Thermodynamics Problems Final Exam Type

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Engineering and Computer Science Concordia University Montréal, Québec H3G 1M8 1 Tutorial 10 Tutorial Leader: Andrew Dafoe Final Tutorial

description

Typical Final Exam type problems in for engineering thermodynamics. Cycle problems with solutions

Transcript of Thermodynamics Problems Final Exam Type

Page 1: Thermodynamics Problems Final Exam Type

Engineering and Computer ScienceConcordia UniversityMontréal, Québec H3G 1M8 1

Tutorial 10

Tutorial Leader: Andrew Dafoe

Final Tutorial

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Problem 8.1

C

𝑇 𝑠𝑢𝑟𝑟=17oC=290K

𝑇 1=25oC=298K

𝑇 2=25oC=298K

�̇�=−25 𝑘𝑊

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Problem 8.1

Δ �̇�=�̇�−�̇�+�̇� (h1−h2)+Δ 𝑃𝐸+Δ 𝐾𝐸

Δ �̇�=�̇�−�̇�+�̇�𝑐𝑝 (𝑇1−𝑇 2 )𝑠𝑖𝑛𝑐𝑒 Δ𝑇=0

Δ �̇�=�̇�−�̇�=0

�̇�=�̇�=−25𝑘𝑊

Δ �̇�air=�̇�air

𝑇 𝑠𝑦𝑠

=−25𝑘𝑊

298𝐾=−0.08389𝑘𝑊 /𝐾

𝑑𝑜𝑒𝑠 h𝑡 𝑖𝑠𝑝𝑟𝑜𝑐𝑒𝑠𝑠𝑣𝑖𝑜𝑙𝑎𝑡𝑒 h𝑡 𝑒 𝑠𝑒𝑐𝑜𝑛𝑑𝑙𝑎𝑤 ?𝑆𝑒𝑐𝑜𝑛𝑑 𝑙𝑎𝑤 :Δ𝑆𝑡𝑜𝑡𝑎𝑙≥0

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Problem 8.1

Δ �̇�air=�̇�air

𝑇 𝑠𝑦𝑠

=−25𝑘𝑊

298𝐾=−0.08389𝑘𝑊 /𝐾

NOPE

𝛥�̇�𝑡𝑜𝑡𝑎𝑙=𝛥�̇�𝑎𝑖𝑟+𝛥�̇�𝑠𝑢𝑟𝑟𝑜𝑢𝑛𝑑𝑖𝑛𝑔𝑠=�̇�𝑎𝑖𝑟

𝑇𝑠𝑦𝑠

+�̇�𝑠𝑢𝑟𝑟

𝑇 𝑠𝑢𝑟𝑟

𝛥�̇�𝑡𝑜𝑡𝑎𝑙=−25𝑘𝑊

298𝐾+

25𝑘𝑊290𝐾

=0.00231𝑘𝑊 /𝐾

𝑛𝑜𝑡𝑒 : h𝑡 𝑖𝑠𝑝𝑟𝑜𝑏𝑙𝑒𝑚𝑖𝑙𝑙𝑢𝑠𝑡𝑟𝑎𝑡𝑒𝑠𝑎𝑟𝑒𝑣𝑒𝑟𝑠𝑖𝑏𝑙𝑒 𝑠𝑦𝑠𝑡𝑒𝑚

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Problem 8.2

𝑣𝑎𝑐

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Problem 8.2

𝑣𝑎𝑐

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Problem 8.2

𝑣𝑎𝑐

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Problem 8.3

𝑆=𝑐𝑜𝑛𝑠𝑡

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Problem 8.3

Δ h=h2−h1=−807.4𝑘𝐽 /𝑘𝑔

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Problem 8.4

𝑉 2=𝑉 1

2

𝑃𝑜𝑙𝑦𝑡𝑟𝑜𝑝𝑖𝑐 (𝑛=1.3)

𝐺𝑖𝑏𝑏𝑠𝑒𝑞 :𝑇𝑑𝑠=𝑑𝑢+𝑃𝑑𝑣

𝑁𝑒𝑒𝑑𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒𝑎𝑡 h𝑏𝑜𝑡 𝑠𝑡𝑎𝑡𝑒𝑠 𝑓𝑖𝑟𝑠𝑡…

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Problem 8.4

𝑉 2=𝑉 1

2

𝑃𝑜𝑙𝑦𝑡𝑟𝑜𝑝𝑖𝑐 (𝑛=1.3)

