ENGR 2213 ThermodynamicsENGR 2213 Thermodynamics

F. C. Lai

School of Aerospace and Mechanical

Engineering

University of Oklahoma

Exergy (Availability)Exergy (Availability)

- Work potential, maximum useful work

The term “availability” was made popular in the Statesby MIT School of Engineering in the 1940s.

An equivalent work “exergy” was introduced in Europein the 1950s.

Dead StateA system is said to be in the dead state when it is inthermodynamic equilibrium with its surroundings.

Exergy Exergy

At the dead state,1. A system is at the same temperature and pressure of its surroundings (in thermal and mechanical equilibrium).

2. There are no unbalanced magnetic, electrical and surface tension effects between the system and its surroundings (mechanical equilibrium).

3. It has zero kinetic and potential energy relative to its surroundings (zero velocity and zero elevation above a reference level).

Exergy Exergy

At the dead state,

4. It does not react with the surroundings (chemically inert).

Work = f (initial state, process path, final state)

To maximize the work output

► the process to be executed is reversible.► the final state is a dead state.

Exergy Exergy

A system will deliver the maximum possible work as it undergoes a reversible process from the specified initial state to the state of its surroundings,i.e., the dead state.

Exergy does not represent the amount of work thata work-producing device will actually deliver uponinstallation. Rather, it represents the upper limit onthe amount of work a device can deliver withoutviolating any thermodynamic laws.

Exergy Exergy

The difference between exergy and the actual workdelivered by a device represents the room Engineers have for improvement.

Exergy of a system at a specified state depends onthe conditions of the surroundings (the dead state)as well as the properties of the system.

Exergy is a property of the combined system (thesystem and its surroundings) and not of the systemalone.

Vapor Power CyclesVapor Power Cycles

Working FluidsHigh temperature: sodium, potassium, mercury.Low temperature: benzene, freon.Water: low cost, availability, high enthalpy of vaporization.

Fuel Types

Coal, natural gas, nuclear, geothermal.

Vapor Power CyclesVapor Power Cycles

Carnot Power Cycle

T

S

1 2

34

1. Process 2-3The quality of steam atthe turbine exit may betoo low.

2. Process 4-1The pump has to dealwith two-phase flows.

Vapor Power CyclesVapor Power Cycles

Carnot Power Cycle

T

S

1 2

34

1. Process 1-2Isothermal heat transferat variable pressure.

Vapor Power CyclesVapor Power Cycles

Rankine Cycle

Rankine cycle is the idealcycle for vapor power plants.

T

S

1

2

3

4State 1: saturated liquidState 2: compressed liquidState 3: superheated vaporState 4: saturated mixture

Ideal Rankine CyclesIdeal Rankine Cycles

T

S

1

2

3

4

Process 1-2: isentropic compression in a pumpProcess 2-3: constant-pressure heat addition in a boilerProcess 3-4: isentropic expansion in a turbineProcess 4-1: constant-pressure heat rejection in a condenser

Turbine

Boiler

Condenser

Pump

1

2

3

4

Ideal Rankine CyclesIdeal Rankine Cycles

net out

in in

w q1

q q 4 1

3 2

h h1

h h

T

S

1

2

3

4

wp = h2 – h1 = v(p2 – p1)

qin = h3 – h2

wt = h3 – h4

qout = h4 – h1

p 2 1

t 3 4

w h hback work ratio

w h h

Example 1Example 1

Consider a steam power plant operating on the simple ideal Rankine cycle. The steam entersthe turbine at 3 MPa and 350 ºC and is condensedin the condenser at a pressure of 75 kPa. Determine(a) the thermal efficiency of this cycle.(b) the back work ratio of this cycle.

