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

Size: px
Start display at page:

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

Transcription

1 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 no cange wit time. Ketic and potential energy cange are negligible. Air i an ideal ga. 4 e proce volve no ternal irreveribilitie uc a riction, and tu it i an iotermal, ternally reverible proce. Propertie Notg tat () or ideal gae, we ave ce 5. Analyi We take te compreor a te ytem. Notg tat te entalpy o air rema contant, te energy balance or ti teady-low ytem can be expreed te rate orm a (teady) Δytem Rate o net energy traner Rate o cange ternal, ketic, by eat, work, and ma potential, etc.energie ereore, Q W kw Notg tat te proce i aumed to be an iotermal and ternally reverible proce, te rate o entropy cange o air i determed to be,air kw ΔS air kw/k 98 K y Q P AIR cont. P kw

2 An ulated cylder i itially illed wit aturated R-4a vapor at a peciied preure. e rerigerant expand a reverible manner until te preure drop to a peciied value. e al temperature te cylder and te work done by te rerigerant are to be determed. Aumption e ketic and potential energy cange are negligible. e cylder i well-ulated and tu eat traner i negligible. e termal energy tored te cylder itel i negligible. 4 e proce i tated to be reverible. Analyi (a) i i a reverible adiabatic (i.e., ientropic) proce, and tu. From te rerigerant table (able A- troug A-), Alo, v v g P 0.8 MPa u u at. MPa g@0.8 MPa g@0.8 MPa m /kg kj/kg kj/kg K and V 0.05 m m.95 kg v m /kg P 0.4 MPa x u u g + x u g ( )( 7.45).9 kj/kg R-4a 0.8 MPa 0.05 m at@0.4 MPa 8.9 (b) We take te content o te cylder a te ytem. i i a cloed ytem ce no ma enter or leave. e energy balance or ti adiabatic cloed ytem can be expreed a 44 Net energy traner by eat, work, and ma W W b, b, ange ternal, ketic, potential, etc.energie ΔU m( u u ) Δytem 44 Subtitutg, te work done durg ti ientropic proce i determed to be W ( ) (.95 kg)( ) 7.09 kj m u kj/kg b, u PROPRIARY MARIAL. 008 e McGraw-Hill ompanie, Inc. Limited ditribution permitted only to teacer and educator or coure preparation. I you are a tudent ug ti Manual, you are ug it wit permiion.

3 Saturated Rerigerant-4a vapor at 60 i compreed teadily by an adiabatic compreor. e mimum power put to te compreor i to be determed. Aumption i i a teady-low proce ce tere i no cange wit time. Ketic and potential energy cange are negligible. e device i adiabatic and tu eat traner i negligible. Analyi e power put to an adiabatic compreor will be a mimum wen te compreion proce i reverible. For te reverible adiabatic proce we ave. From te rerigerant table (able A- troug A-), Alo, v v g P 60 at. vapor P 900 V m g@60 g@ kj/kg 0.48 m /kg 4. kj/kg kj/kg K m /m 6.0 kg/m 0.7 kg/ 0.48 m /kg R-4a ere i only one let and one exit, and tu. We take te compreor a te ytem, wic i a control volume ce ma croe te boundary. e energy balance or ti teady-low ytem can be expreed te rate orm a 44 (teady) Δytem 0 Rate o net energy traner by eat, work, and ma Rate o cange ternal, ketic, potential, etc.energie m m (ce Q Δke Δpe 0) ( ) Subtitutg, te mimum power upplied to te compreor i determed to be W ( 0.7 kg/)( ) 9.7 kw kj/kg PROPRIARY MARIAL. 008 e McGraw-Hill ompanie, Inc. Limited ditribution permitted only to teacer and educator or coure preparation. I you are a tudent ug ti Manual, you are ug it wit permiion.

