (ST EAM, s = s s,7, P = P 7 ) ; (59) Starea [7]: h 7 = h 5 η steam,turb (h 5 h s,7 ) ; Randamentul turbinei  Definition of steam turbine efficiency (60) T 7 = T (ST EAM, P = P 7, h = h 7 ) ; s 7 = s (ST EAM, P = P 7, h = h 7 ) ; (61) SSSF conservation of energy for the steam turbine: adiabatic, neglect ke and pe h 5 = w steam,turb + y h 6 + (1 y) h 7 ; (62) Steam Condenser analysis (1 y) h 7 = q out + (1 y) h 1 ; SSSF conservation of energy for the Condenser per unit mass (63) Q out = ṁ steam q out ; (64) Cycle Statistics MassRatio gastosteam = ṁ gas /ṁ steam ; Ẇ net = ṁ gas (w gas,turb w gas,comp )+ṁ steam (w steam,turb w steam,pumps ) ; (65) definitia puterii nete a ciclului  net η th = Ẇnet/ Q in 100 [%] ; Cycle thermal efficiency, in percent (67) Bwr = ṁgas w gas,comp + ṁ steam w steam,pumps ṁ gas w gas,turb + ṁ steam w steam,turb ; Back work ratio (68) Ẇ net,steam = ṁ steam (w steam,turb w steam,pumps ) ; (69) Ẇ net,gas = ṁ gas (w gas,turb w gas,comp ) ; (70) NetW orkratio gastosteam = Ẇnet,gas/Ẇnet,steam; (71) Data$ = Date$; (72) 5
6 Solution Bwr = Data$ = T 5,11 = 40 [K] η comp = 1 η gas,turb = 1 η pump = 1 η steam,turb = 1 η th = 62.4 [%] MassRatio gastosteam = ṁ gas = [kg g as/s] ṁ steam = [kg s team/s] NetW orkratio gastosteam = P ratio,gas = 14 Q in = [kw] Q out = [kw] q in = 877 [kj/kg g as] q out = 1523 [kj/kg s team] v1 = [m 3 /kg] v3 = [m 3 /kg] Ẇ net = [kw] Ẇ net,gas = [kw] Ẇ net,steam = [kw] w gas,comp = [kj/kg g as] w gas,comp,isen = [kj/kg g as] w gas,turb = [kj/kg g as] w gas,turb,isen = [kj/kg g as] w pump1 = [kj/kg s team] w pump1,s = [kj/kg s team] w pump2 = [kj/kg s team] w pump2,s = [kj/kg s team] w steam,pumps = [kj/kg s team] w steam,turb = [kj/kg s team] y = Arrays Row P i T s,i T i h s,i h i s s,i s i [kpa] [K] [K] [kj/kg] [kj/kg] [kj/kgk] [kj/kgk]
7 Ts Plot all(pratiog as) 7
8 etath(p [6]) etath vs eta 8
Equations P $UnitSystem K kpa. F luid$ = Air (1) Input data for fluid. $If Fluid$= Air. C P = [kj/kg K] ; k = 1.
P09169 Equations Thermodynamics  An Engineering Approach (5th Ed)  Cengel, Boles  McgrawHill (2006)  pg. 549 Ciclul Brayton cu regenerare (aer, heliu) cu comprimare si destindere in trepte, c p =ct.
More informationEquations P Se va nota cu y fractia din debitul masic care intra in turbina care e extrasa din turbina pentru preincalzitorul inchis
P10110 Equations Thermodynamics  An Engineering Approach (5th Ed)  Cengel, Boles  McgrawHill (2006)  pg. 602 Ciclul Rankine cu regenerare cu preincalzitoare deschise multiple  Regeneration using
More informationEquations. P v2
P10108v2 Equations Thermodynamics  An Engineering Approach (5th Ed)  Cengel, Boles  McgrawHill (2006)  pg. 602 Problem 10.108 (996) Effect of Number of Reheat Stages on Rankine Cycle Using EES
More informationECE309 THERMODYNAMICS & HEAT TRANSFER MIDTERM EXAMINATION. Instructor: R. Culham. Name: Student ID Number:
ECE309 THERMODYNAMICS & HEAT TRANSFER MIDTERM EXAMINATION June 19, 2015 2:30 pm  4:30 pm Instructor: R. Culham Name: Student ID Number: Instructions 1. This is a 2 hour, closedbook examination. 2. Permitted
More informationI. (20%) Answer the following True (T) or False (F). If false, explain why for full credit.
