( )( ) 7 MPa q in = = 10 kpa q out. 1 h. = s. Thus, and = 38.9% (b) (c) The rate of heat rejection to the cooling water and its temperature rise are
|
|
- Imogene Gordon
- 6 years ago
- Views:
Transcription
1 . A team poer plant operate on a imple ideal Ranke cycle beteen te peciied preure limit. e termal eiciency o te cycle, te ma lo rate o te team, and te temperature rie o te coolg ater are to be determed. Aumption Steady operatg condition exit. Ketic and potential energy cange are negligible. Analyi (a From te team table (able A-, A-, and A-,.8 kj u, and (b v v 0 0 kpa v( P P ( m (,000 0 kpa.0 kj + p, P MPa 00 C m kj.8000 kj K kj kpa m kj P 0 kpa x x.8 + t W& m & kj..8.8 kj kj 0. kj 8.%. kj,000 kj/ 0. kj. 0 kg/ ( 0.80(.. kj (c e rate o eat rejection to te coolg ater and it temperature rie are m & (.8 kg/(.8 kj 0,8 kj/ 0,8 kj/ Δcoolgater ( mc & 000 kg/.8 kj C 8. C coolgater ( ( MPa 0 kpa
2 . A team poer plant tat operate on te ideal reeat Ranke cycle i conidered. e turbe ork put and te termal eiciency o te cycle are to be determed. Aumption Steady operatg condition exit. Ketic and potential energy cange are negligible. Analyi From te team table (able A-, A-, and 0 kpa. kj v 0 kpa m, ( p v P P kj ( m ( kpa MPa kpa m.08 kj + p, P MPa 00 C P MPa kj 8. kj. kj K 0.0 kj 0 kpa P MPa 00 C 8. kj. kj K P 0 kpa x.0 + x. + ( 0.800(.. kj e turbe ork put and te termal eiciency are determed rom, ( + ( kj and u, t ( ( kj, p, 0 kj 8 kj.08 0 kj 0.8.8%
3 . A team poer plant tat operate on an ideal reeat Ranke cycle beteen te peciied preure limit i conidered. e preure at ic reeatg take place, te total rate o eat put te boiler, and te termal eiciency o te cycle are to be determed. Aumption Steady operatg condition exit. Ketic and potential energy cange are negligible. Analyi (a From te team table (able A-, A-, and A-, v v p, at@ 0 kpa at@ 0 kpa v ( P P ( m (,000 0 kpa. kj + p, P MPa 00 C P 0 kpa.8 kj m kj kpa m kj 0.8 kj.80 kj K + x 00 C P 0 kpa. kj + x ( 0.0(. ( 0.0(. ( te reeat preure. kj.88 kj K 0 kpa P. MPa (b e rate o eat upply i Q & m& + 8. kj [( ( ] ( kg/( ,0 kw (c e termal eiciency i determed rom u, t m& ( ( kj/(.,8 kj/.%,0 kj/ kj.8 kj,8 kj/
4 . A team poer plant operate on an ideal regenerative Ranke cycle it to open eedater eater. e poer put o te poer plant and te termal eiciency o te cycle are to be determed. Aumption Steady operatg condition exit. Ketic and potential energy cange are negligible. Analyi Boiler II P III P II I 8 P I urbe 0 Condener 0 MPa 0. MPa 0. MPa kpa y 8 - y - y - z 0 (a From te team table (able A-, A-, and A-, v v pi, kpa. kj m ( P P ( m ( 00 kpa + pi, P 0. MPa at.liuid v v pii, kj ( P P ( m ( kpa 0. kj P 0. MPa at.liuid v v piii, v v + pii, kj kpa m ( P P ( m ( 0, kpa 0. kj + piii, P 0 MPa 00 C P8 0. MPa 8 8 P 0. MPa x 0. kj m MPa.8 kj.0 kj K 8.8 kj 0.8 kj m kj + x kj kpa m kj kpa m 0.0 kj ( 0.0( kj
5 . An ideal regenerative Ranke cycle it a cloed eedater eater i conidered. e ork produced by te turbe, te ork conumed by te pump, and te eat added te boiler are to be determed. Aumption Steady operatg condition exit. Ketic and potential energy cange are negligible. Analyi From te team table (able A-, A-, and A-, v v 0 0 kpa. kj m v ( P P kj (0.000 m (000 0kPa kpa m.0 kj kj p, P 000 kpa. kj 0 C.0 kj K 8 Cloed P 000 kpa 8. kj Pump P 0 kpa x x. + (0.8(.. kj For an ideal cloed eedater eater, te eedater i eated to te exit temperature o te extracted team, ic ideally leave te eater a a aturated liuid at te extraction preure. P 000 kpa. kj x 0. C MPa 8. kj P 000 kpa MPa y. kj 0. C An energy balance on te eat excanger give te 8 0 kpa -y raction o team extracted rom te turbe ( m & / m& or cloed eedater eater: m& ii m& ee m& + m& m& + m& y + + y Rearrangg,.. y en, + ( y( ( 0.( kj Alo,, P,.0 kj.. kj, P, t 0..8 kj Boiler urbe Condener
Special Topic: Binary Vapor Cycles
0- Special opic: Bary Vapor ycle 0- Bary poer cycle i a cycle ic i actually a combation o to cycle; one te ig temperature region, and te oter te lo temperature region. It purpoe i to creae termal eiciency.