4 Steam i expanded an ientropic turbe. e work produced i to be determed. Aumption i i a teady-low proce ce tere i no cange wit time. e proce i ientropic (i.e., reverible-adiabatic). Analyi ere i one let and two exit. We take te turbe a te ytem, wic i a control volume ce ma croe te boundary. e energy balance or ti teady-low ytem can be expreed te rate orm a 44 (teady) Δytem 0 Rate o net energy traner by eat, work, and ma From a ma balance, Rate o cange ternal, ketic, potential, etc.energie (0.05)(5 kg/) 0.5 kg/ (0.95)(5 kg/) 4.75 kg/ Notg tat te expanion proce i ientropic, te entalpie at tree tate are determed a ollow: P kj/kg kj/kg K (able A - 6) 4 MPa 5 kg/ MPa 0.7 MPa 50 Steam turbe P 4 MPa kj/kg K 979. kj/kg (able A - 6) P Subtitutg, W kj/kg K 68 kw 09. kj/kg (able A - 6) (5 kg/)(979. kj/kg) (0.5 kg/)(09. kj/kg) (4.75 kg/)(68.4 kj/kg) PROPRIARY MARIAL. 008 e McGraw-Hill ompanie, Inc. Limited ditribution permitted only to teacer and educator or coure preparation. I you are a tudent ug ti Manual, you are ug it wit permiion.

5 An adiabatic pump i ued to compre aturated liquid water a reverible manner. e work put i to be determed by dierent approace. Aumption Steady operatg condition exit. Ketic and potential energy cange are negligible. Heat traner to or rom te luid i negligible. Analyi e propertie o water at te let and exit o te pump are (able A-4 troug A-6) 9.8 kj/kg P kj/kg x 0 v m /kg P 5 MPa v kj/kg m (a) Ug te entropy data rom te compreed liquid water table w P /kg kj/kg (b) Ug let peciic volume and preure value w v ( P P ) ( m /kg)(5,000 0) 5.4 kj/kg P rror 0.% (b) Ug average peciic volume and preure value [ / ( ) m /kg](5,000 0) 5.0 kj/kg w v ( P P ) P avg rror 0% 0 5 MPa pump Dicuion e reult ow tat any o te metod may be ued to calculate reverible pump work. PROPRIARY MARIAL. 008 e McGraw-Hill ompanie, Inc. Limited ditribution permitted only to teacer and educator or coure preparation. I you are a tudent ug ti Manual, you are ug it wit permiion.

6 Nitro i compreed an adiabatic compreor. e mimum work put i to be determed. Aumption i i a teady-low proce ce tere i no cange wit time. e proce i adiabatic, and tu tere i no eat traner. Nitro i an ideal ga wit contant peciic eat. Propertie e propertie o nitro at an anticipated average temperature o 400 K are c p.044 kj/kg K and k.97 (able A-b). Analyi ere i only one let and one exit, and tu m. We take te compreor a te ytem, wic i a control volume ce ma croe te boundary. e energy balance or ti teadylow ytem can be expreed te rate orm a 44 (teady) Δytem 0 Rate o net energy traner by eat, work, and ma m + Rate o cange ternal, ketic, potential, etc.energie m ( ) For te mimum work put to te compreor, te proce mut be reverible a well a adiabatic (i.e., ientropic). i beg te cae, te exit temperature will be 600 Nitro compreor ( k ) / k 0.97 /.97 P 600 (0 K) P K 0 Subtitutg to te energy balance equation give w p c ( ) (.044 kj/kg K)(479 0)K 84 kj/kg PROPRIARY MARIAL. 008 e McGraw-Hill ompanie, Inc. Limited ditribution permitted only to teacer and educator or coure preparation. I you are a tudent ug ti Manual, you are ug it wit permiion.