I. (20%) Answer the following True (T) or False (F). If false, explain why for full credit. Both the Kelvin and Fahrenheit scales are absolute temperature scales. Specific volume, v, is an intensive property,
More informationChapter 5. Mass and Energy Analysis of Control Volumes. by Asst. Prof. Dr.Woranee Paengjuntuek and Asst. Prof. Dr.Worarattana Pattaraprakorn
Chapter 5 Mass and Energy Analysis of Control Volumes by Asst. Prof. Dr.Woranee Paengjuntuek and Asst. Prof. Dr.Worarattana Pattaraprakorn Reference: Cengel, Yunus A. and Michael A. Boles, Thermodynamics:
More informationME Thermodynamics I
HW22 (25 points) Given: 1 A gas power cycle with initial properties as listed on the EFD. The compressor pressure ratio is 25:1 Find: 1 Sketch all the processes on a ph diagram and calculate the enthalpy,
More informationMAE 11. Homework 8: Solutions 11/30/2018
MAE 11 Homework 8: Solutions 11/30/2018 MAE 11 Fall 2018 HW #8 Due: Friday, November 30 (beginning of class at 12:00p) Requirements:: Include T s diagram for all cycles. Also include p v diagrams for Ch
More informationME 354 THERMODYNAMICS 2 MIDTERM EXAMINATION. Instructor: R. Culham. Name: Student ID Number: Instructions
ME 354 THERMODYNAMICS 2 MIDTERM EXAMINATION February 14, 2011 5:30 pm  7:30 pm Instructor: R. Culham Name: Student ID Number: Instructions 1. This is a 2 hour, closedbook examination. 2. Answer all questions
More informationChapter 7. Entropy. by Asst.Prof. Dr.Woranee Paengjuntuek and Asst. Prof. Dr.Worarattana Pattaraprakorn
Chapter 7 Entropy by Asst.Prof. Dr.Woranee Paengjuntuek and Asst. Prof. Dr.Worarattana Pattaraprakorn Reference: Cengel, Yunus A. and Michael A. Boles, Thermodynamics: An Engineering Approach, 5th ed.,
More informationME Thermodynamics I
Homework  Week 01 HW01 (25 points) Given: 5 Schematic of the solar cell/solar panel Find: 5 Identify the system and the heat/work interactions associated with it. Show the direction of the interactions.
More informationCHAPTER 5 MASS AND ENERGY ANALYSIS OF CONTROL VOLUMES
Thermodynamics: An Engineering Approach 8th Edition in SI Units Yunus A. Çengel, Michael A. Boles McGrawHill, 2015 CHAPTER 5 MASS AND ENERGY ANALYSIS OF CONTROL VOLUMES Lecture slides by Dr. Fawzi Elfghi
More information+ m B1 = 1. u A1. u B1.  m B1 = V A. /v A = , u B1 + V B. = 5.5 kg => = V tot. Table B.1.
5.6 A rigid tank is divided into two rooms by a membrane, both containing water, shown in Fig. P5.6. Room A is at 200 kpa, v = 0.5 m3/kg, VA = m3, and room B contains 3.5 kg at 0.5 MPa, 400 C. The membrane
More informationc Dr. Md. Zahurul Haq (BUET) Thermodynamic Processes & Efficiency ME 6101 (2017) 2 / 25 T145 = Q + W cv + i h 2 = h (V2 1 V 2 2)
Thermodynamic Processes & Isentropic Efficiency Dr. Md. Zahurul Haq Professor Department of Mechanical Engineering Bangladesh University of Engineering & Technology (BUET Dhaka1000, Bangladesh zahurul@me.buet.ac.bd
More informationECE309 INTRODUCTION TO THERMODYNAMICS & HEAT TRANSFER. 20 June 2005
ECE309 INTRODUCTION TO THERMODYNAMICS & HEAT TRANSFER 20 June 2005 Midterm Examination R. Culham & M. Bahrami This is a 90 minute, closedbook examination. You are permitted to use one 8.5 in. 11 in. crib
More informationAvailability and Irreversibility
Availability and Irreversibility 1.0 Overview A critical application of thermodynamics is finding the maximum amount of work that can be extracted from a given energy resource. This calculation forms the
More informationLecture 35: Vapor power systems, Rankine cycle
ME 00 Thermodynamics I Spring 015 Lecture 35: Vapor power systems, Rankine cycle Yong Li Shanghai Jiao Tong University Institute of Refrigeration and Cryogenics 800 Dong Chuan Road Shanghai, 0040, P. R.
