Thermodynamics Final cheat sheet PDF

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Summary

Macroscopic char. of a system (eg. Mass, vol, energy, pressure, temp.) need knowledge of prev. state Steady State properties change Extensive prop. can add up Intensive may vary from place to place, add Equilibrium balanced, change if isolated at that point in time Work work done work done on system...

Description

Property- Macroscopic char. of a system (eg. Mass, vol, energy, pressure, temp.) *Don’t need knowledge of prev. state Steady State – properties don’t change w/time Extensive prop. – can add up Intensive – may vary from place to place, can’t add Equilibrium – balanced, wouldn’t change if isolated at that point in time

Work W>0 – work done by system; W0 – heat into system; Q 1, γ max = = TH W cycle Q out −Q ¿

Quality- Two Phase, Liquid-Vapor Mixture Works for h, too

m vapor mliquid +m vapor ν =x ν g +(1−x )ν f ν− νf x= ν g −ν f x=

Enthalpy

H =U + pV ↔ h=u+ pν Units are Joules Kelvin Plank Statement It is impossible for any system to operate in a thermodynamic cycle and deliver a net amount of energy by work to its surroundings while receiving energy by heat transfer from a single thermal reservoir

W cycle Q¿

=

T Q ¿ −Qout 95%) 2

η=

ν /2 (ν 2 /2) S

Compressor/Pump (typically 75 to 85 %)

ηC =

´ CV / m) ´ S h2 S−h1 (−W = ´ CV / m h 2−h1 −W ´

Rankine Cycle

h 2−h3 h2 − h1

Isentropic Process of an Ideal Gas, specific heats constant ( k−1) / k

( ) ( ) ()

T2 p2 = T1 p1 T2 v = 1 T1 v2 p2 v1 = p1 v2 cp k= cv

2

( δQT ) +σ

(k−1 )

k

Q dS =∑ j +σ´ Tj dt j

1

b

Control Volume

´ d Scv Q ´ i si −∑ m =∑ +∑ m T dt

´ W´ CV Q = cv +h1−h 2 m ´ m ´

If Int. Rev.

´ CV W m ´

IntRev

´ CV W m ´

IntRev

Rankine Cycle

s=x s g +(1−x)s f Closed Systems

S 2−S1=∫

Heat Transfer One inlet, one exit, SS, excluding KE, PE

( ) ( )

Entropy

2

=∫ TdS 1

=−∫ vdp

Polytropic Processes n

p v =constant W´ CV −n = (p v − p m ´ IntRev n−1 2 2 1 p W´ CV =− p1 v 1 ln 2 , m ´ IntRev p1 p W´ CV =−RT ln 2 , id p1 m ´ IntRev

( ) ( ) ( )

Rankine Cycle Green Highlighter (area under whole curve):

´¿ Q m ´

( ) ( )

Blue Highlighter (area under bottom):

Red Pen (area enclosed):

´ Q out m ´

h1 −h2−(h4 −h3 ) h2−h3 h 1−h4 η= =1− h1−h 4 Back work ratio – ratio of work input to work output

bwr=

´ P/ m ´ h4 −h3 W = ´ /m W ´ h 1−h2 t

1→ 2

:

Turbine

W´ t =h1 −h2 m ´

Isentropic exp. from sat. vap to condensor pressure

3 → 4 : Pump W´ P =h 4−h3 m ´ Isentropic comp. to state for in comp. liq. region

2→ 3

:

Condensor

´ Q out =h2−h3 m ´

Heat transfer from working fluid at const. P., exits as sat. liq.

4 → 1 : Boiler ´¿ Q =h1−h4 m ´ Heat transfer at const. P.

´ net Q

W net W´ P ´ = m ´ m ´ = Irreversibilities

4

( )

IntRev

=−∫ 3

Internal – see pump, turbine External – heat transfer from hot combustion gases to working fluid Engines

MEP=

net work for one cycl displacement volum

For 2 engines of equal disp. Vol., the one with hgiher MEP produces more work Air Standard: IG working fluid, CM, HT, no exhaust/intake, int. rev. Cold Air Standard: const. spec. hts.