Example 1 (continued)Example 1 (continued)

State 1: saturated liquid at p1 = 75 kPa

State 2: compressed liquid at p2 = 3 MPa

Table A-5 h1 = hf = 384.39 kJ/kg v1 = vf = 0.001037 m3/kg

wp = v(p2 – p1) = (0.001037)(3000-75) = 3.03 kJ/kgh2 = h1 + wp = 384.39 + 3.03 = 387.42 kJ/kg

Example 1 (continued)Example 1 (continued)

State 3: superheated vapor at p3 = 3 MPa and T3 = 350 ºC

State 4: saturated mixture at p4 = 75 kPa

Table A-6 h3 = 3115.3 kJ/kg s3 = 6.7428 kJ/kg·K

s4 = s3 = sf + x4sfg

4 f4

fg

s s 6.7428 1.213x 0.886

s 6.2434

Example 1 (continued)Example 1 (continued)

State 4: saturated mixture at p4 = 75 kPah4 = hf + x4hfg

= 384.39 + 0.886(2278.6) = 2403.2 kJ/kgqin = h3 – h2 = 3115.3 – 387.42 = 2727.88 kJ/kgqout = h4 – h1 = 2403.2 – 384.39 = 2018.81 kJ/kg

out

in

q 2018.811 1 0.26

q 2727.88

Example 1 (continued)Example 1 (continued)

p

t

w 3.03bwr 0.004

w 712.1

net

in

w 709.070.26

q 2727.88

wt = h3 – h4 = 3115.3 – 2403.2 = 712.1 kJ/kg

wnet = wt – wp = 712.1 – 3.03 = 709.07 kJ/kg

LCarnot

H

T (91.78 273)1 1 0.414

T (350 273)

Real Rankine CyclesReal Rankine Cycles

T

S

1

2

3

4

Two most common sourcesof irreversibilities

► Fluid friction► Undesired heat loss to the surroundings

Real Rankine CyclesReal Rankine Cycles

T

S

1

2

3

4

2’

3’

4’

Process 1-2’Irreversibilities in the pump

Process 2’-3’Pressure drop due to friction in the boiler

Process 3’-4’Irreversibilities in the turbine

Process 4’-1’Pressure drop due to friction in the condenser

Real Rankine CyclesReal Rankine Cycles

s 2 1p

a 2' 1

w h h

w h h

Efficiency of Pump

Efficiency of Turbine

a 3 4't

s 3 4

w h h

w h h

h2’ = (h2 – h1)/ηp + h1

h4’ = h3 – ηp(h3 – h4)

T

S

1

2

3

4

2’

4’

Example 2Example 2

Consider a steam power plant operating on the simple ideal Rankine cycle. The steam entersthe turbine at 3 MPa and 350 ºC and is condensedin the condenser at a pressure of 75 kPa. Giventhat ηp = ηt = 0.85, determine the thermal efficiency of this cycle.

Example 2 (continued)Example 2 (continued)

State 1: saturated liquid at p1 = 75 kPa

State 2: compressed liquid at p2 = 3 MPa

Table A-5 h1 = hf = 384.39 kJ/kg v1 = vf = 0.001037 m3/kg

ws = v(p2 – p1) = (0.001037)(3000-75) = 3.03 kJ/kgh2’ = h1 + wp/ηp = 384.39 + 3.03/0.85 = 387.95 kJ/kg

Example 2 (continued)Example 2 (continued)

State 3: superheated vapor at p3 = 3 MPa and T3 = 350 ºC

State 4: saturated mixture at p4 = 75 kPa

Table A-6 h3 = 3115.3 kJ/kg s3 = 6.7428 kJ/kg·K

ws = 712.1 kJ/kg

h4’ = h3 – ηtws = 3115.3 – 0.85(712.1) = 2510 kJ/kg

Example 2 (continued)Example 2 (continued)

qin = h3 – h2’ = 3115.3 – 387.95 = 2727.35 kJ/kg

qout = h4’ – h1 = 2510 – 384.39 = 2125.61 kJ/kg

out

in

q 2125.611 1 0.22

q 2727.35

Increase the Efficiency of a Rankine CycleIncrease the Efficiency of a Rankine Cycle