7 Steam i expanded an adiabatic turbe wit an ientropic eiciency o 0.9. e power put o te turbe i to be determed. Aumption i i a teady-low proce ce tere i no cange wit time. Ketic and potential energy cange are negligible. e device i adiabatic and tu eat traner i negligible. Analyi ere i only one let and one exit, and tu m. We take te actual turbe a te ytem, wic i a control volume ce ma croe te boundary. e energy balance or ti teadylow ytem can be expreed te rate orm a 44 (teady) Δytem 0 Rate o net energy traner by eat, work, and ma a, Rate o cange ternal, ketic, potential, etc.energie m a, ( + m From te team table (able A-4 troug A-6), P MPa 400 P 0.7 kj/kg x 6.95 kj/kg K g + x ) (ce Δke Δpe 0) (0.876)(5.) 5.7 kj/kg g e actual power put may be determed by multiplyg te ientropic power put wit te ientropic eiciency. en, a, η η, ( (0.9)( kg/)(.7 5.7)kJ/kg 649 kw ) P MPa 400 Steam turbe η 9% P 0 PROPRIARY MARIAL. 008 e McGraw-Hill ompanie, Inc. Limited ditribution permitted only to teacer and educator or coure preparation. I you are a tudent ug ti Manual, you are ug it wit permiion.

8 Steam enter an adiabatic turbe at a peciied tate, and leave at a peciied tate. e ma low rate o te team and te ientropic eiciency are to be determed. Aumption i i a teady-low proce ce tere i no cange wit time. Potential energy cange are negligible. e device i adiabatic and tu eat traner i negligible. Analyi (a) From te team table (able A-4 and A-6), P 7 MPa 600 P kj/kg kj/kg K a 780. kj/kg ere i only one let and one exit, and tu. We take te actual turbe a te ytem, wic i a control volume ce ma croe te boundary. e energy balance or ti teady-low ytem can be expreed te rate orm a 44 (teady) Δytem 0 Rate o net energy traner by eat, work, and ma ( + V Rate o cange ternal, ketic, potential, etc.energie / ) a, 4444 a, + ( + V V + /) V (ce Δpe 0) Subtitutg, te ma low rate o te team i determed to be (40 m/) 6000 kj/ kg/ (80 m/) kj/kg 000 m / (b) e ientropic exit entalpy o te team and te power put o te ientropic turbe are and P 50,, x 0.98 g x g ( + {( V V )/ } ) ( 6.95 kg/) 874 kw ( 0.98)( 04.7) 467. kj/kg (40 m/) (80 m/) en te ientropic eiciency o te turbe become a 6000 kw η % 874 kw kj/kg 000 m / H O 6 MW PROPRIARY MARIAL. 008 e McGraw-Hill ompanie, Inc. Limited ditribution permitted only to teacer and educator or coure preparation. I you are a tudent ug ti Manual, you are ug it wit permiion.

9 D S Rerigerant-4a enter an adiabatic compreor wit an ientropic eiciency o 0.80 at a peciied tate wit a peciied volume low rate, and leave at a peciied preure. e compreor exit temperature and power put to te compreor are to be determed. Aumption i i a teady-low proce ce tere i no cange wit time. Ketic and potential energy cange are negligible. e device i adiabatic and tu eat traner i negligible. Analyi (a) From te rerigerant table (able A- troug A-), P 0 g at. vapor v v P MPa g@0 8. kj/kg From te ientropic eiciency relation, u, η P a a a a MPa 9.6 kj/kg a 6.97 kj/kg kj/kg K m /kg (b) e ma low rate o te rerigerant i determed rom V m v ( )/η ( ) 0./60 m / kg/ 0.6 m /kg / kj/kg ere i only one let and one exit, and tu. We take te actual compreor a te ytem, wic i a control volume ce ma croe te boundary. e energy balance or ti teady-low ytem can be expreed a 44 (teady) Δytem 0 Rate o net energy traner by eat, work, and ma a, a, Rate o cange ternal, ketic, potential, etc.energie + m m ( 4444 (ce Δke Δpe 0) ) Subtitutg, te power put to te compreor become, W a, ( kg/)( ).70 kw kj/kg R-4a η 80% 0. m /m PROPRIARY MARIAL. 008 e McGraw-Hill ompanie, Inc. Limited ditribution permitted only to teacer and educator or coure preparation. I you are a tudent ug ti Manual, you are ug it wit permiion.

