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Chap4 Closed System

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1 OBJECTIVES DESIGNED AND PREPARED BY : MOHD KAMAL ARIFFIN/2003 CHAPTER 4 : ENERGY ANALYSIS OF CLOSED SYSTEMS DESIGNED AND PREPARED BY : MOHD KAMAL ARIFFIN/2003 BOUNDARY WORK 1 2 One form of mechanical work frequently encountered in practice is associated with the expansion or compression of a gas in a piston-cylinder device. Example : Automotive engines and air compressors The real piston moves at very high speeds, therefore, the boundary work in real engines or compressor/engines is determined by direct measurements For our analysis, the process is assumed as a quasi-static process, the piston moves at low velocities. V 1 p V V 2 p 1 p 2 1 2 p dV dA=pdV P is the initial pressure of the gas V is the total cylinder volume A is the cross-sectional areaof piston The differential work done during the process δW b = Fds = pAds = pdV Thus, the total boundary work done, kJ pdV W W 2 1 12 b = = The total work done during the process is equal to the area under the process curve on a p-V diagram ds CHAPTER 4 : ENERGY ANALYSIS OF CLOSED SYSTEMS
Transcript
Page 1: Chap4 Closed System

1

OBJECTIVES

DESIGNED AND PREPARED BY : MOHD KAMAL ARIFFIN/2003

CHAPTER 4 : ENERGY ANALYSIS OF CLOSED SYSTEMS

DESIGNED AND PREPARED BY : MOHD KAMAL ARIFFIN/2003

BOUNDARY WORK

1 2

• One form of mechanical work frequently encountered in practice is associated with the expansion or compression of a gas in a piston-cylinder device.

• Example : Automotive engines and air compressors• The real piston moves at very high speeds, therefore, the

boundary work in real engines or compressor/engines is determined by direct measurements

• For our analysis, the process is assumed as a quasi-static process, the piston moves at low velocities.

V1

p

VV2

p1

p2

1

2

p

dV

dA=pdV

P is the initial pressure of the gasV is the total cylinder volumeA is the cross-sectional areaof piston

The differential work done during the processδWb = Fds = pAds = pdV

Thus, the total boundary work done,

kJ pdVWW 2112b ∫==

The total work done during the process is equal to the area under the process curve on a p-V diagram

ds

CHAPTER 4 : ENERGY ANALYSIS OF CLOSED SYSTEMS

Page 2: Chap4 Closed System

2

DESIGNED AND PREPARED BY : MOHD KAMAL ARIFFIN/2003

MY EXAMPLE 4-1A gas in a closed tank undergone an expansion process according to pV1.3

= constant. Derive the boundary work for the process.

Substituting p = c/V1.3 into

∫=2

1 1.312 VCdV W

0.3VpVp

11.3

VVpVVp

2211

11.31

1.311

11.32

1.322

−=

+−−

=+−+−

2

2

13.1

13.1V C ⎥

⎤⎢⎣

⎡+−

=+−

∫=2

112 pdV W

CHAPTER 4 : ENERGY ANALYSIS OF CLOSED SYSTEMS

Solution

DESIGNED AND PREPARED BY : MOHD KAMAL ARIFFIN/2003

MY EXAMPLE 4-2A working fluid undergone a process according to the function of p = 2V2 – V. The initial and the final volume are 0.5 m3 and 0.05 m3 respectively. Determine the work done during the process.

∫=2

112 pdV W

2

1

23

2V

3V2

⎥⎦

⎤⎢⎣

⎡−= ⎟⎟

⎞⎜⎜⎝

⎛−−⎟⎟

⎞⎜⎜⎝

⎛−=

2V

3V2

2V

3V2 2

131

22

32

( ) ( ) ( ) ( ) kJ -1.1325.0

35.02

205.0

305.02W

2323

12 =⎟⎟⎠

⎞⎜⎜⎝

⎛−−⎟⎟

⎞⎜⎜⎝

⎛−=

Substituting V1 = 0.5 m3 and V2 = 0.05 m3, thus

( )dV V2V W 21

212 ∫ −=

CHAPTER 4 : ENERGY ANALYSIS OF CLOSED SYSTEMS

Solution

Substituting p = 2V2 – V into

Page 3: Chap4 Closed System

3

DESIGNED AND PREPARED BY : MOHD KAMAL ARIFFIN/2003

BOUNDARY WORK FOR A CONSTANT VOLUME PROCESS

Normally this process occurs when a working fluid is contained in a rigid tank which has a fixed boundary.

