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Chapter#2

Zeroth law,

FirstLawofThermodynamics,A li ti i l tApplicationinclosesystem

Zeroth lawofthermodynamics

If each of two given systems 1 and 2 are in thermal equilibriumwith a third system 3, then the two systems 1 and 2 are inthermal equilibrium with each other.q

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Joulespaddlewheelexperiments

Workdoneonsystem(water)byloweringthemass Temperatureincreaseofwaterafterperformingwork

(Heattransfer?) Systemrestoredtotheinitialstatewithanotherwaterbath. Heattransfertowaterbath?

Joule performed many experiments. In some experiments, thepaddle wheel was replaced by an electrical resistor, suppliedwith an electric current by using a reversible cell.

Joule found, during a cycle change net work done on thesystem was always proportional to net energy removed fromsystem was always proportional to net energy removed fromthe system as heat irrespective of the type of workinteraction.

Mathematically,inSIunits, or

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Firstlawofthermodynamics

Whenever the system undergoes a cyclic change, howevercomplex the cycle may be, the algebraic sum of work transfers isequal to the algebraic sum of energy transfers as heat.

Consider a reversible cycle,

Applying first law to the cycle,

Subtractingequation(2)fromequation(1),

dQ anddW pathdependent,(dQdW)pathindependent

Define(dQdW)=Energychangeofsystem=dEWhere,E=K.E.+P.E.+U=propertyofsystem

dE =d(K.E.+P.E.+U)=dQ dWor,dU =dQ dW

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Example:A fluid system undergoes a non-flow frictionless processfollowing the pressure-volume relation as p = 5/V + 1.5 where pis in bar and V is in m3. During the process the volume changesfrom 0.15 m3 to 0.05 m3 and the system rejects 45 kJ of heat.D t i Ch i i t lDetermine : Change in internal energy.

Solution:Given, V1 = 0.15 m3, V2 = 0.05 m3 ,Heat rejected Q = 45 kJ

Now work done = PdV = [ (5/V + 1 5)dV ] 105 N-mNow, work done = PdV = [ (5/V + 1.5)dV ] 10 N-m= - 5.64 105 N-m = - 564 kJ

Applying the first law energy equation, U = Q W = 519 kJ

Perpetual motion machine of first kind (PMMI)

A cyclically operating device which would produce workcontinuously without absorbing energy from surrounding

(Impossible)(Impossible)

From first law for cyclic process,

If then W = 0

Think: Expansion of a high pressure gas in a piston cylinderassembly to a certain extent, without absorbing any heat(PMMI?)

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Firstlawanalysisforfewelementaryprocesses

Isochoricprocess

Isobaricprocessp

Isothermalprocess

Adiabaticprocess

Polytropic process

Constantinternalenergyprocess

ConstantVolumeProcess(Isochoric)

ConsideragasconfinedinrigidvesselofvolumeVasthesystem.

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Applying the first law, dU = dQ dW

Here, dW = 0

So dU = dQ or Q = U USo, dU = dQ, or Q = U2 U1

heat interaction of system =Change in the internal energy of the system

Specific heat at constant volume,Specific heat at constant volume,CV = (dq/dT)V = (u/T)V

IfCV isconstantthen,

Also,ifCV isvarieswithtemperaturethen,

Where,meanspecificheat=

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Example:An insulated rigid tank initially contains 0.7 kg ofhelium at 30C and 364 kPa. A paddle wheel with apower rating of 0.02 hp is operated within the tank for30 min Determine:30 min. Determine:(a) the final temperature(b) the final pressure of the helium gas(Given, CV = 3.1156 kJ/kg.C)

Solution: Consider He gas in tank as closed system

Given, m = 0.7 kg, T1 = 30C, P1 = 364 kPa

Here, no work done by system but, work is done ontosystem by paddle wheel action.

So, work done in process (amount of paddle wheelwork done on the system)work done on the system),W = - (power)(time) = - (0.02)(3060) = - 26.9 kJ

System is adiabatic, heat transfer Q = 0

Applying first law,U = Q W,U = W

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mCV(T2 T1) = WPut values,(0.7)(3.1156)( T2 300) = - (- 26.9)T2 = 39.3C AnsT2 39.3 C Ans

Also, for He (ideal gas), at constant volume,P1/ T1= P2/T2Put values, 364/300 = P2/312.3,P2 = 378.9 kPa Ans

ConstantPressureProcess(Isobaric)

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Consider heating of a gas at constant pressure andreversibly.

Work done by system from state 1 to state 2,2

Applying first law of thermodynamics,dU = dQ dW

= dQ PdV

2 11

( )W P dV P V V= =

Q= dQ d(PV)

Or, dQ = dU + d(PV)dQ = d(U + PV)

Put U + PV = H, So, dQ = dH, Q = H2 H1

Heat transfer at constant P = Total enthalpy change

Here, H = enthalpy of systemProperty of system, point functionMeasured in KJSpecific enthalpy = h = H/m = u + PvMolar enthalpy = = H/nMolar enthalpy H/n

Specific heat at constant pressure (CP):CP = (dq/dT)P = (h/T)P

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Total heat interaction

If Cp is constant, Q = mCPT = mCP(T2 T1)

2

1

T

PT

Q m C d T= Q mCPT mCP(T2 T1)

If Cp is not constant,

2 1( )pQ mC T T= 2TPC d T

Where, mean specific heat = 12 1

TPC T T

=

Example:1 kg of gaseous CO2 contained in a closed systemundergoes a reversible process at constant pressure.During this process 42 kJ of internal energy isdecreased Determine the work done during the processdecreased. Determine the work done during the process.(Take CP = 840 J/kgC and CV = 600 J/kgC)