10 old water i eated by ot water a eat excanger. e rate o eat traner and te rate o entropy eration wit te eat excanger are to be determed. Aumption Steady operatg condition exit. e eat excanger i well-ulated o tat eat lo to te urroundg i negligible and tu eat traner rom te ot luid i equal to te eat traner to te cold luid. ange te ketic and potential energie o luid tream are negligible. 4 Fluid propertie are contant. Propertie e peciic eat o cold and ot water are given to be 4.8 and 4.9 kj/kg., repectively. Analyi We take te cold water tube a te ytem, wic i a control volume. e energy balance or ti teady-low ytem can be expreed te rate orm a 44 Rate o net energy traner by eat, work, and ma + m Rate o cange ternal, ketic, potential, etc.energie m mc ( 0 (teady) Δytem 4444 p (ce Δke Δpe 0) ) 0 en te rate o eat traner to te cold water ti eat excanger become [ mc ( )] cold water (0.5 kg/)(4.8 kj/kg. )(45 5 ).5 kw p Notg tat eat ga by te cold water i equal to te eat lo by te ot water, te let temperature o te ot water i determed to be.5 kw [ mc p( )] ot water mc ( kg/)(4.9 kj/kg. ) (b) e rate o entropy eration wit te eat excanger i determed by applyg te rate orm o te entropy balance on te entire eat excanger: 0 (teady) S S + S ΔS 44 ytem { 4444 cold Rate o net entropy traner by eat and ma + ot cold Rate o entropy eration + + S 4 ot + S 4 S 0 0 p Rate o cange o entropy (ce Q 0) cold ( ) + ot ( 4 ) Notg tat bot luid tream are liquid (compreible ubtance), te rate o entropy eration i determed to be S cold cp ln + ot 4 cp ln Hot water 00 kg/ (0.5 kg/)(4.8 kj/kg.k)ln + ( kg/)(4.9 kj/kg.k)ln kw/k old water kg/ PROPRIARY MARIAL. 008 e McGraw-Hill ompanie, Inc. Limited ditribution permitted only to teacer and educator or coure preparation. I you are a tudent ug ti Manual, you are ug it wit permiion.

11 In an ice-makg plant, water i rozen by evaporatg aturated R-4a liquid. e rate o entropy eration i to be determed. Aumption Steady operatg condition exit. Ketic and potential energy cange are negligible. Analyi We take te control volume ormed by te R-4a evaporator wit a gle let and gle exit a te ytem. e rate o entropy eration wit ti evaporator durg ti proce can be determed by applyg te rate orm o te entropy balance on te ytem. e entropy balance or ti teady-low ytem can be expreed a S S 44 Rate o net entropy traner by eat and ma + + w S { Rate o entropy eration + S S S 0 (teady) S Δ ytem Rate o cange o entropy R ( ) w R g e propertie o te rerigerant are (able A-) kj/kg kj/kg K 0 w e rate o tat mut be removed rom te water order to reeze it at a rate o 4000 kg/ i Q w (4000 / 600 kg/)(.7 kj/kg) 70.8 kw i were te eat o uion o water at atm i.7 kj/kg. e ma low rate o R-4a i 70.8 kj/ R.800 kg/ kj/kg g R-4a 0 Q 0 at. vapor Subtitutg, S R g w 70.8 kw (.800 kg/)(0.786 kj/kg K) kw/k 7 K PROPRIARY MARIAL. 008 e McGraw-Hill ompanie, Inc. Limited ditribution permitted only to teacer and educator or coure preparation. I you are a tudent ug ti Manual, you are ug it wit permiion.

