University:
The Pennsylvania State University
Course:
CHE 220 | Introduction to Chemical Engineering Thermodynamics
Academic year:

2020
Views:

355

Pages:

9
Author:

study_genius

Topics for Today - Reversible adiabatic proud ( compression ) expansion ) * Reading 3.8 Examples with ideal gases . shmmary EXAMPLEZ C From P 15 ) Lecture ↳ +125.3 2 133 | bar 1bar < , 400k \300K " Process Heat ( hsymd z z z l and work receives 2068.6 z Process ) -> - 0 - 9567 831.4 2900 - veleares ( Jlmol ) 0 956.7 -2068.6 125.3 heat Work ( 1254mA W 2068.6 0 total v * Q On ) alittle engine that can tnvnheatintowork ttthisproussoperatesasacyde .AT#ieend,thestateotthlworkingfluidisves+orektoitsinitia1statebu+asaVesvltoftheprucess,anetamovn+ofheat has converted operates to work . in reverse The opposite direction . would happen if been cycle Reversible Mechanical Need the Tv or compression First ;i÷ ;; Law iidT=,fd¥ ideal gas : the closed in expansion path of : Process Adiabatic . process dU=dQ+dw dU= adt C irg ) dQ= dW= 0 C adiabatic ( reversible - PDV aisle - . Rte PV=RT~>kPHnV=1nR+1hT 4,1+181=1,1 system 2 is cYd¥+aisl¥=-R¥ aist¥= . ( ) .BA#wmefTaYIaph path take substitute ) diﬀerential into )¥ aistr =p is on i ( PN ) ReversibleAtiabatitpalh l¥=§÷.d=V define pz P :y= : , gislais .vP=p,vy a- ( P ,v ) -w( Pz ,K ) intearatefom , Int 'p÷= ln(v÷Y~> - Reversibkatinbaticpathon ( D ,v ) graph . term : Rertttiabatic lfcompressim The pathofthetev . adiabatic onthepvgraphof idealgas looks like Process but is steeper . an isotherm Por . Adiabatic cY¥=t¥} ¥ From 2 ¥= d1v= ohEs¥ . : cjsdtg working having to with = R¥+RtIp - d¥ integrate tom t¥= ( T÷)$YR In ( Tip ) on i From * Path ideal calculate V gases , . ( IP ) , tsp ) to : More useful bk it is always relates initial a CRT & good final it ) of states idea to . avoid EXAMPLE I : Vitro gen 1bar from and the solution compressed by is , 20°C final \ , to K P, Vz ✓ V, Tz=T the Determine . . ( Ppt )M4 work =29J/my{ ) ( aegis 's , process 465.03k = , balance : ,z=¢twnW= on 1 bar temperature l÷,=(I÷§s%2 Energy 5 revtatiabatic civstz - Ti ) = 3555.4 mto , Example Solution ( 1 The Gmt 'd ) W Adiabatic this fact peat for 5bar procedure Tz=Ti * Ze = bar I is the our = air leads produce as before : 184.8k = ( Tz to from expansion . same ( By,M$ expansion to to reversible - T , ) = cooling refrigeration 2241.3 - . We . can : 01 use 20 ' C Example Nitrogen refrigerator temperature of 200k 1bar @ : @ 20°C , To avoid bar 1 cool . vacuum work from Work for 2 PFI bar Pz=PH 3 Pz=Pz 4 R=1 return to 1- T, , Tz= const Determination of PH P must : Pz=P4( They )9K Determination of Tz = : C heating to wehied to to zooc K adiabatic expansion ) . such be 74=200 cool , bar . 200k . PH >1 . ( adiabatic compression ) ( cmst P cooling ) 293.15k T4= atmosphere equipment heating , design low K ? Tz= , bar by 293.15 = to : some Crmpressim State 1 ; to a resources no must minimize To Use this . vacuum expand compression leads we , Because . 2006 , Available . mechanical ; 1bar capable of reaching Small a Solution available is that the expansion in 3 4 produces . 379.5k adiabatic compression from P, ,T , : )l%= } Tz=Ti( ¥ 429.7k EXAMPLEZ ( 'd ) -1=293.15 Tz= P ,= Pz=3 -79.5 1bar 0U|z= Qiz = adiabatic const .P )= 2824.35 0412=2824.35 0 0U34=G( -1926.5 Ty .5)= OUSY = -795 y - 1926.9 200k 1bar u Adiabatic const )= DU23=Cv(T3-T DU41=a( W4|= F- Tx opt = 044 3 -2824.3 -3959.5 E 1135.15 1926.9 )= 2701.3 -941=-7745 Ty )= , - , 2 co .P , Q23= 9>(-13-7) W23=0U23 -0,23 QH 34=0 W34= Pz=3 u , T4= Py= -13=293.15 429.7 he Cvtzt W/z= Q Cmt •: t • 4 1 >