[PDF] The First Law of Thermodynamics: Closed Systems Heat Transfer





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The First Law of Thermodynamics: Closed Systems Heat Transfer

Consider two systems: a) the entire oven (including the heater) and b) only the air in the oven (without the heater) see Fig 3-3. Solution: The energy content 



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Thermodynamic Systems and Processes. 18. Define isolated system closed system



First Law of Thermodynamics: Closed Systems

Problem 3-73 A 0.3-m3 tank contains oxygen initially at 100kPa and 27°C. A Neglect the energy stored in the paddle wheel. Solution: Step 1: Draw a ...





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problem and some choices of system boundries are more difficult than others. One ... (b). Free flight



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Closed thermodynamic systems: application of first law of thermodynamics Some time may then be spent on the solution of closed system problems that involve ...



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Thermodynamics. Solution. Here hig=h₂-h=85.3-264 = 58.9 BTU/lb. The answer is (d). P. 11-6. A nonflow (closed) system contains 1 pound of an ideal gas (C 



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Thermodynamic Properties and calculation

SOLUTIONS: ○ In suche case take the system as 1 mol of air contained in an mechanically reversible processes in a closed system: ○ (a) Heating at ...



The First Law of Thermodynamics: Closed Systems Heat Transfer

Solution: The energy content of the oven is increased during this process. a) The energy transfer to the oven is not caused by a temperature difference between 



First Law of Thermodynamics: Closed Systems

Determine the heat transfer for this process. Solution: Step 1: Draw a schematic diagram to represent the problem. Step 2: What to determine?



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First Law of Thermodynamics: Closed Systems

First Law of Thermodynamics: Closed Systems Problem 3-73 A 0 3-m3 tank contains oxygen initially at 100kPa and 27°C A paddle wheel within the tank is rotated until the pressure inside rise to 150kPa During the process 2KJ of heat is lost to the surroundings Determine the paddle-wheel work done



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Searches related to thermodynamics closed system problems and solutions PDF

Solution: Step 1: Draw a schematic diagram to represent the problem Step 2: What to determine? The heat transfer between the system and the surroundings Q Step 3: The information given in the problem statement 1 Argon in the cylinder: m=5kg P=400kPa and T=30V; 2 A boundary work done by the system W b=15KJ; 3

What is a closed system theory?

Closed systems were defined as a limit case: as systems for which the environment has no significance or is significant only through specified channels. The theory concerned itself with open systems. This question was posted on Quora.com: “What are some examples of open and closed systems?”

What is a closed system?

In nonrelativistic classical mechanics, a closed system is a physical system that doesn't exchange any matter with its surroundings, and isn't subject to any net force whose source is external to the system. A closed system in classical mechanics would be equivalent to an isolated system in thermodynamics .

What is closed system in physics?

A closed system is a physical system that does not allow transfer of matter in or out of the system, though, in different contexts, such as physics, chemistry or engineering, the transfer of energy is or is not allowed.

M.BahramiENSC388(F09)1

st

LawofThermodynamics:ClosedSystems1

duringaninteraction acrossthesystemboundary. motionsofatomsandmolecules.HeatTransfer atemperaturedifference. systemsthatareatthesame temperature. systemboundary. perunittime. action.

Notation:

-Q(kJ)amountofheattransfer -Q°(kW)rateofheattransfer(power) -q(kJ/kg)Ͳheattransferperunitmass -q°(kW/kg)Ͳpowerperunitmass

Signconvention

M.BahramiENSC388(F09)1

st

LawofThermodynamics:ClosedSystems2

ModesofHeatTransfer

Conduction

andtheenergytransportbyfreeelectrons.

Convection

fluidmotion(advection). fanor

Radiation

molecules. Work

Notation:

-W(kJ)amountofworktransfer -W°(kW)power

Q=5kJQ=Ͳ5kJ

HeatinHeatout

System

M.BahramiENSC388(F09)1

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LawofThermodynamics:ClosedSystems3

-w(kJ/kg)Ͳworkperunitmass -w°(kW/kg)Ͳpowerperunitmass

Signconvention

negative.

