MAE110AHomework7:Solutions

11/21/2017

1

H7.1:Aturbochargerbooststheinletairpressuretoanautomobileengine(therebyincreasingenginepoweroutput).Itconsistsofanexhaustgasdriventurbinedirectlyconnectedtoanaircompressor.Assumethatboththeturbineandthecompressorareinternallyreversibleandadiabatichavingalsothesamemassflowrate.Given Model𝑇" = 30°CAiristheworkingfluidandidealgas𝑇( = 650°CIgnorepotentialandkineticenergyeffects𝑝" = 100kPaSteadystate,adiabaticoperationandinternallyreversible𝑝( = 170kPaTurbineandCompressoraredirectlyconnected𝑝2 = 100kPa𝑚 = 0.1kg/sDeterminea) 𝑇2,turbineexittemperatureand𝑊9:;,poweroutputoftheturbineb) 𝑃>,compressorexitpressureand𝑇>,compressorexittemperaturec) Repeat(a)and(b)assumingisentropicefficienciesof80%and75%forthecompressor

andturbine,respectivelyBasicEquations𝑑𝑚@A

𝑑𝑡 = 𝑚CC

− 𝑚EE

𝑑𝑆@A𝑑𝑡 =

𝑄H𝑇H+ 𝑚C𝑠C − 𝑚E𝑠E + 𝜎LM

𝑑𝐸@A𝑑𝑡 = 𝑄CO +𝑊CO + 𝑚C ℎC +

𝑉C>

2 + 𝑔𝑧CC

− 𝑄9:; −𝑊9:; − 𝑚E ℎE +𝑉E>

2 + 𝑔𝑧E9

𝑠> − 𝑠" = 𝑠9 𝑇> − 𝑠9 𝑇" − 𝑅 ln𝑝>𝑝"

𝜂L =𝑊/𝑚

Y

𝑊/𝑚

𝜂; =𝑊/𝑚𝑊/𝑚

Y

2

Analysis𝑑𝑚@A

𝑑𝑡 = 𝑚CC

− 𝑚EE

= 0 ⇒ 𝑚( = 𝑚2 = 𝑚also𝑚" = 𝑚> = 𝑚

𝑑𝑆@A𝑑𝑡 =

𝑄H𝑇H+ 𝑚C𝑠C − 𝑚E𝑠E + 𝜎LM = 0 ⇒ 𝑚"𝑠" − 𝑚>𝑠> = 0 ⇒ 𝑚 𝑠" − 𝑠> = 0

⇒ 𝑠1 = 𝑠2also𝑠3 = 𝑠4a) Turbine:

𝑠2 − 𝑠( = 𝑠9 𝑇2 − 𝑠9 𝑇( − 𝑅 ln𝑝2𝑝(= 0 ⇒ 𝑠9 𝑇2 = 𝑠9 𝑇( + 𝑅 ln

𝑝2𝑝(

Determine𝑠9 𝑇( using𝑇( = 650°ConTableA-22𝑠9 𝑇( = 2.87688kJ/kg∙K

𝑠9 𝑇2 = 2.87688kJ/kg∙K +8.314J/mol∙K29.0g/mol ln

100kPa170kPa = 2.72475kJ/kg∙K

Determine𝑇2using𝑠9 𝑇2 = 2.72475kJ/kg∙KonTableA-22𝑻𝟒𝒔 = 𝟖𝟎𝟓𝐊𝑑𝐸@A𝑑𝑡 = 𝑄CO +𝑊CO + 𝑚C ℎC +

