( ) Given: In a constant pressure combustor. C4H10 and theoretical air burns at P1 = 0.2 MPa, T1 = 600K. Products exit at P2 = 0.

Size: px
Start display at page:

Download "( ) Given: In a constant pressure combustor. C4H10 and theoretical air burns at P1 = 0.2 MPa, T1 = 600K. Products exit at P2 = 0."

Transcription

1 (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. The average pecific heat at contant preure for butane i c = 1.71 / K in the temperature range 3 K 6 K. p ( ) (a) Determine the equilibrium compoition in the product tream neglecting O and NO (thi will require 4 atom balance and 1 equilibrium equation). (b) Now determine the mole fraction of O and NO in the product tream uing two additional equilibrium equation and the mole fraction you determined in part (a). (c) Auming the reactant are premixed and enter the combution chamber at 6 K,. MPa, determine the heat tranfer to or from the combution chamber in per of n-butane. (Hint: the concentration of O and NO will be much lower than the other concentration and do not need to be included in the thermochemitry calculation). Given: In a contant preure combutor C4H1 and theoretical air burn at P1 =. MPa, T1 = 6K Product exit at P =. MPa, T = K Find: (a) y,n, y,co, y,h O, y,co, y,h (b) y,o, y,no (c) Q CV Sytem ketch:

2 Aumption: (1) Ideal gae () Steady tate, uniform flow (3) Contant preure combution (4) Neglect KE, PE (5) W CV = Baic equation: de dt CV ( ) = = Q W CV CV + n h n h k 1, k 1, k, k, k k ( ) ( ) Q = n h n h = H H h T = h + h T K T CV, k, k 1, k 1, k 1 i f,98, i, i 98 k k C D yc y D p = A B ya y B p ref C + D A B Solution: (a) Proce Equation Stoichiometric equation (a toich = x + y 4 ) : C 4 H (O N ) 5H O + 4CO N Reaction equation with 85% theoretical air (a =.85a toich = 5.55) : C 4 H (O N ) bco + cco + dh O + eh + fn + go + hno Neglecting NO and O, the reaction equation become C 4 H (O N ) bco + cco + dh O + eh N Atom balance: C: 4=b+c c=4-b H: 1=d+e e=5-d O: 5.55=b+c+d=4+b+d d=7.5-b

3 N: = Chemical equilibrium Equilibrium equation and contant from Table A-7: CO + H H O + CO K eq = = 4.59 at T = K From our reaction equation, Solve the quadratic equation, It give, K eq = y CO 1 y H O 1 y 1 CO y 1 ( P ) = cd (4 b)(7.5 b) = H P ref be b( b) (4 b)(7.5 b) b( b) = 4.59 b =.5877 and 3.88 Chooe the poitive value. Then, c = 1.418, d = 4.468, and e = Thu, the reaction equation i CH (O N ).5877 CO CO H O H +.774N The equilibrium compoition i.5877 CO CO H O H +.774N At tate, n,tot n,co + n,co + n,h o + n,h + n,n = = / Therefore, mole fraction are y CO = n,co n,tot = =.8691 y CO = n,co = n,tot =.4743 y N = n,n =.774 n,tot =.6967

4 y H = n,h =.5377 n,tot =.186 y H O = n,h O n,tot = = y CO =.8691 y CO =.4743 y N =.6967 y H =.186 y H O = (b) In order to olve equilibrium decompoition for NO and O imultaneouly by adding two additional equilibrium equation, we need to etablih a reaction equation a decribed below,.774n CO CO H O H xco + yco + zo + pn + qno H O H Note: For NO, 1 O + 1 N NO i required. For O, there are two option: CO CO + 1 O and H O H + 1 O. But, we only need one more equilibrium equation a the problem tatement. Then, we have to chooe CO CO + 1 O baed on a higher equilibrium contant: K CO CO+ 1 O = > K H O H + 1 O = Herein, it i aumed that the effect of H O and H i negligible and the gae exit a an inert ga. Atom balance: C: =x + y y = 4 x O: = x + y + z + q z = ( x y q)/ N:.774 =p+q p =.774.5q With equilibrium CO CO + 1 O 1 K eq (T ) = y CO 1 y O ( P ) = =.136 at T = K y 1 CO P ref yz.5 x ( kpa (x + y + z + p + q )11.35 kpa ).5 =.136 yz.5 x(x + y + z + p + q ).5 = =.996

