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10/03/2014 1 Midterm Review Assumptions Steady state = 0 =0 Adiabatic / perfectly insulated = 0 =0 Isentropic / reversible adiabatic = No change in kinetic energy = No change in potential energy = Ideal gas = Polytropic = Cold air = =

MAAE3400 Midterm Review

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Page 1: MAAE3400 Midterm Review

10/03/2014

1

Midterm Review

Assumptions

Steady state πœ•

πœ•π‘‘= 0 π‘Žπ‘›π‘‘

𝑑

𝑑𝑑= 0

Adiabatic / perfectly insulated 𝑄 = 0 π‘Žπ‘›π‘‘ 𝑄 = 0 Isentropic / reversible adiabatic 𝑠𝑖 = 𝑠𝑒

No change in kinetic energy 𝑉𝑖= 𝑉

𝑒

No change in potential energy 𝑧𝑖 = 𝑧𝑒 Ideal gas 𝑝𝑣 = 𝑅𝑇 Polytropic 𝑝𝑣𝑛 = π‘π‘œπ‘›π‘ π‘‘ Cold air 𝑐𝑝 = π‘π‘œπ‘›π‘ π‘‘ π‘Žπ‘›π‘‘ 𝑐𝑣 = π‘π‘œπ‘›π‘ π‘‘

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Ideal gas processes (state 1 to state 2)

𝑝𝑣 = 𝑅𝑇

𝑅 =𝑅

𝑀 get M from table A1 𝑅 = 8.314

π‘˜π½

π‘˜π‘šπ‘œπ‘™ 𝐾 π‘…π‘Žπ‘–π‘Ÿ = 287

𝐽

π‘˜π‘” 𝐾= 53.3

π‘“π‘‘βˆ™π‘™π‘π‘“

π‘™π‘π‘šβˆ™Β°π‘…

Constant temperature 𝑝1

𝑝2=

𝑣1

𝑣2

βˆ’1

Constant volume 𝑝1

𝑝2=

𝑇1

𝑇2

Constant pressure 𝑣1

𝑣2=

𝑇1

𝑇2

For isentropic and ideal gas 𝑝2𝑝1 𝑆=π‘π‘œπ‘›π‘ π‘‘

=π‘π‘Ÿ2π‘ƒπ‘Ÿ1

π‘π‘Ÿ 𝑇 = expπ‘ π‘œ 𝑇

𝑅

𝑣2𝑣1 𝑆=π‘π‘œπ‘›π‘ π‘‘

=π‘£π‘Ÿ2π‘£π‘Ÿ1

Polytropic process (state 1 to state 2)

𝑝𝑣𝑛 = π‘π‘œπ‘›π‘ π‘‘ = 𝑝1𝑣1𝑛 = 𝑝2𝑣2

𝑛

Ideal gas 𝑝2

𝑝1=

𝑣2

𝑣1

βˆ’π‘›

𝑇2

𝑇1=

𝑣2

𝑣1

1βˆ’π‘›

𝑇2

𝑇1=

𝑝2

𝑝1

π‘›βˆ’1

𝑛

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Steam cycles β€’ Need two properties to fix state β€’ In two phase region both known properties

should not be p and T Vapour Quality, x:

𝑣 = 𝑣𝑓 + π‘₯ 𝑣𝑔 βˆ’ 𝑣𝑓

𝑒 = 𝑒𝑓 + π‘₯ 𝑒𝑔 βˆ’ 𝑒𝑓

β„Ž = β„Žπ‘“ + π‘₯ β„Žπ‘” βˆ’ β„Žπ‘“ = β„Žπ‘“ + π‘₯β„Žπ‘“π‘”

Cycles with open systems

Each component is treated as a separate control volume 𝑑𝐸𝑐𝑣𝑑𝑑

= 𝑄 𝑐𝑣 βˆ’π‘Š 𝑐𝑣

+π‘š β„Žπ‘– βˆ’ β„Žπ‘’ +𝑽𝑖2 βˆ’ 𝑽𝑒

2

2+ 𝑔 𝑧𝑖 βˆ’ 𝑧𝑒

Steam Turbines β€’ Designed to produce a given pressure drop β€’ Given state 1 and P2 or pressure drop State2s