More informationSolutions to Homework #10
Solution to Homeork #0 0-6 A teady-lo Carnot enge it ater a te orkg luid operate at peciied condition. e termal eiciency, te preure at te turbe let, and te ork put are to be determed. Aumption Steady operatg
More informationT Turbine 8. Boiler fwh fwh I Condenser 4 3 P II P I P III. (a) From the steam tables (Tables A-4, A-5, and A-6), = = 10 MPa. = 0.
- - A team poer plant operate on an ideal regenerative anke cycle it to open feedater eater. e poer put of te poer plant and te termal efficiency of te cycle are to be determed. Aumption Steady operatg
More informationv v = = Mixing chamber: = 30 or, = s6 Then, and = 52.4% Turbine Boiler process heater Condenser 7 MPa Q in 0.6 MPa Q proces 10 kpa Q out
0-0- A cogeneration plant i to generate poer and proce eat. art o te team extracted rom te turbe at a relatively ig preure i ued or proce eatg. e poer produced and te utilization actor o te plant are to
More information8-4 P 2. = 12 kw. AIR T = const. Therefore, Q &
8-4 8-4 Air i compreed teadily by a compreor. e air temperature i mataed contant by eat rejection to te urroundg. e rate o entropy cange o air i to be determed. Aumption i i a teady-low proce ce tere i
More information1 = The rate at which the entropy of the high temperature reservoir changes, according to the definition of the entropy, is
8-7 8-9 A reverible eat um wit eciied reervoir temerature i conidered. e entroy cange o two reervoir i to be calculated and it i to be determed i ti eat um atiie te creae entroy rcile. Aumtion e eat um
More informationProblem 1 The turbine is an open system. We identify the steam contained the turbine as the control volume. dt + + =
ME Fall 8 HW olution Problem he turbe i an open ytem. We identiy the team contaed the turbe a the control volume. Ma conervation: t law o thermodynamic: Aumption: dm m m m dt + + de V V V m h + + gz +
More informationThermodynamics Lecture Series
Termodynamics Lecture Series Ideal Ranke Cycle Te Practical Cycle Applied Sciences Education Researc Group (ASERG) Faculty of Applied Sciences Universiti Teknologi MARA email: drjjlanita@otmail.com ttp://www5.uitm.edu.my/faculties/fsg/drjj1.tml
More informationSIMPLE RANKINE CYCLE. 3 expander. boiler. pump. condenser 1 W Q. cycle cycle. net. shaft
SIMPLE RANKINE CYCLE um boiler exander condener Steady Flow, Oen Sytem - region ace Steady Flow Energy Equation for Procee m (u Pum Proce,, Boiler Proce,, V ρg) 0, 0, Exanion Proce,, 0, Condener Proce,,
More informationsince (Q H /T H ) = (Q L /T L ) for reversible cycles. Also, since Q diff is a positive quantity. Thus,
7-5 7-84 he alidity o the Clai eqality i to be demontrated g a reerible and an irreerible heat enge operatg between the ame temperatre limit. Analyi Conider two heat enge, one reerible and one irreerible,
More information6-5. H 2 O 200 kpa 200 C Q. Entropy Changes of Pure Substances
Canges f ure Substances 6-0C Yes, because an ternally reversible, adiabatic prcess vlves n irreversibilities r eat transfer. 6- e radiatr f a steam eatg system is itially filled wit supereated steam. e
More informationMAE320-HW7A. 1b). The entropy of an isolated system increases during a process. A). sometimes B). always C). never D).