𝑉C>

2 + 𝑔𝑧CC

− 𝑄9:; −𝑊9:; − 𝑚E ℎE +𝑉E>

2 + 𝑔𝑧E9

= 0

⇒ 𝑚(ℎ( −𝑊9:;jk − 𝑚2ℎ2 = 0 ⇒ 𝑊9:;jk = 𝑚 ℎ( − ℎ2 Determineℎ(andℎ2using𝑇( = 650°Cand𝑇2 = 805K,respectively,onTableA-22ℎ( = 958.76kJ/kg,ℎ2 = 827.46kJ/kg𝑊9:;jk = 0.1kg/s 958.76kJ/kg − 827.46kJ/kg = 𝟏𝟑. 𝟏𝟑kJ/sb) Compressor:𝑊COop = 𝑊9:;jk = 13.13kJ/s𝑑𝐸@A𝑑𝑡 = 𝑄CO +𝑊CO + 𝑚C ℎC +

𝑉C>

2 + 𝑔𝑧CC

− 𝑄9:; −𝑊9:; − 𝑚E ℎE +𝑉E>

2 + 𝑔𝑧E9

= 0

⇒ 𝑊COop + 𝑚"ℎ" − 𝑚>ℎ> = 0 ⇒ ℎ> =𝑊COop𝑚 + ℎ"

Determineℎ"using𝑇" = 30°ConTableA-22ℎ" = 303.21kJ/kg

ℎ> =13.13kJ/s0.1kg/s + 303.21kJ/kg = 434.51kJ/kg

Determine𝑇>usingℎ> = 434.51kJ/kgonTableA-22𝑻𝟐𝒔 = 𝟒𝟑𝟑𝐊

𝑠> − 𝑠" = 𝑠9 𝑇> − 𝑠9 𝑇" − 𝑅 ln𝑝>𝑝"= 0 ⇒ 𝑝> = 𝑝"𝑒

Ys tp uYs tov

Determine𝑠9 𝑇" and𝑠9 𝑇> using𝑇"and𝑇>,respectively,usingTableA-22𝑠9 𝑇" = 1.71200kJ/kg∙K,𝑠9 𝑇> = 2.07240kJ/kg∙K

𝑝> = 100kPa 𝑒

>.wx>2wkJ/kg∙Ku".x">wwkJ/kg∙Ky.("2J/mol∙K>z.wg/mol = 𝟑𝟓𝟏. 𝟓𝟐kPa

3

c)NewActualValuesfortheTurbine:

𝜂; =𝑊/𝑚𝑊/𝑚

Y

⇒ 𝑊9:;jk = 𝜂;𝑊9:;jk Y= 0.8 13.13kJ/s = 𝟏𝟎. 𝟓𝟎kJ/s

𝑊9:;jk = 𝑚 ℎ( − ℎ2 ⇒ ℎ2 = ℎ( −𝑊9:;jk𝑚 = 958.76kJ/kg −

10.50kJ/s0.1kg/s = 853.76kJ/kg

Determine𝑇2usingℎ2 = 853.76kJ/kgonTableA-22𝑻𝟒 = 𝟖𝟐𝟖. 𝟖𝟓KNewActualandIsentropicValuesfortheCompressor:𝑊COop = 𝑊9:;jk = 10.50kJ/s

ℎ> =10.50kJ/s0.1kg/s + 303.21kJ/kg = 408.21kJ/kg

Determine𝑇>usingℎ> = 408.21kJ/kgonTableA-22𝑻𝟐 = 𝟒𝟎𝟕. 𝟏𝟑𝐊

𝜂L =𝑊/𝑚

Y

𝑊/𝑚⇒ 𝑊COop Y

= 𝜂L𝑊COop = 0.75 10.50kJ/s = 7.875kJ/s

ℎ>Y =7.875kJ/s0.1kg/s + 303.21kJ/kg = 381.96kJ/kg

Determine𝑇>Yusingℎ>Y = 381.96kJ/kgonTableA-22𝑇>Y = 381.18K

𝑠> − 𝑠" = 𝑠9 𝑇>Y − 𝑠9 𝑇" − 𝑅 ln𝑝>𝑝"= 0 ⇒ 𝑝> = 𝑝"𝑒

Ys tp uYs tov

Determine𝑠9 𝑇" and𝑠9 𝑇>Y using𝑇"and𝑇>Y,respectively,onTableA-22𝑠9 𝑇" = 1.71200kJ/kg∙K,𝑠9 𝑇>Y = 1.9431kJ/kg∙K