12 Steam expand a turbe rom a peciied tate to anoter peciied tate. e rate o entropy eration durg ti proce i to be determed. Aumption i i a teady-low proce ce tere i no cange wit time. Ketic and potential energy cange are negligible. Propertie From te team table (able A-4 troug 6) P 6 MPa 450 P 0 at. vapor 0.9 kj/kg 6.79 kj/kg K kj/kg kj/kg K Analyi ere i only one let and one exit, and tu. We take te turbe a te ytem, wic i a control volume ce ma croe te boundary. e energy balance or ti teady-low ytem can be expreed te rate orm a Subtitutg, 44 (teady) Δytem 0 Rate o net energy traner by eat, work, and ma m Rate o cange ternal, ketic, potential, etc.energie m ( ) Q (5,000/600 kg/)( )kJ/kg 4000 kj/ 89. kj/ e rate o total entropy eration durg ti proce i determed by applyg te entropy balance on an extended ytem tat clude te turbe and it immediate urroundg o tat te boundary temperature o te extended ytem i 5 at all time. It give 0 S S + S ΔS ytem 0 44 { 44 Rate o net entropy traner by eat and ma m m Rate o entropy eration b,urr + S Rate o cange o entropy 0 P 6 MPa 450 SAM URBIN P 0 at. vapor 4 MW Subtitutg, te rate o entropy eration durg ti proce i determed to be S ( ) + ( 5,000/600 kg/)( ) b,urr 89. kw kj/kg K + 98 K.0 kw/k PROPRIARY MARIAL. 008 e McGraw-Hill ompanie, Inc. Limited ditribution permitted only to teacer and educator or coure preparation. I you are a tudent ug ti Manual, you are ug it wit permiion.

13 Liquid water i eated a camber by mixg it wit upereated team. For a peciied mixg temperature, te ma low rate o te team and te rate o entropy eration are to be determed. Aumption i i a teady-low proce ce tere i no cange wit time. Ketic and potential energy cange are negligible. ere are no work teraction. Propertie Notg tat < 00 0., te cold water and te exit mixture tream exit a a compreed liquid, wic can be approximated a a aturated liquid at te given temperature. From able A-4 troug A-6, 00 kj/m P kj/kg 0 P P kj/kg 7.80 kj/kg kj/kg K o 5.8 kj/kg 0.8 kj/kg K Analyi (a) We take te mixg camber a te ytem, wic i a control volume. e ma and energy balance or ti teady-low ytem can be expreed te rate orm a (teady) Ma balance: m Δ ytem 0 + nergy balance: 44 (teady) Δytem 0 Rate o net energy traner by eat, work, and ma + Rate o cange ternal, ketic, potential, etc.energie ombg te two relation give + ( + ) ( ) + ( ) Solvg or m and ubtitutg, te ma low rate o te upereated team i determed to be Alo, ( ) (00/60kJ/) (.5 kg/)( ) m kg/ ( )kJ/kg kj/kg 0.66 kg/ (b) e rate o total entropy eration durg ti proce i determed by applyg te entropy balance on an extended ytem tat clude te mixg camber and it immediate urroundg o tat te boundary temperature o te extended ytem i 5 at all time. It give 0 S S + S ΔS ytem 0 44 { 44 Rate o net entropy traner by eat and ma + Rate o entropy eration b,urr + S Rate o cange o entropy 0 Subtitutg, te rate o entropy eration durg ti proce i determed to be S + b,urr (.666 kg/)( 0.8 kj/kg K) ( 0.66 kg/)( 7.80 kj/kg K) (00/ 60 kj/) (.5 kg/)( kj/kg K) + 98 K 0. kw/k 0.5 kg/ 50 MIXING HAMBR PROPRIARY MARIAL. 008 e McGraw-Hill ompanie, Inc. Limited ditribution permitted only to teacer and educator or coure preparation. I you are a tudent ug ti Manual, you are ug it wit permiion.

