TERMODİNAMİK ÇALIŞMA SORULARI

Transkript

1 TRMODİNAMİK ÇALIŞMA SORULARI SÜRKLİ AKIŞLI AÇIK SİSTMLR LÜLLR YAYICILAR 5-8 Sürekli akışlı adyabatik bir lülede, ava lüleye 00 kpa basınç, 00 o C sıcaklık ve 45 m/s ızla girmekte, 00 kpa basınç ve 0 m/s ızla çıkmaktadır.lülen giriş kesit alanı 80 cm²'dir. (a) Lüleden akan avanın kütlesel debisi, (b) Havanın lüleden çıkış sıcaklığını, (c) Lülen çıkış kesit alanını esaplayınız. çözüm: (a).09 kg/s (b) 85 o C (c) 79.9 cm² kpa basınç ve 400 o C sıcaklıktaki su buarı, sürekli akışlı, adyabatik bir lüleye 0 m/s ızla girmekte, 00 kpa basınç 00 o C sıcaklıkta çıkmakadır, bu sıra 5 kw ısı kaybı meydana gelmektedir.çıkış ızını ve buarın lüleye çıkışındaki acimsel debisi esaplayınız. çözüm: 606 m/s.74 m³/s 5- Hava sürekli akışlı adyabatik bir lüleye 90 kpa basınç, -0 o C sıcaklık ve 80 m/s ızla girmekte, 00 kpa basınçta giriş ızından düşük çıkmaktadır. (a) Havanın çıkış sıcaklığını, (b) Havanın çıkış ızını esaplayınız. TÜRBİNLR KOMPRSÖRLR 5-46 Su buarı sürekli akışlı adyabatik türbe 6 MPa basınç, 400 o C sıcaklık ve 80 m/s ızla girmekte, 40 kpa basınç ve yüzde 9 kuruluk derecesde, 50 m/s ızla çıkmaktadır.buarın kütle debisi 0 kg/s olduğuna göre, (a) Akışın ketik enerjisdeki değişimi, (b) Türbde üretilen gücü, (c) Türb giriş kesit alanını esaplayınız. çözüm: (a) -.95 kj/kg, (b) 4.6 MW (c) 0.09 m² 5-48 Su buarı sürekli akışlı bir adyabatik türbe 0 MPa basınç ve 500 o C sıcaklıkta girmekte, 0 kpa basınç ve yüzde 90 kuruluk derecesiyle çıkmaktadır. Ketik ve potansiyel enerji değişimleri imal ederek, 5 MW güç üretilebilmesi iç gerekli kütle debisi esaplayınız. çözüm: 4.85 kg/s 5-50 Adyabatik bir ava kompresörü 0 L/s debi ile ava 0 kpa basınç ve 0 o C sıcaklıktan 000 kpa basınç ve 00 o C sıcaklığa sıkıştırmaktadır. (a) Kompresör iç gerekli işi (b) Kompresörü çalıştırmak iç gerekli gücü esaplayınız. 5-5 Karbon dioksit sürekli akışlı adyabatik bir kompresöre 00 kpa basınç ve 00 K sıcaklıkta, 0.5 kg/s debiyle girmekte, 600 kpa basınç ve 450 K sıcaklıkta çıkmaktadır.ketik enerji değişimleri imal ederek, (a) Kompresör girişde karbon dioksit acimsel debisi, (b) Kompresörü çalıştırmak iç gerekli gücü esaplayınız. çözüm: (a) 0.8 m³/s (b) 68.8 kw

2 5-0 Air is accelerated a nozzle from 45 m/s to 80 m/s. Te mass flow rate, te exit temperature, and te exit area of te nozzle are to be determed. Assumptions Tis is a steady-flow process sce tere is no cange wit time. Air is an ideal gas wit constant specific eats. Potential energy canges are negligible. 4 Te device is adiabatic and tus eat transfer is negligible. 5 Tere are no work teractions. Properties Te gas constant of air is 0.87 kpa.m /kg.k (Table A-). Te specific eat of air at te anticipated average temperature of 450 K is c p.0 kj/kg. C (Table A-). Analysis (a) Tere is only one let and one exit, and tus m m m. Usg te ideal gas relation, te specific volume and te mass flow rate of air are determed to be RT ( 0.87 kpa m /kg K)(47 K).455 m P 00 kpa m A (0.00 m )(45 m/s).094 kg/s m /kg /kg P 0 kpa T 0 C 45 m/s A cm AIR P 0 kpa 80 m/s ( b) We take nozzle as te system, wic is a control volume sce mass crosses te boundary. Te energy balance for tis steady-flow system can be expressed te rate form as system Rate of net energy transfer Rate of cange ternal, ketic, potential, etc. energies 5-5 Substitutg, m( / ) m( 0 /) (sce Q W pe 0 c p, ave T T 0 (.0 kj/kg K)( T 00 (80 m/s) C) (45 m/s) kj/kg 000 m /s It yields T 85. C (c) Te specific volume of air at te nozzle exit is RT ( 0.87 kpa m /kg K)(85. 7 K).5 m P 00 kpa m A.094 kg/s A 80 m/s A m 79.9 cm.5 m /kg /kg PROPRITARY MATRIAL. 0 Te McGraw-Hill Companies, Inc. Limited distribution permitted only to teacers and educators for course preparation. If you are a student usg tis Manual, you are usg it wit permission.

