Lecture 17

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ME 200 L17: Conservation Laws Prime Movers

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HW 7 Posted Due in One Week:

This week, do some HomeWork 7 everyday!

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Nozzles

Common Steady-flow Energy Devices

Diffusers Water, Steam, Gas Turbines

Heat Exchangers and Mixers Compressors

Pump

Throttles

Rotating Machinery

A turbine is a steady-flow device used to produce mechanical work (W) by reducing the internal &/or kinetic &/or potential energy of the working fluid.

• For gas turbines, the fluid drives rotating blades while the υ increases from inlet to exit as the working fluid expands (or the p drops).

Turbines

Turbines

Fluid expanding as it passes the blades imparts a force on the blades, which in turn produces a torque on the shaft causing rotation.

Rotating Machinery

A compressor is a steady-flow device that consumes shaft work from rotating blades and in turn compresses the fluid.

υ decreases from inlet to exit as the working fluid is compressed and the pressure rises .

Rotating Machinery

A pump is a steady-flow device that consumes shaft work from rotating blades that compress the fluid.

• Compressors are used for gas systems, pumps for liquids so operating assumptions are similar.

Liquid specific volume does not change significantly with pressure

Real Hardware

GE LM5000 gas turbine: 6.2 m long, 12.5 tons, 3600 rpm, produces 55.2 MW

Real Hardware: F-15 Engine

Real Hardware: PW-4000

D ~ 8 ft

L ~ 12 ft

62,000 lbf

– Boeing 747-400

Airbus A300-610

Real Hardware: GE-90

122,000 lbf

– Boeing 777, etc.

Rotating Machinery

 cv

  cv

  e

 e

 h e

V e

2

2

 gz e

  i

 i

 h i

V i

2

2

 gz i

Given in problem statement.

Any rotating shafts, electricity, or moving boundaries? Certainly! W ≠ 0

• The Δke is often small compared to the Δh, so Δke

≈ 0.

No appreciable elevation change, so Δpe ≈ 0.

Rotating Machinery

For only one inlet and one outlet.

 cv

 cv

  cv

 e h e

 m

 h

1

 h

2

 i h i

• Analysis works for aircraft gas turbines, stationary utility power gas turbines, steam turbines etc.

Gas (air) Turbine Example

 i

 m

1

 dm cv dt

 cv

  i

 i

  e

 e

  cv

  e

V

 i

V

1

Air

T

1

=410 K

P

1

=2.3 bars

A

1

=200 cm 2

 cv

 e

 h e e

 m

2

V e

2

2

2

 

1

W

 cv

T

2

=290 K

P

2

=1.0 bars

A

2

=280 cm 2

V

2 v

2

A

2

V

2 v

2

A

1

V

1 v

1

 gz e

  i

• Relatively low pressure ratio of 2.8 and relatively low temperature T

1 may just indicate that this is the last stage of expansion.

 i

 h i

V i

2

2

 gz i

Gas (air) Turbine Example

 i

 m

1

V

 i

V

1

T

1

=410 K

P

1

A

1

=2.3 bars

=200 cm 2

Air

 cv

28.33

kW

 cv

 e

T

2

=290 K

P

2

=1.0 bars

A

2

=280 cm 2

2

1

2

Table A-22

State

V

2 v

2

T, K

290

410

A

2

V

2 v

2

A

1

V

1 v

1 u, kJ/kg h, kJ/kg

206.91

290.16

293.43

411.12

V

2 v

2

V

2

T

2

P

2

R

300

60

100

( 290 )(.

287 ) m

3

( min

/ min

) sec

KPa

( K )( KJ / kg

K )

6 .

007 kg / s

Q

Turbine

W

Turbine

1

V

1

2

2

 h

2

V

2

2

2

1700

W

Turbine

60

W

Turbine

153.48

2

2000

    

673.23

kW

Q

Turbine

( in

 h exit

)

 

 m (

V

2 in

2000

)

 m (

V

2 out

2000

)

290.16

178.6

2

2000

Steam Turbine Example

6 Kg/s of steam at an inlet velocity of 75 m/s enter a turbine stage at 3 MPa and

400 o C and exit at a velocity of 125 m/s at a pressure of 2 Mpa at 360 o C. Find: (a) the power developed by this steam turbine stage, (b) the percentage change in the steam density across the turbine and (c) the percentage change in the flow area. The heat loss through the casing is 33 kW.