14 An ulated rigid tank i connected to a piton-cylder device wit zero clearance tat i mataed at contant preure. A valve i opened, and ome team te tank i allowed to low to te cylder. e al temperature te tank and te cylder are to be determed. Aumption Bot te tank and cylder are well-ulated and tu eat traner i negligible. e water tat rema te tank underwent a reverible adiabatic proce. e termal energy tored te tank and cylder temelve i negligible. 4 e ytem i tationary and tu ketic and potential energy cange are negligible. Analyi (a) e team tank A undergoe a reverible, adiabatic proce, and tu. From te team table (able A-4 troug A-6), v v g P 500 u u at. vapor P 50 v ( at. mixture), A g@500 v u, g@500 + x u m /kg kj/kg kj/kg K, A x v + x, A g, A e itial and te al mae tank A are u, m m, A, B V A v m, A, A 0.4 m.067 kg m /kg m, A, A, A kg (b) e boundary work done durg ti proce i at@ (0.905)( ).0789 m u g g ( V, B ) PB m, B B W b, PdV PB 0 v, akg te content o bot te tank and te cylder to be te ytem, te energy balance or ti cloed ytem can be expreed a or, u, 44 Net energy traner by eat, work, and ma P m B, B v W, B Δytem 44 ange ternal, ketic, potential, etc.energie b, ΔU ( ΔU ) A + ( ΔU ) B Wb, + ( ΔU ) A + ( ΔU ) B 0 ( mu mu ) A + ( mu ) B 0 m + ( m u m u ) 0 +, B, B A (0.905)(05. kj/kg) 76.6 kj/kg and m, A V A v ( mu mu ) (.067)( 560.7) ( 0.7)( 76.6), A A, B m, B m 0.7 kg.0789 m /kg kj/kg At 50, 467. and g 69. kj/kg. u at te al tate, te cylder will conta a aturated liquid-vapor mixture ce < < g. ereore,.5, B at@50 /kg Sat. vapor m 50 PROPRIARY MARIAL. 008 e McGraw-Hill ompanie, Inc. Limited ditribution permitted only to teacer and educator or coure preparation. I you are a tudent ug ti Manual, you are ug it wit permiion.

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

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

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

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

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

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

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

( )( ) 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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

KNOWN: Data are provided for a closed system undergoing a process involving work, heat transfer, change in elevation, and change in velocity.

KNOWN: Data are provided for a closed system undergoing a process involving work, heat transfer, change in elevation, and change in velocity. Problem 44 A closed system of mass of 10 kg undergoes a process during which there is energy transfer by work from the system of 0147 kj per kg, an elevation decrease of 50 m, and an increase in velocity

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

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

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

2-21. for gage pressure, the high and low pressures are expressed as. Noting that 1 psi = kpa,

2-21. for gage pressure, the high and low pressures are expressed as. Noting that 1 psi = kpa, - -58E The systolic and diastolic pressures of a healthy person are given in mmhg. These pressures are to be expressed in kpa, psi, and meter water column. Assumptions Both mercury and water are incompressible

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

& 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

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

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

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

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

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

CHAPTER 8 ENTROPY. Blank

CHAPTER 8 ENTROPY. Blank CHAPER 8 ENROPY Blank SONNAG/BORGNAKKE SUDY PROBLEM 8-8. A heat engine efficiency from the inequality of Clausius Consider an actual heat engine with efficiency of η working between reservoirs at and L.

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

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

ECE309 INTRODUCTION TO THERMODYNAMICS & HEAT TRANSFER. 20 June 2005

ECE309 INTRODUCTION TO THERMODYNAMICS & HEAT TRANSFER. 20 June 2005 ECE309 INTRODUCTION TO THERMODYNAMICS & HEAT TRANSFER 20 June 2005 Midterm Examination R. Culham & M. Bahrami This is a 90 minute, closed-book examination. You are permitted to use one 8.5 in. 11 in. crib

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

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

ENERGY TRANSFER BY WORK: Electrical Work: When N Coulombs of electrical charge move through a potential difference V

ENERGY TRANSFER BY WORK: Electrical Work: When N Coulombs of electrical charge move through a potential difference V Weight, W = mg Where m=mass, g=gravitational acceleration ENERGY TRANSFER BY WOR: Sign convention: Work done on a system = (+) Work done by a system = (-) Density, ρ = m V kg m 3 Where m=mass, V =Volume

More information

Compressor 1. Evaporator. Condenser. Expansion valve. CHE 323, October 8, Chemical Engineering Thermodynamics. Tutorial problem 5.

Compressor 1. Evaporator. Condenser. Expansion valve. CHE 323, October 8, Chemical Engineering Thermodynamics. Tutorial problem 5. CHE 33, October 8, 014. Cemical Engineering Termodynamics. Tutorial problem 5. In a simple compression refrigeration process, an adiabatic reversible compressor is used to compress propane, used as a refrigerant.

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

(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 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

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

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

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

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

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

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

PROBLEM 6.3. Using the appropriate table, determine the indicated property. In each case, locate the state on sketches of the T-v and T-s diagrams.

PROBLEM 6.3. Using the appropriate table, determine the indicated property. In each case, locate the state on sketches of the T-v and T-s diagrams. PROBLEM 63 Using the appropriate table, determine the indicated property In each case, locate the state on sketches of the -v and -s diagrams (a) water at p = 040 bar, h = 147714 kj/kg K Find s, in kj/kg

More information

Thermodynamics [ENGR 251] [Lyes KADEM 2007]

Thermodynamics [ENGR 251] [Lyes KADEM 2007] CHAPTER V The first law of thermodynamics is a representation of the conservation of energy. It is a necessary, but not a sufficient, condition for a process to occur. Indeed, no restriction is imposed

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

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

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

CHAPTER. The First Law of Thermodynamics: Closed Systems

CHAPTER. The First Law of Thermodynamics: Closed Systems CHAPTER 3 The First Law of Thermodynamics: Closed Systems Closed system Energy can cross the boundary of a closed system in two forms: Heat and work FIGURE 3-1 Specifying the directions of heat and work.

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

+ 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

HEAT ENGINES AND REFRIGERATORS

HEAT ENGINES AND REFRIGERATORS EA ENGINES AND REFRIGERAORS 9 onceptual uestions 9.. s =. (a) < 0, > 0. ork is done by the system; the area under the curve is positive. s (b) > 0, < 0. ork is done on the system to compress it to a smaller

More 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

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

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

EXAM # 1 ME 300 SP2017

EXAM # 1 ME 300 SP2017 CIRCLE YOUR LECTURE BELOW: 8:3 am :3 am 3:3 pm Prof. Lucht Prof. Chen Prof. Goldenstein EXAM # ME 3 SP7 INSTRUCTIONS. Please place all your electronics, including but not limited to cell phones, computers,

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

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

Name: I have observed the honor code and have neither given nor received aid on this exam.

Name: I have observed the honor code and have neither given nor received aid on this exam. ME 235 FINAL EXAM, ecember 16, 2011 K. Kurabayashi and. Siegel, ME ept. Exam Rules: Open Book and one page of notes allowed. There are 4 problems. Solve each problem on a separate page. Name: I have observed

More information

Today lecture. 1. Entropy change in an isolated system 2. Exergy

Today lecture. 1. Entropy change in an isolated system 2. Exergy Today lecture 1. Entropy change in an isolated system. Exergy - What is exergy? - Reversible Work & Irreversibility - Second-Law Efficiency - Exergy change of a system For a fixed mass For a flow stream

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

MAE 320 THERODYNAMICS FINAL EXAM - Practice. Name: You are allowed three sheets of notes.

MAE 320 THERODYNAMICS FINAL EXAM - Practice. Name: You are allowed three sheets of notes. 50 MAE 320 THERODYNAMICS FINAL EXAM - Practice Name: You are allowed three sheets of notes. 1. Fill in the blanks for each of the two (Carnot) cycles below. (a) 5 a) Heat engine or Heat pump/refrigerator

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

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

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

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

20 m neon m propane. g 20. Problems with solutions:

20 m neon m propane. g 20. Problems with solutions: Problems with solutions:. A -m tank is filled with a gas at room temperature 0 C and pressure 00 Kpa. How much mass is there if the gas is a) Air b) Neon, or c) Propane? Given: T7K; P00KPa; M air 9; M

More information

MAE 110A. Homework 3: Solutions 10/20/2017

MAE 110A. Homework 3: Solutions 10/20/2017 MAE 110A Homework 3: Solutions 10/20/2017 3.10: For H 2O, determine the specified property at the indicated state. Locate the state on a sketch of the T-v diagram. Given a) T 140 C, v 0.5 m 3 kg b) p 30MPa,