Since the volume is constant, dV = 0, therefore, there in no boundary work done during this process (the area under the process curve is zero)

p2

p1

υ1 = υ2

1

2

Area = W12 = 0

∫=2

1

v

v 12 pdVW

0

= 0 υ

p

CHAPTER 4 : ENERGY ANALYSIS OF CLOSED SYSTEMS

DESIGNED AND PREPARED BY : MOHD KAMAL ARIFFIN/2003

EXAMPLE 4-1CHAPTER 4 : ENERGY ANALYSIS OF CLOSED SYSTEMS

Page 4: Chap4 Closed System

4

DESIGNED AND PREPARED BY : MOHD KAMAL ARIFFIN/2003

BOUNDARY WORK FOR A CONSTANT PRESSURE PROCESS

υ2

21p1= p2

υ1

W12 = Area = p1(V2 – V1)

∫=2

1

v

v 12 pdVW ∫=

2

1

v

v dVp ( )12 VVp −=

υ

p

Substituting an ideal gas equation of state, pV = mRT, then for ideal gas,

( )1212 TTmRW −=

The volume of the system increases during constant pressure expansion (+ve W) and decreases during constant volume compression (-ve W). Using boundary work definition, ••

CHAPTER 4 : ENERGY ANALYSIS OF CLOSED SYSTEMS

DESIGNED AND PREPARED BY : MOHD KAMAL ARIFFIN/2003

MY EXAMPLE 4-3CHAPTER 4 : ENERGY ANALYSIS OF CLOSED SYSTEMS

15 kg of water in a piston-cylinder assembly, initially at 10 bar and 350 oC. The water is now cooled at constant pressure to a saturated vapor. Sketch the process on a p- υ and T- υ and determine the work done during the process.

At 10 bar (1000 kPa) Ts =179.91 oC, T>TS shs T1 = 350 oC

10

TS = 179.9 oC12

p = 10 bar

179.9

1

2

350

W12 = mp(υ2 -υ1 )

The water at state 2 is saturated vapor, thus

υ2 = υg pada 10 bar = 0.19436 m3/kg.

= (15)(10 x 102)(0.1936 – 0.2825) = -1321.5 kJ

υ

p

υ

T

From Table A-5 at 10 bar (1 MPa) and 350 oC,

υ1 = 0.2825 m3/kg

Solution

Page 5: Chap4 Closed System

5

A piston-cylinder system contains an air initially at 200 kPa and 30 oC. At this condition, the piston is resting on a set of stops with a cylinder volume of 400 litre. A pressure of 400 kPa is needed to move the piston. The air is then heated until the volume inside the cylinder doubles. By using ideal gas model, determine the final temperature and the total work done during the processes. For the air take cv = 0.718 kJ/kgK and R = 0.287 kJ/kgK

DESIGNED AND PREPARED BY : MOHD KAMAL ARIFFIN/2003

MY EXAMPLE 4-4

υ (m3/kg)

p(bar)

••

400

200

V1= V2 V3

1

2 3

υ =c

p =c

T1 = 30 oC

T2T3

3

33

2

22

1

11TVp

TVp

TVp

==

11

1333 Vp

TVpT =

Wtotal= W12 + W23

kJ 1601000400

1000400 x 2 x 400 =⎥⎦

⎤⎢⎣

⎡−⎟

⎠⎞

⎜⎝⎛=

Using

= p(V3 - V2)

( )( )( )( )( )K 1212

V200273302V400

1

1

=

+=

V3 =2V2 and (V1 = V2)

0

CHAPTER 4 : ENERGY ANALYSIS OF CLOSED SYSTEMS

Solution

Air200 kPa

30oC

Piston

Cylinder

Stops

DESIGNED AND PREPARED BY : MOHD KAMAL ARIFFIN/2003

BOUNDARY WORK FOR A CONSTANT TEMPERATURE PROCESS (ISOTHERMAL)

To maintain the temperature of a system, heat has to be supplied continuously during isothermal expansion and heat has to be rejected during isothermal compression.