MAE0-W7A The homework i due Monday, November 4, 06. Each problem i worth the point indicated. Copying o the olution rom another i not acceptable. (). Multiple choice (0 point) a). Which tatement i invalid
More informationKNOWN: Air undergoes a polytropic process in a piston-cylinder 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 information( ) Given: In a constant pressure combustor. C4H10 and theoretical air burns at P1 = 0.2 MPa, T1 = 600K. Products exit at P2 = 0.
(SP 9) N-butane (C4H1) i burned with 85 percent theoretical air, and the product of combution, an equilibrium mixture containing only O, CO, CO, H, HO, N, and NO, exit from a combution chamber at K,. MPa.
More informationLecture 38: Vapor-compression refrigeration systems
ME 200 Termodynamics I Lecture 38: Vapor-compression refrigeration systems Yong Li Sangai Jiao Tong University Institute of Refrigeration and Cryogenics 800 Dong Cuan Road Sangai, 200240, P. R. Cina Email
More informationDelft University of Technology DEPARTMENT OF AEROSPACE ENGINEERING
Delft University of Technology DEPRTMENT OF EROSPCE ENGINEERING Course: Physics I (E-04) Course year: Date: 7-0-0 Time: 4:00-7:00 Student name and itials (capital letters): Student number:. You have attended
More informationChapter 8 EXERGY A MEASURE OF WORK POTENTIAL
8- Chapter 8 EXERGY A MEAURE OF ORK POENIAL Exergy, Irreveribility, Reverible ork, and econd-law Efficiency 8-C Reverible work differ from the ueful work by irreveribilitie. For reverible procee both are
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 information300 kpa 77 C. (d) If we neglect kinetic energy in the calculation of energy transport by mass
6-6- Air flows steadily a ie at a secified state. The diameter of the ie, the rate of flow energy, and the rate of energy transort by mass are to be determed. Also, the error oled the determation of energy
More informationGiven A gas turbine power plant operating with air-standard Brayton cycle
ME-200 Fall 2017 HW-38 1/4 Given A ga turbine power plant operating with air-tandard Brayton cycle Find For ientropic compreion and expanion: (a) Net power (kw) produced by the power plant (b) Thermal
More informationSENSITIVITY ANALYSIS FOR COUNTER FLOW COOLING TOWER- PART I, EXIT COLD WATER TEMPERATURE
SENSITIVITY ANALYSIS FOR COUNTER FLOW COOLING TOWER- PART I, EXIT COLD WATER TEMPERATURE *Citranjan Agaral Department of Mecanical Engineering, College of Tecnology and Engineering, Maarana Pratap Univerity
More informationm = P 1V 1 RT 1 P 2 = P 1. P 2 V 2 T 2 = P 1V 1 T 1 V 1 2 V 1 T 2 = T 1. {z} T 2 = T 1 1W 2 = PdV = P 1 (V 2 V 1 ). Z T2 (c vo + αt)dt.
NAME: SOLUTION AME 0 Thermodynamic Examination Prof J M Power March 00 Happy 56th birthday, Sir Dugald Clerk, inventor of the two-troke engine, b March 85 (5) A calorically imperfect ideal ga, with ga
More informationChapter 12 Radiation Heat Transfer. Special Topic: Heat Transfer from the Human Body
Chapter 1 Radiation Heat ranfer Special opic: Heat ranfer from the Human Body 1-7C Ye, roughly one-third of the metabolic heat generated by a peron who i reting or doing light work i diipated to the environment
More informationENGINEERING OF NUCLEAR REACTORS. Tuesday, October 9 th, 2014, 1:00 2:30 p.m.