Otto Cycle

Otto Cycle

1→ 2 : W 12 =u2 −u m Isentropic comp.

Otto Cycle

W cycle

2→ 3 : Q 23 =u3 −u2 m

W 34 W 12 =u3−u 4− (u2− − m W net u4 −u1 η= =1− Q 23 u3 − u2 =

Const. Vol. HT from ext. source

3→4 : W 34 =u3−u m

Compression Ratio

4→1 : Q 41 =u 4−u1 m

r=

For isentropic, cold air:

Isentropic exp.

Const. Vol. Heat rejection *sign convention rules not obeyed here

Diesel

cp T 2 k−1 =r , k= T1 cv 1 η=1− k−1 r Diesel

Diesel

1→ 2 : W 12 =u2 −u m

2→ 3 : Q 23 =h3 −h2 m W 23 3 =∫ pdv m 2

Isentropic comp.

Q 41 W cycle u 4−u1 m m =1− =1− η= Q 23 Q 23 h3−h 2 m m k r −1 v 1 η=1− k−1 c , rc= 3 v k ( r c −1 ) r 2

Area enclosed:

W cycle

=

Q cycle

(

Const. Pressure. HT from ext. source

3→4 : W 34 =u3−u m Isentropic exp.

4→1 : Q 41 =u 4−u1 m Const. Vol. Heat rejection

1→ 2 W 12 =u2 −u m

Blue:

Red:

W¿ m Q¿ m

;

Purple:

W out ; m

Green:

Q out m

2→ 3 Q 23 =u3 −u2 m

:

Isentropic comp.

3→4 : W 34 =p 3 (v 4 m Q 34 =h4−h3 m

)

Cold-Air Standard

Dual

Dual

v1 v 4 = v2 v3

:

;

( ) ( ) ( )

T2 v = 1 T1 v2

k−1

T4 v3 = T 3 v4

k−1

=r k−1 rc = r

k−1

Gas Turbine Power Plants Brayton

Const. Vol. Heat addition

4→5 : W 45 =u4 −u m Isentropic exp.

Q 51 u5 −u1 m η=1− =1− Q 23 Q 34 u 3−u2 +h 4−h3 + m m

Const. P. Heat add

5 →1 : Q 51 =u5−u 1 m Const. vol. ht. rejection Brayton

1→ 2

:

Compressor

W´ C =h2−h 1 m ´

Isentropic exp. from sat. vap to condensor pressure

Brayton (ideal)

2→ 3 ´¿ Q =h3−h2 m ´ Heat transfer from working fluid at const. P., exits as sat. liq.

Brayton

´ W´ T W − C h1 −h2−(h4 −h3 ) m ´ m ´ η= = ´Q ¿ h3−h2 ´ m ´ c/ m ´ h2−h1 W = bwr= W´ t / m ´ h3 −h4

Cold-Air

3→4 Turbine

:

4→1

W´ C ; =h3 −h4 =h4 −h m ´ m ´ m ´ Heat transfer at Isentropic comp. ´ out Q to state for in const. P. Green: comp. liq. region m ´ W´ T

´ out Q

Refrigeration/Heat Pump

1→ 2 s 2→ 3

: isentropic comp. Ht. transfer from refrigerant,

exits as liquid

3→4

: throttling to 2-phase liquid-

vapor mix

4 → 1 : HT. to refrigerant at constant p through evaporator

Red:

Yellow:

W´ T m ´

;

Blue:

´¿ Q m ´

;

T 2 =T 1 η=1−

p2 p1

( k−1 )/ k

( )

,T 4 =T 3

T1 1 =1− ( ) T2 p2 k−1 k p1

p4 p3

( )

( )

Note Brayton and Rankine the same Refrigeration

´¿ Q =h1−h4 m ´

capacity

, Qin is refrigeration

W´ C =h2−h 1 m ´ ´ Q out =h2−h3 , h3=h4 m ´ h1−h 4 Q β= ¿ = W cycle h2− h1

(k −1...