𝑠2𝑠 = 𝑠1 β„Ž2𝑠(𝑠2𝑠, 𝑝2)

π‘₯2𝑠 =𝑠2π‘ βˆ’π‘ π‘“2𝑠

𝑠𝑔2π‘ βˆ’π‘ π‘“2𝑠

β„Ž2𝑠 = β„Žπ‘“2𝑠 + π‘₯ βˆ™ β„Žπ‘“π‘”2𝑠

State 2

β„Ž2 = β„Ž1 + β„Ž2𝑠 βˆ’ β„Ž1 πœ‚π‘‘ 𝑠2(β„Ž2, 𝑝2)

Work

π‘€π‘œπ‘’π‘‘ =π‘Š π‘œπ‘’π‘‘π‘š

= β„Ž1 βˆ’ β„Ž2

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Steam turbine with re-heat Given either T3 or heat input

𝑝3 = 𝑝2

β„Ž3 𝑝3, 𝑇3 or β„Ž3𝑄

π‘š

, β„Ž2

𝑠3 𝑝3, β„Ž3 Work

π‘€π‘œπ‘’π‘‘ =π‘Š π‘œπ‘’π‘‘π‘š

= β„Ž1 βˆ’ β„Ž2 + (β„Ž3 βˆ’ β„Ž4)

Steam condenser Constant pressure process 𝑝3 = 𝑝2

β„Ž2 π‘₯2, 𝑝2

𝑠2 π‘₯2, 𝑝2

𝑣2 π‘₯2, 𝑝2 Assume x3 = 0

β„Ž3 = β„Žπ‘“3 𝑝3

𝑠3 = 𝑠𝑓3 𝑝3

𝑣3 = 𝑣𝑓3 𝑝3

Heat Loss

π‘žπ‘œπ‘’π‘‘ =𝑄 π‘œπ‘’π‘‘

π‘š π‘ π‘‘π‘’π‘Žπ‘š= β„Ž2 βˆ’ β„Ž3

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Liquid Pump Isentropic and incompressible β„Ž4𝑠 = β„Ž3 + 𝑣3 𝑝4 βˆ’ 𝑝3 Isentropic

β„Ž4𝑠 𝑠3, 𝑝4

β„Ž4 =β„Ž4𝑠 βˆ’ β„Ž3

πœ‚π‘+ β„Ž3

Work

𝑀𝑖𝑛 =π‘Š π‘–π‘›π‘š

= β„Ž4 βˆ’ β„Ž3

Steam Generator

π‘žπ‘–π‘› =𝑄 π‘–π‘›π‘š

= (β„Ž1 βˆ’ β„Ž4)

Thermal Efficiency

πœ‚ =𝑛𝑒𝑑 π‘€π‘œπ‘Ÿπ‘˜ π‘œπ‘’π‘‘

β„Žπ‘’π‘Žπ‘‘ 𝑖𝑛𝑝𝑒𝑑=(π‘Š π‘œπ‘’π‘‘/π‘š ) βˆ’ (π‘Š 𝑖𝑛/π‘š )

(𝑄 𝑖𝑛/π‘š )

πœ‚ =β„Ž1 βˆ’ β„Ž2 βˆ’ β„Ž4 βˆ’ β„Ž3

β„Ž1 βˆ’ β„Ž4

Back Work Ratio

π‘π‘€π‘Ÿ =π‘€π‘œπ‘Ÿπ‘˜ 𝑖𝑛𝑝𝑒𝑑 π‘π‘’π‘šπ‘

π‘€π‘œπ‘Ÿπ‘˜ π‘œπ‘’π‘‘π‘π‘’π‘‘ π‘‘π‘’π‘Ÿπ‘π‘–π‘›π‘’

π‘π‘€π‘Ÿ =β„Ž4 βˆ’ β„Ž3β„Ž1 βˆ’ β„Ž2

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Open Feedwater Heaters (OFH) Condensed flows from 2 and 5 mixes, exits OFH at state 6. If not specified, x=0 at 4 and 6 Apply the first law to find y:

𝑑𝐸𝑐𝑣𝑑𝑑

= 𝑄 𝑐𝑣 βˆ’π‘Š 𝑐𝑣

+ π‘šπ‘– β„Žπ‘– +𝑽𝑖2

2

+ 𝑔 𝑧𝑖 βˆ’ π‘šπ‘œ β„Žπ‘œ +π‘½π‘œ2

2+ 𝑔 π‘§π‘œ

0 = 0 βˆ’ 0 +π‘š 2β„Ž2 +π‘š 5β„Ž5 βˆ’π‘š 6β„Ž6

0 = π‘¦π‘š 1β„Ž2 + (1 βˆ’ 𝑦)π‘š 1β„Ž5 βˆ’π‘š 1β„Ž6

Closed Feedwater Heater (two outlets) Two inlets and two outlets If not specified, assume terminal temperature difference is 0 (T6=T7) For the steam trap h7=h8 If not specified, x=0 at 4 and 7 Apply the first law to find y: 0 = 0 βˆ’ 0 +π‘š 2β„Ž2 +π‘š 5β„Ž5 βˆ’π‘š 6β„Ž6 βˆ’π‘š 7β„Ž7

0 = π‘¦π‘š 1β„Ž2 +π‘š 1β„Ž5 βˆ’π‘š 1β„Ž6 βˆ’ π‘¦π‘š 1β„Ž7

𝑦 =β„Ž6 βˆ’ β„Ž5β„Ž2 βˆ’ β„Ž7

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Closed Feedwater Heater

Thermal Efficiency

πœ‚ =𝑛𝑒𝑑 π‘€π‘œπ‘Ÿπ‘˜ π‘œπ‘’π‘‘

β„Žπ‘’π‘Žπ‘‘ 𝑖𝑛𝑝𝑒𝑑= (π‘Š π‘œπ‘’π‘‘/π‘š 1) βˆ’ (π‘Š 𝑖𝑛/π‘š 1)

(𝑄 𝑖𝑛/π‘š 1)

𝑄 π‘œπ‘’π‘‘ = 1 βˆ’ 𝑦 π‘š 1 β„Ž3 βˆ’ β„Ž4 + π‘¦π‘š 1 β„Ž8 βˆ’ β„Ž4

𝑄 π‘œπ‘’π‘‘ = 1 βˆ’ 𝑦 π‘š 1β„Ž3 + π‘¦π‘š 1β„Ž8 βˆ’π‘š 1β„Ž4

Air Cycles

Air standard Cold air standard

π‘π‘Ÿ2 =𝑝2𝑝1π‘π‘Ÿ1

π‘£π‘Ÿ2 =𝑣2𝑣1π‘£π‘Ÿ1

π‘π‘Ÿ4 =𝑝4𝑝3

π‘π‘Ÿ3

π‘£π‘Ÿ4 =𝑣4𝑣3

π‘£π‘Ÿ3

Get enthalpies and internal energies from

the pr and vr values using table A-22

For expansion or compression of Otto and

Diesel cycles

𝑇2𝑇1

=𝑝2𝑝1

π‘˜βˆ’1π‘˜

=𝑣1𝑣2

π‘˜βˆ’1

𝑇4𝑇3

=𝑝4𝑝3

π‘˜βˆ’1π‘˜

=𝑣3𝑣4

π‘˜βˆ’1

βˆ†β„Ž = π‘π‘βˆ†π‘‡

βˆ†π‘’ = π‘π‘£βˆ†π‘‡

cv and cp assumed to be constant

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Otto Cycle Compression Ratio

π‘Ÿ =𝑉1𝑉2

=𝑉4𝑉3

=𝑣1𝑣2

=𝑣4𝑣3

States 1 and 3 are usually given

π‘Šπ‘–π‘›

π‘š= 𝑒2 βˆ’ 𝑒1

π‘„π‘–π‘›π‘š

= 𝑒3 βˆ’ 𝑒2

π‘Šπ‘œπ‘’π‘‘

π‘š= 𝑒3 βˆ’ 𝑒4

π‘„π‘œπ‘’π‘‘π‘š

= 𝑒4 βˆ’ 𝑒1

Otto Cycle Only internal energies needed at each state - Given state 1, state 3 and r - For states 1 and 3, use given temp to find u from table

State 2: π‘£π‘Ÿ2 =1

π‘Ÿπ‘£π‘Ÿ1

get u2 (π‘£π‘Ÿ2) using table State 4: π‘£π‘Ÿ4 = π‘Ÿπ‘£π‘Ÿ3 get u4(π‘£π‘Ÿ4) using table For cold air standard