𝑃𝑜𝑙𝑦𝑡𝑟𝑜𝑝𝑖𝑐 𝑟𝑒𝑙𝑎𝑡𝑖𝑜𝑛 :𝑇2=𝑇1( 𝑣1

𝑣2)𝑛−1

=310×20.3=381.7𝐾

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Problem 8.6

𝑊𝑛𝑒𝑡=700 𝑘𝐽

h𝑖𝑛𝑡 :ΔE=Δ𝑈=Q−W

𝑄=0

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Problem 8.6

𝑊𝑛𝑒𝑡=700 𝑘𝐽

𝑄=0

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Problem 8.6

𝑊𝑛𝑒𝑡=700 𝑘𝐽

𝑓𝑟𝑜𝑚𝑡𝑎𝑏𝑙𝑒𝑠 : 𝑠2=6.5215𝑘𝐽 /𝑘𝑔𝐾

𝑄=0

Δ 𝑠=𝑠2−𝑠1=6.5215−6.7593=−0.238𝑘𝐽 /𝑘𝑔𝐾

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Isentropic Efficiencies

• The isentropic efficiency of a compressor and a turbine is a great source of confusion for students.

• Where for adiabatic devices

• The easiest way to remember is to realize that the isentropic work represents the ideal scenario:– The ideal pump requires less work to operate – The ideal turbine produces the most work

• Simply remember that efficiency is expressed as a percentage (and is never greater than 100%)

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Problem 8.10

𝑄=0

𝜂𝑇=𝑊 𝑟𝑒𝑎𝑙

𝑊 𝑖𝑠𝑒𝑛𝑡𝑟𝑜𝑝𝑖𝑐

𝑘=1.395

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Problem 8.10

𝑄=0

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Problem 8.11

𝑄=0

𝜂𝐶=𝑊 𝑖𝑠𝑒𝑛𝑡𝑟𝑜𝑝𝑖𝑐

𝑊 𝑟𝑒𝑎𝑙

𝑘=1.4

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Problem 8.11

𝑄=0

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Problem 8.13

TC

HE

HE

1

2 3

4

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Problem 8.13

TC

HE

HE

1

2 3

4

𝑓𝑖𝑟𝑠𝑡 𝑠𝑡𝑒𝑝 :𝑐𝑜𝑚𝑝𝑙𝑒𝑡𝑒 h𝑡 𝑒𝑐𝑦𝑐𝑙𝑒𝑢𝑠𝑖𝑛𝑔𝑖𝑠𝑒𝑛𝑡𝑟𝑜𝑝𝑖𝑐 𝑎𝑠𝑠𝑢𝑚𝑝𝑡𝑖𝑜𝑛

𝑇 2𝑠=𝑇1( 𝑃2

𝑃1)𝑘− 1𝑘 =288 (12 )0.286=585.8𝐾

𝑇 4 𝑠=𝑇3( 𝑃4

𝑃3)𝑘−1𝑘 =873( 1

12 )0.286

=429.2𝐾

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Problem 8.13

TC

HE

HE

1

2 3

4

𝑁𝑒𝑥𝑡 , 𝑓𝑖𝑛𝑑 h𝑡 𝑒𝑖𝑠𝑒𝑛𝑡𝑟𝑜𝑝𝑖𝑐𝑤𝑜𝑟𝑘𝑣𝑎𝑙𝑢𝑒𝑠 :

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Problem 8.13

TC

HE

HE

1

2 3

4

𝑖𝑑𝑒𝑎𝑙𝑏𝑎𝑐𝑘𝑤𝑜𝑟𝑘 :𝑤𝐶 (Δ 𝑠=0 )

𝑤𝑇 (Δ 𝑠=0 )

=0.6711

𝑏𝑎𝑑𝑐𝑜𝑚𝑝 .𝑏𝑎𝑐𝑘𝑤𝑜𝑟𝑘 :𝑤𝐶 (𝜂=90 % )

𝑤𝑇 (Δ𝑠=0)

=0.7457

𝑏𝑎𝑑𝑡𝑢𝑟𝑏 .𝑏𝑎𝑐𝑘𝑤𝑜𝑟𝑘 :𝑤𝐶 (Δ𝑠=0)

𝑤𝑇 (𝜂=90 % )

=0.7456

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Engineering and Computer ScienceConcordia UniversityMontréal, Québec H3G 1M8 24

Best of luck on your final(s) next week!

Please feel free to e-mail me with any last minute questions