𝑝> = 100kPa 𝑒

".z2("kJ/kg∙Ku".x">wwkJ/kg∙Ky.("2J/mol∙K>z.wg/mol = 𝟐𝟐𝟑. 𝟗𝟏kPa

4

H7.2:Airentersasteady-statecompressorat100kPaand17°Catarateof4kg/min.Theexitpressureis600kPa.Theprocessisinternallyreversibleandisothermal.Changesinkineticandpotentialenergyarenegligible.Given Model𝑇" = 17°CAiristheworkingfluidandanidealgas𝑝" = 100kPaSteadystate,internallyreversible,andsingle-inlet,single-outlet𝑝> = 600kPaIgnorepotentialandkineticenergyeffects𝑚 = 4kg/minIsothermaloperation,𝑇" = 𝑇> = 𝑇Determine𝑄,rateofheattransfer𝑊CO,powerinputinkWBasicEquations𝑑𝑚@A

𝑑𝑡 = 𝑚CC

− 𝑚EE

𝑄OE;CO𝑚

intrev

= 𝑇𝑑𝑠>

"

𝑊OE;9:;𝑚

CO;�EM

= − 𝑣𝑑𝑝>

"+𝑉"> − 𝑉>>

2 + 𝑔 𝑧" − 𝑧>

𝑠> − 𝑠" = 𝑠9 𝑇> − 𝑠9 𝑇" − 𝑅 ln𝑝>𝑝"

𝑑𝐸@A𝑑𝑡 = 𝑄CO +𝑊CO + 𝑚C ℎC +

𝑉C>

2 + 𝑔𝑧CC

− 𝑄9:; −𝑊9:; − 𝑚E ℎE +𝑉E>

2 + 𝑔𝑧E9

Analysis 𝑑𝑚@A

𝑑𝑡 = 𝑚CC

− 𝑚EE

= 0 ⇒ 𝑚" = 𝑚> = 𝑚

𝑑𝐸@A𝑑𝑡 = 𝑄CO +𝑊CO + 𝑚C ℎC +

𝑉C>

2 + 𝑔𝑧CC

− 𝑄9:; −𝑊9:; − 𝑚E ℎE +𝑉E>

2 + 𝑔𝑧E9

= 0

⇒ 𝑊CO + 𝑚"ℎ" − 𝑄9:; − 𝑚>ℎ> ⇒ 𝑊CO = 𝑄9:; + 𝑚 ℎ> − ℎ" Idealgas ∴ ℎ = ℎ 𝑇 ,isothermal ∴ 𝑇" = 𝑇> ⇒ ℎ" 𝑇" = ℎ> 𝑇> ⇒ 𝑊CO = 𝑄9:; 𝑄OE;CO𝑚

intrev

= 𝑇𝑑𝑠>

"⇒ −

𝑄9:;𝑚 = 𝑇 𝑠> − 𝑠" ⇒ 𝑄9:; = 𝑚𝑇 𝑠" − 𝑠>

5

𝑠> − 𝑠" = 𝑠9 𝑇> − 𝑠9 𝑇" − 𝑅 ln𝑝>𝑝"⇒ 𝑠" − 𝑠> = 𝑅 ln

𝑝>𝑝"

⇒ 𝑄9:; = 𝑚𝑇 𝑅 ln𝑝>𝑝"

𝑄9:; = 4kg/minmin60s 17°C + 273K

8.314J/mol∙K29.0g/mol 1000g/kg ln

600kPa100kPa

= 9931.13kJ/s 𝑄9:; = 𝟗. 𝟗𝟑𝟏kW 𝑊CO = 𝑄9:; = 𝟗. 𝟗𝟑𝟏kW Alternative: Solve for power 𝑊OE;

