5-1 Reference State

T(℃) 0 100 500

H(J/mol) 0 2919 15060

ˆ

CO(g, 0℃. 1atm) CO(g, 100℃. 1atm)

5. 열역학 데이터 표

CO enthalpy change

molJHHH CC /919,20^

100^^

=−=∆ °° , reference state : o℃ , 1 atm

- : 절대값 측정 불가

- : 측정 가능 UΔHΔ ˆ,ˆUH ˆ,ˆ

State property (상태특성)

• A property of a system component whose value depends only on the state of the system such as T, P, mole fractions.

• A state property does not depend on how the system reached that state.

• 예 :

UH ˆ,ˆ

•예제 7.5-1 : 다음 쪽

Example 7.5-1 Use of Tabulated Enthalpy Data

• The following entries taken from a data table for saturated methyl chloride:

1. What reference state was used to generate the given enthalpies? 2. Calculate and for the transition of saturated methyl chloride vapor from

50~0 ˚F . 3. What assumption did you make in solving question 2 regarding the effect of

pressure on specific enthalpy?

Ĥ∆ Û∆

SOLUTION 1. Liquid at –40˚F and 6.878 psia.

2.

3. was assumed independent of P.

0ˆ =H

m

3m

3m

initialinitialfinalfinal

m

Btu/lb4.96psiaft10.73

Btu1.987psia/lbft920)](51.99)(1..969)(4[(18.90)

Btu/lb6.05)ˆˆ(ˆˆˆˆ

Btu/lb05.6)28.20223.196( ) F50(ˆ－ ) F0(ˆˆ

−=⋅

⋅−−

−=−−∆=∆−∆=∆

−=−==∆

VPVPHVPHU

H HH

Ĥ

5-2 Steam Tables

•Subcooled liquid : VLE(vapor-liquid equilibrium curve) 위 부분 •Saturated liquid/ saturated steam/both mixture : VLE 선상 •Superheated steam : VLE 아래 부분

• Reference state of water :

- 0.01℃ and 0.00611 bar (triple point)

- Û ≡ 0

•Tables B.5 : 온도에 따른 액상 물과 포화수증기의

•Tables B.6 : 압력에 따른 액상 물과 포화수증기의

•Tables B.7 : 온도와 압력에 따른 과열수증기의

•예제 7.5-2 : 스팀 도표 이용 연습, 다음 쪽

UHV ˆ,ˆ,ˆ

UHV ˆ,ˆ,ˆ

UHV ˆ,ˆ,ˆ

1. Determine the vapor pressure , specific internal energy, and specific enthalpy of saturated steam 133.5℃ .

2. Show that water at 400℃ and 10 bar is superheated steam and determine its specific volume, specific internal energy, and specific enthalpy relative to liquid water at the triple point, and its dew point.

3. Show that for superheated steam depend strongly on temperature and relatively slightly on pressure.

Example 7.5-2 The Steam Tables

HU ˆˆ and

1. From table B.6

2. From table B.7, [T=400℃, P = 10 bar]

3. At constant P, T : 400℃→450℃ , change by about 3%. ( )

at constant T(400℃), P : 10 →20 bar, change by much less than 1%.

SOLUTION

kJ/kg7.2724ˆ,kJ/kg0.2543ˆ,/kgm606.0ˆ,bar0.3 3* ==== HUVp

℃TUVH 9.179,kJ/kg2958ˆ,/kgm307.0ˆ kJ/kg,3264ˆ dp3 ====

HU ˆˆ andkJ/kg30412958:ˆkJ/kg,33713264:ˆ →→ UH

HU ˆˆ and

예제7.5-3) Energy Balance on s Steam Turbine Steam at 10 bar absolute with 190℃ of superheat is

fed to a turbine at a rate m = 2000 kg/h. The turbine operation is adiabatic, and the effluent

is saturated steam at 1 bar. Calculate the work output of the turbine in kilowatts.

Neglecting kinetic and potential energy changes.