5 With equilibrium 1 O + 1 N NO K eq (T ) = y NO 1 y O 1 1 y N ( P ) = =. P ref q z.5 =. p.5 at T = K Solve 5 equation and 5 unknown (EES code for thi i attached). Then, x =.587 y = z =.8619 p =.77 q =.8463 Therefore, the equilibrium compoition i.587 CO CO O +.77 N NO H O H Conequently, the reaction become C 4 H (O N ).587 CO CO O +.77 N NO H O H The mole fraction for are y CO =.869 y CO =.4746 y N =.6977 y H =.186 y H O =.1499 y NO = y O = EES code "5 equation" y=4-x z=( x-y-q)/ p= *q (y*z^(.5))/x/((x+y+z+p+q )^(.5))=.996 q/(z^(.5))/(p^(.5))=.

6 (c) Uing Table A-3 and A-5, Reactant T = 6 K, p = bar h c 1 1 f,98, C H 4 1 = 15,51 = 1.71 c = = 99.4 K K K p, C H p, C H h, C ( ) ( ) ( ) ( ) 4H 6 K h 1, C 98 4H K = c 1 p, C H T = 1 K K h, C ( 6 ) 4H K h ( ) 1, C4H 98 K = 3, 1 hc ( ) ( ) ( ) 4H 6 K = h 1 f,98, C4H + h 1, C 6 4H K h 1, C 98 95, 49 4H K = 1 ho ( 6 K ) = h f,98,o + h ( ) ( ),O 6 K h,o 98 K = 17,99 8, 68 = 9, 47 hn ( 6 K ) = h f,98, N + h ( ) ( ), N 6 K h, N 98 K = 17,563 8, 669 = 8,894 H 1 = n 1,k h 1,k = (1 ) ( 95,49 ) + (5.55 ) (9,47 C k 4 H 1 ) O + (.774 ) (8,894 ) N = 14,36

7 Product T = K, p = bar 1 hco ( K ) = h ( ) ( ) f,98, CO + h, CO K h, CO 98 K = 393,5 + 1,84 9,364= 3, 8 hco ( K ) = hf,98, CO + h, CO ( K ) h, CO ( 98 K ) = 11, , 48 8, 669 = 5379 hh ( ) O K = hf,98, H ( ) ( ) O + h, HO K h, H 98 41,8 8,593 9, O K = + = hh ( K ) = h f,98, H + h ( ) ( ), H K h, H 98 K = 5,97 hn ( K ) = h f,98, N + h ( ) ( ), N K h, N 98 K = 64,81 8, 669 = 56,14 H = n,k k h,k = (.5877 ) ( 3,8 ) CO + (1.418 ) ( 53,79 ) CO + (4.468 ) ( 169,13 ) H O + (.774 ) (56,14 ) N = 417,635 Q CV = H H 1 = 417,635 14,36 = 557,995 + (.5377 ) (5,97 ) H Thu, of n butane 1 Q CV 1 of n butane per econd in the reaction = 557,995 = 557,995 of n butane

8 (SP3 & SP31 Canceled) (SP3) A imple ideal Rankine cycle with water a the working fluid operate between the preure limit of 15 MPa in the boiler and 1 kpa in the condener. Saturated team enter the turbine. a. Determine the work produced by the turbine per unit ma, the heat tranferred in the boiler per unit ma, and the thermal efficiency of the cycle. b. Now conider that irreveribilitie in the turbine caue the team quality at the outlet of the turbine to be.7. Determine the ientropic efficiency of the turbine and the thermal efficiency of the cycle. Given: A Rankine cycle with water operate at Pcond = 1 kpa, Pboiler = 15 MPa the team quality at the turbine inlet i 1. Find: (a) W turbine/m, Q bolier/m, and η th (b) η ien, η th,irrev Sytem ketch: None Aumption: 1. Steady tate, teady flow. KE and PE negligible 3. Internally reverible pump 4. No work at the boiler and the condener 5. Adiabatic at the turbine and the pump 6. Water enter the pump a a aturated liquid 7. Negligible change of pecific volume in the pump

9 Baic equation: de cv dt = = Q cv W cv + m i (h i + v i i + gz i) m e (h e + v e e + gz e) Solution: (a) T- diagram for the ideal Rankine cycle i Q in/m Q out/m State 1-: pump State -3: boiler State 3-4: turbine State 4-1: condener For bolier and condenor: Q cv/m + h in h out = For turbine and pump: W cv/m + h in h out = Steam propertie can be found from the team table, Table A- and A-3 h 1 = h f,1kpa = v 1 = v f,1kpa =.143 m3 Work done by an internally reverible adiabatic pump (refer Chapter 8..): W pump m = v 1 (P P 1 ) =.143 m3 (15, 1)kPa = h = h 1 + W pump m = = 433.

