Special Topic: Binary Vapor Cycles

Size: px
Start display at page:

Download "Special Topic: Binary Vapor Cycles"

Transcription

1 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. 0- onider te eat excanger o a bary poer cycle. e orkg luid o te toppg cycle (cycle A enter te eat excanger at tate and leave at tate. e orkg luid o te bottomg cycle (cycle B enter at tate and leave at tate. Neglectg any cange ketic and potential energie, and aumg te eat excanger i ell-ulated, te teady-lo energy balance relation yield u, A E 0 (teady E Eytem 0 + E e B A B e E + A i i B or A ( B ( 0- Steam i not an ideal luid or vapor poer cycle becaue it critical temperature i lo, it aturation dome reemble an verted V, and it condener preure i too lo. 0- Becaue mercury a a ig critical temperature, relatively lo critical preure, but a very lo condener preure. It i alo toxic, expenive, and a a lo entalpy o vaporization. 0- In bary vapor poer cycle, bot cycle are vapor cycle. In te combed ga-team poer cycle, one o te cycle i a ga cycle.

2 0- Revie roblem 0- It i to be demontrated tat te termal eiciency o a combed ga-team poer plant cc can be expreed a cc g + g ere g W g / i n an d W / g, are te termal eiciencie o te ga and team cycle, repectively, and te eiciency o a combed cycle i to be obtaed. Analyi e termal eiciencie o ga, team, and combed cycle can be expreed a cc Wtotal Wg g g, W g, g, ere i te eat upplied to te ga cycle, ere i te eat rejected by te team cycle, and ere g, i te eat rejected rom te ga cycle and upplied to te team cycle. Ug te relation above, te expreion g + g cc g, g, g + g + g, g, g, g, g, can be expreed a ereore, te proo i complete. Ug te relation above, te termal eiciency o te given combed cycle i determed to be cc g + g g, 0- e termal eiciency o a combed ga-team poer plant cc can be expreed term o te termal eiciencie o te ga and te team turbe cycle a +. It i to be on tat te value o cc i greater tan eiter o or. Analyi By actorg term, te relation cc g + g g + ( g > g g oitive ce < or cc g + g + g ( > g oitive ce < cc g + g cc g g can be expreed a u e conclude tat te combed cycle i more eicient tan eiter o te ga turbe or team turbe cycle alone.

3 0-0- A team poer plant operatg on te ideal Ranke cycle it reeatg i conidered. e reeat preure o te cycle are to be determed or te cae o gle and double reeat. Aumption Steady operatg condition exit. Ketic and potential energy cange are negligible. Analyi (a Sgle Reeat: From te team table (able A-, A-, and A-, x 0 ka 0. + x + x 00 0 ka (b Double Reeat : Ma 00 x and x 00 g g kj/kg K ( 0.(. ( 0.(. SINGLE Ma 0 ka. kj/kg. kj/kg K 00 DOUBL Ma 0 ka Any preure x elected beteen te limit o Ma and. Ma ill atiy te reuirement, and can be ued or te double reeat preure.

4 E A geotermal poer plant operatg on te imple Ranke cycle ug an organic luid a te orkg luid i conidered. e exit temperature o te geotermal ater rom te vaporizer, te rate o eat rejection rom te orkg luid te condener, te ma lo rate o geotermal ater at te preeater, and te termal eiciency o te Level I cycle o ti plant are to be determed. Aumption Steady operatg condition exit. Ketic and potential energy cange are negligible. Analyi (a e exit temperature o geotermal ater rom te vaporizer i determed rom te teadylo energy balance on te geotermal ater (bre, bre,0,000 Btu/ brec p ( (, lbm/(.0 Btu/lbm F(. F F (b e rate o eat rejection rom te orkg luid to te air te condener i determed rom te teady-lo energy balance on air, air airc p ( (,,00 lbm/( 0. Btu/lbm F(. MBtu/. F (c e ma lo rate o geotermal ater at te preeater i determed rom te teady-lo energy balance on te geotermal ater, geo,0,000 Btu/ geo (d e rate o eat put i and geo geo,0 c p ( (.0 Btu/lbm F( lbm/.0. F +, 0, ,, 000 vaporizer reeater, 0, 000 Btu / en, W W t 00 0 kw 0 kw. Btu 0.%,0,000 Btu/ kw