Solution) Energy Balance on s Steam Turbine 주입 과열 스팀의 온도=10 bar에서의 포화온도+과열온도 =180+190 =370℃ 주입 과열 스팀의 비엔탈피(10 bar, 370℃) = 3201 kJ/kg 배출 스팀(1bar, saturated) = 2675 kJ/kg 에너지 수지식에 대입: = - (2000 kg/h)(2675-3201)(kJ/kg)(1 h/3600 s) = 292 kJ/s = 292 kW

)(^^

inoutS HHmHW −−=∆−=

)(^^

inoutS HHmHW −−=∆−=

물의 상태를 흐름도에 표기 : 액상(l), 기상(v) Determine the flow rates of all stream components

Determine the specific enthalpies of each stream component

Write the appropriate form of the energy balance equation

6. 에너지수지 진행순서

예제 7.6-1 : Energy Balance on a One-Component Process: 주입열량 구하기

120 kg H2O(l)/min

30℃, H=125.7 kJ/kg ^

175 kg H2O(l)/min

30℃, H=271.9 kJ/kg ^

295 kg H2O( )/min

17 bar saturated

H = 2793 kJ/kg

6 cm ID pipe

^

Heat Q (kJ/min)

υ

Q = ∆H + ∆Ek

∑ ∑−=∆out in

i

^

ii

^

i HmHm H

= (295 kg/min)(2793 kJ/kg)-(120 kg/min)(125.7 kJ/kg)- (175 kg/min)(271.9 kJ/kg)

=7.61 ·105 kJ/min

1.

2. kE∆

•Table B.6 : 포화증기 비용(17 bar) = 0.1166 m3/kg

•6 cm ID 단면적= =(3.14)(32 cm2)(1 m2/104 cm2)

= 2.83 · 10-3 m2

•스팀유속, u(m/s) =

2Rπ

)(/)/( 23 mAsmV

=(295 kg/min)(1 min/60 s)(0.1166 m3/kg)/(2.83x10-3 m2 ) = 202 m/s • 운동에너지 : 주입 부분=무시 = {(296 kg/min)/2}(2022 m2/s2)(1 N/1 kg•m/s2)(1 kJ/103 Nm) = 6.02 •103 kJ/min • 주입 열량 계산 : = 7.61 •105 kJ/min + 6.02 •103 kJ/min = 7.67 •105 kJ/min • 운동 에너지 : 전체 열량 중 0.8% 비중 차지 — 많은 온도 차, 상변화, 화학반응 시 무시 가능

2um 2 /EE outlet k,k ==∆

Q = ∆H + ∆Ek

• 공정 흐름 상 여러 성분 포함 경우 : 성분 별 비엔탈피 구함. • 분자 구조가 유사한 기체, 액체 혼합물 : 주어진 온도와 압력 하의 순수한 성분의 비엔탈피 사용 가능.

• 에너지 수지와 물질 수지를 동시에 해결해야 될 문제 많음.

• A gas stream (60.0wt% ethane and 40.0wt% n-butane) is to be heated from 150K to 250K at a pressure of 5 bar.

Calculate the required heat input per kilogram of the

mixture, neglecting potential and kinetic energy changes, using tabulated enthalpy data for ethane and n-butane and assuming ideal gas behavior.

Example 7.6-2 Energy Balance on a Two-Component Process

• Basis : 1kg/s Mixture

• Energy balance

SOLUTION

0,0,0 pkS

pkS

=∆=∆=∆=⇒

∆+∆+∆=−

EEWHQ

EEHWQ

kgkJ478

kJ/s00.1kJ/s478kJ/s478kJ/s)]0.30)(400.0()3.314)(600.0[(

kgskJ0.237400.0

kgsJk3.973600.0

ˆˆ

10462

componentsinlet

componentsoutlet

=⇒=+−

+=

−=∆= ∑∑

HCkgHCkg

HmHmHQ iiii

• 예제 7.6-3 : 과열 수증기 m2량은? 필요한 m1의 부피유속은?

^

m1 kg H2O(v)/hr, 1atm

400℃, H=3278 kJ/kg

superheated

^

m2 kg H2O( v )/hr

300℃, 1 atm superheated

H = 3074 kJ/kg ^

1150 kg H2O(v)/hr, 1 atm

100℃, H=2676 kJ/kg

saturated

• 에너지 수지와 물질 수지를 동시에 해결해야 될 문제 많음.