State 2: 𝑇2 = 𝑇1𝑣1

𝑣2

π‘˜βˆ’1= 𝑇1π‘Ÿ

π‘˜βˆ’1

State 4: 𝑇4 = 𝑇3𝑣3

𝑣4

π‘˜βˆ’1= 𝑇3

1

π‘Ÿ

π‘˜βˆ’1

𝑒𝑛 = 𝑐𝑣𝑇𝑛

Find cv (k) from Table A20

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Diesel Cycle Compression Ratio

π‘Ÿ =𝑉1𝑉2

=𝑣1𝑣2

Cut-off Ratio

π‘Ÿπ‘ =𝑉3𝑉2

=𝑣3𝑣2

π‘Šπ‘–π‘›

π‘š= 𝑒2 βˆ’ 𝑒1

π‘„π‘–π‘›π‘š

= β„Ž3 βˆ’ β„Ž2

π‘Šπ‘œπ‘’π‘‘

π‘š= 𝑒3 βˆ’ 𝑒4

π‘„π‘œπ‘’π‘‘π‘š

= 𝑒4 βˆ’ 𝑒1

Diesel Cycle Given state 1, r and rc

State 1: 𝑒1(𝑇1) and π‘£π‘Ÿ1 𝑇1 use table A22

State 2: π‘£π‘Ÿ2 =1

π‘Ÿπ‘£π‘Ÿ1 get T2, u2 and h2

State 3: 𝑇3 = 𝑇2π‘Ÿπ‘ get vr3, u3 and h3

State 4: π‘£π‘Ÿ4

π‘£π‘Ÿ3=

π‘Ÿ

π‘Ÿπ‘ get u4

For cold air standard State 2: 𝑇2 = 𝑇1π‘Ÿ

π‘˜βˆ’1 State 3: 𝑇3 = 𝑇2π‘Ÿπ‘

State 4: 𝑇4 = 𝑇3𝑣3

𝑣4

π‘˜βˆ’1= 𝑇3

π‘Ÿπ‘

π‘Ÿ

π‘˜βˆ’1

𝑒𝑛 = 𝑐𝑣𝑇𝑛 and β„Žπ‘› = 𝑐𝑝𝑇𝑛

Find cv (k) and cp (k) from Table A20

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Thermal Efficiency

πœ‚π‘π‘¦π‘π‘™π‘’ =π‘Šπ‘π‘¦π‘π‘™π‘’/π‘š

𝑄𝑖𝑛/π‘š=π‘Šπ‘œπ‘’π‘‘/π‘š βˆ’π‘Šπ‘–π‘›/π‘š

𝑄𝑖𝑛/π‘š=𝑄𝑖𝑛/π‘š βˆ’π‘„π‘œπ‘’π‘‘/π‘š

𝑄𝑖𝑛/π‘š

πœ‚π‘‚π‘‘π‘‘π‘œ = 1 βˆ’π‘’4 βˆ’ 𝑒1𝑒3 βˆ’ 𝑒2

πœ‚ π‘‚π‘‘π‘‘π‘œπΆπ‘œπ‘›π‘ π‘‘ 𝑐𝑣

= 1 βˆ’π‘‡1𝑇2

= 1 βˆ’1

π‘Ÿπ‘˜βˆ’1

πœ‚π·π‘–π‘’π‘ π‘’π‘™ = 1 βˆ’π‘’4 βˆ’ 𝑒1β„Ž3 βˆ’ β„Ž2

πœ‚ π·π‘–π‘’π‘ π‘’π‘™πΆπ‘œπ‘›π‘ π‘‘ 𝑐𝑣

= 1 βˆ’1

π‘Ÿπ‘˜βˆ’11

π‘˜βˆ™π‘Ÿπ‘π‘˜ βˆ’ 1

π‘Ÿπ‘ βˆ’ 1

Dual Cycle

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Brayton Cycle

Brayton Cycle For reversible air-standard

π‘ƒπ‘Ÿ2 = π‘ƒπ‘Ÿ1𝑃2𝑃1

π‘ƒπ‘Ÿ4 = π‘ƒπ‘Ÿ3𝑃4𝑃3

For reversible cold air standard

𝑇2𝑇1

=𝑃2𝑃1

π‘˜βˆ’1π‘˜

𝑇4𝑇3

=𝑃4𝑃3

π‘˜βˆ’1π‘˜

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Brayton Cycle with regenerator