9:;𝑚

CO;�EM

= − 𝑣𝑑𝑝>

"+𝑉"> − 𝑉>>

2 + 𝑔 𝑧" − 𝑧> ⇒ −𝑊CO

𝑚 = −𝑅𝑇𝑝 𝑑𝑝

>

"⇒ 𝑊CO = 𝑚𝑅𝑇 ln

𝑝>𝑝"

6

H7.3:Considerasteampowerplantthatoperatesatsteadystatewiththetwopressurelimitsof10MPaand20kPa.Waterentersthepumpassaturatedliquid(pumpinletpressureis20kPaandtheexitpressure10MPa)andleavestheturbineassaturatedvapor(turbineinletpressureis10MPaandexitpressureis20kPa).Bothpumpandturbineareconsideredadiabaticwithnegligiblekineticandpotentialenergychanges.Assumealsothatliquidwaterbehavesasanincompressiblesubstance.Given Model𝑝" = 20kPaWateristheworkingfluidandincompressiblewhenliquid𝑝> = 10MPaIgnorepotentialandkineticenergyeffects𝑝( = 10MPaSteadystate,adiabaticoperation𝑝2 = 20kPaDetermine𝑊;

𝑊�,theratiooftheworkproducedbytheturbinetotheworkconsumedbythepumpassuming:

a) Internallyreversibleprocessesb) Isentropicefficienciesoftheturbineandpumpare80%and70%,respectivelyBasicEquations𝑑𝐸@A𝑑𝑡 = 𝑄CO +𝑊CO + 𝑚C ℎC +

𝑉C>

2 + 𝑔𝑧CC

− 𝑄9:; −𝑊9:; − 𝑚E ℎE +𝑉E>

2 + 𝑔𝑧E9

𝑑𝑚@A

𝑑𝑡 = 𝑚CC

− 𝑚EE

𝑊OE;9:;𝑚

CO;�EM

= − 𝑣𝑑𝑝>

"+𝑉"> − 𝑉>>

2 + 𝑔 𝑧" − 𝑧>

𝜂� =𝑊/𝑚

Y

𝑊/𝑚

𝜂; =𝑊/𝑚𝑊/𝑚

Y

7

Analysisa)Pumpandturbineareinternallyreversible∴𝑠" = 𝑠>, 𝑠( = 𝑠2𝑑𝑚@A

𝑑𝑡 = 𝑚CC

− 𝑚EE

= 0 ⇒ 𝑚" = 𝑚> = 𝑚( = 𝑚2 = 𝑚

PumpWork:Assumeliquidwaterisincompressiblesubstance∴ 𝑣 = constant𝑊OE;

9:;𝑚

CO;�EM

= − 𝑣𝑑𝑝>

"+𝑉"> − 𝑉>>

2 + 𝑔 𝑧" − 𝑧> ⇒ −𝑊CO

𝑚 = − 𝑣𝑑𝑝>

"⇒𝑊COop𝑚 = 𝑣 𝑝> − 𝑝"

Determine𝑣using𝑝" = 20kPaonTableA-3𝑣 = 𝑣�" = 1.0172×10u(m3/kg𝑊COop𝑚 = 1.0172×10u(m3/kg 10MPa − 20kPa = 10.15kJ/kg