Fig.2:Signconventionforheatandwork.

Similaritiesbetweenworkandheattransfer:

phenomena).

Systems

state. processaswellastheendstates.

Pathfunctions

otherhand,arepointfunctions )W-nor WW not function,Path (Wfunction)(Point 12122
1122
1

WVVVdV

ElectricalWork

System

Q

W(+)(Ͳ)

M.BahramiENSC388(F09)1

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LawofThermodynamics:ClosedSystems4

)(Was form rate in the explained becan Which )( e kWVIkJVNW e

Example1:Electricalwork

Solution:

a) Theenergytransfertotheovenisnotcausedbyatemperaturedifference b)

Fig.3:Schematicforexample1.

Mechanicalwork

adistance. )(.kJsFWElectric oven air

Heater

HeaterElectric

oven air

Systemboundary

Systemboundary

M.BahramiENSC388(F09)1

st

LawofThermodynamics:ClosedSystems5

2 1 )(.kJdsFW theremustbeaforceactingontheboundary theboundarymustmove workinteraction,W=0.

MovingBoundaryWork

boundarywork. dsin

PdVPAdsdsFW

b differentialwork. willhavea dependonthecyclepath.

M.BahramiENSC388(F09)1

st

LawofThermodynamics:ClosedSystems6

Fig.5:networkdoneduringacycle.

PolytropicProcess

oftenrelatedbyPV n processcanbefound: nVPVPdVCVPdVW n polytopic 1 11222
12 1 Since CVPVP nn 2211
.Foranidealgas(PV=mRT)itbecomes: VP W net 12 V 1 V 2

M.BahramiENSC388(F09)1

st

LawofThermodynamics:ClosedSystems7

)(1,1 12 kJnnTTmRW polytropic 1 V 1 =P 2 V 2 =mRT 0 =C,whichcanbe foundfrom: )(1,ln 12 112
12 1 kJnVVVPdVVCPdVW isothermalb

Sinceforanidealgas,PV=mRT

0 atconstanttemperatureT 0 ,orP=C/V.

Example2:Polytropicwork

3bar,theinitialvolumeis0.1m

3 ,andthefinalvolumeis0.2m 3 .Determinetheworkfor

Solution:

iii)theexpansionispolytropic. a)n=1.5 2 1 1122
1 V V nVPVPPdVW

WeneedP

2 thatcanbefoundfrom nn VPVP 2211
barbarVVPP n

06.12.01.03

5.1 2 1 12 kJmNkJ barmNmbarW6.17.101 1/10

5.111.032.006.1

3253
kJmNkJ barmNmbarWVVVPPdVW

79.201.02.0ln.101

1/101.03ln

325
312
112
1 integralbecomeW=P(V 2 ͲV 1

M.BahramiENSC388(F09)1

st

LawofThermodynamics:ClosedSystems8

Springwork

xkF s wherek s )(21 2 12 2 kJxxkW sspring

NonͲmechanicalformsofwork

dxFW.

Example3:Mechanicalwork

Fig.6:SchematicPͲVdiagramforExample3.

Solution:

Process1Ͳ2isanexpansion(V

2 >V 1 )andthesystemisdoingwork(W 12 >0),thus: W 12 =P 1 (V 2 ͲV 1 )+[0.5(P 1 +P 2 )-P 1 ](V 2 -V 1 =(V 2 -V 1 )(P 1 +P 2 )/2 3 =V 2 ),soW 23
=0

Process3Ͳ1isacompression(V

3 >V 1 ),workisdoneonthesystem,(W 31
<0) 12 3 V V 2 V 1 P 1 P 2 P

M.BahramiENSC388(F09)1

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LawofThermodynamics:ClosedSystems9

W 31
=ͲP 1 (V 2 -V 1 Wquotesdbs_dbs17.pdfusesText_23
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