3 5-4 Heat is lost from te steam flowg a nozzle. Te velocity and te volume flow rate at te nozzle exit are to be determed. Assumptions Tis is a steady-flow process sce tere is no cange wit time. Potential energy cange is negligible. Tere are no work teractions. Analysis We take te steam as te system, wic is a control volume sce mass crosses te boundary. Te energy balance for tis steady-flow system can be expressed te rate form as nergy balance: 400 C 800 kpa 0 m/s STAM Q 00 C 00 kpa 5-9 m Rate of cange ternal, ketic, potential, etc. energies m system Q sce W pe or Q m Te properties of steam at te let and exit are (Table A-6) P 800 kpa T 400 C.849 m /kg 67.7 kj/kg P 0 kpa T 0 C Te mass flow rate of te steam is Substitutg,.6 m /kg 07. kj/kg m (0.08 m )(0 m/s).08 kg/s m /s A (0 m/s) 67.7 kj/kg kj/kg 000 m /s Te volume flow rate at te exit of te nozzle is (.08 kg/s)(.6 m /kg) m 7. kj/kg 606 m/s.74 m /s kj/kg 000 m /s 5 kj/s.08 kg/s PROPRITARY MATRIAL. 0 Te McGraw-Hill Companies, Inc. Limited distribution permitted only to teacers and educators for course preparation. If you are a student usg tis Manual, you are usg it wit permission.

4 5-6 Air is decelerated a diffuser from 600 ft/s to a low velocity. Te exit temperature and te exit velocity of air are to be determed. Assumptions Tis is a steady-flow process sce tere is no cange wit time. Air is an ideal gas wit variable specific eats. Potential energy canges are negligible. 4 Te device is adiabatic and tus eat transfer is negligible. 5 Tere are no work teractions. Properties Te entalpy of air at te let temperature of 50 F is.88 Btu/lbm (Table A-7). Analysis (a) Tere is only one let and one exit, and tus m m m. We take diffuser as te system, wic is a control volume sce mass crosses te boundary. Te energy balance for tis steady-flow system can be expressed te rate form as 5- system Rate of cange ternal, ketic, potential, etc. energies AIR or, m( / ) m( /) (sce Q W pe, Btu/lbm 600 ft/s Btu/lbm 5,07 ft /s 9.07 Btu/lbm From Table A-7, T 540 R (b) Te exit velocity of air is determed from te conservation of mass relation, Tus, A A A RT / P RT / A P AT P (540 R)( psia) (600 ft/s) 4 ft/s A T P 4 (50 R)(4.5 psia) PROPRITARY MATRIAL. 0 Te McGraw-Hill Companies, Inc. Limited distribution permitted only to teacers and educators for course preparation. If you are a student usg tis Manual, you are usg it wit permission.

5 5-49 Steam expands a turbe. Te cange ketic energy, te power put, and te turbe let area are to be determed. Assumptions Tis is a steady-flow process sce tere is no cange wit time. Potential energy canges are negligible. Te device is adiabatic and tus eat transfer is negligible. Properties From te steam tables (Tables A-4 troug 6) and P 6 MPa T 400 C P 40 kpa x m /kg 78. kj/kg x f fg Analysis (a) Te cange ketic energy is determed from kj/kg P 6 MPa T 400 C 80 m/s STAM m kg/s 50 m/s (80 m/s) kj/kg ke.95 kj/kg P 40 kpa 000 m /s x.9 50 m/s (b) Tere is only one let and one exit, and tus m m m. We take te turbe as te system, wic is a control volume sce mass crosses te boundary. Te energy balance for tis steady-flow system can be expressed te rate form as system m( Rate of cange ternal, ketic, potential, etc.energies / ) W m( /) (sce Q pe W m Ten te power put of te turbe is determed by substitution to be W (c) Te let area of te turbe is determed from te mass flow rate relation, (0 kg/s)( )kJ/kg 4,590 kw 4.6 MW W 5- m m A A (0 kg/s)( m /kg) 80 m/s.09m PROPRITARY MATRIAL. 0 Te McGraw-Hill Companies, Inc. Limited distribution permitted only to teacers and educators for course preparation. If you are a student usg tis Manual, you are usg it wit permission.