10 P 3 = 15, kpa, x 3 = 1 h 3 = = K P 4 = 1 kpa, 4 = 4 x 4 = 4 f fg = = / h 4 = h f + x 4 h fg = = Thu, W turbine/m = h 3 h 4 = = / Q bolier/m = h 3 h = = / Q condenor/m = h 4 h 1 = = / The thermal efficiency of the cycle i η th = 1 Q out Q in = 1 Q condenor/m Q bolier/m = =.3139 (b) T- diagram for the Rankine cycle with irreveribilitie at the turbine i Q in/m Q out/m State 1-: pump State -3: boiler State 3-4: turbine State 4-1: condener Except propertie at tate 4, all propertie are identical with part (a).

11 h 1 = h = 433. h 3 = 61.5 h 4 = Becaue the team quality at the outlet of the turbine i.7 P 4 = 1 kpa, x 4 = 1 h 4 = h f + x 4 h fg = = / The ientropic efficiency of the turbine i η ien = h 3 h 4 h 3 h 4 = =.876 Newly calculated Q condenor/m, Q condenor m = h 4 h 1 = = The thermal efficiency of the cycle with irrerveribilitie at the turbine i η th,irrev = 1 Q out Q in = Q condenor/m Q bolier/m = =.741

12 (SP33) A team power plant operate on an ideal reheat Rankine cycle between the preure limit of 15 MPa and 1 kpa. The ma flow rate of team through the cycle i 1 /. Steam enter both tage of the turbine at 5 C. If the moiture content of the team at the exit of the low-preure turbine i not to exceed 1 percent, a. Determine the preure at which reheating take place. b. Determine the total rate of heat input in the boiler. c. Determine the thermal efficiency of the cycle. d. Plot the cycle on a T- diagram with repect to aturation line. Given: An ideal reheat Rankine cycle with water operate at Pmin = 1 kpa, Pmax = 15 MPa Tturb1,inlet =Tturb,inlet = 5 m = 1 / The quality at the exit of the low-preure turbine i targeted.9 Find: (a) Preheat (b) Q in (c) η th (d) T- diagram Sytem ketch: None Aumption: 1. Steady tate, teady flow. KE and PE negligible 3. Internally reverible pump

13 4. No work at the boiler and the condener 5. Adiabatic at the turbine and the pump 6. Water enter the pump a a aturated liquid 7. Negligible change of pecific volume in the pump Baic equation: de cv dt = = Q cv W cv + m i (h i + v i i + gz i) m e (h e + v e e + gz e) Solution: T- diagram for the ideal reheat Rankine cycle i State 1-: pump 15 MPa State -3: boiler State 3-4: 1 t turbine State 4-5: reheat State 5-6: nd turbine State 6-1: condener (a) Steam propertie can be found from the team table, Table A-, A-3, and A-4 h 1 = h f,1kpa = v 1 = v f,1kpa =.11 m3 Work done by an internally reverible adiabatic pump (refer Chapter 8..): W pump m = v 1 (P P 1 ) =.11 m3 h = h 1 + W pump m = = 6.97 (15, 1)kPa = 15.14

14 P3 = 15, kpa, T3 = 5 h3 = / 3 = /K (need double interpolation from the table) From the condition, the quality at the exit of the low-preure turbine i targeted.9, we et x 6 =.9 P6 = 1 kpa h 6 = h f + x 6 h fg = = = f + x 6 fg = ( ) = 7.41 K T5 = 5, 5 = = MPa P 5 MPa 3MPa P 5 =.16 MPa(the reheat preure) = h 5 h = h 5 = / P4 =.16 MPa, 4 = 3 h 4 = 8.55 h 4 = (need double interpolation from the table) h 4 = (b) The rate of heat input i Q in = m [(h 3 h ) + (h 5 h 4 )] = 1 = 45,4.8 [( ) + ( )] (c)

15 The thermal efficiency i determined from Q in and Q out. Q out = m (h 6 h 1 ) = 1 ( ) = 5,84. Thu, η th = 1 Q out Q in (d) = 1 5,84. =.46 or 4.6 % 45,4.8 It i drawn at the initial part of the olution.

