Uploaded by Elisa M. Wéber

Formulas Sheet

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UNIT CONVERSIONS:
1 N = 1 kg ∙ m⁄s 2
1 Pa = 1 N/m2
1J=1N∙m
`
1 lbf = 32.174 lbm ∙ ft⁄s 2
1 W = 1 J⁄s
1 J = 1 Volt ∙ Ampere ∙ s
1 bar = 105 Pa = 0.1 MPa = 100 kPa
𝑇(K) = 𝑇(℃) + 273.15
WORK, HEAT AND ENERGY BALANCE:
𝑡2
𝐸2 − 𝐸1 = 𝑄 − 𝑊
𝑊 = ∫ 𝑊̇ 𝑑𝑡
𝑑𝐸
= 𝑄̇ − 𝑊̇
𝑑𝑡
𝑄 = ∫ 𝑄̇ 𝑑𝑡
𝑄̇𝑥 = −𝜅𝐴
𝑡1
𝑡2
𝑑𝑇
𝑑𝑥
𝑄̇𝑒 = 𝜀𝜎𝐴(𝑇𝑏4 − 𝑇s4 )
𝑡1
𝑠2
Ʋ2
𝑊 = ∫ 𝑝 𝑑Ʋ
𝑄̇𝑐 = ℎ𝐴(𝑇b − 𝑇f )
𝑊 = ∫ 𝐹 𝑑𝑠
𝑠1
Ʋ1
𝑊̇ = −ℰ𝑖
𝑊̇ = 𝐹𝑉 = 𝒯𝜔
ENERGY ANALYSIS OF CYCLES:
𝑄cycle = 𝑊𝑐𝑦𝑐𝑙𝑒
∆𝐸cycle = 𝑄cycle − 𝑊𝑐𝑦𝑐𝑙𝑒 = 0
𝜂power =
𝑊𝑐𝑦𝑐𝑙𝑒
𝑄𝑖𝑛
𝛽ref =
𝑄in
𝑊cycle
𝛾hp =
𝑄out
𝑊cycle
EVALUATION OF PROPERTIES:
𝑣 = 𝑣𝑓 + 𝑥(𝑣g − 𝑣f )
𝑢 = 𝑢f + 𝑥(𝑢g − 𝑢f )
ℎ = ℎf + 𝑥(ℎg − ℎf )
𝑠 = 𝑠f + 𝑥(𝑠g − 𝑠f )
7
INCOMPRESSIBLE SUBSTANCE MODEL:
𝑇2
𝑢2 − 𝑢1 = ∫ 𝑐 (𝑇) 𝑑𝑇
𝑇1
𝑇2
𝑇2
ℎ2 − ℎ1 = ∫ 𝑐 (𝑇) 𝑑𝑇 + 𝑣 (𝑝2 − 𝑝1 )
𝑠2 − 𝑠1 = ∫
𝑇1
𝑇1
𝑐 (𝑇 )
𝑑𝑇
𝑇
IDEAL GAS MODEL:
𝑃𝑉 = 𝑚𝑅𝑇
𝑃𝑣 = 𝑅𝑇
𝑇2
𝑇2
𝑢2 − 𝑢1 = ∫ 𝑐𝑣 (𝑇) 𝑑𝑇
ℎ2 − ℎ1 = ∫ 𝑐𝑝 (𝑇) 𝑑𝑇
𝑇1
𝑇1
𝑇2
𝑠2 − 𝑠1 = ∫ 𝑐𝑣 (𝑇)
𝑇1
𝑇2
𝑑𝑇
𝑣2
+ 𝑅 𝑙𝑛
𝑇
𝑣1
𝑠2 − 𝑠1 = ∫ 𝑐𝑝 (𝑇)
𝑇1
𝑑𝑇
𝑃2
− 𝑅 𝑙𝑛
𝑇
𝑃1
POLYTROPIC PROCESS:
𝑃2
𝑉1 𝑛
=( )
𝑃1