1

1

6 kg

/

 cv

33 s kW

6 kg / s

 cv

Q

Turbine

W

Turbine

1

V

2

1

2 h

2

V

2

2

2

 m e

 m

2

 m

V

2

A V

2 2

AV

1 1 v

2

 v

2 v

1

33

W

Turbine

6 3230.9

 

2

2000

3159.3

2000

2

 

366.6

kW

Turbine

( )

2 2

/ v

2

AV / v

1 1 1

125 A

2

/ 0.141

75 A

1

/ 0.094

A

2

/

  

1

) /

 

( v

1

 v

2

) / v

2

A

1

(0.094 0.141) / 0

(75 / 125) * (0.141 / 0.094)

0.9

Air Compressors

A pump is a steady-flow device that consumes shaft work from rotating blades that compress the fluid.

• Compressors are used for gas systems, pumps for liquids so operating assumptions are similar.

September 17th, 2010 ME 200 17

Example

At steady state, a well-insulated compressor takes in air at 60 ºF, 14.2 psi, with a volumetric flow rate of

1200 ft 3 /min, and compresses it to 500 ºF, 120 psi.

Kinetic and potential energy changes from inlet to exit can be neglected. Determine the compressor power, in hp, and the volumetric flow rate at the exit, in ft 3 /min.

Example

Find

– W cv

– A

2

V

2

= ? in hp

= ? in ft 3 /min

Assumptions

• The control volume is at steady state; the flow is steady

Q, Δke, and Δpe are negligible.

• T he air is an ideal gas.

System ( air flowing through compressor )

1 compressor

2

Basic Equations dm cv dt

  i

 i

  e

 e P

 

RT air

P

1

T

1

= 14.2 psi

= 60 ºF

A

1

V

1

= 1200 ft 3 /min

P

2

T

2

= 120 psi

= 500 ºF dE cv dt

  cv

  cv

  i

 i

 h i

V i

2

2

 gz i

  e

 e

 h e

V e

2

2

 gz e

Example

Solution dm cv dt

  i

 i

  e

 e 1

 m

2

 dE cv dt

 cv

  cv

  cv

  i

 i

 h i

V i

2

2

 m

 h

1

 h

2

 gz i

  e

 e

 h e

V e

2

2

 gz e

 m

AV

 

RT

P m

A

1

V

1

P

1

RT

1

Example

m

14.2

psi

1200 ft

3

 in

1540

 f

520

R

 1 h 1 ft

2

28.9

lb m



R

2

 cv

 m

 h

1

 h

2

From Table A-22E h

1

124 Btu lb m h

2

231 Btu lb m

W cv

 lb

5310 124 231 h m

   Btu lb m

1 hp

2540 Btu h

5310 lb m h

W cv

 

223 hp

Example

A

2

V

2

 

2

A

2

V

2

 m RT

2

P

2

A V

2

5310 lb m h

1540

28.9

lbf

 ft lb m



R

120 psi

960

R

1 ft

2

1 h in

2

144 60 min

A

2

V

2

262 ft

3 min

Combustors

In ME 300 and ME 525 we learn about the actual combustion process including pollutant reduction, efficiency enhancement, fuels, alternate

Fuels etc. For now we will assume that fuel is burned and the energy released in the combustion process is available to us as heat to calculate how much the compressed air can be heated or how much water can be boiled etc.

 cv

  cv

  e

 e

 h e

V e

2

2

 gz e

  i

 i

 h i

V i

2

2

 gz i

 fuel

  air

( h e

 h i

)

Fuel

( H .

V .)

  air

( h e

 h i

)

H. V. means heating value of the fuel

Summary

Control volume energy and mass conservation equations that we learned are applicable to many practical energy devices and equipment.

Important learning comes from application of appropriate assumptions, considering the appropriate working substances and their properties in the proper range of operation to estimate different energy quantities.

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