697.pdf

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여재익, [email protected], (02) 880-9334 - 2006 Spring - 1

Chapter 4 Chapter 4

• 1st Law of Thermodynamics in the Closed System

• In general, total energy E,

E = U (internal energy) + KE + PE

)

( Q dE W

W E

Q

δ

δ = +

Δ + Δ

≡ Δ

Non-perfect differential 불완전 미분 (Path dependent) Non-perfect differential 불완전 미분 (Path dependent)

Perfect differential 완전미분

) (

1 2

2

z z mg PE

m KE

−

=

= ν

• If the system is static, KE = PE = 0

• Conservation of Energy (1) For a cycle process,

W U

Q = Δ + Δ Δ

P

∫

=

+

= W Q

W dU

Q δ δ

δ δ

0

“한 Cycle 동안 System에 주어진 Net Heat는 System이 행한 Net Work와 같다.”

W

net

Q Q

₁

−

₂

=

Q1= Cycle 과정 중 System으로 준 열량.

Q2= Cycle 과정 중 System으로부터 방출된 열량.

Wnet= System이 행한 Net Work.

(2) Energy conservation in Isolated (고립) System.

Æ주위와 열이나 일을 교환하지 않는 밀폐 system Q12= 0, W12= 0

ÆE2-E1= 0 If static, U2-U1= 0 (3) Internal energy

Extensive property:

That is. it depends on the massof the system.

(4) Enthalpy: thermodynamic property

Defined total enthalpy

Or, per unit mass

specific enthalpy

For saturated steam, we know that specific volume specific internal energy specific enthalpy Look up of steam table

pV U H = +

ν p u h = +

One reason for introducing enthalpy at this time is that although the steam tables list values for internal energy, many other tables and charts of thermodynamic properties give values for enthalpy but not for the internal energy.

g

f x

xν ν ν =(1− ) +

g

f xu

u x u=(1− ) +

g

f xh

h x h=(1− ) +

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• Example 1

A cylinder fitted with a piston has a volume of 0.1m³and contains 0.5 kg of steam at 0.4 Mpa. Heat is transferred to the system until the temperature is 300°C, while the pressure remains constant.

Find

W U Q=Δ +Δ Δ

) (

1 1 2 2

1 2 2

1 2

12 1

ν ν p p m

V V p dV p dV p W

−

=

−

=

Δ ∫ ∫

T

(Method 1) Consider the First Law,

(Method 2)

From steam table, internal energy )

(

) (

1 2

1 1 2 2 1 2

12 1 2 12

h h m

p p m u u m

W u u m Q

−

=

− +

−

=

Δ +

−

= Δ

ν ν

Since p = constant Since p = constant

8 .

2 2804

1 1

= +

= u

u x u

u _f _fg

kJ u

u m

W U U Q

1 . 771 0 . 91 ) ( ₂ ₁

12 1 2 12

= +

−

=

Δ +

−

= Δ

Steam table look up, enthalpy Steam table look up, enthalpy

From last time … From last time…

• Recall the 1st Law of Thermodynamics from last time,

With assumptions:

a simple compressible substance & quasi-equilibrium process.

We find that this expression can be evaluated for two separate cases.

(1) Constant Volume

The specific heat ≡The amount of heat required per unit mass to raise the temperature by 1 degree.

(2) Constant Pressure, for constant p.

pdV dU

W dU Q

+

= +

= δ

δ Neglect KE, PE changeNeglect KE, PE change

v v v

v T

u T U m T Q

c m 1( ) ( )

) 1(

∂

= ∂

∂

= ∂

= δ

δ

)

(H U pV

dH

Q= = +

δ

p p p

p T

h T H m T Q

c m 1( ) ( )

) 1(

∂

= ∂

∂

= ∂

= δ

δ

• Some comments about Internal energy, Enthalpy and specific heat of ideal gas.

RT p ν =

) (T f u =

For an ideal gas For an ideal gas, the internal energy is:

We will explain this later We will explain this later We had

v

T

c ( u )

∂

= ∂

But since “u”But since “u”depends only on Tdepends only on T

dT c

_v

= du

or du=c_vdT ……….(1)

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From the definition of enthalpy,

for an ideal gas

Now, from , we can also write

Or ---(2)

) (T f h =

p

T

c ( h )

∂

= ∂

dT c

_p

= dh

dT c dh =

_p

Pressure-Volume (p-v) diagram for ideal gas

Since u(T) and h(T), these are also lines of constant u or h or T.

For all processes 1Æ2, 1Æ2’, 1Æ2’’,

The changes in internal energy and enthalpy are the same.

Since u(T) and h(T), these are also lines of constant u or h or T.

For all processes 1Æ2, 1Æ2’, 1Æ2’’,

The changes in internal energy and enthalpy are the same.

• Further, for ideal gasc_v = f(T),c_p = f(T) RdT

du dh= +

T d R T d c dT

c_p = _v + R c c_p− _v =

R c_v

1 1

= −

γ ^v

p

c

=c γ

R c_p

−1

=γ γ Æ

This says, the difference between the constant pressure and constant volume specific heats of

an ideal gas is always constant.

This says, the difference between the constant pressure and constant volume specific heats of

an ideal gas is always constant.

Æ Using

Diatomic gas (N2, O2, CO, etc)

9/7=1.29 9/2 R

7/2 R 극고온

7/5=1.4 7/2 R

5/2 R 상온

General

5/3=1.67 5/2 R

3/2 R 극저온

5/3=1.67 5/2 R

3/2 R Monatomic gas

(He, Ar, etc)

γ=cp/cv

cp

cv

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• Adiabatic System

pdv du or

pdV dU Q

+

= +

=

= 0 0

δ 0 Where (ideal gas), du=c_vdT ν p=RT

so + ν =0

ν ^d dT RT c_v

=0

+ ν

νc ^d R T dT

v

; c_v R

1 1

= −

γ

Take the integral,

C v T

or

C c v

T R

v

=

− +

= +

ln ) 1 ( ln

ln ln

γ

Then for adiabatic ( ) process, we findδQ=0

1

C

Tv

^γ⁻

=

^;

v p = RT

or

v

¹

C

₁

R

pv

^γ⁻

=

^;

pv

^γ

= C

2

or

3 1

C Tp =

− − γ γ

or

γ γ γ

γ

1

1 2 1 2

2 1 1 2

1 1 2 1 2

) (

) ( ) (

−

=

p p T T

v v p p

v v T T

For ideal gas in Adiabatic System For ideal gas in Adiabatic System

• In general, for polytropic gas (i.e. ) Consider

n=0 : Isobaric Process (등압 과정)

n= : Isochoric Process (등적 과정)

γ>1 n=γ : Adiabatic process (단열 과정) n=1 : Isothermal process (등온 과정)

c pv

ⁿ

= c

pv

ⁿ

=

∞

±

p

v

n V p V p n

v p v m p

n v cm v

v cm dv

v dv m c

dv p m dV p W

n n n n

−

= −

−

= −

−

= −

=

−

∫

1 1

) (

) 1 (

) (

1 1 2 2 1 1 2 2

1 1 1 2 2 1 2 1

2 1 2

12 1

Note,

c v p v p

c pv

n n

n

=

2 2 1 1

If ideal gas,

n T T W mR

−

= − 1

) (

₂ ₁

12

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• Homework Set #3

• due 3/30

• 4-1, 4-3, 4-6, 4-13, 4-20, 4-23, 4-26, 4-28