5 0-0- A team poer plant operate on te imple ideal Ranke cycle. e turbe let temperature, te poer put, te termal eiciency, and te mimum ma lo rate o te coolg ater reuired 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-, (b and v v u, p, ka v ( ( m /kg( 000. ka.0 ka + p, ka ka. kj/kg m /kg kj/kg.0 kj/kg.0 kj/kg Ma. kj/kg 0. t W kj ka m... kj/kg kj/kg. 0.. kj/kg. kj/kg.%. kj/kg ( 0.( 0,000, kj/ kj/ t Ma. ka (c e ma lo rate o te coolg ater ill be mimum en it i eated to te temperature o te team te condener, ic i 0., cool c W 0,000, 0, kj/ 0, kj/ (. kj/kg ( 0.. kg/

6 0-0- A team poer plant operatg on an ideal Ranke cycle it to tage o reeat i conidered. e termal eiciency o te cycle and te ma lo rate o te team 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-, en, u, (b v v ka v( ( m /kg(,000 ka.0 ka + p, Ma 00 Ma Ma Ma 00 ka x. kj/kg m /kg kj ka m kj/kg Ma 00 t 0. kj/kg.0 kj/kg K 00. kj/kg. kj/kg. kj/kg K. kj/kg. kj/kg. kj/kg K g. + x. + ( + ( + ( g.. 0. kj/kg kj/kg ka ( 0.0(.0. kj/kg Ma Ma kj/kg W. kj/kg.% 0. kj/kg 0,000 kj/. kj/kg. kg/ Ma

7 0-0- An 0-MW team poer plant operatg on a regenerative Ranke cycle it an open eedater eater i conidered. e ma lo rate o team troug te boiler, te termal eiciency o te cycle, and te irreveribility aociated it te regeneration proce are to be determed. Aumption Steady operatg condition exit. Ketic and potential energy cange are negligible. Analyi Boiler II Open urbe -y y ondener I 0 Ma 0. Ma y 0 ka -y (a From te team table (able A-, A-, and A-, v v ka.kj/kg v( / p kj ( 0.000m /kg( 00 0ka / ( 0. pi, 0. Ma at. liuid v v m /kg ka m 0.kJ/kg kJ/kg Ma 0.0 kj/kg 0.000m /kg v ( / p kj ( m /kg( 0, ka /( 0. ka m 0. kj/kg kj/kg pii, 0 Ma 00 x 0. Ma. kj/kg. kj/kg K g.0 + x g ( 0.( 0.0. kj/kg ( ( 0.0(.. kj/kg..

8 0- x 0 ka g. + x g.+ ( 0.(. 0. kj/kg ( ( 0.0(.. kj/kg. 0. e raction o team extracted i determed rom te teady-lo energy balance euation applied to te eedater eater. Notg tat W ke pe 0, E E E E E 0 (teady ytem 0 ( y ( i i mee m + m m y + ere y i te raction o team extracted rom te turbe ( /. Solvg or y, en, and y kj/kg ( y( ( 0.(.. W m... kj/kg 0,000 kj/. kg/. kj/kg (b e termal eiciency i determed rom Alo, t 0. Ma. 0. Ma. kj/kg.%. 0 ka. kj/kg K.0 kj/kg K 0. kj/kg K. kj/kg en te irreveribility (or exergy detruction aociated it ti regeneration proce i i regen 0gen 0 + mee mii 0 urr L [ y ( y ] ( 0 K [.0 ( 0.(. ( 0.( 0. ]. kj/kg 0

9 0-0- An 0-MW team poer plant operatg on an ideal regenerative Ranke cycle it an open eedater eater i conidered. e ma lo rate o team troug te boiler, te termal eiciency o te cycle, and te irreveribility aociated it te regeneration proce are to be determed. Aumption Steady operatg condition exit. Ketic and potential energy cange are negligible. Analyi Boiler II Open urbe -y y ondener I 0 Ma 0. Ma y 0 ka -y (a From te team table (able A-, A-, and A-, v v 0 0 ka. kj/kg m v ( ( m /kg( 00 0 ka + pi, 0. Ma at.liuid v v pii, /kg kj ka m Ma 0.0 kj/kg m ( ( m /kg( 0, ka v + pii, 0 Ma Ma x 0 ka x. kj/kg. kj/kg K 0.0 kj/kg /kg kj ka m kj/kg g.0 + x ( 0.( g. + x.+ g g 0. kj/kg. kj/kg ( 0.(. 0. kj/kg

10 0- e raction o team extracted i determed rom te teady-lo energy euation applied to te eedater eater. Notg tat W ke pe 0, E E E 0 (teady ytem 0 E ( y ( i i mee m + m m y + E ere y i te raction o team extracted rom te turbe ( /. Solvg or y, en, and y kj/kg ( y( ( 0.( 0.. W m...0 kj/kg 0,000 kj/. kj/kg. 0 kg/ (b e termal eiciency i determed rom Alo, t. kj/kg.0%. kj/kg. kj/kg 0. 0 ka.0 kj/kg K 0. kj/kg K. kj/kg en te irreveribility (or exergy detruction aociated it ti regeneration proce i i regen 0gen 0 + mee mii 0 urr L [ y ( y ] ( 0 K [.0 ( 0.(. ( 0.( 0. ].0 kj/kg 0

11 0-0- An ideal reeat-regenerative Ranke cycle it one open eedater eater i conidered. e raction o team extracted or regeneration 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 ka. kj/kg v ( ( m /kg( 00 ka kj ka m 0. kj/kg kj/kg pi, 0. Ma at. liuid v v Ma /kg 0. kj/kg m /kg v( ( m /kg( 0, ka pii, kj ka m 0. kj/kg kj/kg pii, 0 Ma 00.0 Ma.0 Ma 00. kj/kg. kj/kg K. kj/kg. kj/kg. kj/kg K 0. Ma 0. kj/kg. 0. ka 0. x g. + x. + g Boiler II ( 0.(.. kj/kg Open 0 Ma 0. Ma ka I y Ma urbe -y onden. e raction o team extracted i determed rom te teady-lo energy balance euation applied to te eedater eater. Notg tat W ke pe 0, E E E 0 (teady ytem 0 E ( y ( i i mee m + m m y + E ere y i te raction o team extracted rom te turbe ( /. Solvg or y, y (b e termal eiciency i determed rom ( ( ( ( ( y( ( 0.0(... kj/kg. kj/kg and t.%. kj/kg. kj/kg

12 0-0- A nonideal reeat-regenerative Ranke cycle it one open eedater eater i conidered. e raction o team extracted or regeneration 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 urbe II Open I y -y ondener y -y (a From te team table (able A-, A-, and A-, v ka. kj/kg m v ( ( m /kg( 00 ka pi, 0. Ma at. liuid v 0. Ma /kg pi, kj ka m 0. kj/kg Ma 0. kj/kg m /kg ( ( m /kg( 0, ka pii, v kj ka m 0. kj/kg kj/kg pii, 0 Ma 00.0 Ma. kj/kg. kj/kg K. kj/kg ( ( 0.(.. kj/kg..

13 0-.0 Ma Ma ka x. kj/kg. kj/kg K 0. kj/kg. kj/kg (. ( 0.( ( 0.(.. 0. (. ( 0.(.. g + x g. kj/kg. kj/kg e raction o team extracted i determed rom te teady-lo energy balance euation applied to te eedater eater. Notg tat W ke pe 0, E E E E E 0 (teady ytem 0 ( y ( i i mee m + m m y + ere y i te raction o team extracted rom te turbe ( /. Solvg or y, y (b e termal eiciency i determed rom and t ( + ( ( (... kj/kg ( y( ( 0.(.. 0. kj/kg 0. kj/kg.%. kj/kg

14 A team poer plant operate on an ideal reeat-regenerative Ranke cycle it one reeater and to eedater eater, one open and one cloed. e raction o team extracted rom te turbe or te open eedater eater, te termal eiciency o te cycle, and te poer put are to be determed. Aumption Steady operatg condition exit. Ketic and potential energy cange are negligible. Analyi Boiler loed II Hig- urbe (a From te team table (able A-, A-, and A-, v v pi, 0 Open I y. kj/kg z I m /kg v ( ( m /kg( 00 ka Lo- urbe -y-z ondener kj ka m 0.0 kj/kg ka pi, 0. kj/kg Ma at.liuid v 0. Ma m /kg pii, v( kj ( m /kg(, ka ka m.0 kj/kg kj/kg pii, 0. Ma at.liuid v v 0 0 Ma 00.0 Ma.0 Ma Ma. 0. Ma 0. kj/kg 0. kj/kg m /kg Ma 0. Ma ( + ( m /kg(, ka. kj/kg. kj/kg K v. kj/kg. kj/kg K Ma y 0 0. Ma z ka - y - z kj ka m

15 0-0. Ma 0 0. Ma 0 ka 0 x 0. kj/kg 000. kj/kg 0.0 g + x. +. kj/kg. 0.. g ( 0.0(.0 e raction o team extracted i determed rom te teady-lo energy balance euation applied to te eedater eater. Notg tat W ke pe 0, E 0 (teady E Eytem 0 E i i E e e ( ( y( ( ere y i te raction o team extracted rom te turbe ( /. Solvg or y, y For te open FWH, y E + E E + ( y z + z ( i i E E (teady ytem e e ere z i te raction o team extracted rom te turbe ( / at te econd tage. Solvg or z, z 0 ( y( 0.. ( 0.0( (b ( + ( 0 (.. + (. 0.. kj/kg ( y z( ( (.0. and t. 0.. kj/kg 0. kj/kg.%. kj/kg (c W m ( kg/(. kj/kg, kw kj/kg

16 0-0- A cogeneration poer plant i modiied it reeat and tat produce MW o poer and upplie MW o proce eat. e rate o eat put te boiler and te raction o team extracted or proce eatg 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 0 Ma 00. kj/kg 0. kj/kg. kj/kg. kj/kg K Boiler roce eater urbe Ma. kj/kg II I Ma. kj/kg 00. kj/kg K. 0. x 0. g. ka Ma + x g. + ( 0.(. Ma. kj/kg e ma lo rate troug te proce eater i proce,000 kj/. kg/ ka (. 0. kj/kg Alo, W ( + ( ( + (.( or, 000 kj/ ( ( ( It yield. kg/ and m.. 0. kg/ Mixg camber: E E E 0 (teady 0 or, en, ytem E E m m m m + m i i e e + ( 0.(. + (.( 0.. ( + ( (. kg/(. 0.0 kj/kg + ( 0. kg/(,00 kw (b e raction o team extracted or proce eatg i. kg/ y.%. kg/ total 0.0 kj/kg.. kj/kg ondener

( )( ) 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

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

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

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

1 = The rate at which the entropy of the high temperature reservoir changes, according to the definition of the entropy, is

1 = 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 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

Thermodynamics Lecture Series

Thermodynamics 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 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

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

Chapter 8 EXERGY A MEASURE OF WORK POTENTIAL

Chapter 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

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

Chapter 12 Radiation Heat Transfer. Special Topic: Heat Transfer from the Human Body

Chapter 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 information

since (Q H /T H ) = (Q L /T L ) for reversible cycles. Also, since Q diff is a positive quantity. Thus,

since (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 information

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

6-5. H 2 O 200 kpa 200 C Q. Entropy Changes of Pure Substances

6-5. 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 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

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

SENSITIVITY 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 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 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

Then the amount of water that flows through the pipe during a differential time interval dt is (1) 4

Then 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 information

& out. R-134a 34 C

& 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 information

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

( ) 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 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

Carnot Factor of a Vapour Power Cycle with Regenerative Extraction

Carnot 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 information

Unit 12 Refrigeration and air standard cycles November 30, 2010

Unit 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 information

ENGINEERING OF NUCLEAR REACTORS. Tuesday, October 9 th, 2014, 1:00 2:30 p.m.

ENGINEERING 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

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

4-93 RT RT. He PV n = C = = Then the boundary work for this polytropic process can be determined from. n =

4-93 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 information

Chapters 19 & 20 Heat and the First Law of Thermodynamics

Chapters 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 information

Design of Robust PI Controller for Counter-Current Tubular Heat Exchangers

Design 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 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

Delft University of Technology DEPARTMENT OF AEROSPACE ENGINEERING

Delft 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 information

Chapter 5 MASS AND ENERGY ANALYSIS OF CONTROL VOLUMES

Chapter 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 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

Section 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

Section 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 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

The average velocity of water in the tube and the Reynolds number are Hot R-134a

The 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 information

(b) The heat transfer can be determined from an energy balance on the system

(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 information

I. (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. 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 information

Main components of the above cycle are: 1) Boiler (steam generator) heat exchanger 2) Turbine generates work 3) Condenser heat exchanger 4) Pump

Main 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 information

Chapter 7: 17, 20, 24, 25, 32, 35, 37, 40, 47, 66 and 79.