From ideal gas equation, pV = mRT : the value of mR is constant and for isothermal process, T is constant, thus pV = constant = C

p1V1 = p2V2 = p3V3 = .............= pnVn

∫=2

1

v

v12 pdVW

⎟⎟⎠

⎞⎜⎜⎝

⎛=

1

2

VV lnpV

p2

υ2υ1

p1

2

1

pυ = C

pV = C p = C/V

∫=2

1

v

v12 V

dVCW

⎟⎟⎠

⎞⎜⎜⎝

⎛=

2

1

pp lnpV

⎟⎟⎠

⎞⎜⎜⎝

⎛=

1

212 V

V ln RTmW

Substituting pV = mRT

2

1

vv V ]ln[ C=

Area

υ

p

CHAPTER 4 : ENERGY ANALYSIS OF CLOSED SYSTEMS

Page 6: Chap4 Closed System

6

DESIGNED AND PREPARED BY : MOHD KAMAL ARIFFIN/2003

EXAMPLE 4-3CHAPTER 4 : ENERGY ANALYSIS OF CLOSED SYSTEMS

0.5 kg of gas (R = 0.1890 kJ/kgK) initially at 1.5 bar and 20 oC is contained in a closed tank. Gas is now compressed isothermally to a pressure of 15 bar. By assuming the gas is an ideal gas, determine the work done during the process.

DESIGNED AND PREPARED BY : MOHD KAMAL ARIFFIN/2003

MY EXAMPLE 4-5

⎟⎟⎠

⎞⎜⎜⎝

⎛=

2

112 p

p ln mRTW ( )( )( ) kJ 63.76 -151.5 ln 273200.189050. =⎟

⎠⎞

⎜⎝⎛+=

( )( ) kJ 76.63 1846.0

01846.0ln01846.010 x 15VV ln VpW 2

1

22212 −=⎟

⎠⎞

⎜⎝⎛=⎟⎟

⎞⎜⎜⎝

⎛=

32

11 m 1846.0

10 x 5.1)27320)(1890.0)(5.0(

pmRTV =

+==

32

22 m 01846.0

10 x 15)27320)(1890.0)(5.0(

pmRTV =

+==

m = 0.5 kgp1 = 1.5 barT1 = 20 oC

p2 = 15 barT2 = T1 = 20 oC

Compression

pV = C (T = C)

or ⎟⎟⎠

⎞⎜⎜⎝

⎛=

1

22212 V

V ln VpW

Solution

CHAPTER 4 : ENERGY ANALYSIS OF CLOSED SYSTEMS

Page 7: Chap4 Closed System

7

BOUNDARY WORK FOR A POLYTROPIC PROCESS

DESIGNED AND PREPARED BY : MOHD KAMAL ARIFFIN/2003

A polytropic process is process which the pressure and volume of the system are related by pV n = Constant where n is polytropic index.

∫= 2

1

V

V n12 VdV CW

2

1

V

V

1+n-

1+n-V C ⎥

⎤⎢⎣

⎡= ⎥

⎤⎢⎣

⎡=

1-n V- V pV

1+n-2

1+n-1n kJ

1-nVp - Vp 2211=

( ) kJ 1-n

T - TmR W 2112 =

p2

υ2υ1

p1

2

1

pυn = Constant

nVCp =

Substituting pV = mRT, for an ideal gas

Area

Or per unit mass, kJ/kg 1-np - p w 2211

12υυ

=

Or per unit mass,

( ) kJ/kg 1-nT - TR w 21

12 =

υ

p

CHAPTER 4 : ENERGY ANALYSIS OF CLOSED SYSTEMS

p-v-T RELATIONS FOR POLYTROPIC PROCESS

DESIGNED AND PREPARED BY : MOHD KAMAL ARIFFIN/2003

pυn = C

n

2

1

1

2

pp

⎟⎟⎠

⎞⎜⎜⎝

⎛υυ

=

p = p n22

n11 υυ

υRT = p

CRT n =υ⎟⎠⎞

⎜⎝⎛υ

CRT 1n =υ −

1n

2

1

1

2

TT

⎟⎟⎠

⎞⎜⎜⎝

⎛υυ

=

pRT = υ

Cp

RTpn

=⎟⎠

⎞⎜⎝

⎛ CpRT

1n

n=−

n1n

1

2

1

2

pp

TT

⎟⎟⎠

⎞⎜⎜⎝

⎛=

1n22

1n11 TT −− υ=υ 1n

2

n2

1n1

n1

pT

pT

−− =

CHAPTER 4 : ENERGY ANALYSIS OF CLOSED SYSTEMS

Page 8: Chap4 Closed System

8

EXAMPLE 4-4

DESIGNED AND PREPARED BY : MOHD KAMAL ARIFFIN/2003

CHAPTER 4 : ENERGY ANALYSIS OF CLOSED SYSTEMS

A piston-cylinder device contains 0.05 m3 of a gas initially at 200 kPa. At this state, a linear spring that has a spring constant of 150 kN/m is touching the piston but exerting no force on it. Now heat is transferred to the gas, causing the piston to rise and to compress the spring until the volume inside the cylinder doubles. If the cross-sectional area of the piston is 0.25 m2, determine

a) the final pressure inside the cylinder b) the total work done by the gasc) The fraction of this work done against the spring to compress it

Solution

EXAMPLE 4-4 cont..