KNOWN: Air undergoes a polytropic process in a piston-cylinder assembly. The work is known.

KNOWN: 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 A gas turbine power plant operating with air-standard Brayton cycle

Given 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 information

2b m 1b: Sat liq C, h = kj/kg tot 3a: 1 MPa, s = s 3 -> h 3a = kj/kg, T 3b

2b 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

MAE320-HW7A. 1b). The entropy of an isolated system increases during a process. A). sometimes B). always C). never D).

MAE320-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 information

SOLUTION MANUAL ENGLISH UNIT PROBLEMS CHAPTER 11 SONNTAG BORGNAKKE VAN WYLEN. FUNDAMENTALS of. Thermodynamics. Sixth Edition

SOLUTION 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 information

T 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.

T 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 information

v 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

v 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 information

Problem 1 The turbine is an open system. We identify the steam contained the turbine as the control volume. dt + + =

Problem 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 information

SOLUTION MANUAL CHAPTER 12

SOLUTION MANUAL CHAPTER 12 SOLUION MANUAL CHAPER CONEN SUBSECION PROB NO. In-ext Concept Quetion a-g Concept problem - Brayton cycle, ga turbine - Regenerator, Intercooler, nonideal cycle 5-9 Ericon cycle 0- Jet engine cycle -5

More information

MAE 11. Homework 8: Solutions 11/30/2018

MAE 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 information

NONISOTHERMAL OPERATION OF IDEAL REACTORS Plug Flow Reactor

NONISOTHERMAL OPERATION OF IDEAL REACTORS Plug Flow Reactor NONISOTHERMAL OPERATION OF IDEAL REACTORS Plug Flow Reactor T o T T o T F o, Q o F T m,q m T m T m T mo Aumption: 1. Homogeneou Sytem 2. Single Reaction 3. Steady State Two type of problem: 1. Given deired

More information

Thermodynamics ENGR360-MEP112 LECTURE 7

Thermodynamics ENGR360-MEP112 LECTURE 7 Thermodynamics ENGR360-MEP11 LECTURE 7 Thermodynamics ENGR360/MEP11 Objectives: 1. Conservation of mass principle.. Conservation of energy principle applied to control volumes (first law of thermodynamics).

More information

Chapter 5. Mass and Energy Analysis of Control Volumes

Chapter 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 information

m = 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.

m = 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 information

Bernoulli s equation may be developed as a special form of the momentum or energy equation.

Bernoulli 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 information

Course: MECH-341 Thermodynamics II Semester: Fall 2006

Course: MECH-341 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: MECH-341 Thermodynamics II Semester: Fall

More information

ENT 254: Applied Thermodynamics

ENT 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 information

Unit Workbook 2 - Level 5 ENG U64 Thermofluids 2018 UniCourse Ltd. All Rights Reserved. Sample

Unit 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 information

FUNDAMENTALS of Thermodynamics

FUNDAMENTALS of Thermodynamics SOLUTION MANUAL SI UNIT PROBLEMS CHAPTER 15 SONNTAG BORGNAKKE VAN WYLEN FUNDAMENTALS of Thermodynamics Sixth Edition CONTENT SUBSECTION PROB NO. Correspondence table Concept-Study Guide Problems 1-20 Equilibrium

More information

ME Thermodynamics I. Lecture Notes and Example Problems

ME 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 information

8-4 P 2. = 12 kw. AIR T = const. Therefore, Q &

8-4 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 information

(SP 1) DLLL. Given: In a closed rigid tank,

(SP 1) DLLL. Given: In a closed rigid tank, (SP 1) Given: In a closed rigid tank, State 1: m 1,ice = 1, m 1,g = 0.05 P1= 0.0381 kpa, T1= -30 o C State 2: the liquid vapor equilibrium line, either saturated liquid or saturated vapor Find: (a) The

More information

The First Law of Thermodynamics. By: Yidnekachew Messele

The 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 information

MAE 101A. Homework 3 Solutions 2/5/2018

MAE 101A. Homework 3 Solutions 2/5/2018 MAE 101A Homework 3 Solution /5/018 Munon 3.6: What preure gradient along the treamline, /d, i required to accelerate water upward in a vertical pipe at a rate of 30 ft/? What i the anwer if the flow i

More information

THERMODYNAMICS, 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, 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 information

Chapter 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 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 information

The exergy of asystemis the maximum useful work possible during a process that brings the system into equilibrium with aheat reservoir. (4.