Alternative:Use1stLaw𝑑𝐸@A𝑑𝑡 = 𝑄CO +𝑊CO + 𝑚C ℎC +

𝑉C>

2 + 𝑔𝑧CC

− 𝑄9:; −𝑊9:; − 𝑚E ℎE +𝑉E>

2 + 𝑔𝑧E9

⇒ 𝑊COop + 𝑚"ℎ" − 𝑚>ℎ> = 0 ⇒𝑊COop𝑚 = ℎ> − ℎ",whereℎ> = ℎ>Y

TurbineWork:𝑑𝐸@A𝑑𝑡 = 𝑄CO +𝑊CO + 𝑚C ℎC +

𝑉C>

2 + 𝑔𝑧CC

− 𝑄9:; −𝑊9:; − 𝑚E ℎE +𝑉E>

2 + 𝑔𝑧E9

⇒ 𝑚(ℎ( −𝑊9:;jk − 𝑚2ℎ2 = 0 ⇒𝑊9:;jk𝑚 = ℎ( − ℎ2

Determineℎ2and𝑠2using𝑝2 = 20kPaonTableA-3ℎ2 = ℎ�2 = 2609.7kJ/kg,𝑠2 = 𝑠�2 = 7.9085kJ/kg∙KAt𝑝( = 10MPaonTableA-3find𝑠�𝑠� = 5.6141kJ/kg∙KKnowingthat𝑠( = 𝑠2 = 7.9085kJ/kg∙K ⇒ 𝑠( > 𝑠�at𝑝( = 10MPa ∴ state 3 issuperheatedvaporDetermineℎ(using𝑝( = 10MPaand𝑠( = 7.9085kJ/kg∙Konprovidedtables:http://newmaeweb.ucsd.edu/courses/MAE110A/FA_2017/table_water_superheat.pdfℎ( = 4711.1kJ/kg𝑊9:;jk𝑚 = 4711.1kJ/kg − 2609.7kJ/kg = 2101.4kJ/kg

𝑊;

𝑊�=𝑊9:;jk/𝑚𝑊COop/𝑚

=2101.4kJ/kg10.15kJ/kg = 𝟐𝟎𝟕. 𝟎𝟑

8

b)

𝜂� =𝑊/𝑚

Y

𝑊/𝑚⇒𝑊COop𝑚 =

𝑊COop/𝑚 Y𝜂�

=10.15kJ/kg

0.8 = 12.69kJ/kg

𝜂; =𝑊/𝑚𝑊/𝑚

Y

⇒𝑊9:;jk𝑚 = 𝜂;

𝑊9:;jk𝑚

Y= 0.7 2101.4kJ/kg = 1470.98kJ/kg

𝑊;

𝑊�=𝑊9:;jk/𝑚𝑊COop/𝑚

=1470.98kJ/kg12.69kJ/kg = 𝟏𝟏𝟓. 𝟗𝟐

9

6.59:Inaheat-treatingprocess,a1-kgmetalpart,initiallyat1075K,isquenchedinatankcontaining100kgofwater,initiallyat295K.Thereisnegligibleheattransferbetweenthecontentsofthetankandtheirsurroundings.Takethespecificheatofthemetalpartandwatertobeconstantat0.5kJ/kgKand4.2kJ/kgK,respectively.Given Model𝑇",�E;�� = 1075KWaterandmetalpartaretheclosedsystem𝑇",��;E� = 295KNegligiblepotentialandkineticenergyeffects𝑚�E;�� = 1kgAdiabaticprocessandnoworkisdone𝑚��;E� = 100kgWaterandmetalpartareincompressiblesubstanceswith𝑐�E;�� = 0.5kJ/kg∙Kconstantspecificheat𝑐��;E� = 4.2kJ/kg∙KDeterminea) 𝑇� ,thefinalequilibriumtemperatureinKb) 𝜎,theamountentropyproducedwithinthetank,inkJ/KBasicEquationsΔ𝐾𝐸 + Δ𝑃𝐸 + Δ𝑈 = 𝑄CO +𝑊CO − 𝑄9:; −𝑊9:;

∆𝑆 =𝛿𝑄𝑇

>

"+ 𝜎

𝑇𝑑𝑠 = 𝑑𝑢 + 𝑝𝑑𝑣Analysisa)Δ𝐾𝐸 + Δ𝑃𝐸 + Δ𝑈 = 𝑄CO +𝑊CO − 𝑄9:; −𝑊9:; ⇒ Δ𝑈 = Δ𝑈�E;�� + Δ𝑈��;E� = 0⇒ 𝑚�E;��𝑐�E;�� 𝑇� − 𝑇",�E;�� + 𝑚��;E�𝑐��;E� 𝑇� − 𝑇",��;E� = 0