DESIGNED AND PREPARED BY : MOHD KAMAL ARIFFIN/2003

CHAPTER 4 : ENERGY ANALYSIS OF CLOSED SYSTEMS

Page 9: Chap4 Closed System

9

kg/x520 2.1

13

11.2

1

2

12 m 0.3540 0.1115 x

pp

=⎟⎠⎞

⎜⎝⎛=⎟⎟

⎞⎜⎜⎝

⎛= υυ

1.5 kg of a pure substance initially at 20 bar and 250 oC undergoes a polytropic expansion process until the pressure is 5 bar. The polytropic index is 1.2. Determine the work done if the pure substance is (a) water and (b) air (R = 0.287 kJ/kgK)

DESIGNED AND PREPARED BY : MOHD KAMAL ARIFFIN/2003

MY EXAMPLE 4-6

m = 1.5 kgp1 = 20 barT1 = 250 oC

p2 = 5 barExpansion

pV1.2 = CW12 = ?

( )1-np - pm

1-nVp - Vp W 22112211

12υυ

==[ ]

kJ 345.011.2

)(0.3540)10 x (5)(0.1115)10 x (20(1.5) 22

=−−

=

At 20 bar, Ts = 212.38 oC, T1>Ts shs υ1 = 0.1115 m3/kg20 bar250 oC

a)

b) /kgm 0.075110 x 20

273) 0(0.287)(25p

RT 32

1

1 =+

==1υ

kg/31.21

11.2

1

2

12 m 0.2384 0.0751 x

520 x

pp

=⎟⎠⎞

⎜⎝⎛=⎟⎟

⎞⎜⎜⎝

⎛= υυ

( ) [ ] kJ 232.511.2

)(0.2384)10 x (5)(0.0751)10 x (20(1.5)1-np - pm W

222211

12 =−−

==υυ

n

2

1

1

2

pp

⎟⎟⎠

⎞⎜⎜⎝

⎛υυ

=

CHAPTER 4 : ENERGY ANALYSIS OF CLOSED SYSTEMS

n

2

1

1

2pp

⎟⎟⎠

⎞⎜⎜⎝

⎛=

υυ

INTERNAL ENERGY, ENTHALPY AND SPECIFIC HEAT

DESIGNED AND PREPARED BY : MOHD KAMAL ARIFFIN/2003

The specific heat of a substance can be defined as the energy required to raise the temperature of a unit mass of a substance by one degree

The values of u and h are depend on temperature of the system ( function of T), thus

dTducv =

dTdhcp =and

du = cv dT and dh = cp dT

integrating

Specific heat terms

Specific Heat atConstant Volume

Tucv δδ

=

Specific Heat atConstant Pressure

Thcp δδ

=

u2 – u1 = cv (T2 – T1) h2 – h1 = cp (T2 – T1)

CHAPTER 4 : ENERGY ANALYSIS OF CLOSED SYSTEMS

Page 10: Chap4 Closed System

10

cp - cv = Rcp = cv + R

INTERNAL ENERGY, ENTHALPI AND SPECIFIC HEAT

DESIGNED AND PREPARED BY : MOHD KAMAL ARIFFIN/2003

From definition of enthalpy and the ideal gas equation of state, h = u + pυ and pυ = RT, we can write h = u + RT or in differential forms

RdTdu

dTdh

+= or

Let introduce another ideal gas property called the SPECIFIC HEAT RATIO, k, defined as

cp cv

v

p

cc

k ratio,heat Specific =

vv

v

v

pcR

cc

cc

=−pp

v

p

pcR

cc

cc

=−

vcR1k =−

pcR

k11 =−

1kRcv −

=1k

kRcp −=

k1/k

cp - cv = R cp - cv = R

Deviding by cv Deviding by cp

CHAPTER 4 : ENERGY ANALYSIS OF CLOSED SYSTEMS

dh = du +pdv and dh = du + RdT or

ENERGY BALANCE FOR CLOSED SYSTEM

DESIGNED AND PREPARED BY : MOHD KAMAL ARIFFIN/2003

Change in total properties is zero

From the first law of thermodynamics, energy balance for closed system can be expressed as