The 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 information

CHAPTER 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 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 information

c 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)

c 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 information

Lecture 38: Vapor-compression refrigeration systems

Lecture 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 information

Chapter 7. Entropy: A Measure of Disorder

Chapter 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 information

FINAL EXAM. ME 200 Thermodynamics I, Spring 2013 CIRCLE YOUR LECTURE BELOW:

FINAL 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 information

Chapter 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 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 information

first law of ThermodyNamics

first 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 information

( )( ) 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

( )( ) 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 . 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.

More information

m 5 ME-200 Fall 2017 HW-19 1/2 Given Diffuser and two-stage compressor with intercooling in a turbojet engine

m 5 ME-200 Fall 2017 HW-19 1/2 Given Diffuser and two-stage compressor with intercooling in a turbojet engine ME-00 Fall 07 HW-9 / Gven Dffuer and two-tage compreor wth ntercoolng n a turbojet engne Fnd (a) Temperature (K) of ar at the dffuer ext (b) Ma flow rate (/) of ar through the engne (c) Total power nput

More information

(1)5. Which of the following equations is always valid for a fixed mass system undergoing an irreversible or reversible process:

(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 information

Chapter 10. Closed-Loop Control Systems

Chapter 10. Closed-Loop Control Systems hapter 0 loed-loop ontrol Sytem ontrol Diagram of a Typical ontrol Loop Actuator Sytem F F 2 T T 2 ontroller T Senor Sytem T TT omponent and Signal of a Typical ontrol Loop F F 2 T Air 3-5 pig 4-20 ma

More information

ME 2322 Thermodynamics I PRE-LECTURE Lesson 23 Complete the items below Name:

ME 2322 Thermodynamics I PRE-LECTURE 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: 1 L H 2. (10 pt) Can the thermal efficiency of an actual engine ever exceed that of an equivalent Carnot

More information

Availability and Irreversibility

Availability 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 information

AAE COMBUSTION AND THERMOCHEMISTRY

AAE COMBUSTION AND THERMOCHEMISTRY 5. COMBUSTIO AD THERMOCHEMISTRY Ch5 1 Overview Definition & mathematical determination of chemical equilibrium, Definition/determination of adiabatic flame temperature, Prediction of composition and temperature

More information

CHAPTER 5 MASS AND ENERGY ANALYSIS OF CONTROL VOLUMES

CHAPTER 5 MASS AND ENERGY ANALYSIS OF CONTROL VOLUMES Thermodynamics: An Engineering Approach 8th Edition in SI Units Yunus A. Çengel, Michael A. Boles McGraw-Hill, 2015 CHAPTER 5 MASS AND ENERGY ANALYSIS OF CONTROL VOLUMES Lecture slides by Dr. Fawzi Elfghi

More information

4. The following graphs were prepared from experimental data for a reactant, A. What is the correct order of A?

4. The following graphs were prepared from experimental data for a reactant, A. What is the correct order of A? Diebolt Summer 010 CHM 15 HOUR EXAM I KEY (15 pt) Part One: Multiple Choice. 1. Conider the reaction IO 3 + 5I + 6H + 3I + 3H O. If the rate of diappearance of I i 9.010-3 M/, what i the rate of appearance

More information

Lecture 35: Vapor power systems, Rankine cycle

Lecture 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 information

Department 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 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 information

1 year chemistry n0tes new st CHAPTER 8 CHEMICAL EQUILIBRIUM MCQs Q.1 A reaction is reversible because (a) reactants are reactive (b) products are

1 year chemistry n0tes new st CHAPTER 8 CHEMICAL EQUILIBRIUM MCQs Q.1 A reaction is reversible because (a) reactants are reactive (b) products are year chemitry n0te new t CHAPTER 8 CHEMICAL EQUILIBRIUM MCQ Q.1 A reaction i reverible becaue reactant are reactive (b) product are reactive (c) product are table (d) reactant are table Q.2 A large value

More information

ECE309 INTRODUCTION TO THERMODYNAMICS & HEAT TRANSFER. 12 June 2006

ECE309 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, closed-book examination. You are permitted to use one 8.5 in. 11 in. crib sheet (one side

More information

EP225 Note No. 5 Mechanical Waves

EP225 Note No. 5 Mechanical Waves EP5 Note No. 5 Mechanical Wave 5. Introduction Cacade connection of many ma-pring unit conitute a medium for mechanical wave which require that medium tore both kinetic energy aociated with inertia (ma)