⇒ 𝑇� =𝑚�E;��𝑐�E;��𝑇",�E;�� + 𝑚��;E�𝑐��;E�𝑇",��;E�

𝑚�E;��𝑐�E;�� + 𝑚��;E�𝑐��;E�

𝑇� =1kg 0.5kJ/kg∙K 1075K + 100kg 4.2kJ/kg∙K 295K

1kg 0.5kJ/kg∙K + 100kg 4.2kJ/kg∙K = 𝟐𝟗𝟓. 𝟗K

b)

∆𝑆 =𝛿𝑄𝑇

>

"+ 𝜎 ⇒ 𝜎 = Δ𝑆�E;�� + Δ𝑆��;E� = 𝑚�E;��Δ𝑠�E;�� + 𝑚��;E�Δ𝑠��;E�

𝑇𝑑𝑠 = 𝑑𝑢 + 𝑝𝑑𝑣 ⇒ 𝑑𝑠 =𝑐𝑇 𝑑𝑇 ⇒ 𝑠> − 𝑠" = Δ𝑠 = 𝑐 ln

𝑇>𝑇"

⇒ 𝜎 = 𝑚�E;��𝑐�E;�� ln𝑇�

𝑇",�E;��+ 𝑚��;E�𝑐��;E� ln

𝑇�𝑇",��;E�

𝜎 = 1kg 0.5kJ/kg∙K ln295.9K1075K + 100kg 4.2kJ/kg∙K ln

295.9K295K

𝜎 = 𝟎. 𝟔𝟑𝟒𝟑𝟕𝐤𝐉/𝐊

10

6.95: The figure provides steady-state test data for a control volume in which two entering steams of air mix to form a single exiting stream. Stray heat transfer and kinetic energy and potential energy effects are negligible. A hard-read photocopy of the data sheet indicates that the pressure of the exiting stream is either 1.0 MPa or 1.8 MPa. Assuming the ideal gas model for air with cp = 1.02 kJ/kg K. Given Model𝑇" = 800KThecontrolvolumeisgivenandisatsteadystate𝑇> = 650KNegligiblepotentialandkineticenergyeffects𝑝" = 1.8MPaAdiabaticoperationandnoworkisdone𝑝> = 1.0MPaAirisanidealgaswithconstantspecificheat𝑚" = 1.5kg/s𝑚> = 2.0kg/s𝑐� = 1.02kJ/kg∙K DetermineIfeitherorbothoftheoutputpressurevalues,𝑝( = 1.0MPaand𝑝( = 1.8MPa,arepossibleBasicEquations𝑑𝐸@A𝑑𝑡 = 𝑄CO +𝑊CO + 𝑚C ℎC +