Q

W

SystemdE

E = U+KE+PE

δQ - δW = dE dU + dKE + dPE = δQ - δWdU = δQ - δWU2 - U1 = Q12 - W12

m(u2 – u1) = Q12 – W12 u2- u1 = q12- w12

∫∫ δδ W = QFor cyclicProcess

Total energy entering the system

Ein

Total energyLeaving the system

Eout

Change in the totalEnergy of the system

∆Esystem

- =

dU = 0

CHAPTER 4 : ENERGY ANALYSIS OF CLOSED SYSTEMS

In the rate form, kW dtdEEE systemoutin =− &&

Page 11: Chap4 Closed System

11

MY EXAMPLE 4-7

DESIGNED AND PREPARED BY : MOHD KAMAL ARIFFIN/2003

A closed system undergoes a cyclic process through paths 1-2-1. During the process 1-2 through path A, the system produces 50kJ of work and at the same time rejects 20 kJ of heat. While during the process 2-1 through path B the system is supplied with 30 kJ of work. If the energy of the system at state 1, E1 is 35 kJ, determine the heat transfer during the process 2-1, Q21.

For process A, E2 - E1 = Q12 - W12

E2 = Q12 - W12 + E1 = -20 - (50) + 35= -35 kJ

Q21 = E1 - E2 + W21 = 35 - (-35) + (-30)= 40 kJ

For process B E1 - E2 = Q21 - W21

Q21 = W12 + W21 - Q12 = 50 + (-30) - (-20)= 40 kJ

Q12 + Q21 = W12 + W21

Or by considering a cyclic process1

A

B

W12 = 50 kJQ12 = -20 kJ

W21 = -30 kJQ21 = ?

E1 = 35 kJ

2

Prop I

PROP

II

CHAPTER 4 : ENERGY ANALYSIS OF CLOSED SYSTEMS

DESIGNED AND PREPARED BY : MOHD KAMAL ARIFFIN/2003

MY EXAMPLE 4-8An electric heater with a rating of 240 V and 10 A is used to heat a fluid in a rigid tank for 15 second. The fluid is stirred by a paddle wheel with a work amount to 20 kJ. During this process the internal energy of the fluid increases in the amount of 30 kJ. Determine the amount of heat transfer during the process.

Since no change in volume (regid tank), the boundary work is zero. Thus, the work involve during the process are the work by the heater and the paddle wheel.

∆U = Q – (Wb + We + Wpaddle)

Electric heater work,We = VI x t = = 36 kJ

Electric Motor

Electric heater

Paddle wheel

Q = ∆U + (We + Wpaddle)= 30 + (-36 – 20) = -26 kJ

0

Solution

CHAPTER 4 : ENERGY ANALYSIS OF CLOSED SYSTEMS

Page 12: Chap4 Closed System

12

DESIGNED AND PREPARED BY : MOHD KAMAL ARIFFIN/2003

A completely insulated closed system with an elastic wall contains an air at a temperature of 30 oC. The air is heated by a 1 kW electric heater for 30 second causing the change in system volume and the temperature increases to 60 oC. Determine the boundary work during the process. (cv for air = 0.718 kJ/kgK)

Air0.5 kg30 oC

MY EXAMPLE 4-9

Generally, U2- U1 = Q12 - (Wb + We)

With,We = x t = 1 kW x 30 s = 30 kJ

Wb = -mcv(T2 – T1) – We= -0.5 x 0.718 x (60 – 30) – (-30)= 19.23 kJ

For an ideal gas, U2 - U1 = mcv(T2 - T1)

CHAPTER 4 : ENERGY ANALYSIS OF CLOSED SYSTEMS

Solution

EXAMPLE 4-5

DESIGNED AND PREPARED BY : MOHD KAMAL ARIFFIN/2003

CHAPTER 4 : ENERGY ANALYSIS OF CLOSED SYSTEMS

A piston-cylinder device contains 25 g of saturated water vapor that is maintained at a constant pressure of 300 kPa. A resistance heater within the cylinder is turned on and passes a current of 0.2 A for 5 min from a 120V source. At the same time, a heat loss of 3.7 kJ occurs. (a) Show that for a closed system the boundary work, Wb and the change in internal energy, ∆U in the first law relation can be combined into one term ∆H, for a constant pressure process. (b) Determine the final temperature of the steam