6 5-5 Steam expands a turbe. Te mass flow rate of steam for a power put of 5 MW is to be determed. Assumptions Tis is a steady-flow process sce tere is no cange wit time. Ketic and potential energy canges are negligible. Te device is adiabatic and tus eat transfer is negligible. Properties From te steam tables (Tables A-4 troug 6) P MPa T 500 C 75. kj/kg 5- P kpa x.90 x f fg kj/kg H O Analysis Tere is only one let and one exit, and tus m m m. We take te turbe as te sy stem, wic is a control volume sce mass crosses te boundary. Te energy balance for tis steady-flow system can be expressed te rate form as system Rate of cange ternal, ketic, potential, etc.energies W m( ) m W m (sce Q ke pe Substitutg, te required mass flow rate of te steam is determed to be 5000 kj/s m( ) kj/kg m 4.85 kg/s PROPRITARY MATRIAL. 0 Te McGraw-Hill Companies, Inc. Limited distribution permitted only to teacers and educators for course preparation. If you are a student usg tis Manual, you are usg it wit permission.

7 5-5 Air is compressed at a rate of 0 L/s by a compressor. Te work required per unit mass and te power required are to be determed. Assumptions Tis is a steady-flow process sce tere is no cange wit time. Ketic and potential energy canges are negligible. Air is an ideal gas wit constant specific eats. Properties Te constant pressure specific eat of air at te average temperature of (000)/60 C4 K is c p.08 kj/kg K (Table A-b). Te gas constant of air is R.87 kpa m /kg K (Table A-). Analysis (a) Tere is only one let and one exit, and tus m m m. We take te compressor as te system, wic is a control volume sce mass crosses te boundary. Te energy balance for tis steady-flow system can be expressed te rate form as 5-5 system by eat, work, and mass Tus, W m W w p Rate of cange ternal, ketic, potential, etc.energies m m( (sce ke pe ) mc p ( T T ) c ( T T ) (.08 kj/kg K)(00 0)K 85.0 kj/kg MPa 00 C Compressor 0 kpa 0 C 0 L/s (b) Te specific volume of air at te let and te mass flow rate are ( 0.87 kpa m /kg K)(0 7 K) RT m /kg 0 kpa P m 0.00 m /s.047 kg/s m /kg Ten te power put is determed from te energy balance equation to be W mc ( T T ) (0.047 kg/s)(.08 kj/kg K)(00 0)K kw p PROPRITARY MATRIAL. 0 Te McGraw-Hill Companies, Inc. Limited distribution permitted only to teacers and educators for course preparation. If you are a student usg tis Manual, you are usg it wit permission.

8 5-56 CO is compressed by a compressor. Te volume flow rate of CO at te compressor let and te power put to te compressor are to be determed. Assumptions Tis is a steady-flow process sce tere is no cange wit time. Ketic and potential energy canges are negligible. Helium is an ideal gas wit variable specific eats. 4 Te device is adiabatic and tus eat transfer is negligible. Properties Te gas constant of CO is R.889 kpa.m /kg.k, and its molar mass is M 44 kg/kmol (Table A-). Te let and exit entalpies of CO are (Table A-0) 5-8 T T 0 K 9, 4 kj / kmol 450 K 5, 48 kj / kmol Analysis (a) Tere is only one let and one exit, and tus specific volume of air and its volume flow rate are m m m. Te let CO RT P kpa m /kg K 00 K.5667 m /kg 00 kpa m (0.5 kg/s)( m /kg) (b) We take te compressor as te system, wic is a control volume sce mass crosses te boundary. Te energy balance for tis steady- flow system can be expressed te rate form as Substitutg W system Rate of cange ternal, ketic, potential, etc. energies 0.8 m W m m (sce Q ke pe W m( ) m( ) / M 0.5 kg/s 5,48 9,4 kj/kmol 44 kg/kmol /s 68.8 kw PROPRITARY MATRIAL. 0 Te McGraw-Hill Companies, Inc. Limited distribution permitted only to teacers and educators for course preparation. If you are a student usg tis Manual, you are usg it wit permission.