𝑄 π‘–π‘›π‘š

= β„Ž3 βˆ’ β„Žπ‘₯

πœ‚π‘Ÿπ‘’π‘” =π‘Žπ‘π‘‘π‘’π‘Žπ‘™

max π‘‘β„Žπ‘’π‘œπ‘Ÿ.=β„Žπ‘₯ βˆ’ β„Ž2β„Ž4 βˆ’ β„Ž2

Brayton Cycle with reheat

π‘Š π‘‘π‘’π‘Ÿπ‘π‘–π‘›π‘’ = π‘š β„Ž3 βˆ’ β„Žπ‘Ž + β„Žπ‘ βˆ’ β„Ž4

𝑄 𝑖𝑛 = π‘š β„Ž3 βˆ’ β„Ž2 + β„Žπ‘ βˆ’ β„Žπ‘Ž

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Brayton Cycle with intercooling

π‘Š π‘π‘œπ‘šπ‘ = π‘š β„Žπ‘ βˆ’ β„Ž1 + β„Ž2 βˆ’ β„Žπ‘‘

𝑄 π‘œπ‘’π‘‘ = π‘š β„Žπ‘ βˆ’ β„Žπ‘‘ + β„Ž(𝑐𝑦𝑐𝑙𝑒 π‘œπ‘’π‘‘) βˆ’ β„Ž1

Brayton Cycle

π‘Š 𝑐𝑦𝑐𝑙𝑒 = π‘Š π‘‘π‘’π‘Ÿπ‘π‘–π‘›π‘’ βˆ’π‘Š π‘π‘œπ‘šπ‘

πœ‚π‘‘β„Žπ‘’π‘Ÿπ‘šπ‘Žπ‘™ = 1 βˆ’π‘„ π‘œπ‘’π‘‘

𝑄 𝑖𝑛

π‘π‘€π‘Ÿ =π‘Š π‘π‘œπ‘šπ‘

π‘Š π‘‘π‘’π‘Ÿπ‘π‘–π‘›π‘’

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Aircraft Propulsion

π‘Š 𝑐𝑦𝑐𝑙𝑒 = π‘Š π‘‘π‘’π‘Ÿπ‘π‘–π‘›π‘’ βˆ’π‘Š π‘π‘œπ‘šπ‘ = 0

β„Žπ‘–π‘› +𝑉𝑖𝑛2

2= β„Žπ‘œπ‘’π‘‘ +

π‘‰π‘œπ‘’π‘‘2

2

𝐹𝑑 = π‘š (𝑉5 βˆ’ π‘‰π‘Ž)

Aircraft Propulsion Relative to the engine v1=0 and v4 = 0.

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Refrigeration:

𝑄 π‘–π‘›π‘š

= β„Ž1 βˆ’ β„Ž4

π‘Š π‘π‘š

= β„Ž2 βˆ’ β„Ž1

𝑄 π‘œπ‘’π‘‘π‘š

= β„Ž2 βˆ’ β„Ž3

β„Ž3 = β„Ž4

𝑄 π‘œπ‘’π‘‘ = 𝑄 𝑖𝑛 +π‘Š 𝑐

𝛽 =β„Ž1 βˆ’ β„Ž4β„Ž2 βˆ’ β„Ž1

Cascade:

For intermediate heat exchanger

π‘š 𝐡 β„Ž8 βˆ’ β„Ž5 = π‘š 𝐴 β„Ž3 βˆ’ β„Ž2 For COP

𝛽 =π‘š 𝐴 β„Ž1 βˆ’ β„Ž4

π‘š 𝐴 β„Ž2 βˆ’ β„Ž1 +π‘š 𝐡 β„Ž6 βˆ’ β„Ž5

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Multistage Compression with Intercooling

For COP

𝛽 =(1 βˆ’ π‘₯)(β„Ž1 βˆ’ β„Ž8)

1 βˆ’ π‘₯ β„Ž2 βˆ’ β„Ž1 + (β„Ž4 βˆ’ β„Ž3)


Midterm review

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