Solution: Given, m = 1 kg, U = 42 kJ = 42 103 J

Now, U = mCV(T2 T1)Put values, 42 103 = 1600(T2 T1)So, T2 T1 = 70C

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Now, Heat transfer in isobaric process,Q = mCP(T2 T1)Put values, Q = 1840( 70)Q = 58800 J or 58.8 kJ

Applying first law, Q = U + WPut values 58.8 = 42 + WW = 16.8 kJWork done on the system during process = 16 8 kJWork done on the system during process 16.8 kJ

(Ans)

ConstantTemperatureProcess(Isothermal)

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System is allowed to undergo reversible expansion.

The process path can be drawn on PV diagram.

A l i fi t l f th Applying first law of thermo.,dU = dQ dW = dQ PdV

Assume gas as ideal one, U = U(T),So, dU = CVdT = 0 at constant temp.

So, We have, dQ = dW, Heat transfer = Work transferdQ = PdV

Integrate, Q = W =

Put P = nRT/V for ideal gas,

2

1

V

V

P d V

22ln

V nRT VQ W dV nRT = = =

PV = constant (process equation)

11

lnV

Q W dV nRTV V

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ReversibleAdiabaticProcess(Isentropic)

Gas is allowed to expand reversibly and adiabatically (dQ= 0), Isentropic process

Applying first law of thermo.,dU = 0 dW, W = UW = nCV(T)= nCV(T2 T1)= nR(T2 T1)/( 1)= (P1V1 P2V2)/( 1)

Again, dU = - dWnCVdT = - PdVnCVdT = - nRTdV/V (For ideal gas P = nRT/V)nRdT/( 1) = - nRTdV/V (For CV = R/( 1))

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Rearranging, dT/T = - ( 1) dV/VIntegrate,ln(T2/T1) = - ( 1) ln(V2/V1)(T2/T1) = (V1/V2)( 1)

T1V1( 1) = T2V2( 1)

TV( 1) = Constant

Using PV = nRT We can also haveUsing PV nRT, We can also have,PV = constant, TP(1-) = constant

PolytropicProcess

When a process undergoes a reversible process inwhich heat is transferred and

Frequently, a plot of lnP versus lnV yields a straightline. (Polytropic process)

For such a process, PVn = constantWhere n = index of expansionWhere, n = index of expansion

We know for any reversible process,Work done = W = PdV

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Putp=C/Vn

PutCasP1V1n andP2V2n,

2 1 11 2

1

( )1

V n n

nV

C dV V VW CV n

+ += = =1 1 2 2

1 11 1 1 2 2 2

1 1 2 2

( . . )( 1)

( )( 1)

n n n nPV V PV VWn

PV PVW

+ += =

N:Numberofmoles

1 2

( 1)( )( 1)

nR T TW Nn

=

Heat transfer during polytropic process:Q = U + W= NCV(T2 T1) + NR(T1 T2)/(n1)

Put value of CV,

= NR(T2 T1)/(1) + NR(T1 T2)/(n1)= NR(T1 T2)[ n+1 + 1]/(1) (n1)= NR[(T1 T2)( n)/(1) (n1)

Q = ( n)/(1) . W

Index n depends on work and heat interactions ofprocess.

Some processes are special cases of polytropic.

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n = 0; PV0 = constant; P = constant; isobaric process

1 PV1 I h l n = 1; PV1 = constant; Isothermal process

n = ; PV = constant; Adiabatic process

n = ; PV = constant; V = constant; Isochoric process

Example: Equation u = 3.64 pv + 90 gives the internal energyof a certain substance where u is in kJ/kg, p is in kPa and v isin m3/kg. A system composed of 3.5 kg of this substanceexpands from an initial pressure of 500 kPa and a volume of0.25 m3 to a final pressure 100 kPa in a process in whichpressure and volume are related by pv1 25 = constantpressure and volume are related by pv1.25 = constant.

(i) If the expansion is quasi-static, find Q, U and W for theprocess.(ii) In another process, the same system expands according tothe same pressure-volume relationship as in part (i) and fromthe same pressure volume relationship as in part (i), and fromthe same initial state to the same final state as in part (i), butthe heat transfer in this case is 32 kJ. Find the work transfer forthis process.(iii) Explain the difference in work transfer in parts (i) and (ii).

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Solution: Given, Internal energy equation : u = 3.64 pv + 90, V1

= 0.25 m3, p1 = 500 kPa, p2 = 100 kPa, pv1.25 = const.

u = u2 u1 = 3.64 (p2v2 p1v1) ... per kg2 1 (p2 2 p1 1) p g U = 3.64 (p2V2 p1V1) ...for 3.5 kg

Using, p1V11.25 = p2V21.25V2 = 0.906 m3and U = 125 2 kJ (Ans)and U 125.2 kJ (Ans)

For a quasi-static polytropic process,W = (p1V1 p2V2)/(n 1)= 137.6 KJ (Ans)

Applying first law,Q = U + W= 12.4 kJ (Ans)

In second case U would remain as 125 2 kJ and Q In second case, U would remain as 125.2 kJ and Q = 32 KJ (given)

Applying first law again,W = Q U= 32 ( 125 2) 32 ( 125.2)= 157.2 kJ (Ans.)

In second case, process is not quasi static and work is not pdV