1. Lowering the condenser pressure T

S

2. Superheating the steam to a higher temperature

3. Increasing the boiler pressure

Example 3Example 3

Consider a steam power plant operating on the simple ideal Rankine cycle. The steam entersthe turbine at 3 MPa and 350 ºC and is condensedin the condenser at a pressure of 75 kPa. Determine the thermal efficiency of this cycle,(a) if the condenser pressure is lowered to 10 kPa(b) in addition to the change in (a) if the steam is superheated to 600 ºC(c) in addition to the change in (b) if the boiler pressure is raised to 15 MPa

Example 3 (continued)Example 3 (continued)

(a) State 1: saturated liquid at p1 = 10 kPa

State 2: compressed liquid at p2 = 3 MPa

Table A-5 h1 = hf = 191.83 kJ/kg v1 = vf = 0.001008 m3/kg

wp = v(p2 – p1) = (0.001008)(3000-10) = 3.01 kJ/kgh2 = h1 + wp = 191.83 + 3.01 = 194.84 kJ/kg

Example 3 (continued)Example 3 (continued)

State 3: superheated vapor at p3 = 3 MPa and T3 = 350 ºC

State 4: saturated mixture at p4 = 10 kPa

Table A-6 h3 = 3115.3 kJ/kg s3 = 6.7428 kJ/kg·K

4 f4

fg

s s 6.7428 0.6493x 0.812

s 7.5009

s4 = s3 = sf + x4sfg

Example 3 (continued)Example 3 (continued)

(a) State 4: saturated mixture at p4 = 10 kPah4 = hf + x4hfg

= 191.83 + 0.812(2392.8) = 2134.8 kJ/kgqin = h3 – h2 = 3115.3 – 194.84 = 2920.46 kJ/kgqout = h4 – h1 = 2134.8 – 191.83 = 1942.97 kJ/kg

out

in

q 1942.971 1 0.335

q 2920.46

Example 3 (continued)Example 3 (continued)

(b) State 1 and State 2 remain the same

State 3: superheated vapor at p3 = 3 MPa and T3 = 600 ºC

Table A-6 h3 = 3682.3 kJ/kg s3 = 7.5085 kJ/kg·K

State 4: saturated mixture at p4 = 10 kPa

4 f4

fg

s s 7.5085 0.6493x 0.914

s 7.5009

s4 = s3 = sf + x4sfg

Example 3 (continued)Example 3 (continued)

(b) State 4: saturated mixture at p4 = 10 kPah4 = hf + x4hfg

= 191.83 + 0.914(2392.8) = 2378.8 kJ/kgqin = h3 – h2 = 3682.3 – 194.84 = 3487.46 kJ/kgqout = h4 – h1 = 2378.8 – 191.83 = 2186.97 kJ/kg

out

in

q 2186.971 1 0.373

q 3487.46

Example 3 (continued)Example 3 (continued)

(c) State 1 remains the same

State 2: compressed liquid at p2 = 15 MPa

wp = v(p2 – p1) = (0.001008)(15000-10) = 15.11 kJ/kgh2 = h1 + wp = 191.83 + 15.11 = 206.94 kJ/kg

State 3: superheated vapor at p3 = 15 MPa and T3 = 600 ºC

Example 3 (continued)Example 3 (continued)

State 3: superheated vapor at p3 = 15 MPa and T3 = 600 ºC

Table A-6 h3 = 3582.3 kJ/kg s3 = 6.6776 kJ/kg·K

State 4: saturated mixture at p4 = 10 kPa

4 f4

fg

s s 6.6776 0.6493x 0.804

s 7.5009

s4 = s3 = sf + x4sfg

Example 3 (continued)Example 3 (continued)

(c) State 4: saturated mixture at p4 = 10 kPah4 = hf + x4hfg

= 191.83 + 0.804(2392.8) = 2115.7 kJ/kgqin = h3 – h2 = 3582.3 – 206.94 = 3375.36 kJ/kgqout = h4 – h1 = 2115.7 – 191.83 = 1923.87 kJ/kg

out

in

q 1923.871 1 0.43

q 3375.36