𝑉C>

2 + 𝑔𝑧CC

− 𝑄9:; −𝑊9:; − 𝑚E ℎE +𝑉E>

2 + 𝑔𝑧E9

𝑑𝑆@A𝑑𝑡 =

𝑄H𝑇H+ 𝑚C𝑠C − 𝑚E𝑠E + 𝜎LM

𝑑𝑚@A

𝑑𝑡 = 𝑚CC

− 𝑚EE

𝑇𝑑𝑠 = 𝑑ℎ − 𝑣𝑑𝑝𝑑ℎ = 𝑐�𝑑𝑇𝑝𝑣 = 𝑅𝑇Analysis𝑑𝑚@A

𝑑𝑡 = 𝑚CC

− 𝑚EE

= 0 ⇒ 𝑚" +𝑚> = 𝑚(

𝑑𝐸@A𝑑𝑡 = 𝑄CO +𝑊CO + 𝑚C ℎC +

𝑉C>

2 + 𝑔𝑧CC

− 𝑄9:; −𝑊9:; − 𝑚E ℎE +𝑉E>

2 + 𝑔𝑧E9

= 0

⇒ 𝑚"ℎ" + 𝑚>ℎ> − 𝑚(ℎ( = 0 ⇒ 𝑚" ℎ" − ℎ( + 𝑚> ℎ> − ℎ( = 0

⇒ 𝑚"𝑐� 𝑇" − 𝑇( + 𝑚>𝑐� 𝑇> − 𝑇( = 0 ⇒ 𝑇( =𝑚"𝑐�𝑇" + 𝑚>𝑐�𝑇>𝑚"𝑐� + 𝑚>𝑐�

=𝑚"𝑇" + 𝑚>𝑇>𝑚" + 𝑚>

𝑇( =1.0kg/s 800K + 2.0kg/s 650K

1.0kg/s + 2.0kg/s = 700K

𝑑𝑆@A𝑑𝑡 =

𝑄H𝑇H+ 𝑚C𝑠C − 𝑚E𝑠E + 𝜎LM = 0 ⇒ 𝑚1𝑠1 + 𝑚2𝑠2 − 𝑚3𝑠3 + 𝜎LM = 0

⇒ 𝑚" 𝑠" − 𝑠( + 𝑚> 𝑠> − 𝑠( + 𝜎𝑐𝑣 = 0⇒ 𝜎𝑐𝑣 = 𝑚" 𝑠( − 𝑠" + 𝑚> 𝑠( − 𝑠>

11

𝑇𝑑𝑠 = 𝑑ℎ − 𝑣𝑑𝑝 ⇒ 𝑑𝑠 =𝑐�𝑇𝑑𝑇 −

𝑅𝑇𝑝𝑑𝑝 ⇒ 𝑠> − 𝑠" = 𝑐� ln

𝑇>𝑇"− 𝑅 ln

𝑝>𝑝"

⇒ 𝜎𝑐𝑣 = 𝑚" 𝑐� ln𝑇(𝑇"− 𝑅 ln

𝑝(𝑝"

+ 𝑚> 𝑐� ln𝑇(𝑇>− 𝑅 ln

𝑝(𝑝>

Check𝑝( = 1.0MPa:

𝜎LM = 1.0kg/s 1.02kJ/kg∙K ln700K800K

−8.314J/mol∙K29.0g/mol

1000g/kg ln1.0MPa1.8MPa

+ 2.0kg/s 1.02kJ/kg∙K ln700K650K

−8.314J/mol∙K29.0g/mol

1000g/kg ln1.0MPa1.0MPa

𝝈𝒄𝒗 = 𝟎. 𝟏𝟖𝟐kJ/K∙sEntropyproductionisgreaterthanzerothereforethisprocessispossiblegiventheadiabaticcondition.Check𝑝( = 1.8MPa:

𝜎LM = 1.0kg/s 1.02kJ/kg∙K ln700K800K

−8.314J/mol∙K29.0g/mol

1000g/kg ln1.8MPa1.8MPa

+ 2.0kg/s 1.02kJ/kg∙K ln700K650K

−8.314J/mol∙K29.0g/mol

1000g/kg ln1.8MPa1.0MPa

𝝈𝒄𝒗 = −𝟎. 𝟑𝟐𝟔kJ/K∙sEntropyproductionislessthanzerothereforethisprocessisnotpossiblegiventheadiabaticcondition.

12

6.103: Refrigerant134aiscompressedfrom2bar,saturatedvapor,to10bar,90°C,inacompressoroperatingatsteadystate.Themassflowrateofrefrigerantenteringthecompressoris7kg/min,andthepowerinputis10.85kW.Kineticandpotentialenergyeffectscanbeneglected.Given Model𝑝" = 2barCompressorandR134aarethecontrolvolume𝑝> = 10barNegligiblepotentialandkineticenergyeffects𝑇> = 90°CSteadystate𝑊CO = 10.85kW𝑚 = 7kg/minDeterminea) 𝑄,therateofheattransferinkWb) Iftheheattransferoccursat𝑇§ = 50°Cfind𝜎LM ,therateofentropyproductioninkW/Kc) Ifthecontrolvolumeisenlargedtoincludetheimmediatesurroundingssuchthatthe

heattransferoccursat𝑇§ = 300Kfind𝜎LM ,therateofentropyproductioninkW/K.Discusstheresults.