.31 ENGINEERING OF NUCLEAR REACTORS Tuesday, October 9 th, 014, 1:00 :30 p.m. OEN BOOK QUIZ 1 (solutions) roblem 1 (50%) Loss o condensate pump transient in a LWR condenser i) Consider the seaater in the
More information& out. R-134a 34 C
5-9 5-76 Saturated refrigerant-4a vapor at a saturation temperature of T sat 4 C condenses side a tube. Te rate of eat transfer from te refrigerant for te condensate exit temperatures of 4 C and 0 C are
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 informationSection A 01. (12 M) (s 2 s 3 ) = 313 s 2 = s 1, h 3 = h 4 (s 1 s 3 ) = kj/kgk. = kj/kgk. 313 (s 3 s 4f ) = ln
0. (a) Sol: Section A A refrigerator macine uses R- as te working fluid. Te temperature of R- in te evaporator coil is 5C, and te gas leaves te compressor as dry saturated at a temperature of 40C. Te mean
More informationStep 1: Draw a diagram to represent the system. Draw a T-s process diagram to better visualize the processes occurring during the cycle.
ENSC 61 Tutorial, eek#8 Ga Refrigeration Cycle A refrigeration yte ug a the workg fluid, conit of an ideal Brayton cycle run revere with a teperature and preure at the let of the copreor of 37C and 100
More informationTHERMODYNAMICS PRACTICE PROBLEMS
HERMODYNAMICS PRACICE PROBLEMS. A Carot refrigerator has a coefficiet of performace of 0. If the refrigerator s terior is to be kept at 45 C, the temperature of the refrigerator s temperature reservoir
More informationThe average velocity of water in the tube and the Reynolds number are Hot R-134a
hater 0:, 8, 4, 47, 50, 5, 55, 7, 75, 77, 8 and 85. 0- Refrigerant-4a is cooled by water a double-ie heat exchanger. he overall heat transfer coefficient is to be determed. Assumtions he thermal resistance
More informationUnit 12 Refrigeration and air standard cycles November 30, 2010
Unit Rerigeration and air tandard cycle Noveber 0, 00 Unit Twelve Rerigeration and Air Standard Cycle Mechanical Engineering 70 Therodynaic Larry Caretto Noveber 0, 00 Outline Two oewhat related topic
More informationCarnot Factor of a Vapour Power Cycle with Regenerative Extraction
Journal of Modern Pysics, 2017, 8, 1795-1808 ttp://www.scirp.org/journal/jmp ISSN Online: 2153-120X ISSN Print: 2153-1196 arnot Factor of a Vapour Power ycle wit Regenerative Extraction Duparquet Alain
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 informationME 200 Thermodynamics 1 Fall 2017 Exam 3
ME 200 hermodynamics 1 Fall 2017 Exam Circle your structor s last name Division 1: Naik Division : Wassgren Division 6: Braun Division 2: Sojka Division 4: Goldenste Division 7: Buckius Division 8: Meyer
More informationAssumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. hl = IOkPa = 191.
~ Chapter 9 Vapor and Combined Power Cycles 9-16 A steam power plant that operates on a simple ideal Rankine cycle is considered. he quality of the steam at the turbine exit, the thermal efficiency of
More informationThermodynamics Lecture Series
Thermodynamics Lecture Series Second Law uality of Energy Applied Sciences Education Research Group (ASERG) Faculty of Applied Sciences Universiti Teknologi MARA email: drjjlanita@hotmail.com http://www.uitm.edu.my/faculties/fsg/drjj.html
More informationThen the amount of water that flows through the pipe during a differential time interval dt is (1) 4
5-98 Review Problems 5-45 A tank oen to te atmosere is itially filled wit. e tank discarges to te atmosere troug a long ie connected to a valve. e itial discarge velocity from te tank and te time required
More informationMAHALAKSHMI ENGINEERING COLLEGE
MAHALAKSHMI ENGINEERING COLLEGE TIRUCHIRAPALLI-621213. Department: Mechanical Subject Code: ME2202 U N IT - 1 Semester: III Subject Name: ENGG. THERMODYNAMICS 1. 1 kg of gas at 1.1 bar, 27 o C is compressed
More informationPsychrometrics. PV = N R u T (9.01) PV = N M R T (9.02) Pv = R T (9.03) PV = m R T (9.04)
Pycrometric Abtract. Ti capter include baic coverage of pycrometric propertie and pycrometric procee. Empai i upon propertie and procee relative to te environment and to proceing of biological material.
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 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 019-4747351 04-9798679 Chapter
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 informationChapter 5 MASS AND ENERGY ANALYSIS OF CONTROL VOLUMES
5- Chapter 5 MASS AND ENERGY ANALYSIS OF CONTROL OLUMES Conseration of Mass 5-C Mass, energy, momentum, and electric charge are consered, and olume and entropy are not consered durg a process. 5-C Mass
More informationDepartment of Civil Engineering & Applied Mechanics McGill University, Montreal, Quebec Canada
Department f Ciil ngeerg Applied Mechanics McGill Uniersity, Mntreal, Quebec Canada CI 90 THRMODYNAMICS HAT TRANSFR Assignment #4 SOLUTIONS. A 68-kg man whse aerage bdy temperature is 9 C drks L f cld
More informationChapters 19 & 20 Heat and the First Law of Thermodynamics
Capters 19 & 20 Heat and te First Law of Termodynamics Te Zerot Law of Termodynamics Te First Law of Termodynamics Termal Processes Te Second Law of Termodynamics Heat Engines and te Carnot Cycle Refrigerators,
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 informationLecture 12 - Non-isolated DC-DC Buck Converter
ecture 12 - Non-iolated DC-DC Buck Converter Step-Down or Buck converter deliver DC power from a higher voltage DC level ( d ) to a lower load voltage o. d o ene ref + o v c Controller Figure 12.1 The
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 informationBernoulli s equation may be developed as a special form of the momentum or energy equation.
BERNOULLI S EQUATION Bernoulli equation may be developed a a pecial form of the momentum or energy equation. Here, we will develop it a pecial cae of momentum equation. Conider a teady incompreible flow
More informationHEAT ENGINES AND REFRIGERATORS
EA ENGINES AND REFRIGERAORS 9 onceptual uestions 9.. s =. (a) < 0, > 0. ork is done by the system; the area under the curve is positive. s (b) > 0, < 0. ork is done on the system to compress it to a smaller
More informationDesign of Robust PI Controller for Counter-Current Tubular Heat Exchangers
Deign of Robut PI Controller for Counter-Current Tubular Heat Excanger Jana Závacká Monika Bakošová Intitute of Information Engineering Automation Matematic Faculty of Cemical Food Tecnology STU in Bratilava
More informationChapter 7 ENTROPY. 7-3C The entropy change will be the same for both cases since entropy is a property and it has a fixed value at a fixed state.
7- Chapter 7 ENROY Entropy and the Increae of Entropy rciple 7-C No. he δ Q repreent the net heat tranfer durg a cycle, which could be poitive. 7-C No. A ytem may produce more (or le) work than it receive
More informationThermodynamic Cycles
Thermodynamic Cycles Content Thermodynamic Cycles Carnot Cycle Otto Cycle Rankine Cycle Refrigeration Cycle Thermodynamic Cycles Carnot Cycle Derivation of the Carnot Cycle Efficiency Otto Cycle Otto Cycle
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 informationCHAPTER 8 OBSERVER BASED REDUCED ORDER CONTROLLER DESIGN FOR LARGE SCALE LINEAR DISCRETE-TIME CONTROL SYSTEMS
CHAPTER 8 OBSERVER BASED REDUCED ORDER CONTROLLER DESIGN FOR LARGE SCALE LINEAR DISCRETE-TIME CONTROL SYSTEMS 8.1 INTRODUCTION 8.2 REDUCED ORDER MODEL DESIGN FOR LINEAR DISCRETE-TIME CONTROL SYSTEMS 8.3
More informationEVAPORATION. Robert evaporator. Balance equations Material balance (total) Component balance. Heat balance
EAPORATION Roert eorator Balance equation Material alance (total) S S=Solution =aor =Steam K=Steam condenate S Comonent alance S S Heat alance S S v Merkel lot can e ued for otaining entaly data Heat ower
More informationNumber of extra papers used if any
Last Name: First Name: Thermo no. ME 200 Thermodynamics 1 Fall 2018 Exam Circle your structor s last name Division 1 (7:0): Naik Division (1:0): Wassgren Division 6 (11:0): Sojka Division 2 (9:0): Choi
More information4-93 RT RT. He PV n = C = = Then the boundary work for this polytropic process can be determined from. n =
- A cylder i itially filled it eliu ga at a ecified tate. Heliu i creed lytrically t a ecified teerature and reure. e eat tranfer durg te rce i t be detered. Autin Heliu i an ideal ga it cntant ecific
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 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 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: Pressure, temperature, and velocity of steam entering a 1.6-cm-diameter pipe.
4.3 Steam enters a.6-cm-diameter 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.6-cm-diameter
More informationJournal of Chemical and Pharmaceutical Research, 2013, 5(12): Research Article
Available online.jocpr.com Journal of emical and Parmaceutical Researc, 013, 5(1):55-531 Researc Article ISSN : 0975-7384 ODEN(USA) : JPR5 Performance and empirical models of a eat pump ater eater system
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 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 informationρ water = 1000 kg/m 3 = 1.94 slugs/ft 3 γ water = 9810 N/m 3 = 62.4 lbs/ft 3
CEE 34 Aut 004 Midterm # Anwer all quetion. Some data that might be ueful are a follow: ρ water = 1000 kg/m 3 = 1.94 lug/ft 3 water = 9810 N/m 3 = 6.4 lb/ft 3 1 kw = 1000 N-m/ 1. (10) A 1-in. and a 4-in.
More informationMAE 320 Thermodynamics HW 5 Assignment
MAE 0 Therodynaics HW 5 Assignent The hoework is due Wednesday, October 9, 06. Each proble is worth the pots dicated. Copyg o the solution ro another is not acceptable.. Multiple choices questions. There
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 Dhaka-1000, Bangladesh zahurul@me.buet.ac.bd
More informationME Thermodynamics I
HW-6 (5 points) Given: Carbon dioxide goes through an adiabatic process in a piston-cylinder assembly. provided. Find: Calculate the entropy change for each case: State data is a) Constant specific heats
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, closed-book examination. 2. Permitted
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 informationBME-A PREVIOUS YEAR QUESTIONS
BME-A PREVIOUS YEAR QUESTIONS CREDITS CHANGE ACCHA HAI TEAM UNIT-1 Introduction: Introduction to Thermodynamics, Concepts of systems, control volume, state, properties, equilibrium, quasi-static process,
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 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 informationHydraulic validation of the LHC cold mass heat exchanger tube.
Hydraulic validation o te LHC cold mass eat excanger tube. LHC Project Note 155 1998-07-22 (pilippe.provenaz@cern.c) Pilippe PROVENAZ / LHC-ACR Division Summary Te knowledge o te elium mass low vs. te
More informationSOLUTION MANUAL ENGLISH UNIT PROBLEMS CHAPTER 11 SONNTAG BORGNAKKE VAN WYLEN. FUNDAMENTALS of. Thermodynamics. Sixth Edition
SOLUION MANUAL ENGLISH UNI PROBLEMS CHAPER SONNAG BORGNAKKE VAN WYLEN FUNDAMENALS of hermodynamic Sixth Edition CHAPER SUBSECION PROB NO. Rankine Cycle 67-8 Brayton Cycle 8-87 Otto, Dieel, Stirling and
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 information(b) The heat transfer can be determined from an energy balance on the system
8-5 Heat is transferred to a iston-cylinder device wit a set of stos. e work done, te eat transfer, te exergy destroyed, and te second-law efficiency are to be deterined. Assutions e device is stationary
More informationNUCLEAR THERMAL-HYDRAULIC FUNDAMENTALS
NUCLEAR THERMAL-HYDRAULIC FUNDAMENTALS Dr. J. Micael Doster Departent of Nuclear Engineering Nort Carolina State University Raleig, NC Copyrigted POER CYCLES Te analysis of Terodynaic Cycles is based alost
More informationExam 3. Feb 12, Problem 1 / 32 Problem 2 / 33 Problem 3 / 32 Total / 100
ROSE-HULMAN Institute of Technology Sophomore Engineering Curriculum ES201 Conservation & Accounting Principles Winter 2014-2015 Section [1 pt]: 01 (1 st period) Name [1 pt] 02 (2 nd period) CM [1 pt]
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 Qur-ān In many engineering applications,
More informationUnsteady State Simulation of Vapor Compression Heat Pump Systems With Modular Analysis (This paper will not be presented.)
Purdue University Purdue e-pubs nternational Refrigeration and ir Conditioning Conference Scool of Mecanical Engineering 0 Unsteady State Simulation of Vapor Compression Heat Pump Systems Wit Modular nalysis
More informationthe first derivative with respect to time is obtained by carefully applying the chain rule ( surf init ) T Tinit
.005 ermal Fluids Engineering I Fall`08 roblem Set 8 Solutions roblem ( ( a e -D eat equation is α t x d erfc( u du π x, 4αt te first derivative wit respect to time is obtained by carefully applying te
More informationME Thermodynamics I
Homework - Week 01 HW-01 (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 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 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 informationSPC 407 Sheet 2 - Solution Compressible Flow - Governing Equations
SPC 407 Sheet 2 - Solution Compressible Flow - Governing Equations 1. Is it possible to accelerate a gas to a supersonic velocity in a converging nozzle? Explain. No, it is not possible. The only way to
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 informationMain components of the above cycle are: 1) Boiler (steam generator) heat exchanger 2) Turbine generates work 3) Condenser heat exchanger 4) Pump
Introducton to Terodynacs, Lecture -5 Pro. G. Cccarell (0 Applcaton o Control olue Energy Analyss Most terodynac devces consst o a seres o coponents operatng n a cycle, e.g., stea power plant Man coponents
More informationDetermination of heat transfer intensity between free streaming water film and rigid surface using thermography
IV Conferencia Panamericana de END Buenos Aires Octubre 2007 Determination of eat transfer intensity between free ing water film and rigid surface using termograpy Ivanka Boras and Srecko Svaic Faculty
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 informationChapter 7: 17, 20, 24, 25, 32, 35, 37, 40, 47, 66 and 79.
hapter 7: 17, 0,, 5,, 5, 7, 0, 7, 66 and 79. 77 A power tranitor mounted on the wall diipate 0.18 W. he urface temperature of the tranitor i to be determined. Aumption 1 Steady operating condition exit.
More informationKelvin Planck Statement of the Second Law. Clausius Statement of the Second Law
Kelv Planck Statement of te Second aw It is imossible to construct an enge wic, oeratg a cycle, will roduce no oter effect tan te extraction of eat from a sgle reservoir and te erformance of an equivalent
More informationMinimizing and maximizing compressor and turbine work respectively
Minimizing and maximizing compressor and turbine ork respectively Reversible steady-flo ork In Chapter 3, Work Done during a rocess as found to be W b dv Work Done during a rocess It depends on the path
More informationChapter 1: 20, 23, 35, 41, 68, 71, 76, 77, 80, 85, 90, 101, 103 and 104.
Chapter 1: 0, 3, 35, 1, 68, 71, 76, 77, 80, 85, 90, 101, 103 and 10. 1-0 The filament of a 150 W incandescent lamp is 5 cm long and has a diameter of 0.5 mm. The heat flux on the surface of the filament,
More informationd dt T R m n p 1. (A) 4. (4) Carnot engine T Refrigerating effect W COPref. = 1 4 kw 5. (A)
. (A). (C) 5. (C) 7. ( to 5) 9. (C) 6. (C). (C). (D) 6. (A) 8. (0.6 to 0.66) 50. (D) 6. (C). (A) 5. (C) 7. (A) 9. (C) 5. (D) 6. (C). () 6. (C) 8. (600) 0. (D) 5. (B) 6. (D) 5. (A) 7. (A) 9. (D). (C) 5.
More informationME 200 Thermodynamics 1 Spring Exam 2
Last Name: First Name: Thermo no. ME 200 Thermodynamics 1 Sprg 2017 - Exam 2 Circle your structor s last name Ardekani Fisher Hess Naik Sojka (onle and on campus) INSTRUCTIONS This is a closed book and
More informationFirst Name Last Name CIRCLE YOUR LECTURE BELOW: Div. 1 10:30 am Div. 2 2:30 pm Div. 3 4:30 pm Prof. Gore Prof. Udupa Prof. Chen
CIRCLE YOUR LECURE BELOW: Div. 1 10:30 am Div. :30 m Div. 3 4:30 m Prof. Gore Prof. Udua Prof. Chen EXAM # 3 INSRUCIONS 1. his is a closed book examination. All needed roerty tables are rovided.. Do not
More informationTwo mark questions and answers UNIT II SECOND LAW 1. Define Clausius statement. It is impossible for a self-acting machine working in a cyclic process, to transfer heat from a body at lower temperature
More informationLECTURE NOTE THERMODYNAMICS (GEC 221)
LETURE NOTE ON THERMODYNAMIS (GE ) Thermodynamics is the branch of science that treats the arious phenomena of energy and related properties of matter especially the relationship between heat, work and
More information