Thermo Final Equation Sheet PDF

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Summary

Final exam formula sheet, covers from work and power to thermodynamic cycles. ...

Description

Interpolation

x −x a ) y = y a +( y b− y a)( x b− x a Work and Power

V2 W e =VI =I R= R W m =Fd W s =T ∅ ´ m =Fv ´ W W s =Tω 2

Energy

1 2 P 1 2 e mch= + v +gz = ρv + v +gz ρ 2 2 P 1 2 e mchflow =u+ + v + gz ρ 2 1 2 e mchflow =u+ ρv + v + gz 2 Efficiency

Desired Required ηtot =η1+ η2+⋯ η=

Energy and Heat Transfer

m ´ =ρ V´ = ρAV Q out =−¿ Q ¿ =+ ¿ Phase Descriptions

V T sat → SHV Pressure at Known Temperature

P> Psat → SCL P< Psat → SHV Quality

m vap m tot V −V f X= V f −V g X=

Enthalpy

h=u+ pv H=U + PV h ≈ hf @T +v f @ t ( Psys + Psat @ t ) Pressure |¿|−P sys

Patm =P¿ sys+ P atm P¿

|¿|=P

Energy Balance

∆ E=∑ E¿ −∑ E out

∆ u+ ∆ KE+∆ PE=Q ¿ +W ¿ + m´ ¿ e Ideal Gas Law

PV =mRT Pv = RT

Conservation of Mass

V´ ´ =ρ V avg A c =ρ V´ = m v ´ =V avg A c V

Isobaric (P=constant)

V1 T1

=

V2 T2

One Inlet One Exit

ρ 1 V 1 A1 = ρ 2 V 2 A 2 V 1 A1 V 2 A2 = v1 v2

Isothermal (T=cons)

P1 V 1=P 2 V 2 Isochoric (V=cons)

P1 P2 = T1 T 2

Turbines, Compressors, Pumps, and Fans

Compressibility Factor

Pv RT P Pr = Pcr Z=

Pv =ZRT T r=

T T cr

Using Ideal Gas Law

Pr ≪1 IGL T r >2∧Pr ≯1 T r >2∧Pr ≯3 T r ≈ 1∧P r ≈ 1

IGL Comp Chart Property Tables

Specific Heats

∆ u=CVavg ∆ T ∆ h =C Pavg ∆ T C p =C v =C h=u+ P ∆ v ∆ h=C Vavg ∆ T +v ∆ P Boundary Work

W B =Fds= PAds =P ∆ V Polytropic

P 2 V 2−P1 V 1

1−n n n P1 V 1 =P2 V 2 mR ( T 2−T 1) 1−n

Conservation of Energy for Steady Flow

( (

)

´ −W´ =m´out h− 1 v 2−gz Q 2 out 1 2 −m´ ¿ h− v +gz 2 ¿ V 1 A1 V 2 A 2 = m´ 1=m´2= v2 v1

)

[

1 2 2 ( v −v ) 2 1 2 1 2 2 If Q=0 :h2 −h1= ( v 1−v 2 ) 2 IF Q=0∧Ideal Gas : 1 2 2 C p (T 2−T 1) = (V 1 −V 2) 2 Q out=m ( h 1−h2 )+

]

Heat Engines (Power Plants)

W out Q out QL =1− =1− Q¿ Q¿ QH Q ¿ +W ¿=Q out −W out W output =Q ¿ −Q out ηth=

Ideal Gas

WB=

Turb :W out =+¿ Comp , P , F :W ¿=−¿ ´ W´ =m Q= ´ ( ∆ h +∆ KE + ∆ PE ) ∆ h=open sys ∆ u=closed sys

Energy Balance for Nozzles and Diffusers

Solids and Liquids

WB=

∆ U + P ∆ V =Q−W other ∆ H=Q−W other

where n ≠ 1

Isothermal Ideal Gas where n=1

v W B =mRT ln ( 2 ) v1 First Law for Closed Systems

∆ E =∆ U + ∆ KE + ∆ PE ∆ E=Q ¿ +W ¿ −Q out −W out Isochoric (V=constant)

∆ U =Q−W other Isobaric (P=constant)

∆ U =Q− P ∆ V −W other First Law Energy Balance

∆ U =Q− P ∆ V −W other

Refrigerators (Air Conditioners)

[

QL Q H Q −1= H −1 = W¿ W ¿ QL Q H =W ¿ +Q L CO PR =

]

−1

Heat Pump

[

Q Q Q CO P HP= H =1+ L = 1− L W¿ W¿ QH

−1

]

Carnot Cycle (most efficient) Four totally reversible processes

1→ 2 ¿ Isothermal heat addition (T 1 = 2→ 3 ¿ Isentropic expansion(S2 =S 3)

3 → 4 ¿ Isothermal heat rejection (T 4 → 1 ¿ Is entropic compression(S 4= W T ηth = net =1− L Q¿ TH Carnot Heat Pump

[ ] [ ]

Q T CO P HP= H = 1 − L W¿ TH Carnot Refrigerator

CO PR =

QL T H = −1 W¿ T L

−1

−1

Approximate Analysis (constant specific heats) A-1/A-2

T2 v2 +R ln T1 v1

( ) ( ( ) (

∆ s=s 2−s1 ≈C vavg ln

T P ∆ s=s 2−s1 ≈C pavg ln 2 −R ln P T1 Guess T 2 , find Cvavg ∨C pavg , calcula Exact Analysis (variable specific heats) A-17 o

o

∆ s=s 2−s1= s2 −s1 −R ln

( ) P2 P1

∑ m´ ¿ h=∑ m´out h where Q=0 Single Fluid System

Q=∑ m´out h−∑ m´ ¿ h Q B → A=m ( h A 2−h A 1 )

Throttling Valve (Decrease Pressure)

h1=h2 If ideal gas :h1=h2 ∧T 1=T 2 Mixing Chamber

m 1 h1 + m 2 h2=m 3 h3 m 1+ m2 =m3 ∑ m¿ =∑ mout Entropy 2

∆ s=s 2−s2=∫ 1

∆ s=

dQ T

Q To

Q +s T o gen ∆ s>0 irreversible ∆ s=0 isentropic ∆ s...