Page 13: Chap4 Closed System

13

EXAMPLE 4-5 cont…

DESIGNED AND PREPARED BY : MOHD KAMAL ARIFFIN/2003

CHAPTER 4 : ENERGY ANALYSIS OF CLOSED SYSTEMS

DESIGNED AND PREPARED BY : MOHD KAMAL ARIFFIN/2003

HEAT TRANSFER DURING THERMODYNAMIC PROCESSES

Q12 = U2 - U1= m(u2 – u1)

Gas unggulQ12 = mcv (T2 - T1)

Q12 = U2 - U1 + W12= U2 - U1 + p ( V2 - V1)= (U2 + pV2) - (U1 + pV1)

Q12 = H2 - H1 = m(h2 – h1)

Gas unggul h2 – h1 = cp(T2 – T1)

Q12 = mcp (T2 - T1)

dT = 0 dU = 0 (dU = mcvdT)

δQ = δW

dU = δQ - δW Q12 = U2 - U1 + W12

Proses PolitropikNext Slide

Proses SeisipaduW12 = 0

ProsesSetekanan

ProsesSesuhu

H2 H1

CHAPTER 4 : ENERGY ANALYSIS OF CLOSED SYSTEMS

Page 14: Chap4 Closed System

14

DESIGNED AND PREPARED BY : MOHD KAMAL ARIFFIN/2003

HEAT TRANSFER DURING THERMODYANAMICS PROCESSES

( ) ( )1-n

T - TmR + T - TmcQ 2112v12 =

1Rcv −

( ) ( ) 1 -

T - TmR - 1 - n

T - TmR Q 212112 γ

=

( ) 1 - n

T - TmR 1 - n 21 ⎟

⎠⎞

⎜⎝⎛⎟

⎞⎜⎝

⎛ −=

γγ

12Q

Proses Politropik

1nVpVpW 2211

12 −−

=

Q12 = (U2 - U1) + W12

m(u2 – u1) = mcv(T2 – T1)

1n)TT(mRW 21

12 −−

=

pV = mRT

1212 W*1nQ ⎟⎠

⎞⎜⎝

⎛−γ−γ

=

W12

CHAPTER 4 : ENERGY ANALYSIS OF CLOSED SYSTEMS

DESIGNED AND PREPARED BY : MOHD KAMAL ARIFFIN/2003

EXAMPLE 4-9

CHAPTER 4 : ENERGY ANALYSIS OF CLOSED SYSTEMS

A piston-cylinder device initially contains 0.5 m3 of nitrogen gas at 400 kPa and 27oC. An electric heater within the device is turned on and is allowed to pass a current of 2 A for 5 min from a 120 V source. Nitrogen expands at constant pressure and a heat loss of 2800 J occurs during the process. Determine the final temperature of nitrogen.

Assumptions : Nitrogen is an ideal gas, KE and PE is zero, specific heat is constant

Solution Qout

Wb

Page 15: Chap4 Closed System

15

DESIGNED AND PREPARED BY : MOHD KAMAL ARIFFIN/2003

EXAMPLE 4-9 Cont…

CHAPTER 4 : ENERGY ANALYSIS OF CLOSED SYSTEMS

We – Qout = ∆U + Wb = mcv(T2 – T1) + mR(T2 – T1)

( ) C56.6K 329.643000.2968)32.245(0.74

2.872TRcm

QWT o1

v

oute2 ==+

+−

=++

−=

Wb = p(V2 – V1) = mR(T2 – T1)

DESIGNED AND PREPARED BY : MOHD KAMAL ARIFFIN/2003

EXAMPLE 4-10A piston cylinder device innitially contains air at 150 kPa and 27oC. At this state, the piston is resting on a pair of stops and the enclosed volume is 400 L. The mass of the piston is such that a 350 kPa pressure is required to move it. The air is now heated until its volume has doubled. Determine (a) the final temperature (b) the work done by the air and (c) the total heat transferred to the air.

Air150 kPa

27oC

piston

cylinder

stops( )( )( )

( )( ) K 1400V150

273272V350Vp

TVpT1

1

11

1333 =

+==

kJ 140 1000400

1000400 x 2 x 350 =⎥⎦

⎤⎢⎣

⎡−⎟

⎠⎞

⎜⎝⎛=

Qtotal = Q12 + