1

( )

kgkJ

kJ/kg/secm10002

(m/sec)KE lb

lbft32.22

(ft/sec)KE

Wρgh)2

Vpv(umQ

W)ρgh2

Vvpm(u)ρgh2

Vvpm(uQ

ρgh2

V(T)PEKE(T)E

vp mvp mWWWWWvp mW

vp mVpVVppdVW

LawFirstWΔEQ

22

2

METRIC

2

ENGLISH

shaft

2

shaft1

2

2221

2

111

22211shaftout flowin flowshaft

22out flow

1111final 1initial 11in flow

=×

=−

=×

=

++++∆×=

++++−+++=

++=++=

−+=++==

==−==

+=

∫

UU

Steady Flow Energy EquationOpen, steady flow thermodynamic system - a region in space

QshaftW

p1p2

v2

v1 h2

V

V

h1

2

EquationEnergy FlowSteady Wρgh)2

V(hmQ shaft

2

+++∆=

( )outin hhmW −=∆==

=∆∆hmΔHW

0Q Elevation, Velocity,

( )outin hhmQ −===

=≅≅h mΔΔHQ

0 Work0,ΔElevation 0,ΔVelocity

2Vh

2Vh

22

2

21

1 +=+

==≅ 0W0,Q0,ΔElevation

Steady Flow Processes Devices

Turbine, Compressor, Pump

Boiler, Condenser, Heat Exchanger

Diffuser, Nozzle

outin hh =

==

====

outin HH0ΔH

0W0,Q0,ΔElevation0,ΔVelocityValve - throttling process

3measured becan quality 100% to87% .87,x

kJ/kg 2038.8kJ/kg 731.95kJ/kg 2505.6

hhhx

kJ/kg 2038.8h kJ/kg 731.95h

kPa 850 @kJ/kg 2505.6quality measurable maximum @h

kPa) 100.@(Phquality measurable maximum @h)P(Thh

0Q 0, W0,ΔPE 0,ΔKE

Wρgh)2gVΔ(hQ

EquationEnergy FlowSteady system odynamicopen therm

1fg

1f2

fg

f

2

barometerg2

barometer2,21

shaft

2

=

−=

−=

==

=

==

=====

+++=

device? this with meaaured be canquality steam kPa 850 of rangeWhat

T

v

100 kPa

850 kPa

1

2

1

2

4C53.97kPa 15 @ TT

%quality 29 .29,2372.3

225.94924.79h

hhx

kJ/kg 2373.3h kJ/kg, 225.94h kPa 15at kJ/kg 924.79h

hkg 700kJ/kg 2608.8kg 200kJ/kg 251.18kg 500kJ/kg 2608.8C@60hhkJ/kg 251.18C@60hh

hmhmhmconstanthm

0,PE 0,KE 0, W0,Q

Wρgh)2gV(hmQ

EquationEnergy FlowSteady kg/sec 200kg/sec 500m

mmm Balance Masssystem odynamicopen therm

Osaturation

fg

f

fgf

c

c

Ovb

Ofa

ccbbaa

shaft

2

c

bac

==

=−

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===

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?conditionsexit the areWhat 15kPa. of pressure aat tank a in steam saturated

C60 kg/sec 200 with mixed iswater C60 of kg/sec 500 OO

water60kg/sec 500

O

steam60kg/sec 200

O

kPa 15

ab

c

5

An adiabatic air compressor is to be powered by a direct coupled adiabatic steam turbine that is also driving a generator. Steam enters the turbine at 12.5 MPa and 500 C at a rate of 25 kg/sec and exits at 10 kPa and a quality of .92. Air enters the compressor at 98 kpa and 295 K at a rate of 10 kg/sec and exits at 1 MPa. Determine the net power delivered to the generator by the turbine.

6

An adiabatic air compressor is to be powered by a direct coupled adiabatic steam turbine that is also driving a generator. Steam enters the turbine at 12.5 MPa and 500 C at a rate of 25 kg/sec and exits at 10 kPa and a quality of .92. Air enters the compressor at 98 kpa and 295 K at a rate of 10 kg/sec and exits at 1 MPa and 620 K. Determine the net power delivered to the generator by the turbine.

airsteamp p

v v

( ) ( )

( ) ( )

kJ/sec 27,106.5W

295.17628.07seckg102392.53343.6

seckg25W

hhmhhmWWW

net

net

c2c1ct2t1tcompressorturbinenet

+=

−−−=

−−−=−=

g2392.5kJ/k2392.1.92191.81h

kPa @10xh@10kPahhkJ/kg 3343.6Mpa) 12.5p500,(Tsuperheat@h

kJ/kg 628.07620)Tairtable@(h17-A Table kJ/kg 295.17295)Tairtable@(h

t2

fgft2

t1

c2

c1

=×+=

+=====

======

12.5 kPa500 C

10 kPax=.92

98 kPa295 K

1 Mpa620 K

netW

compressor turbinegenerator

7

Steam at 3Mpa and 400 C enters an adiabatic nozzle steadily with a velocity of 40 m/sec and leaves at 2.5 MPa and 300 m/sec. Determine (a) the exit temperature and (b) the ratio of inlet to exit area.