• 수증기 량 물질 수지 : 1150 + m1 =m2 (1) • 에너지 수지 :

– 왜냐하면 1) 단열, 2) 샤프트 일=0, 3) 운동, 위치에너지 =0

– 물질량 계산 :

• (1150 kg/h)(2676 kJ/kg) + m1(3278 kJ/kg) = m2(3074 kJ/kg) (2) • 식 (1), (2)를 연립 방정식으로 풀이: • m1 = 2240 kg/h , m2 = 3390 kg/h

– m1의 부피 유속 계산:

• 400 도, 1 기압(1 bar 값과 유사) 상태의 비용 = 3.11 m3/kg • (2240 kg/h)(3.11 m3/kg) = 6980 m3/h

∑ ∑−=out in

i^

ii^

i HmHm HΔ

7. 기계적 에너지수지

Incompressible fluid : 식 7.4-12 식의 변형

Mechanical energy balance

F : friction loss

Negligible potential and kinetic energy : 상변화, 조성변화, 온도차가 큰 경우

Q = ∆ U (closed system) Q = ∆ H (open system)

mWmQUzgg

g2uP

scc

2

/)/(^

−=−∆+∆+∆

+∆ρ

mW

Fzgg

g2uP s

cc

2

−=+∆+

∆+

∆ ˆρ

Bernoulli equation

0zgg

g2uP

cc

2

=∆+∆

+∆ρ

예제 7.7-1) Bernoulli equation

Position 1

Position 2 1 cm ID P2 = 1 atm z2 = 50 m

0.5 cm ID 20 L H2O/min P1 = ? z1 = 0

mzzzsmgmkg

smuuu

smu

smsm

cmcmL

mLu

50/81.9,/1000

/271

/24.4

,/1760min110

)25.0(1

101

min20

12

23

2221

22

22

2

24

223

3

1

=−=∆==

−=−=∆

=

==

ρ

π

Bernoulli Equation 적용:

0Nsmkg1

mzsmgNsmkg12

smumkgmNP

2

2

2

222

3

2

=⋅

∆+

⋅⋅∆

+∆

]/)/[()()/(

]/)/[()/(

)/()/(

ρ

주어진 값 대입:

barPa

mNatmPmNP

kgmNkgmNmkg

PP

56.41056.4

)/1001325.11(/1056.4

0/490/5.135/1000

5

252

251

312

=×=

×==×=

=⋅+⋅−−

예제 7.7-2) Siphoning

- friction loss = 0.80 ft-lbf / lbm

- 5 gal gasoline 이송에 필요한 시간?

mW

Fzgg

g2uP s

cc

2

−=+∆+

∆+

∆ ˆρ

•압력차=0, g = 32.174 ft/s2, 높이차= -2.5 ft, 샤프트일=0

•마찰손실 = 0.80 ft-lbf / lbm , 22

2 uu ≈∆

•u2 구하기:

0/80.02

2

2

22

/174.32

)1)(5.2)(/174.32(

/174.322

)1( =+−+•

mfsftlb

lbftsft

sftlb

lbu lblbftm

f

m

f

u2 = 10.5 ft/s

•부피 유속 및 이송 시간 구하기 :

•부피유속=(10.5 ft/s)(3.124)(0.1252 in2)(1 ft2/144 in2)

= 3.58•10-3 ft3/s

•이송시간=(5 gal)(0.1337 ft3/gal)/(3.58•10-3 ft3/s)=187 s=3.1 min

• Water flows from an elevated reservoir through a conduit a turbine at a lower level and out of the turbine through a similar conduit. At a point 100m above the turbine the pressure is 207 kPa, and at a point 3m below the turbine the pressure is 124 kPa. What must the water flow rate be if the turbine output is 1.00 MW?

Example 7.7-3 Hydraulic Power Generation

• From Equation 7.7-2, ① = ②

SOLUTION

kg/s915m/kgN)101083(m/sN1000.1

m/kgN1010m/skg1s

N1m103m9.81m103

m/s9.81

/83m/skg1kg/m101.00

N1/mN1083

N/1083kPa83kPa207)(124

m/sN101.00MW1.00

g

g

6

22

2

233

23

23

6

=⋅−−⋅×−

=

⋅−=⋅

−=∆

−=∆=

⋅−=⋅×

×−=

∆

×−=−=−=∆

⋅×==

∆+∆

−=⇒

−=∆+

∆

m

zg

zg

kgmNP

mP

W

zPWm

mWzP

s

s

s

ρ

ρ

ρ

•http://cis.kepco.co.kr/cis/customer/electric_info/statistical_sum.jsp •한국전력공사, 전력통계

•단위: MW

Key notation

Q : heat

Ws : shaft work

Wf : flow work

Ek : kinetic energy

Ep : potential energy

U : internal energy

H : enthalpy

4절 학습효과 증진을 위한 학습방안 및 활용

문제파악을 위해 흐름도를 작성함.

흐름도에 모든 산술자료를 표시함.

에너지입량, 에너지출량, 일, 열 등의 에너지를 분류함.

각 변수 단위의 일관성을 유지함.