More informationME Thermodynamics I. Lecture Notes and Example Problems
ME 227.3 Thermodynamics I Lecture Notes and Example Problems James D. Bugg September 2018 Department of Mechanical Engineering Introduction Part I: Lecture Notes This part contains handout versions of
More informationECE309 INTRODUCTION TO THERMODYNAMICS & HEAT TRANSFER. 13 June 2007
ECE309 INTRODUCTION TO THERMODYNAMICS & HEAT TRANSFER 13 June 2007 Midterm Examination R. Culham This is a 2 hour, openbook examination. You are permitted to use: course text book calculator There are
More informationChapter 5. Mass and Energy Analysis of Control Volumes
Chapter 5 Mass and Energy Analysis of Control Volumes Conservation Principles for Control volumes The conservation of mass and the conservation of energy principles for open systems (or control volumes)
More informationECE309 INTRODUCTION TO THERMODYNAMICS & HEAT TRANSFER. 12 June 2006
ECE309 INTRODUCTION TO THERMODYNAMICS & HEAT TRANSFER 1 June 006 Midterm Examination R. Culham This is a hour, closedbook examination. You are permitted to use one 8.5 in. 11 in. crib sheet (one side
More informationDepartment of Mechanical Engineering ME 322 Mechanical Engineering Thermodynamics. Lecture 26. Use of Regeneration in Vapor Power Cycles
Department of Mechanical Engineering ME 322 Mechanical Engineering Thermodynamics Lecture 2 Use of Regeneration in Vapor Power Cycles What is Regeneration? Goal of regeneration Reduce the fuel input requirements
More information1 st Law Analysis of Control Volume (open system) Chapter 6
1 st Law Analysis of Control Volume (open system) Chapter 6 In chapter 5, we did 1st law analysis for a control mass (closed system). In this chapter the analysis of the 1st law will be on a control volume
More informationDishwasher. Heater. Homework Solutions ME Thermodynamics I Spring HW1 (25 points)
HW1 (25 points) (a) Given: 1 for writing given, find, EFD, etc., Schematic of a household piping system Find: Identify system and location on the system boundary where the system interacts with the environment
More informationLecture 38: Vaporcompression refrigeration systems
ME 200 Termodynamics I Lecture 38: Vaporcompression refrigeration systems Yong Li Sangai Jiao Tong University Institute of Refrigeration and Cryogenics 800 Dong Cuan Road Sangai, 200240, P. R. Cina Email
More informationCourse: MECH341 Thermodynamics II Semester: Fall 2006
FINAL EXAM Date: Thursday, December 21, 2006, 9 am 12 am Examiner: Prof. E. Timofeev Associate Examiner: Prof. D. Frost READ CAREFULLY BEFORE YOU PROCEED: Course: MECH341 Thermodynamics II Semester: Fall
More informationThermodynamics is the Science of Energy and Entropy
Definition of Thermodynamics: Thermodynamics is the Science of Energy and Entropy  Some definitions.  The zeroth law.  Properties of pure substances.  Ideal gas law.  Entropy and the second law. Some
More informationTHERMODYNAMICS, FLUID AND PLANT PROCESSES. The tutorials are drawn from other subjects so the solutions are identified by the appropriate tutorial.
THERMODYNAMICS, FLUID AND PLANT PROCESSES The tutorials are drawn from other subjects so the solutions are identified by the appropriate tutorial. THERMODYNAMICS TUTORIAL 2 THERMODYNAMIC PRINCIPLES SAE
More information374 Exergy Analysis. sys (u u 0 ) + P 0 (v v 0 ) T 0 (s s 0 ) where. e sys = u + ν 2 /2 + gz.
374 Exergy Analysis The value of the exergy of the system depends only on its initial and final state, which is set by the conditions of the environment The term T 0 P S is always positive, and it does
More informationReadings for this homework assignment and upcoming lectures
Homework #3 (group) Tuesday, February 13 by 4:00 pm 5290 exercises (individual) Thursday, February 15 by 4:00 pm extra credit (individual) Thursday, February 15 by 4:00 pm Readings for this homework assignment
More informationME 200 Final Exam December 14, :00 a.m. to 10:00 a.m.
CIRCLE YOUR LECTURE BELOW: First Name Last Name 7:30 a.m. 8:30 a.m. 10:30 a.m. 11:30 a.m. Boregowda Boregowda Braun Bae 2:30 p.m. 3:30 p.m. 4:30 p.m. Meyer Naik Hess ME 200 Final Exam December 14, 2015
More information(1)5. Which of the following equations is always valid for a fixed mass system undergoing an irreversible or reversible process:
Last Name First Name ME 300 Engineering Thermodynamics Exam #2 Spring 2008 March 28, 2008 Form A Note : (i) (ii) (iii) (iv) Closed book, closed notes; one 8.5 x 11 sheet allowed. 60 points total; 60 minutes;
More informationUnit Workbook 2  Level 5 ENG U64 Thermofluids 2018 UniCourse Ltd. All Rights Reserved. Sample
Pearson BTEC Level 5 Higher Nationals in Engineering (RQF) Unit 64: Thermofluids Unit Workbook 2 in a series of 4 for this unit Learning Outcome 2 Vapour Power Cycles Page 1 of 26 2.1 Power Cycles Unit
More informationCHAPTER 7 ENTROPY. Copyright Hany A. AlAnsary and S. I. AbdelKhalik (2014) 1
CHAPTER 7 ENTROPY S. I. AbdelKhalik (2014) 1 ENTROPY The Clausius Inequality The Clausius inequality states that for for all cycles, reversible or irreversible, engines or refrigerators: For internallyreversible
More informationThe exergy of asystemis the maximum useful work possible during a process that brings the system into equilibrium with aheat reservoir. (4.
Energy Equation Entropy equation in Chapter 4: control mass approach The second law of thermodynamics Availability (exergy) The exergy of asystemis the maximum useful work possible during a process that
More informationME Thermodynamics I
HW6 (5 points) Given: Carbon dioxide goes through an adiabatic process in a pistoncylinder assembly. provided. Find: Calculate the entropy change for each case: State data is a) Constant specific heats
More informationPrevious lecture. Today lecture
Previous lecture ds relations (derive from steady energy balance) Gibb s equations Entropy change in liquid and solid Equations of & v, & P, and P & for steady isentropic process of ideal gas Isentropic
More informationTHE FIRST LAW APPLIED TO STEADY FLOW PROCESSES
Chapter 10 THE FIRST LAW APPLIED TO STEADY FLOW PROCESSES It is not the sun to overtake the moon, nor doth the night outstrip theday.theyfloateachinanorbit. The Holy Qurān In many engineering applications,
More informationIII. Evaluating Properties. III. Evaluating Properties
F. Property Tables 1. What s in the tables and why specific volumes, v (m /kg) (as v, v i, v f, v g ) pressure, P (kpa) temperature, T (C) internal energy, u (kj/kg) (as u, u i, u f, u g, u ig, u fg )
More informationENT 254: Applied Thermodynamics
ENT 54: Applied Thermodynamics Mr. Azizul bin Mohamad Mechanical Engineering Program School of Mechatronic Engineering Universiti Malaysia Perlis (UniMAP) azizul@unimap.edu.my 0194747351 049798679 Chapter
More informationExisting Resources: Supplemental/reference for students with thermodynamics background and interests:
Existing Resources: Masters, G. (1991) Introduction to Environmental Engineering and Science (Prentice Hall: NJ), pages 15 29. [ Masters_1991_Energy.pdf] Supplemental/reference for students with thermodynamics
More informationfirst law of ThermodyNamics
first law of ThermodyNamics First law of thermodynamics  Principle of conservation of energy  Energy can be neither created nor destroyed Basic statement When any closed system is taken through a cycle,
More informationME 201 Thermodynamics
ME 0 Thermodynamics Solutions First Law Practice Problems. Consider a balloon that has been blown up inside a building and has been allowed to come to equilibrium with the inside temperature of 5 C and
More informationFundamentals of Thermodynamics. Chapter 8. Exergy
Fundamentals of Thermodynamics Chapter 8 Exergy Exergy Availability, available energy Anergy Unavailable energy Irreversible energy, reversible work, and irreversibility Exergy analysis : Pure Thermodynamics
More information5/6/ :41 PM. Chapter 6. Using Entropy. Dr. Mohammad Abuhaiba, PE
Chapter 6 Using Entropy 1 2 Chapter Objective Means are introduced for analyzing systems from the 2 nd law perspective as they undergo processes that are not necessarily cycles. Objective: introduce entropy
More informationIntroduction to Engineering thermodynamics 2 nd Edition, Sonntag and Borgnakke. Solution manual
Introduction to Engineering thermodynamics 2 nd Edition, Sonntag and Borgnakke Solution manual Chapter 6 Claus Borgnakke The picture is a false color thermal image of the space shuttle s main engine. The
More informationTeaching schedule *15 18
Teaching schedule Session *15 18 19 21 22 24 Topics 5. Gas power cycles Basic considerations in the analysis of power cycle; Carnot cycle; Air standard cycle; Reciprocating engines; Otto cycle; Diesel
More informationUNIT I Basic concepts and Work & Heat Transfer
SIDDHARTH GROUP OF INSTITUTIONS :: PUTTUR Siddharth Nagar, Narayanavanam Road 517583 QUESTION BANK (DESCRIPTIVE) Subject with Code: Engineering Thermodynamics (16ME307) Year & Sem: IIB. Tech & IISem
More informationThe First Law of Thermodynamics. By: Yidnekachew Messele
The First Law of Thermodynamics By: Yidnekachew Messele It is the law that relates the various forms of energies for system of different types. It is simply the expression of the conservation of energy
More information2b m 1b: Sat liq C, h = kj/kg tot 3a: 1 MPa, s = s 3 > h 3a = kj/kg, T 3b
.6 A upercritical team power plant ha a high preure of 0 Ma and an exit condener temperature of 50 C. he maximum temperature in the boiler i 000 C and the turbine exhaut i aturated vapor here i one open
More information: 2017,, : 2017, < Part I > Zero exergy line < Part II > : hs
: : 217,, : 217, < Part I > 1. 2. 3. 4. 5. Zero exergy line 6. 7. < Part II > : hs : 217, 5, : 217,  1 / 24   2 / 24  : : 111 = 1 Q = m C ΔT 919 = 1 = + T P A = + = + Q rev = T ΔS δq ds T rev (=
More informationFINAL EXAM. ME 200 Thermodynamics I, Spring 2013 CIRCLE YOUR LECTURE BELOW:
ME 200 Thermodynamics I, Spring 2013 CIRCLE YOUR LECTURE BELOW: Div. 5 7:30 am Div. 2 10:30 am Div. 4 12:30 am Prof. Naik Prof. Braun Prof. Bae Div. 3 2:30 pm Div. 1 4:30 pm Div. 6 4:30 pm Prof. Chen Prof.
More informationThermodynamics: An Engineering Approach Seventh Edition in SI Units Yunus A. Cengel, Michael A. Boles McGrawHill, 2011.
Thermodynamics: An Engineering Approach Seventh Edition in SI Units Yunus A. Cengel, Michael A. Boles McGrawHill, 2011 Chapter 7 ENTROPY Mehmet Kanoglu University of Gaziantep Copyright The McGrawHill
More informationChapter 7. Entropy: A Measure of Disorder
Chapter 7 Entropy: A Measure of Disorder Entropy and the Clausius Inequality The second law of thermodynamics leads to the definition of a new property called entropy, a quantitative measure of microscopic
More informationIn the next lecture...
16 1 In the next lecture... Solve problems from Entropy Carnot cycle Exergy Second law efficiency 2 Problem 1 A heat engine receives reversibly 420 kj/cycle of heat from a source at 327 o C and rejects
More informationBrown Hills College of Engineering & Technology
UNIT 4 Flow Through Nozzles Velocity and heat drop, Mass discharge through a nozzle, Critical pressure ratio and its significance, Effect of friction, Nozzle efficiency, Supersaturated flow, Design pressure
More informationThermodynamics: An Engineering Approach Seventh Edition Yunus A. Cengel, Michael A. Boles McGrawHill, Chapter 7 ENTROPY
Thermodynamics: An Engineering Approach Seventh Edition Yunus A. Cengel, Michael A. Boles McGrawHill, 2011 Chapter 7 ENTROPY Copyright The McGrawHill Companies, Inc. Permission required for reproduction
More informationc Dr. Md. Zahurul Haq (BUET) Entropy ME 203 (2017) 2 / 27 T037
onsequences of Second Law of hermodynamics Dr. Md. Zahurul Haq Professor Department of Mechanical Engineering Bangladesh University of Engineering & echnology BUE Dhaka000, Bangladesh zahurul@me.buet.ac.bd
More informationEngineering Thermodynamics Solutions Manual
Engineering Thermodynamics Solutions Manual Prof. T.T. AlShemmeri Download free books at Prof. T.T. AlShemmeri Engineering Thermodynamics Solutions Manual 2 2012 Prof. T.T. AlShemmeri & bookboon.com
More information10 minutes reading time is allowed for this paper.
EGT1 ENGINEERING TRIPOS PART IB Tuesday 31 May 2016 2 to 4 Paper 4 THERMOFLUID MECHANICS Answer not more than four questions. Answer not more than two questions from each section. All questions carry the
More informationENTROPY. Chapter 7. Mehmet Kanoglu. Thermodynamics: An Engineering Approach, 6 th Edition. Yunus A. Cengel, Michael A. Boles.
Thermodynamics: An Engineering Approach, 6 th Edition Yunus A. Cengel, Michael A. Boles McGrawHill, 2008 Chapter 7 ENTROPY Mehmet Kanoglu Copyright The McGrawHill Companies, Inc. Permission required
More informationThermodynamics ENGR360MEP112 LECTURE 7
Thermodynamics ENGR360MEP11 LECTURE 7 Thermodynamics ENGR360/MEP11 Objectives: 1. Conservation of mass principle.. Conservation of energy principle applied to control volumes (first law of thermodynamics).
More informationChapter Two. Basic Thermodynamics, Fluid Mechanics: Definitions of Efficiency. Laith Batarseh
Chapter Two Basic Thermodynamics, Fluid Mechanics: Definitions of Efficiency Laith Batarseh The equation of continuity Most analyses in this book are limited to onedimensional steady flows where the velocity
More informationApplied Thermodynamics. Gas Power Cycles
Applied Thermodynamics Gas Power Cycles By: Mohd Yusof Taib Faculty of Mechanical Engineering myusof@ump.edu.my Chapter Description Aims To identify and recognized ideal thermodynamics cycle. To analyze
More informationChemical Engineering Thermodynamics Spring 2002
10.213 Chemical Engineering Thermodynamics Spring 2002 Test 2 Solution Problem 1 (35 points) High pressure steam (stream 1) at a rate of 1000 kg/h initially at 3.5 MPa and 350 ºC is expanded in a turbine
More informationBMEA PREVIOUS YEAR QUESTIONS
BMEA PREVIOUS YEAR QUESTIONS CREDITS CHANGE ACCHA HAI TEAM UNIT1 Introduction: Introduction to Thermodynamics, Concepts of systems, control volume, state, properties, equilibrium, quasistatic process,
More informationFUNDAMENTALS OF THERMODYNAMICS
FUNDAMENTALS OF THERMODYNAMICS SEVENTH EDITION CLAUS BORGNAKKE RICHARD E. SONNTAG University of Michigan John Wiley & Sons, Inc. PUBLISHER ASSOCIATE PUBLISHER ACQUISITIONS EDITOR SENIOR PRODUCTION EDITOR
More informationTo receive full credit all work must be clearly provided. Please use units in all answers.
Exam is Open Textbook, Open Class Notes, Computers can be used (Computer limited to class notes, lectures, homework, book material, calculator, conversion utilities, etc. No searching for similar problems
More informationWhere F1 is the force and dl1 is the infinitesimal displacement, but F1 = p1a1
In order to force the fluid to flow across the boundary of the system against a pressure p1, work is done on the boundary of the system. The amount of work done is dw =  F1.dl1, Where F1 is the force
More informationR13. II B. Tech I Semester Regular Examinations, Jan THERMODYNAMICS (Com. to ME, AE, AME) PART A
SET  1 II B. Tech I Semester Regular Examinations, Jan  2015 THERMODYNAMICS (Com. to ME, AE, AME) Time: 3 hours Max. Marks: 70 Note 1. Question Paper consists of two parts (PartA and PartB) 2. Answer
More informationConsequences of Second Law of Thermodynamics. Entropy. Clausius Inequity
onsequences of Second Law of hermodynamics Dr. Md. Zahurul Haq Professor Department of Mechanical Engineering Bangladesh University of Engineering & echnology BUE Dhaka000, Bangladesh zahurul@me.buet.ac.bd
More informationCHAPTER INTRODUCTION AND BASIC PRINCIPLES. (Tutorial). Determine if the following properties of the system are intensive or extensive properties: Property Intensive Extensive Volume Density Conductivity
More informationUsing the Entropy Rate Balance to Determine the Heat Transfer and Work in an Internally Reversible, Polytropic, Steady State Flow Process
Undergraduate Journal of Mathematical Modeling: One + Two Volume 8 08 Spring 08 Issue Article Using the Entropy Rate Balance to Determine the Heat Transfer and Work in an Internally Reversible, Polytropic,
More informationES 202 Fluid and Thermal Systems
ES Fluid and Thermal Systems Lecture : Power Cycles (/4/) Power cycle Road Map of Lecture use Rankine cycle as an example the ideal Rankine cycle representation on a Ts diagram divergence of constant
More information1. (10) Calorically perfect ideal air at 300 K, 100 kpa, 1000 m/s, is brought to rest isentropically. Determine its final temperature.
AME 5053 Intermediate Thermodynamics Examination Prof J M Powers 30 September 0 0 Calorically perfect ideal air at 300 K, 00 kpa, 000 m/s, is brought to rest isentropically Determine its final temperature
More informationThermal Energy Final Exam Fall 2002
16.050 Thermal Energy Final Exam Fall 2002 Do all eight problems. All problems count the same. 1. A system undergoes a reversible cycle while exchanging heat with three thermal reservoirs, as shown below.
More informationME 300 Thermodynamics II Spring 2015 Exam 3. Son Jain Lucht 8:30AM 11:30AM 2:30PM
NAME: PUID#: ME 300 Thermodynamics II Spring 05 Exam 3 Circle your section (5 points for not circling correct section): Son Jain Lucht 8:30AM :30AM :30PM Instructions: This is a closed book/note exam.
More informationLecture 44: Review Thermodynamics I
ME 00 Thermodynamics I Lecture 44: Review Thermodynamics I Yong Li Shanghai Jiao Tong University Institute of Refrigeration and Cryogenics 800 Dong Chuan Road Shanghai, 0040, P. R. China Email : liyo@sjtu.edu.cn
More informationChapter 5: The First Law of Thermodynamics: Closed Systems
Chapter 5: The First Law of Thermodynamics: Closed Systems The first law of thermodynamics can be simply stated as follows: during an interaction between a system and its surroundings, the amount of energy
More informationT098. c Dr. Md. Zahurul Haq (BUET) First Law of Thermodynamics ME 201 (2012) 2 / 26
Conservation of Energy for a Closed System Dr. Md. Zahurul Haq Professor Department of Mechanical Engineering Bangladesh University of Engineering & Technology (BUET Dhaka, Bangladesh zahurul@me.buet.ac.bd
More informationME 2322 Thermodynamics I PRELECTURE Lesson 23 Complete the items below Name:
Lesson 23 1. (10 pt) Write the equation for the thermal efficiency of a Carnot heat engine below: T η = T 1 L H 2. (10 pt) Can the thermal efficiency of an actual engine ever exceed that of an equivalent
More informationESO 201A Thermodynamics
ESO 201A Thermodynamics Instructor: Sameer Khandekar Tutorial 9 [727] A completely reversible heat pump produces heat at arate of 300 kw to warm a house maintained at 24 C. Theexterior air, which is at
More informationT222 T194. c Dr. Md. Zahurul Haq (BUET) Gas Power Cycles ME 6101 (2017) 2 / 20 T225 T226
The Carnot Gas Power Cycle Gas Power Cycles 1 2 : Reversible, isothermal expansion at T H 2 3 : Reversible, adiabatic expansion from T H to T L 3 4 : Reversible, isothermal compression at T L Dr. Md. Zahurul
More informationConsequences of Second Law of Thermodynamics. Entropy. Clausius Inequity
onsequences of Second Law of hermodynamics Dr. Md. Zahurul Haq Professor Department of Mechanical Engineering Bangladesh University of Engineering & echnology BUE Dhaka000, Bangladesh zahurul@me.buet.ac.bd
More informationKNOWN: Pressure, temperature, and velocity of steam entering a 1.6cmdiameter pipe.
4.3 Steam enters a.6cmdiameter pipe at 80 bar and 600 o C with a velocity of 50 m/s. Determine the mass flow rate, in kg/s. KNOWN: Pressure, temperature, and velocity of steam entering a.6cmdiameter
More informationExergy and the Dead State
EXERGY The energy content of the universe is constant, just as its mass content is. Yet at times of crisis we are bombarded with speeches and articles on how to conserve energy. As engineers, we know that
More informationSpring_#8. Thermodynamics. Youngsuk Nam
Spring_#8 Thermodynamics Youngsuk Nam ysnam1@khu.ac.krac kr Ch.7: Entropy Apply the second law of thermodynamics to processes. Define a new property called entropy to quantify the secondlaw effects. Establish
More information20 m neon m propane. g 20. Problems with solutions:
Problems with solutions:. A m tank is filled with a gas at room temperature 0 C and pressure 00 Kpa. How much mass is there if the gas is a) Air b) Neon, or c) Propane? Given: T7K; P00KPa; M air 9; M
More informationFind: a) Mass of the air, in kg, b) final temperature of the air, in K, and c) amount of entropy produced, in kj/k.
PROBLEM 6.25 Three m 3 of air in a rigid, insulated container fitted with a paddle wheel is initially at 295 K, 200 kpa. The air receives 1546 kj of work from the paddle wheel. Assuming the ideal gas model,
More informationKNOWN: Air undergoes a polytropic process in a pistoncylinder assembly. The work is known.
PROBLEM.7 A hown in Fig. P.7, 0 ft of air at T = 00 o R, 00 lbf/in. undergoe a polytropic expanion to a final preure of 5.4 lbf/in. The proce follow pv. = contant. The work i W = 94.4 Btu. Auming ideal
More informationEVALUATION OF THE BEHAVIOUR OF STEAM EXPANDED IN A SET OF NOZZLES, IN A GIVEN TEMPERATURE
Equatorial Journal of Engineering (2018) 913 Journal Homepage: www.erjournals.com ISSN: 01847937 EVALUATION OF THE BEHAVIOUR OF STEAM EXPANDED IN A SET OF NOZZLES, IN A GIVEN TEMPERATURE Kingsley Ejikeme
More informationENGR Thermodynamics
ENGR 224  hermodynamics #1  Diagram for a Cascade VCR Cycle (21 ts) Baratuci Final 13Jun11 On a full sheet of paper, construct a complete Diagram for the cascade cascade vaporcompression refrigeration
More informationME 200 Final Exam December 12, :00 a.m. to 10:00 a.m.
CIRCLE YOUR LECTURE BELOW: First Name Last Name 7:30 a.m. 8:30 a.m. 10:30 a.m. 1:30 p.m. 3:30 p.m. Mongia Abraham Sojka Bae Naik ME 200 Final Exam December 12, 2011 8:00 a.m. to 10:00 a.m. INSTRUCTIONS
More informationIsentropic Efficiency in Engineering Thermodynamics
June 21, 2010 Isentropic Efficiency in Engineering Thermodynamics Introduction This article is a summary of selected parts of chapters 4, 5 and 6 in the textbook by Moran and Shapiro (2008. The intent
More informationUBMCC11  THERMODYNAMICS. B.E (Marine Engineering) B 16 BASIC CONCEPTS AND FIRST LAW PART A
UBMCC11  THERMODYNAMICS B.E (Marine Engineering) B 16 UNIT I BASIC CONCEPTS AND FIRST LAW PART A 1. What do you understand by pure substance? 2. Define thermodynamic system. 3. Name the different types
More informationESO201A: Thermodynamics
ESO201A: Thermodynamics First Semester 20152016 MidSemester Examination Instructor: Sameer Khandekar Time: 120 mins Marks: 250 Solve subparts of a question serially. Question #1 (60 marks): One kmol
More informationThermodynamics Lecture Series
Thermodynamics ecture Series Reference: Chap 0 Halliday & Resnick Fundamental of Physics 6 th edition Kinetic Theory of Gases Microscopic Thermodynamics Applied Sciences Education Research Group (ASERG)
More informationERRATA SHEET Thermodynamics: An Engineering Approach 8th Edition Yunus A. Çengel, Michael A. Boles McGrawHill, 2015
ERRATA SHEET Thermodynamics: An Engineering Approach 8th Edition Yunus A. Çengel, Michael A. Boles McGrawHill, 2015 December 2015 This errata includes all corrections since the first printing of the book.
More informationR13 SET  1 '' ''' '' ' '''' Code No RT21033
SET  1 II B. Tech I Semester Supplementary Examinations, June  2015 THERMODYNAMICS (Com. to ME, AE, AME) Time: 3 hours Max. Marks: 70 Note: 1. Question Paper consists of two parts (PartA and PartB)
More informationThermodynamics II. Week 9
hermodynamics II Week 9 Example Oxygen gas in a piston cylinder at 300K, 00 kpa with volume o. m 3 is compressed in a reversible adiabatic process to a final temperature of 700K. Find the final pressure
More information