More information

ME 354 Tutorial, Week#13 Reacting Mixtures

ME 354 Tutorial, Week#13 Reacting Mixtures ME 354 Tutorial, Week#13 Reacting Mixtures Question 1: Determine the mole fractions of the products of combustion when octane, C 8 H 18, is burned with 200% theoretical air. Also, determine the air-fuel

More information

Previous lecture. Today lecture

Previous 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 information

ME 300 Thermodynamics II

ME 300 Thermodynamics II ME 300 Thermodynamics II Prof. S. H. Frankel Fall 2006 ME 300 Thermodynamics II 1 Week 1 Introduction/Motivation Review Unsteady analysis NEW! ME 300 Thermodynamics II 2 Today s Outline Introductions/motivations

More information

Stoichiometry, Energy Balances, Heat Transfer, Chemical Equilibrium, and Adiabatic Flame Temperatures. Geof Silcox

Stoichiometry, Energy Balances, Heat Transfer, Chemical Equilibrium, and Adiabatic Flame Temperatures. Geof Silcox Stoichiometry, Energy Balances, Heat ransfer, Chemical Equilibrium, and Adiabatic Flame emperatures Geof Silcox geoff@che.utah.edu (80)58-880 University of Utah Chemical Engineering Salt Lake City, Utah

More information

Thermodynamics II. Week 9

Thermodynamics 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

SOLUTIONS

SOLUTIONS SOLUTIONS Topic-2 RAOULT S LAW, ALICATIONS AND NUMERICALS VERY SHORT ANSWER QUESTIONS 1. Define vapour preure? An: When a liquid i in equilibrium with it own vapour the preure exerted by the vapour on

More information

1 year n0tes chemistry new st CHAPTER 7 THERMOCHEMISTRY MCQs Q.1 Which of the following statements is contrary to the first law of thermodynamics?

1 year n0tes chemistry new st CHAPTER 7 THERMOCHEMISTRY MCQs Q.1 Which of the following statements is contrary to the first law of thermodynamics? year n0te chemitry new t CHAPTER 7 THERMOCHEMISTRY MCQ Q.1 Which of the following tatement i contrary to the firt law of thermodynamic? (a) energy can neither be created nor detroyed (b) one form of energy

More information

TRIPLE SOLUTIONS FOR THE ONE-DIMENSIONAL

TRIPLE SOLUTIONS FOR THE ONE-DIMENSIONAL GLASNIK MATEMATIČKI Vol. 38583, 73 84 TRIPLE SOLUTIONS FOR THE ONE-DIMENSIONAL p-laplacian Haihen Lü, Donal O Regan and Ravi P. Agarwal Academy of Mathematic and Sytem Science, Beijing, China, National

More information

III. Evaluating Properties. III. Evaluating Properties

III. 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 information

Brown Hills College of Engineering & Technology

Brown 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 information

Lecture 44: Review Thermodynamics I

Lecture 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 information

Chapter 16. In Chapter 15 we analyzed combustion processes under CHEMICAL AND PHASE EQUILIBRIUM. Objectives

Chapter 16. In Chapter 15 we analyzed combustion processes under CHEMICAL AND PHASE EQUILIBRIUM. Objectives Chapter 16 CHEMICAL AND PHASE EQUILIBRIUM In Chapter 15 we analyzed combustion processes under the assumption that combustion is complete when there is sufficient time and oxygen. Often this is not the

More information

Solutions to Homework #10

Solutions 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 information

Chapter 5: The First Law of Thermodynamics: Closed Systems

Chapter 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 information

ME 375 FINAL EXAM Wednesday, May 6, 2009

ME 375 FINAL EXAM Wednesday, May 6, 2009 ME 375 FINAL EXAM Wedneday, May 6, 9 Diviion Meckl :3 / Adam :3 (circle one) Name_ Intruction () Thi i a cloed book examination, but you are allowed three ingle-ided 8.5 crib heet. A calculator i NOT allowed.

