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ME 200 L23: Clausius Inequality andControl Volume Example Problems

Kim See’s Office ME Gatewood Wing Room 2172 Please check your HW Grades on Blackboard

Please return all previously graded Homework(s) today with HW8

https://engineering.purdue.edu/ME200/ThermoMentor© Program Launched Quiz 2 performance suggests it is making a

difference, Let Examination 2 prove that too.

Spring 2014 MWF 1030-1120 AMJ. P. Gore

gore@purdue.eduGatewood Wing 3166, 765 494 0061

Office Hours: MWF 1130-1230TAs: Robert Kapaku rkapaku@purdue.edu

Dong Han han193@purdue.edu

Clausius Inequality►The Clausius inequality is developed from

the Kelvin-Planck as:

cycleb

T

Q∫ (Eq. 5.13)

cycle = 0 no irreversibilities present within the system

cycle > 0 irreversibilities present within the system

cycle < 0 impossible

Eq.5.14

Example: Use of Clausius Inequality

QH=1000 kJ, TH=500 K and QL=600 kJ at (a) 200

K, (b) 300 K, (c) 400 K. Find if each cycle is reversible, irreversible or ideal.

Solution: Use the given QH, QL values to find work and ensure that the work produced does not result in a negative value for cycle

cycleC

out

H

in T

Q

T

Q

b

T

Q∫

Example: Use of Clausius Inequality

(b) kJ/K 0K 300

kJ 600

K 500

kJ 1000cycle cycle = 0 kJ/K = 0

(a) kJ/K 1K 200

kJ 600

K 500

kJ 1000cycle cycle = +1 kJ/K > 0

Irreversibilities present within system

No irreversibilities present within system

(c) kJ/K 5.0K 400

kJ 600

K 500

kJ 1000cycle cycle = –0.5 kJ/K < 0

Impossible

Review for Examination 2

5

– Control Mass and Control Volume– Simple Compressible Substance: State Principle– Conservation of Mass– First Law of Thermodynamics or Conservation of Energy– Property Relations

• Subcooled or Saturated Solid, Subcooled or compressed liquid and Saturated Liquid, Saturated Liquid Vapor Mixture, Superheated Vapor, Ideal Gases

• p-V-T, p-v-T, and p-V-Z-T relations• Internal Energy, Enthalpy, Entropy

– SI and British System of Units. Make Sure lbm versus lbf is managed with the 32 ft/s2 factor properly

– Boundary Work versus Shaft Work– Heat Transfer and Entropy relationship– Reversible processes: Internally reversible, Externally

reversible

Review for Examination 2

6

– High Temperature Reservoir, Low Temperature Reservoir– Second Law of Thermodynamics– Efficiency and Coefficient of Performance– Carnot Engine and Carnot Heat Pump– Control Volumes

• Nozzles• Diffusers• Compressors• Pumps• Turbines• Heaters• Heat Exchangers

– Integrated Control Volumes such as a Pump feeding into a Boiler which feeds into a Turbine which feeds into a Condenser

Example Problem: NozzleExample Problem: Nozzle

7

Given: Air at 800 K expands to an exit temperature of 660 K. The inletvelocity is sufficiently low to not contribute significantly to the total energy.

Find: The exit velocity.

Assumptions: Change in PE neglected, No heat transfer, No work doneother than flow work, Steady state, Steady flow, Mass is conserved.

2 2

2

2 2

2

2000 2000 821 9 607 02 655

CVCV CV i i e e

H ,L B,S I E

CVi e

I E

ei e

e i e

dE V VQ W m ( h gZ ) m ( h gZ ) bb

dt

dmm m

dt

Vh h

V ( h h ) ( . . ) m / s

Example Problem: DiffuserExample Problem: Diffuser

8

Given: Steam at 100oC, 1 bar is pressurized through a diffuserto 1.5 bars, 120oC and negligible velocity. Find the inlet velocity.

Find: The inlet velocity.

Assumptions: Change in PE neglected, No heat transfer, No work doneother than flow work, Steady state, Steady flow, Mass is conserved.

2 2

2

2 2

2

2000 2711 4 2676 2 265 33

CVCV CV i i e e

H ,L B,S I E

CVi e

I E

ie i

i

dE V VQ W m ( h gZ ) m ( h gZ )

dt

dmm m

dt

Vh h

V ( . . ) . m / s

Example Problem: CompressorExample Problem: Compressor

9

Given: Air is compressed from 1 bar, 300 K to 10 bars, 800 K by a compressorusing 550 kJ/kg of electrical work input.

Find: Heat transferred to the cooling fluid.

Assumptions: Change in PE, KE neglected, Steady state, Steady flow, Mass is conserved.

2 2

2 2

550 821 95 300 19 28 24

CVCV CV i i e e

H ,L B,S I E

CVi e

I E

C C e i C C e i

C C e i

dE V VQ W m ( h gZ ) m ( h gZ )

dt

dmm m

dt

Q W m( h h ) Q / m W / m ( h h )

q w ( h h ) ( . . ) . kJ

Example Problem: Heater/CombustorExample Problem: Heater/Combustor

10

Given: Air at 800 K and 10 bars is heated to 1400 K by heat addition from acombustor.

Find: Find the heat added by the combustor.

Assumptions: Change in PE, KE neglected, Steady state, Steady flow, Mass is conserved, No work done.

2 2

2 2

1515 42 821 95 693 47

CVCV CV i i e e

H ,L B,S I E

CVi e

I E

Comb. e i Comb. e i

Comb. e i

dE V VQ W m ( h gZ ) m ( h gZ )

dt

dmm m

dt

Q m( h h ) Q / m ( h h )

q ( h h ) . . . kJ / kg

Example Problem: TurbineExample Problem: Turbine

11

Given: Air at 1400 K and 10 bars is expanded to 900 K by a turbine.

Find: Find the work output per unit mass of air if the process is adiabatic.

Assumptions: Change in PE, KE neglected, Steady state, Steady flow, Mass is conserved, Heat Transfer is negligible.

2 2

2 2

1515 42 932 93 582 49

CVCV CV i i e e

H ,L B,S I E

CVi e

I E

T i e T i e

T i e

dE V VQ W m ( h gZ ) m ( h gZ )

dt

dmm m

dt

W m( h h ) W / m ( h h )

w ( h h ) . . . kJ / kg

Consider the compressor, combustor, and turbine on slides 11, 12, and 13as a system: Net work = 582 49 521 76 60 73

693 47

60 73 693 47 8 76

T c

Comb.

w w . . . kJ / kg

q . kJ / kg

. / . . %

Example Problem: Heat ExchangerExample Problem: Heat Exchanger

12

Given: 0.6 kg/s of air at 2000 K flows through a counter-flow HX and exits at 1000 K. On the other side of the HX, 1 kg/s of air is heated from 800 K to 1400 K.

Find: The heat loss to the surroundings.

Assumptions: Change in PE, KE neglected, Steady state, Steady flow, Mass is conserved, No work done.

2 2

2000 1000 1400 800

2 2

0 6 2252 1 1046 04 1 1515 42 821 95

30 166 693 47 4 35

CVCV CV i i e e

H ,L B,S I E

CVi e

I E

Loss hot cold

dE V VQ W m ( h gZ ) m ( h gZ )

dt

dmm m

dt

Q m ( h h ) m ( h h )

. ( . . ) ( . . )

. kW out of . kW or . %