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

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

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

2-18. (a) For mercury, (b) For water,

2-18. (a) For mercury, (b) For water, -8-5 CD EES Bt a gage and a manmeter are attaced t a gas t measure its pressure. Fr a specified reading f gage pressure, te difference between te fluid levels f te tw arms f te manmeter is t be determined

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

Simulation and verification of a plate heat exchanger with a built-in tap water accumulator

Simulation and verification of a plate heat exchanger with a built-in tap water accumulator Simulation and verification of a plate eat excanger wit a built-in tap water accumulator Anders Eriksson Abstract In order to test and verify a compact brazed eat excanger (CBE wit a built-in accumulation

More information

ME 201 Thermodynamics

ME 201 Thermodynamics ME 0 Thermodynamics Solutions First Law Practice Problems. Consider a balloon that has been blown up inside a building and has been allowed to come to equilibrium with the inside temperature of 5 C and

More 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

Circle your instructor s last name

Circle your instructor s last name ME 00 Thermodynamics Fall 07 Exam Circle your structor s last name Division : Naik Division : Wassgren Division 6: Braun Division : Sojka Division 4: Goldenste Division 7: Buckius Division 8: Meyer INSTRUCTIONS

More information

ESO 201A Thermodynamics

ESO 201A Thermodynamics ESO 201A Thermodynamics Instructor: Sameer Khandekar Tutorial 9 [7-27] A completely reversible heat pump produces heat at arate of 300 kw to warm a house maintained at 24 C. Theexterior air, which is at

More information

ENGR Thermodynamics

ENGR Thermodynamics ENGR 224 - hermodynamics #1 - Diagram for a Cascade VCR Cycle (21 ts) Baratuci Final 13-Jun-11 On a full sheet of paper, construct a complete Diagram for the cascade cascade vapor-compression refrigeration

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

In the next lecture...

In the next lecture... 16 1 In the next lecture... Solve problems from Entropy Carnot cycle Exergy Second law efficiency 2 Problem 1 A heat engine receives reversibly 420 kj/cycle of heat from a source at 327 o C and rejects

More 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

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

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

Fundamentals of Heat Transfer Muhammad Rashid Usman

Fundamentals of Heat Transfer Muhammad Rashid Usman Fundamentals of Heat Transfer Muammad Rasid Usman Institute of Cemical Engineering and Tecnology University of te Punjab, Laore. Figure taken from: ttp://eatexcanger-design.com/20/0/06/eat-excangers-6/

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

Given: Hot fluid oil, Cold fluid - water (T 1, T 2 ) (t 1, t 2 ) Water

Given: Hot fluid oil, Cold fluid - water (T 1, T 2 ) (t 1, t 2 ) Water . In a counter flow double pipe eat excanger, oil is cooled fro 85 to 55 by water entering at 5. Te ass flow rate of oil is 9,800 kg/ and specific eat f oil is 000 J/kg K. Te ass flow rate of water is

More information

Chapter 4. Energy Analysis of Closed Systems

Chapter 4. Energy Analysis of Closed Systems Chapter 4 Energy Analysis of Closed Systems The first law of thermodynamics is an expression of the conservation of energy principle. Energy can cross the boundaries of a closed system in the form of heat

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

EF 152 Exam #3, Fall, 2012 Page 1 of 6

EF 152 Exam #3, Fall, 2012 Page 1 of 6 EF 5 Exam #3, Fall, 0 Page of 6 Name: Setion: Guidelines: ssume 3 signifiant figures for all given numbers. Sow all of your work no work, no redit Write your final answer in te box provided - inlude units

More information

MAE 110A. Homework 6: Solutions 11/9/2017

MAE 110A. Homework 6: Solutions 11/9/2017 MAE 110A Hoework 6: Solutions 11/9/2017 H6.1: Two kg of H2O contained in a piston-cylinder assebly, initially at 1.0 bar and 140 C undergoes an internally ersible, isotheral copression to 25 bar. Given

More information