熱工基礎(chǔ)英chapter 5第五章

Example1:MassofAirinaRoom

第五章

Determinethemassoftheairinaroomdimensionsare4m×5m×6mat100kPaand25℃Solution:V=4×5×6=120m3

pV 100103120

m 140.3kg

RT 287(27325)

Example2(初終狀態(tài)的參數(shù)的計算)

0.112m3ofgashasapressureof138Pa.Itiscompressedto690PaaccordingtothelawpV14=C.Determinethenewvolumeofthegas.

Solution:

SincethegasiscompressedaccordingtothelawpV14=C,then,

11 22

pV14pV14

p V14 V p1/14

12 or 21

p2 V1 V1 p2

fromwhich,

1/14

138

1/14

VVp1 0.112 0.035m3

2 1

p2

690

Example2-2(初終狀態(tài)的參數(shù)的計算)

Agaswhoseoriginalpressureandtemperaturewere300kPaand25℃,respectively,iscompressedaccordingtothelawpV14=Cuntilitstemperaturebecomes180℃.Determinethenewpressureofthegas.(1.299MPa)

Example2-3(初終狀態(tài)的參數(shù)的計算)

Agaswhoseoriginalvolumeandtemperaturewere0.015m3and285℃,respectively,isecxpandedaccordingtothelawpV135=Cuntilitsvolumeis0.09m3.Determinethenewtemperatureofthegas.(25.4℃)

Example2-4(初終狀態(tài)的參數(shù)的計算)

0.675kgofgasat1.4MPaand280℃isexpandedtofourtimestheoriginalvolumeaccordingtothelawpV13=C.Determine:

theoriginalandfinalvolumeofthegas,(0.0765m3,0.306m3)

thefinalpressureofthegas,(231kPa)

thefinaltemperatureofthegas.(92℃)

Example3

0.014m3gasatapressureof2070kPaexpandstoapressureof207kPaaccordingtothelawpV135=C,Determinetheworkdonebythegasduringtheexpansion.

Solution:

p 2070 1

VV(1)1/n0.014( )1/1350.014101350.077m3

p

2

2 1 207

pVpV 20701030.0142071030.077

W11 22 37.3kJn1 1.351

WtnW1.3537.350.4kJ

Example4

Agasiscompressedisothermallyfromapressureandvolumeof100kPaand0.056m3,respectively,toavolumeof0.007m3.Determinethefinalpressureandtheworkdoneonthegas.Solution:

t

WWpVlnV21001030.056ln0.00711.64kJ

V1 0.056

Example5(理想氣體的熱量計算)

2kgofgas,occupying0.7m3,hadanoriginaltemperatureof15℃.Itwasthenheatedatconstantvolumeuntilitstemperaturebecame135℃.Howmuchheatwastransferredtothegasandwhatwasitsfinalpressure?Take,cv=0.72kJ/kg-KandR=0.29kJ/kg-K.

Solution:

Qmcv(T2T1)20.72(13515)20.72120172.8kJ

Nowp1v1=mRT1,andT127315288K

1

pmRT120.29288238.6kPa

V1 0.7

sincethevolumeremainsconstant,then

p1p2pTp1(273135)238.6338.02kPaTT 22T 288

1 2 1

Example5-1(理想氣體的熱量計算)

Agaswhosepressure,volumeandtemperatureare275kPa,0.009m3and185℃,respectively,hasitsstatechangedatconstantpressureuntilitstemperaturebecomes15℃.Howmuchheatistransferredfromthegasandhowmuchworkisdoneonthegasduringtheprocess?TakeR=0.29kJ/kg-K,cp=1.005kJ/kg-K(-31.78kJ)

Example5-2(理想氣體熱力過程的計算)

0.25kgofairatapressureof140kPaoccupies0.15m3andfromthisconditionitiscompressedto

1.4MPaaccordingtothelawpV125=C.Determine:

thechangeofinternalenergyoftheair,(30.73kJ)

theworkdoneonorbytheair,(-49.1kJ)

theheatreceivedorrejectedbytheair.(-18.37kJ)Example6(綜合)

Aquantityofgashasaninitialpressure,volumeandtemperatureof0.14MPa,0.14m3and25℃,respectively.Itiscompressedtoapressureof1.4MPaaccordingtothelawpV1.25=constant.

Determine:

thefinalvolume,temperatureofthegasandthechangeofentropy;

theheattransferredbythegasandthevolumeworkofthepolytropicprocess.Takecp=1.041kJ/kgK,cv=0.743kJ/kgK.

Solution:

R=cp-cv=1.041-0.743=0.298kJ/kgK.p1V1=mRT1andT1=25+273=298K

∴m=p1V1/(RT1)=0.221kg

For1kgofgas,

ssclnV2clnp2

V p

2 p v

1 1

Also,pVnpVn

11 22

1/125

VVp11/n0.140.14

0.0222m3

p

2 1

2

0.0222

1.4

1.4

ss1.041ln 0.743ln 0.207kJ/kgK

1 2 0.14 0.14

Butthereis0.221kgofgas

Changeofentropy=-0.2210.207=-0.0457kJ/K

n1

V

TTV1

1

pV2

2986.31025472.3K

W 1

1p2V245.92

n1

Qmcv(T2T1)W17.29kJ

Example6-1

0.3kgofairatapressureof350kN/m2andatemperatureof35℃receivesheatenergyatconstantvolumeuntilitspressurebecomes700kN/m2.Itthenreceivesheatenergyatconstantpressureuntilitsvolumebecomes0.2289m3.Determinethechangeofentropyduringeachprocess.(1.491kJ/K;0.332kJ/K)

Takecp=1.006kJ/kgK,cv=0.717kJ/kgK.Example6-2

A quantityofgasaninitialpressure,volumeandtemperatureof0.13MPa,0.224m3and21℃,

respectively.Itiscompressedtoavolumeof0.028m3accordingtothelawpV13=constant.Determinethechangeofentropyandstatewhetheritisanincreaseofdecrease.

TakeR=0.287kJ/kgK,cv=0.717kJ/kgK.(-0.0518kJ/K)

Example6-3

1kgofairhasavolumeof56litresandatemperatureof190,Theairthenreceivesheatatconstantpressureuntilitstemperaturebecomes500.Fromthisstatetheair