More information

External Flow: Flow over Bluff Objects (Cylinders, Spheres, Packed Beds) and Impinging Jets

External Flow: Flow over Bluff Objects (Cylinders, Spheres, Packed Beds) and Impinging Jets External Flow: Flow over Bluff Object (Cylinder, Sphere, Packed Bed) and Impinging Jet he Cylinder in Cro Flow - Condition depend on pecial feature of boundary layer development, including onet at a tagnation

More information

SIMPLE RANKINE CYCLE. 3 expander. boiler. pump. condenser 1 W Q. cycle cycle. net. shaft

SIMPLE 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 information

Exergy and the Dead State

Exergy 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 information

ECE309 THERMODYNAMICS & HEAT TRANSFER MIDTERM EXAMINATION. Instructor: R. Culham. Name: Student ID Number:

ECE309 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 information

Special Topic: Binary Vapor Cycles

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 information

To receive full credit all work must be clearly provided. Please use units in all answers.

To 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 information

ME Thermodynamics I

ME 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 information

Correction for Simple System Example and Notes on Laplace Transforms / Deviation Variables ECHE 550 Fall 2002

Correction for Simple System Example and Notes on Laplace Transforms / Deviation Variables ECHE 550 Fall 2002 Correction for Simple Sytem Example and Note on Laplace Tranform / Deviation Variable ECHE 55 Fall 22 Conider a tank draining from an initial height of h o at time t =. With no flow into the tank (F in

More information

CHAPTER 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 information

ME 354 THERMODYNAMICS 2 MIDTERM EXAMINATION. Instructor: R. Culham. Name: Student ID Number: Instructions

ME 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, closed-book examination. 2. Answer all questions

More information

Lecture Notes II. As the reactor is well-mixed, the outlet stream concentration and temperature are identical with those in the tank.

Lecture Notes II. As the reactor is well-mixed, the outlet stream concentration and temperature are identical with those in the tank. Lecture Note II Example 6 Continuou Stirred-Tank Reactor (CSTR) Chemical reactor together with ma tranfer procee contitute an important part of chemical technologie. From a control point of view, reactor

More information

5/6/ :41 PM. Chapter 6. Using Entropy. Dr. Mohammad Abuhaiba, PE

5/6/ :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 information

Step 1: Draw a diagram to represent the system. Draw a T-s process diagram to better visualize the processes occurring during the cycle.

Step 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 information

Energy and Energy Balances

Energy and Energy Balances Energy and Energy Balances help us account for the total energy required for a process to run Minimizing wasted energy is crucial in Energy, like mass, is. This is the Components of Total Energy energy

More information

Chapter 4 Total Entropy Cannot Decrease 2012/5/6

Chapter 4 Total Entropy Cannot Decrease 2012/5/6 Chapter 4 Total Entropy Cannot Decreae "Ice melting" - a claic example of entropy increaing It happen all the time! Nothing you can do About it. Spontaneou and irreverible. It i approaching to an equilibrium

More information

Find: a) Mass of the air, in kg, b) final temperature of the air, in K, and c) amount of entropy produced, in kj/k.

Find: 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 information

SCHOOL OF COMPUTING, ENGINEERING AND MATHEMATICS SEMESTER 2 EXAMINATIONS 2014/2015 ME258. Thermodynamics

SCHOOL OF COMPUTING, ENGINEERING AND MATHEMATICS SEMESTER 2 EXAMINATIONS 2014/2015 ME258. Thermodynamics s SCHOOL OF COMPUING, ENGINEERING AND MAHEMAICS SEMESER EXAMINAIONS 04/05 ME58 hermodynamics ime allowed: WO hours Answer: Any FOUR Questions Items permitted: Any approved calculator Items supplied: Steam

More information

376 CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD. D(s) = we get the compensated system with :

376 CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD. D(s) = we get the compensated system with : 376 CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD Therefore by applying the lead compenator with ome gain adjutment : D() =.12 4.5 +1 9 +1 we get the compenated ytem with : PM =65, ω c = 22 rad/ec, o

More information

Control Systems Engineering ( Chapter 7. Steady-State Errors ) Prof. Kwang-Chun Ho Tel: Fax:

Control Systems Engineering ( Chapter 7. Steady-State Errors ) Prof. Kwang-Chun Ho Tel: Fax: Control Sytem Engineering ( Chapter 7. Steady-State Error Prof. Kwang-Chun Ho kwangho@hanung.ac.kr Tel: 0-760-453 Fax:0-760-4435 Introduction In thi leon, you will learn the following : How to find the

