Thermo final cheat sheet 1 PDF

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equations for thermodynamics...

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Unit Conversions - Temperature 𝑇(℃) = 𝑇(𝐾) − 273.15 𝑇(℉) = 1.8 ∗ 𝑇(℃) + 32 T(ºR) = T(ºF) + 459.67 - Pressure 1 bar = 105 Pa = 100 kPa Patm = 101 kPa = 14.7 lbf/in^2 6895 Pa = 1 psi - Energy + Power 1 Btu = 778 ft*lbf 1 hp = 550 ft*lbf/s = 2545 Btu/hr N = kg*m/s^2 - Volume 1 m = 3.28084 ft - Fluids 𝜌𝐻2𝑂 = 1000 kg/m^3 = 1.94 slug/ft^3 𝛾𝐻2 𝑂 = 9800 N/m^3 = 62.4 lbf/ft^3 𝛾𝐻𝑔 = 133,200 N/m^3 - Mass Slug = 32.174 lbmass = lbforce*s^2/ft Mass Conservation

𝑑𝑀𝑎𝑠𝑠𝑐𝑣 = 𝑑𝑡

∑ 𝑚𝑖 󰇗 − ∑ 𝑚󰇗𝑒

𝑚󰇗= 𝜌𝐴|𝑉| =

𝐴|𝑉 | 𝑣

Volumetric flow rate = 𝐴|𝑉| Flow work = (𝑃 ∗ 𝐴) ∗ |𝑉 |

Work and Power 𝑊 = ∫ 𝐹 ∙ 𝑑𝑟 - Volume expansion work: 𝑊 = ∫ 𝑃 𝑑𝑉 𝑊 = 𝑚 ∗ ∫ 𝑃 𝑑𝑣 𝑃𝑜𝑤𝑒𝑟 =

𝑑𝑊 𝑑𝑡

Velocity^2 to kJ/kg

𝑚2 𝑠2

∗ 𝑘𝑔∗𝑚 ∗ 𝑁∗𝑠2

𝐽 𝑁∗𝑚

∗

𝑘𝐽

103 𝐽

Adiabatic: Δ𝑄 = Δ𝑠 = 0 Isentropic: Δ𝑠 = 0

Ideal Gas 𝑃𝑣 = 𝑅𝑇 , 𝑅 is universal 𝑃 = 𝜌𝑅𝑇 𝑃𝑉 = 𝑚𝑅𝑇 𝑘𝐽 8.314

Mixtures 𝑥 − 𝑥𝑓 𝑀𝑣𝑎𝑝 (𝑥 = 𝑣, 𝑢, ℎ) = 𝑋= 𝑥𝑔 − 𝑥𝑓 𝑀

{1545

Enthalpy

∆𝑢 = ∫ 𝐶𝑣 (𝑇) 𝑑𝑇 𝑇2

𝑇1

∆ℎ = ∫ 𝐶𝑝 (𝑇) 𝑑𝑇 + 𝑣 (𝑃2 − 𝑃1 ) 𝑇1

𝐶𝑣 ≈

𝑑𝑡

𝐶𝑣 (𝑇1 ) + 𝐶𝑣 (𝑇2 ) 2

𝑇

)

𝑟𝑒𝑣

, 𝑆 = 𝑠∗𝑚

𝑆2 − 𝑆1 = ∫1 ( 𝑇 ) 2 𝛿𝑄

Isentropic Efficiency

Nozzle: 𝜂𝑛 =

= ℎ 1−ℎ 2 ≤ 1 ℎ −ℎ

1 |𝑉 |2 2 2 1 |𝑉 ( 2 2 |2 ) 𝑠

Compressor/Pump: 𝜂𝑐/𝑝 =

𝑄󰇗

1 1 2 2 = 𝑄𝑐𝑣󰇗 − 𝑊𝑐𝑣󰇗 + ∑𝑖 𝑚󰇗 (ℎ 𝑖 𝑖 + 2 |𝑉𝑖 | + 𝑔 ∗ 𝑧𝑖 ) − ∑ 𝑒 𝑚󰇗𝑒(ℎ𝑒 + 2 |𝑉𝑒 | + 𝑔 ∗ 𝑧𝑒 )