𝑉2
𝑃1 𝑉1𝑛 = 𝑃2 𝑉2𝑛
𝑛
𝑃𝑉 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
2
𝑃2 𝑉2 − 𝑃1 𝑉1
∫ 𝑃 𝑑𝑉 =
1−𝑛
1
2
∫ 𝑃 𝑑𝑉 = 𝑃1 𝑉1 ln
𝑛≠1
1
𝑉2
𝑉1
𝑛=1
IDEAL GASS UNDER POLYTROPIC PROCESS:
𝑇2
𝑃2 (𝑛−1)⁄𝑛
𝑉1 𝑛−1
=( )
=( )
𝑇1
𝑃1
𝑉2
2
𝑚𝑅 (𝑇2 − 𝑇1 )
∫ 𝑃 𝑑𝑉 =
1−𝑛
1
2
𝑛≠1
∫ 𝑃 𝑑𝑉 = 𝑚𝑅𝑇 ln
1
𝑉2
𝑉1
𝑛=1
MASS AND ENERGY BALANCE FOR CONTROL VOLUMES:
𝑑𝑚cv
= ∑ 𝑚̇𝑖 − ∑ 𝑚̇𝑒
𝑑𝑡
𝑖
𝑒
𝑚̇ = 𝜌𝐴𝑉 =
𝐴𝑉
𝑣
𝑑𝐸𝑐𝑣
𝑉𝑖2
𝑉𝑒2
̇
̇
= 𝑄cv − 𝑊cv + ∑ 𝑚̇𝑖 (ℎ𝑖 +
+ 𝑔𝑧𝑖 ) − ∑ 𝑚̇𝑒 (ℎ𝑒 +
+ 𝑔𝑧𝑒 )
𝑑𝑡
2
2
𝑖
𝑒
8
MASS AND ENERGY BALANCE FOR CONTROL VOLUMES (TRANSIENT ANALYSIS):
∆𝑚𝑐𝑣 = ∑ 𝑚𝑖 − ∑ 𝑚𝑒
𝑖
𝑒
∆𝑈𝑐𝑣 = 𝑄cv − 𝑊cv + ∑ 𝑚𝑖 ℎ𝑖 − ∑ 𝑚𝑒 ℎ𝑒
𝑖
𝑒
THERMAL EFFICIENCY AND COEFFICIENT OF PERFORMANCE:
𝜂=
𝑊𝑐𝑦𝑐𝑙𝑒
𝑄𝐶
=1−
𝑄𝐻
𝑄𝐻
𝜂max = 1 −
𝛽=
𝑄𝐶
𝑄𝐶
=
𝑊𝑐𝑦𝑐𝑙𝑒 𝑄𝐻 − 𝑄𝐶
𝛽max =
1
𝑇𝐻 ⁄𝑇𝐶 − 1
𝛾=
𝑄𝐻
𝑄𝐻
=
𝑊𝑐𝑦𝑐𝑙𝑒 𝑄𝐻 − 𝑄𝐶
𝛾max =
1
1 − 𝑇C ⁄𝑇H
𝑇C
𝑇H
SECOND LAW OF THERMODYNAMICS:
∮(
𝛿𝑄
) = −𝜎cycle
𝑇 b
2
𝑆2 − 𝑆1 = ∫ (
1
𝛿𝑄
)
𝑇 int
rev
𝑑𝑆 = (
𝛿𝑄
)
𝑇 𝑖𝑛𝑡
𝑟𝑒𝑣
𝑇 𝑑𝑆 = 𝑑𝑈 + 𝑃 𝑑Ʋ
𝑇 𝑑𝑠 = 𝑑𝑢 + 𝑃 𝑑𝑣
𝑇 𝑑𝑆 = 𝑑𝐻 − Ʋ 𝑑𝑃
𝑇 𝑑𝑠 = 𝑑ℎ − 𝑣 𝑑𝑃
2
𝑆2 − 𝑆1 = ∫ (
1
𝛿𝑄
) + 𝜎
𝑇 b
𝑆2 − 𝑆1 =
𝑄
+ 𝜎
𝑇b
𝑄̇𝑗
𝑑𝑆
=∑
+ 𝜎̇
𝑑𝑡
𝑇𝑗
𝑗
9
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