Review System Property State Process Cycle Zeroth Law.

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Review • System • Property • State • Process • Cycle • Zeroth Law
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Transcript of Review System Property State Process Cycle Zeroth Law.

Review

• System

• Property

• State

• Process

• Cycle

• Zeroth Law

Example 2-4

Applied Thermodynamics

Thermodynamics: Energy, Work, Power and Heat

Energy

• Forms of Energy– Kinetic: KE = ½ mv2

– Potential: PE = mgh

• Specific energy:– ke = ½ v2

– pe = gh

• Internal energy, U

• Efficiency, η

Energy

• Total ∆Esystem has three macroscopic quantities:– ∆KE: motion of system– ∆PE: displacement of system in a gravitational

field– ∆U: Internal Energy is an extensive property.

Energy: Forms or Carriers

• Many forms: kinetic, potential, thermal, radiant, elastic, chemical etc….

• Energy is [ex]changed (dynamic) and stored (static)

• It may be better to say energy is carried.

Work: Definition

• Means of energy transfer across a boundary:– Expansion work

Wk = ΣF δx = F ∆x = F (x2 – x1) Wk = Σp A δx = = Σp δV = p ∆V

– Shaft WorkWk = Σ T δ θ = T ∆ θ = T (θ2 – θ1)

P

V

1

2

dV

P

V

1

2

dV

Work: Sign Convention

• W>0: work transfer out of system• W<0: work transfer into system

Caution! Work is path dependent

• dW = p∙dV ----- meaningless because:∫dW = W2 – W1 = ∫p ∙ dV

Implies we can assign values to W1 and W2 .

• Instead we writeδW = p ∙ dV and W = ∫δW = ∫p ∙ dVWhere δW is an inexact differential.

i.e. the left side cannot be integrated and evaluated at the limits.

• Work is not a property.

Example: Work

• CO2 is slowly heated from 50C to 500C in two steps as shown.– p1 = 100 kPa

– p3 = 150 kPa

– T2 = 350C

– m = 0.044kg

• Calculate total Work.

P

V

1

2 3P2=P3

P1

V3V2V1

Example: Work

• Assume quasi-equilibrium• Assume Ideal gas

kJ .J

kPaPam..kPa m..kPa

m .:similarly

m .

kPaPa

kmolkg

kPa

kJJK

KkmolkJ.kg .

m .

kPaPa

kmolkg

kPa

kJJK

KkmolkJ.kg .

33

3

3

3

2022200

1000034500428015002680034502150100

04280

03450

100044150

100027335031480440

02680

100044100

10002735031480440

2

3

2

2

22

1

1

11

23212213

1

W

W

V

V

MpTRm

V

V

MpTRm

V

VVpVVpp

dVpWV

V

P

V

1

2 3P2=P3

P1

V3V2V1

Power

• Energy flow or energy currentPower = dE/dt = IE

• Rate of doing WorkPower = dW/dt = F ∙ dx/dt = F ∙ v

W = ∫F ∙ v dt =∫p ∙ dV

• Involves a flow across a potentialPower = -Δp |dV/dt|

= -Δφ |dq/dt| = V∙Iq

Heat: Definition

• Thermal energy moving across a boundary (not the lay definition)

• Only induced by a temperature difference• Adiabatic process: no transfer of heat• Like work, heat depends on the process, • Heat is not a property• Q>0: heat transfer to system• Q<0: heat transfer from the system

Equivalence of Work and Heat

• Heat and work are both energy transitions• Work can affect a system as if heat had been

transferred. (the opposite is not always true)

Internal Energy

• In a Macroscopic analysis anything not KE or PE is Internal Energy, U.– Specific internal energy, u = U/m, is intensive.– Sensible U – related to temperature – Latent U – associated with phase change

• Microscopically Internal Energy is made of:– Translation, Rotation and Vibration of molecules– Chemical bonds within molecules– Plus: orbital states, nuclear spin and nuclear forces

Work in a Polytropic Process

• pVn = constant

• If n≠1, general polytropic process

• If n = 1, isothermal process

• If n = 0, isobaric process

1

211 V

VVpW ln

nVpVp

W

1

1122

VVpW 2

Example: Polytropic Process

• A gas in a piston–cylinder assembly undergoes a process for which:pVn = constant

Pi = 3 bar, Vi = 0.1 m3, Vf = 0.2 m3

• Determine the Work if a) n=1.5, b) n=1.0, c) n=0

p (b

ar)

V (m3)

1

2a2b

2c3.0

0.1 0.2

pVn=k

Example: Polytropic Process, n=1.5

kJ

mN kJ

bar N/m

.m .bar m .bar .

bar .m .m .bar ,.for

233

.

3

3

617101

110

51110320061

0612010351

11

1

3

5

51

2

112

11221111

1222

11

12

2211

2

1

2

1

.W

VV

ppn

nVpVp

nVVpVVp

W

nVV

kdVVkdVpW

VpVpkpV

n

nnnn

nnV

V n

V

V

nnn

p (b

ar)

V (m3)

1

2a

3.0

0.1 0.2

pVn=k

Example: Polytropic Process, n=1

kJ .mN

kJ bar

mN

m .m .lnm .bar

lnln

2

3

33

.

7920101

110

1020103

3

5

1

211

1

2

221101

2

1

2

1

W

VV

VpVV

kdVVkdVpW

VpVpkpV

V

V

V

V

p (b

ar)

V (m3)

1

2b

3.0

0.1 0.2

pVn=k

Example: Polytropic Process, n=0

kJ mN

kJ bar

mNm .m .bar

233 30

101

110

102033

5

12

210

W

VVpW

ppkpV

pVn=k

p (b

ar)

V (m3)

1

2c3.0

0.1 0.2

Reversibility

• Process are idealized as reversible– The process can be reversed with a return to

the original state.– No dissipative effects– No production of entropy

• Irreversible work– Friction work and viscous work always oppose

mechanical work– Transfer of heat through a finite ∆T