Entropy and Ideal Gas

Δ𝑠 = 𝑠(𝑇2 , 𝑃2 ) − 𝑠(𝑇1 , 𝑃1 ) = ∫𝑇 2

+ 𝑚󰇗(𝑠1 − 𝑠2 ) + 𝜎󰇗𝑐𝑣 𝑇

Refrig: 𝛽𝑟𝑒𝑣 =

2

1

Energy Conservation

+ ∑ 𝑚󰇗𝑖 ∗ 𝑠𝑖 − ∑ 𝑚𝑒󰇗 ∗ 𝑠𝑒 + 𝜎󰇗𝑐𝑣 𝑄󰇗 1 (∑ 𝑚󰇗 𝑇

Power: 𝜂𝑟𝑒𝑣 =

Rev: 1𝑄2 = ∫1 𝑇𝑑𝑠 = 𝑚𝑇(𝑠2 − 𝑠1 ) 𝑊󰇗𝑡 ⁄𝑚󰇗 (𝑊󰇗𝑡 ⁄𝑚󰇗 )𝑠

𝑙𝑏𝑚𝑜𝑙∗°𝑅

Δ𝑠 = 𝑠(𝑇2 , 𝑃2 ) − 𝑠(𝑇1 , 𝑃1 ) =

+ 𝜎󰇗𝑐𝑣 )

𝐶𝑂𝑃𝑟𝑒𝑣,1 = 𝐶𝑂𝑃𝑟𝑒𝑣,2

𝑇 ∗ 𝑑𝑠 = 𝑑ℎ − 𝑣 ∗ 𝑑𝑃

≤1

󰇗 𝑡 ⁄𝑚󰇗) (𝑊 𝑠

(𝑊󰇗𝑡 ⁄𝑚󰇗 )𝑎

𝑊󰇗 𝑄󰇗𝐻

𝑄󰇗𝑖𝑛 𝑊󰇗

Heat Pump: 𝛾𝑟𝑒𝑣 =

2𝑠

≤1

Real gas 𝑃𝑉 𝑧 = 𝑅𝑇 , where z = compressibility factor

Compressed Liquid Approx. 𝑥(𝑇, 𝑃) ≈ 𝑥𝑓 (𝑇) , x = (v,u,h,s)

=

=

𝑄󰇗𝑜𝑢𝑡 𝑊󰇗

𝑄𝐻 −𝑄𝐶 𝑄𝐻 𝑄󰇗𝑖𝑛

=1−

𝑄󰇗𝑜𝑢𝑡 −𝑄󰇗𝑖𝑛

=

=

Entropy Balance (Process) 𝑆2 − 𝑆1 =

2 𝛿𝑄 ∫1 ( 𝑇 ) 𝑏

+𝜎

𝑆2 − 𝑆1 = ∑ + 𝜎 𝑇 𝑑𝑆 𝑑𝑡

=∑

𝑄󰇗

𝑇

𝑄

+ 𝜎󰇗

𝑇 𝐶𝑝 (𝑇) ∫𝑇12 𝑇 𝑑𝑡

+ 𝑅 ∗ ln ( 𝑣2 ) 𝑣

− 𝑅 ∗ ln ( 𝑃

1

2nd Law and Entropy ∮𝑐𝑦𝑐𝑙𝑒 (

𝛿𝑄 𝑇

) ≤0 𝑏

S production: 𝜎 = − ∮ ( ) ≥ 0

𝑇𝑐 𝑇ℎ

𝛿𝑄 𝑇

𝑏

𝜎 > 0 : irreversible cycle 𝜎 = 0 : reversible cycle 𝜎 < 0 : impossible

𝑇𝐶 𝑇𝐻 −𝑇𝐶

𝑄󰇗𝑜𝑢𝑡 = 𝑄󰇗𝑜𝑢𝑡 −𝑄󰇗𝑖𝑛

𝑇 𝐶𝑣 (𝑇) 𝑑𝑡 𝑇 1

Δ𝑠 = 𝑠°(𝑇2 ) − 𝑠°(𝑇1 ) − 𝑅 ∗ ln ( 𝑃2 )