Chapter 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 information

the first derivative with respect to time is obtained by carefully applying the chain rule ( surf init ) T Tinit

the 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 information

NUCLEAR THERMAL-HYDRAULIC FUNDAMENTALS

NUCLEAR 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 information

Teaching schedule *15 18

Teaching 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 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

Department of Mechanical Engineering ME 322 Mechanical Engineering Thermodynamics. Calculation of Entropy Changes. Lecture 19

Department of Mechanical Engineering ME 322 Mechanical Engineering Thermodynamics. Calculation of Entropy Changes. Lecture 19 Department of Mecanical Engineering ME Mecanical Engineering ermodynamics Calculation of Entropy Canges Lecture 9 e Gibbs Equations How are entropy alues calculated? Clausius found tat, dq dq m re re ds

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

300 kpa 77 C. (d) If we neglect kinetic energy in the calculation of energy transport by mass

300 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 information

Journal of Chemical and Pharmaceutical Research, 2013, 5(12): Research Article

Journal 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 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

What lies between Δx E, which represents the steam valve, and ΔP M, which is the mechanical power into the synchronous machine?

What lies between Δx E, which represents the steam valve, and ΔP M, which is the mechanical power into the synchronous machine? A 2.0 Introduction In the lat et of note, we developed a model of the peed governing mechanim, which i given below: xˆ K ( Pˆ ˆ) E () In thee note, we want to extend thi model o that it relate the actual

More information

SPC 407 Sheet 2 - Solution Compressible Flow - Governing Equations

SPC 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 information

ME Thermodynamics I

ME 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 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

ECE 3510 Root Locus Design Examples. PI To eliminate steady-state error (for constant inputs) & perfect rejection of constant disturbances

ECE 3510 Root Locus Design Examples. PI To eliminate steady-state error (for constant inputs) & perfect rejection of constant disturbances ECE 350 Root Locu Deign Example Recall the imple crude ervo from lab G( ) 0 6.64 53.78 σ = = 3 23.473 PI To eliminate teady-tate error (for contant input) & perfect reection of contant diturbance Note:

More information

International Journal of Scientific & Engineering Research, Volume 4, Issue 12, December ISSN

International Journal of Scientific & Engineering Research, Volume 4, Issue 12, December ISSN International Journal of Scientific & Engineering Reearch, Volume 4, Iue 12, December-2013 180 A Novel approach on entropy production rate Behnaz Jalili 1*, Cyru Aghanaafi 2*, Mohammad Khademi 3* Abtract

More information

Developing Transfer Functions from Heat & Material Balances

Developing Transfer Functions from Heat & Material Balances Colorado Sool of Mine CHEN43 Stirred ank Heater Develoing ranfer untion from Heat & Material Balane Examle ranfer untion Stirred ank Heater,,, A,,,,, We will develo te tranfer funtion for a tirred tank

More information

Heat Recovery for next Generation of Hybrid Vehicles: Simulation and Design of a Rankine Cycle System

Heat Recovery for next Generation of Hybrid Vehicles: Simulation and Design of a Rankine Cycle System Page 0440 EVS24 Stavanger, Norway, May 13-16, 2009 Heat Recovery or next Generation o Hybrid Vehicles: Simulation and Design o a Ranke Cycle System 1 IFP, 1-4 avenue de Bois Préau, 92852 Rueil-Malmaison,

More information

EVAPORATION. Robert evaporator. Balance equations Material balance (total) Component balance. Heat balance

EVAPORATION. 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 information

Unsteady State Simulation of Vapor Compression Heat Pump Systems With Modular Analysis (This paper will not be presented.)

Unsteady 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 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

Design By Emulation (Indirect Method)

Design By Emulation (Indirect Method) Deign By Emulation (Indirect Method he baic trategy here i, that Given a continuou tranfer function, it i required to find the bet dicrete equivalent uch that the ignal produced by paing an input ignal

More information

Department of Civil Engineering & Applied Mechanics McGill University, Montreal, Quebec Canada

Department 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 information

Readings for this homework assignment and upcoming lectures

Readings 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 information

KNOWN: Pressure, temperature, and velocity of steam entering a 1.6-cm-diameter pipe.

KNOWN: 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 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

Chapter 3 Thermoelectric Coolers

Chapter 3 Thermoelectric Coolers 3- Capter 3 ermoelectric Coolers Contents Capter 3 ermoelectric Coolers... 3- Contents... 3-3. deal Equations... 3-3. Maximum Parameters... 3-7 3.3 Normalized Parameters... 3-8 Example 3. ermoelectric

More information

= T. (kj/k) (kj/k) 0 (kj/k) int rev. Chapter 6 SUMMARY

= T. (kj/k) (kj/k) 0 (kj/k) int rev. Chapter 6 SUMMARY Capter 6 SUMMARY e second la of termodynamics leads to te definition of a ne property called entropy ic is a quantitative measure of microscopic disorder for a system. e definition of entropy is based

More information

ME 200 Thermodynamics 1 Fall 2017 Exam 3

ME 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 information

Radiation Heat Transfer

Radiation Heat Transfer CM30 ranport I Part II: Heat ranfer Radiation Heat ranfer Profeor Faith Morrion Department of Chemical Engineering Michigan echnological Univerity CM30 ranport Procee and Unit Operation I Part : Heat ranfer

More information

Methods for transfer matrix evaluation applied to thermoacoustics

Methods for transfer matrix evaluation applied to thermoacoustics Proceeding of the Acoutic 2012 Nante Conference 23-27 April 2012, Nante, France Method for tranfer matrix evaluation applied to thermoacoutic F. Bannart, G. Penelet, P. Lotton and J.-P. Dalmont Laboratoire

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

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

Kelvin Planck Statement of the Second Law. Clausius Statement of the Second Law

Kelvin 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 information

Lecture 12 - Non-isolated DC-DC Buck Converter

Lecture 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 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

Simulation and Analysis of Biogas operated Double Effect GAX Absorption Refrigeration System

Simulation and Analysis of Biogas operated Double Effect GAX Absorption Refrigeration System Simulation and Analysis of Biogas operated Double Effect GAX Absorption Refrigeration System Simulation and Analysis of Biogas operated Double Effect GAX Absorption Refrigeration System G Subba Rao, 2

More information

ME 200 Final Exam December 14, :00 a.m. to 10:00 a.m.

ME 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

Conductance from Transmission Probability

Conductance from Transmission Probability Conductance rom Transmission Probability Kelly Ceung Department o Pysics & Astronomy University o Britis Columbia Vancouver, BC. Canada, V6T1Z1 (Dated: November 5, 005). ntroduction For large conductors,

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

Inference for Two Stage Cluster Sampling: Equal SSU per PSU. Projections of SSU Random Variables on Each SSU selection.

Inference for Two Stage Cluster Sampling: Equal SSU per PSU. Projections of SSU Random Variables on Each SSU selection. Inference for Two Stage Cluter Sampling: Equal SSU per PSU Projection of SSU andom Variable on Eac SSU election By Ed Stanek Introduction We review etimating equation for PSU mean in a two tage cluter

More information

Answer Key THERMODYNAMICS TEST (a) 33. (d) 17. (c) 1. (a) 25. (a) 2. (b) 10. (d) 34. (b) 26. (c) 18. (d) 11. (c) 3. (d) 35. (c) 4. (d) 19.

Answer Key THERMODYNAMICS TEST (a) 33. (d) 17. (c) 1. (a) 25. (a) 2. (b) 10. (d) 34. (b) 26. (c) 18. (d) 11. (c) 3. (d) 35. (c) 4. (d) 19. HERMODYNAMICS ES Answer Key. (a) 9. (a) 7. (c) 5. (a). (d). (b) 0. (d) 8. (d) 6. (c) 4. (b). (d). (c) 9. (b) 7. (c) 5. (c) 4. (d). (a) 0. (b) 8. (b) 6. (b) 5. (b). (d). (a) 9. (a) 7. (b) 6. (a) 4. (d).

More information

Physics 207 Lecture 23

Physics 207 Lecture 23 ysics 07 Lecture ysics 07, Lecture 8, Dec. Agenda:. Finis, Start. Ideal gas at te molecular level, Internal Energy Molar Specific Heat ( = m c = n ) Ideal Molar Heat apacity (and U int = + W) onstant :

More information

Chapter A 9.0-V battery is connected to a lightbulb, as shown below. 9.0-V Battery. a. How much power is delivered to the lightbulb?

Chapter A 9.0-V battery is connected to a lightbulb, as shown below. 9.0-V Battery. a. How much power is delivered to the lightbulb? Capter continued carges on te plates were reversed, te droplet would accelerate downward since all forces ten act in te same direction as gravity. 5. A 0.5-F capacitor is able to store 7.0 0 C of carge

More information

Lecture 21. The Lovasz splitting-off lemma Topics in Combinatorial Optimization April 29th, 2004

Lecture 21. The Lovasz splitting-off lemma Topics in Combinatorial Optimization April 29th, 2004 18.997 Topic in Combinatorial Optimization April 29th, 2004 Lecture 21 Lecturer: Michel X. Goeman Scribe: Mohammad Mahdian 1 The Lovaz plitting-off lemma Lovaz plitting-off lemma tate the following. Theorem

More information

Thermodynamics Lecture Series

Thermodynamics 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 information

Nonisothermal Chemical Reactors

Nonisothermal Chemical Reactors he 471 Fall 2014 LEUE 7a Nnithermal hemical eactr S far we have dealt with ithermal chemical reactr and were able, by ug nly a many pecie ma balance a there are dependent react t relate reactr ize, let

More information

CFD Analysis and Optimization of Heat Transfer in Double Pipe Heat Exchanger with Helical-Tap Inserts at Annulus of Inner Pipe

CFD Analysis and Optimization of Heat Transfer in Double Pipe Heat Exchanger with Helical-Tap Inserts at Annulus of Inner Pipe IOR Journal Mecanical and Civil Engineering (IOR-JMCE) e-in: 2278-1684,p-IN: 2320-334X, Volume 13, Issue 3 Ver. VII (May- Jun. 2016), PP 17-22 www.iosrjournals.org CFD Analysis and Optimization Heat Transfer

More information

Finite Difference Formulae for Unequal Sub- Intervals Using Lagrange s Interpolation Formula

Finite Difference Formulae for Unequal Sub- Intervals Using Lagrange s Interpolation Formula Int. Journal o Mat. Analyi, Vol., 9, no. 7, 85-87 Finite Dierence Formulae or Unequal Sub- Interval Uing Lagrange Interpolation Formula Aok K. Sing a and B. S. Badauria b Department o Matematic, Faculty

More information

Number of extra papers used if any

Number 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 information

Hydraulic validation of the LHC cold mass heat exchanger tube.

Hydraulic 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 information

1 Power is transferred through a machine as shown. power input P I machine. power output P O. power loss P L. What is the efficiency of the machine?

1 Power is transferred through a machine as shown. power input P I machine. power output P O. power loss P L. What is the efficiency of the machine? 1 1 Power is transferred troug a macine as sown. power input P I macine power output P O power loss P L Wat is te efficiency of te macine? P I P L P P P O + P L I O P L P O P I 2 ir in a bicycle pump is

More information

This appendix derives Equations (16) and (17) from Equations (12) and (13).

This appendix derives Equations (16) and (17) from Equations (12) and (13). Capital growt pat of te neoclaical growt model Online Supporting Information Ti appendix derive Equation (6) and (7) from Equation () and (3). Equation () and (3) owed te evolution of pyical and uman capital

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

(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

Homework #7 Solution. Solutions: ΔP L Δω. Fig. 1

Homework #7 Solution. Solutions: ΔP L Δω. Fig. 1 Homework #7 Solution Aignment:. through.6 Bergen & Vittal. M Solution: Modified Equation.6 becaue gen. peed not fed back * M (.0rad / MW ec)(00mw) rad /ec peed ( ) (60) 9.55r. p. m. 3600 ( 9.55) 3590.45r.

More information

Velocity or 60 km/h. a labelled vector arrow, v 1

Velocity or 60 km/h. a labelled vector arrow, v 1 11.7 Velocity en you are outide and notice a brik wind blowing, or you are riding in a car at 60 km/, you are imply conidering te peed of motion a calar quantity. ometime, owever, direction i alo important

More information

RP 2.4. SEG/Houston 2005 Annual Meeting 1513

RP 2.4. SEG/Houston 2005 Annual Meeting 1513 P 2.4 Measurement of sear ave velocity of eavy oil De-ua Han, Jiajin Liu, University of Houston Micael Batzle, Colorado Scool of Mines Introduction It is ell knon tat te fluids ave no sear modulus and

More information

Systems Analysis. Prof. Cesar de Prada ISA-UVA

Systems Analysis. Prof. Cesar de Prada ISA-UVA Sytem Analyi Prof. Cear de Prada ISAUVA rada@autom.uva.e Aim Learn how to infer the dynamic behaviour of a cloed loo ytem from it model. Learn how to infer the change in the dynamic of a cloed loo ytem

More information