8

Steam at 3Mpa and 400 C enters an adiabatic nozzle steadily with a velocity of 40 m/sec and leaves at 2.5 MPa and 300 m/sec. Determine (a) the exit temperature and (b) the ratio of inlet to exit area.

p

v

3 MPa400C40m/sec

2.5MPa300m/sec

/kgm 09938.v

kJ/kg 3231.7hMpa) 3.P400.,(T @superheat v&h

31

1

11

=

====

kJ/kg 3187.544.23231.71h/secm 1000

kJ/kg 1secm

2300

240kJ/kgm 3231.71.9h

2V

2Vhh

equationenergy flowsteady 2

Vh2

Vh

2

222

222

2

22

21

12

22

2

21

1

=−=

−+=

−+=

+=+

.187274.53513.951

AA

11626.300 A 2.5

09938.40 A 3

V A V AρAVm b)

11626. v2.5)p3187.5,(h @ v

C, 376.5T2.5)p3187.5,(h @ T a)

1

2

21

2

21

1

11

2

2

2

2

==

=

=

==

===

==

vv

1

2

Open thermodynamic system - a region in space

4.65

9

10

( )if1

1

1

1

1

mmmTuvh

−=1pim

( )

( ) ( )( )

( )( )( ) s h a f t1ifiiff WhmmumumQ +−−−=

++−−−=−−=

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=

=−+=

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+=

∫∫

shaft111ifiiff

11ifflow

1end1if11flow

flow

fff

1ifiii

shaftflowif

shaftflow

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WvpummumumQvpmmW

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pdVW

umEummumEWWEEQ

WWΔEQmmm in, flow that willmass theall

plus tank in the mass initial of consists system

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1f

oo

oi

of

v

p

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shaft

i

Tk T0T arbitrary, is T

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cc

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shaftW

11

( )( )( )( ) ( )

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F167T

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0381.6323.831.125TQ381.670460.243hmm

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of

of

f

1fi

ii

ffff

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=

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=−−==+××=×−

=+××+×==×+×+×=

−−−=

A 200 cubic ft tank contains2. lbm carbon dioxide and .4lbm helium at an initial temperature of 70 F. 3 lbmof air at 14.7 and 70 F are admitted to the tank.What is the final temperature of the tank?

( )if1

1

1

1

mmmTuh

−=

1p

im

12

There initially are 2 kg of liquid water and 8 kg of vapor in an insulated vertical piston cylinder device. A constant pressure of 300 kPa is maintained by the piston weight. Steam at .5 Mpa and 350 C are added until the contents of the cylinder are allvapor. What is the final temperature in the cylinder and themass of steam admitted?

.5 MPa350 C

300 kPa

.5 MPa350 C

300 kPa

work

steam2 kg

8 kg

4-53

13

( )

( )( )

( )

( ) ( )( ) ( )

( )

( ) ( )( )( )

( )( )

kg 9.76kg 10kg 19.76mm

kg 19.76kJ/kg 2724.9kJ/kg 3168.1.7kJ/kg 2292.23kJ/kg 3168.110kgm

hhhhmm

hhmhhm0hmhmhmhm0

hmmhmhm0vpmmumm)vpmu(m)vpmu(m0

vpmvpmvpmmummumum0

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vpmmW

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umEu mmumE

mcylinder, in thefinally mass theis system the

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∫

∫

kJ/kg 3168.1hC350 MPa, @.5

kJ/kg 2292.23h2163.5.8561.43h

hxhhkJ/kg 2724.9hh

kPa 300 @

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14

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ΔEEEconservedEnergy , definedEnergy

212net

1212net

in massout massoutinoutin

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12out massin massoutinoutin

12

12outinoutin

12out massin massoutinoutin

12out massin massoutinoutin

outin

−=−==−=−==

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=−

space in region a SYSTEM OPEN

mass of quantity contained a SYSTEM CLOSED

LawFirst

15

( ) ( ) ( )

( ) ( ) ( )( ) ( ) ( ) ( )( ) ( ) ( ) ( )

( ) ( )( ) ( ) ( ) ( )

( )

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12out massin massoutinoutin

inout

1122out12in12out bin boutin

in12out121122

outoo12inoo12out bin boutin

2outooinooout bin boutin

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umumh mmh mm WWQQUUEEWWQQ UUEEWWQQ

,Q ,with Wumumhmmh mm WWQQ

ummummumum)v(pmm)v(pmmWWQQ

UUVpVpWWQQUUWWWWQQ

0 UUEEWWQQ

−=−−−+−+−−=−+−+−

−=−+−+−

−=−++

−=−−−+−+−−−−+−=−−−+−+−

−=−+−+−−=−+−+−

−=−+−+−

space in regionSYSTEM UNSTEADY

umumh mmW-Q

mor m mass, of quantitySYSTEM UNSTEADY

112212

21