𝐶𝑂𝑃𝑖𝑟𝑟𝑒𝑣 < 𝐶𝑂𝑃𝑟𝑒𝑣 = 𝐶𝑂𝑃𝑚𝑎𝑥

𝑏

𝑉1

𝑉1

2nd Law and Cycle COP’s

𝑇 ∗ 𝑑𝑠 = 𝑑𝑢 + 𝑃 ∗ 𝑑𝑣

Turbine: 𝜂𝑡 =

𝑚𝑅(𝑇2 −𝑇1 )

𝑉1 ) 𝑉2

,𝑛 ≠ 1 (1−𝑛) ={ 𝑉2 𝑚𝑅𝑇 ln ( ) ,𝑛 = 1

Entropy Balance (Cycle)

𝑄󰇗 ∑ 𝑇

=(

𝑛−1

󰇗 + 𝑚󰇗 [(ℎ𝑖 − ℎ𝑒 ) + 1 (|𝑉𝑖 |2 − |𝑉𝑒 |2 ) + 𝑔 (𝑧𝑖 − 𝑧𝑒 )] 1 Inlet, 1 Exit: 0 = 𝑄𝑐𝑣󰇗 − 𝑊𝑐𝑣 2

Δ𝑠 = 𝑠2 − 𝑠1 =

Entropy

𝛿𝑄

1 𝑊2

𝑓𝑡∗𝑙𝑏𝑓

𝑑𝑡

1 in, 1 out: 0 = ∑

𝐶𝑝 = 𝐶𝑣 + 𝑅

𝑑𝑆 = (

=

𝑑𝑆𝑐𝑣

Constant Cv approximation

𝑇2 =( ) 𝑇1 𝑃1

𝑤

𝐾𝑚𝑜𝑙∗𝐾 𝐵𝑡𝑢

𝑑𝐸𝑐𝑣

𝐻 = 𝑈+𝑃∗𝑉 ℎ =𝑢+𝑃∗𝑣

𝑇2

𝑛−1 𝑃2 𝑛

𝑅 = 1.986 𝑙𝑏𝑚𝑜𝑙∗ °𝑅

𝛽 = (1 − 𝑋)𝛽𝑓 + 𝑋 ∗ 𝛽𝑔

Specific Heat (heat capacity) C used when incompressible, otherwise Cv and Cp

𝑅

Polytropic Process 𝑃 ∗ 𝑉 𝑛 = 𝑐 𝑜𝑟 𝑃 ∗ 𝑣 𝑛 = 𝑐 - Volume expansion work (𝑃2 𝑉2−𝑃1 𝑉1 ) ,𝑛 ≠ 1 (1−𝑛) 1𝑊2 = { 𝑉 𝑃1 𝑉1 ln ( 2 ) ,𝑛 = 1

Polytropic Ideal Gas

𝑃𝑣 = 𝑅𝑇 , where 𝑅 = 𝑀

1 Law ∆𝐸 = ∆𝐾𝐸 + ∆𝑃𝐸 + ∆𝑈 = 𝑄 12 − 𝑊 1 2 Rate Form: 𝐸󰇗 = 𝑄󰇗 − 𝑊󰇗 st

𝑇𝐻 𝑇𝐻 −𝑇𝐶

𝐾𝐸 = 𝑚𝑣 2 2 1

Energy

𝑃𝐸 = 𝑚𝑔𝑧

1st Law and Cycles Power: 𝑄𝐻 − 𝑄𝐶 = 𝑊𝑐𝑦𝑐𝑙𝑒 R+HP: 𝑄𝑜𝑢𝑡 − 𝑄𝑖𝑛 = 𝑊𝑐𝑦𝑐𝑙𝑒

1

𝑃2

𝑃1

)

Ideal Rankine Cycle

T: P:

𝑊󰇗𝑡

𝑚󰇗

𝑊󰇗𝑝 𝑚󰇗

= ℎ1 − ℎ2

C:

= ℎ4 − ℎ3

𝑄󰇗𝑖𝑛 𝑚󰇗

B:

𝑄󰇗𝑜𝑢𝑡 𝑚󰇗

Carnot Refrigeration Cycle

= ℎ2 − ℎ3

Com:

ℎ4 = ℎ3 + 𝑣3 (𝑃4 − 𝑃3 )

= ℎ1 − ℎ4

T:

In second P equation above, use v in m3/kg and P in kPa to get h4 directly in kJ/kg 𝜂=

𝑊󰇗𝑡−𝑊󰇗𝑝 = 𝑄󰇗𝑖𝑛

1−

ℎ2 −ℎ3 ℎ1 −ℎ4

𝐵𝑊 =

𝑊󰇗𝑝 𝑊󰇗𝑡

=

ℎ4 −ℎ3 ℎ1 −ℎ2

𝑊󰇗𝑐𝑜𝑚 𝑚󰇗

𝑊󰇗𝑡

𝑚󰇗

= ℎ2 − ℎ1 Con: 𝑄󰇗𝑜𝑢𝑡 = 𝑚󰇗𝑇𝐻 (𝑠3 − 𝑠2 )

= ℎ3 − ℎ4

𝛽

𝑄󰇗 = 𝑊󰇗 𝑖𝑛−𝑊󰇗 𝑐𝑜𝑚 𝑡

Along streamlines: Newton’s Law of Viscosity

h is depth, z is height 𝑑𝑃 = −𝜌𝑔 = −𝛾 𝑑𝑧 𝑃(𝑧) = 𝑃(𝑧0 ) − 𝜌𝑔(𝑧 − 𝑧0 )

No slip condition

𝑃𝑔𝑎𝑢𝑔𝑒 = 𝑃 − 𝑃𝑎

Pascal’s Principle: same depth in same fluid = same pressure Specific Gravity: 𝑆𝐺 =

𝜌

𝜌𝐻2 𝑂 |4℃

=

𝛾

𝛾 𝐻2 𝑂

Archimedes (buoyancy): 𝐹𝐵 = 𝜌𝑔𝑉 = weight of fluid displaced 𝐼

𝑐

yR = distance from free surface to resultant force Ixc = moment of inertia of gate yc = 1/2 the length of the gate. 1

𝑏𝑎3

2

|𝑉| 2

𝜏 = shear stress , 𝜇 = viscosity (Pa*s) , 𝛾󰇗 = rate of shear strain = speed = du/dy

+ 𝑦𝑐 , 𝑦𝑅 = 𝑦 𝑥𝑐 ∗𝐴

12

𝑄󰇗𝑜𝑢𝑡 − 𝑄󰇗𝑖𝑛

=

𝑇𝐶 𝑇𝐻 −𝑇𝐶

+ 𝜌 + 𝑔𝑧 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑃

V: ℎ4 = ℎ3

𝐹B + 𝐹𝑠 =

others same as Carnot

Momentum Conservation

𝜕 ∫𝜌 𝜕𝑡

∗ 𝑣 ∗ 𝑑𝑣 + ∯𝑐.𝑠. 𝜌 ∗ 𝑣 ∗ (𝑣 ∙ 𝑛)𝑑𝐴

𝐹𝑠 is external force or F = P*A (P always acts inward) 𝑛 is unit outward normal with respect to the c.v. 𝑣 ∙ 𝑛 = |𝑣||𝑛| cos(𝜃) = |𝑣|cos(𝜃) Non-uniform flow Solve for u (speed) equation

𝜏 = 𝜇 ∗ 𝛾󰇗

Submerged Gate 𝐹𝑅 = 𝛾 ∗ ℎ𝑐 ∗ 𝐴, hc = depth of center of gate

Rectangle: 𝐼𝑥𝑐 =

𝑄󰇗𝑖𝑛

Bernoulli Equation

Fluid Statics

𝑃(ℎ) = 𝑃(0) + 𝛾 ∗ ℎ

=

E: 𝑄󰇗𝑖𝑛 = 𝑚󰇗𝑇𝑐 (𝑠1 − 𝑠4 )

Ideal Vapor-Comp Refrigeration

Circle: 𝐼𝑥𝑐 = 4 𝑅 4 𝜋

Volumetric flow rate = |𝑉|𝐴 = ∫𝐴 𝑢(𝑥)𝑑𝐴...