BasicEquations𝑑𝐸@A𝑑𝑡 = 𝑄CO +𝑊CO + 𝑚C ℎC +

𝑉C>

2 + 𝑔𝑧CC

− 𝑄9:; −𝑊9:; − 𝑚E ℎE +𝑉E>

2 + 𝑔𝑧E9

𝑑𝑆@A𝑑𝑡 =

𝑄H𝑇H+ 𝑚C𝑠C − 𝑚E𝑠E + 𝜎LM

𝑑𝑚@A

𝑑𝑡 = 𝑚CC

− 𝑚EE

Analysisa)𝑑𝑚@A

𝑑𝑡 = 𝑚CC

− 𝑚EE

= 0 ⇒ 𝑚" = 𝑚> = 𝑚

𝑑𝐸@A𝑑𝑡 = 𝑄CO +𝑊CO + 𝑚C ℎC +

𝑉C>

2 + 𝑔𝑧CC

− 𝑄9:; −𝑊9:; − 𝑚E ℎE +𝑉E>

2 + 𝑔𝑧E9

= 0

⇒ 𝑊CO + 𝑚"ℎ" − 𝑄9:; − 𝑚>ℎ> = 0 ⇒ 𝑄9:; = 𝑊CO + 𝑚 ℎ" − ℎ>

13

Determineℎ"using𝑝" = 2baronTableA-11ℎ" = ℎ� = 241.30kJ/kgDetermineℎ>using𝑝> = 10barand𝑇> = 90°ConTableA-12ℎ> = 324.01kJ/kg

𝑄9:; = 10.85kW + 7kg/minmin60s 241.30kJ/kg − 324.01kJ/kg = 𝟏. 𝟐kW

b)𝑑𝑆@A𝑑𝑡 =

𝑄H𝑇H+ 𝑚C𝑠C − 𝑚E𝑠E + 𝜎LM = 0 ⇒

𝑄OE;¨©𝑇§

+ 𝑚"𝑠" − 𝑚>𝑠> + 𝜎LM = 0

⇒ 𝜎LM =𝑄9:;𝑇§

+ 𝑚 𝑠> − 𝑠"

Determine𝑠"using𝑝" = 2barfromTableA-11𝑠" = 0.9253kJ/kg∙KDetermine𝑠>using𝑝> = 10barand𝑇> = 90°CfromTableA-12𝑠> = 1.0707kJ/kg∙K

𝜎LM =1.2kW

50°C + 273K + 7kg/minmin60s 1.0707kJ/kg∙K − 0.9253kJ/kg∙K

𝝈𝒄𝒗 = 𝟎. 𝟎𝟐𝟎𝟔𝟖kW/Kc)

𝜎LM =𝑄9:;𝑇§

+ 𝑚 𝑠> − 𝑠"

𝜎LM =1.2kW300K + 7kg/min

min60s 1.0707kJ/kg∙K − 0.9253kJ/kg∙K

𝝈𝒄𝒗 = 𝟎. 𝟎𝟐𝟎𝟗𝟔kW/KTheentropyproductionincreaseswhenthecontrolvolumesizeincreasestoincludetheimmediatesurroundingsbecausethereareirreversibilitiesassociatedwiththeheattransfertothesurroundings(heattransferacrossfinitetemperaturedifferenceisasourceofirreversibility)thatarenotaccountedforwhenthecontrolvolumeisjustthecompressor.Theseextrairreversibilitiesyieldanincreaseinentropyproduction.