More information

THE FIRST LAW APPLIED TO STEADY FLOW PROCESSES

THE 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 information

1 st Law Analysis of Control Volume (open system) Chapter 6

1 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 information

CHAPTER 7 ENTROPY. Copyright Hany A. Al-Ansary and S. I. Abdel-Khalik (2014) 1

CHAPTER 7 ENTROPY. Copyright Hany A. Al-Ansary and S. I. Abdel-Khalik (2014) 1 CHAPTER 7 ENTROPY S. I. Abdel-Khalik (2014) 1 ENTROPY The Clausius Inequality The Clausius inequality states that for for all cycles, reversible or irreversible, engines or refrigerators: For internally-reversible

More information

Isentropic Efficiency in Engineering Thermodynamics

Isentropic 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 information

Psychrometrics. 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)

Psychrometrics. 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 information

wb Thermodynamics 2 Lecture 9 Energy Conversion Systems

wb Thermodynamics 2 Lecture 9 Energy Conversion Systems wb1224 - Thermodynamics 2 Lecture 9 Energy Conversion Systems Piero Colonna, Lecturer Prepared with the help of Teus van der Stelt 8-12-2010 Delft University of Technology Challenge the future Content

More information

Green-Kubo formulas with symmetrized correlation functions for quantum systems in steady states: the shear viscosity of a fluid in a steady shear flow

Green-Kubo formulas with symmetrized correlation functions for quantum systems in steady states: the shear viscosity of a fluid in a steady shear flow Green-Kubo formula with ymmetrized correlation function for quantum ytem in teady tate: the hear vicoity of a fluid in a teady hear flow Hirohi Matuoa Department of Phyic, Illinoi State Univerity, Normal,

More information

Chapter 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.

Chapter 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 information

COMBUSTION OF FUEL 12:57:42

COMBUSTION OF FUEL 12:57:42 COMBUSTION OF FUEL The burning of fuel in presence of air is known as combustion. It is a chemical reaction taking place between fuel and oxygen at temperature above ignition temperature. Heat is released

More information

Digital Control System

Digital Control System Digital Control Sytem - A D D A Micro ADC DAC Proceor Correction Element Proce Clock Meaurement A: Analog D: Digital Continuou Controller and Digital Control Rt - c Plant yt Continuou Controller Digital

More information

Dishwasher. Heater. Homework Solutions ME Thermodynamics I Spring HW-1 (25 points)

Dishwasher. Heater. Homework Solutions ME Thermodynamics I Spring HW-1 (25 points) HW-1 (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 information

1 Routh Array: 15 points

1 Routh Array: 15 points EE C28 / ME34 Problem Set 3 Solution Fall 2 Routh Array: 5 point Conider the ytem below, with D() k(+), w(t), G() +2, and H y() 2 ++2 2(+). Find the cloed loop tranfer function Y () R(), and range of k

More information

ECE309 INTRODUCTION TO THERMODYNAMICS & HEAT TRANSFER. 13 June 2007

ECE309 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, open-book examination. You are permitted to use: course text book calculator There are

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.

+ 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 information

Chapter 4. The Laplace Transform Method

Chapter 4. The Laplace Transform Method Chapter 4. The Laplace Tranform Method The Laplace Tranform i a tranformation, meaning that it change a function into a new function. Actually, it i a linear tranformation, becaue it convert a linear combination

More information

ME 300 Thermodynamics II Exam 2 November 13, :00 p.m. 9:00 p.m.

ME 300 Thermodynamics II Exam 2 November 13, :00 p.m. 9:00 p.m. ME 300 Therodynaics II Exa 2 Noveber 3, 202 8:00 p.. 9:00 p.. Nae: Solution Section (Circle One): Sojka Naik :30 a.. :30 p.. Instructions: This is a closed book/notes exa. You ay use a calculator. You

More information

Entropy Minimization in Design of Extractive Distillation System with Internal Heat Exchangers

Entropy Minimization in Design of Extractive Distillation System with Internal Heat Exchangers Entropy Minimization in Deign of Extractive Ditillation Sytem with Internal Heat Exchanger Diego F. Mendoza, Carlo.M. Riaco * Group of Proce Sytem Engineering, Department of Chemical and Environmental

More information

CONTROL SYSTEMS. Chapter 2 : Block Diagram & Signal Flow Graphs GATE Objective & Numerical Type Questions

CONTROL SYSTEMS. Chapter 2 : Block Diagram & Signal Flow Graphs GATE Objective & Numerical Type Questions ONTOL SYSTEMS hapter : Bloc Diagram & Signal Flow Graph GATE Objective & Numerical Type Quetion Quetion 6 [Practice Boo] [GATE E 994 IIT-Kharagpur : 5 Mar] educe the ignal flow graph hown in figure below,

More information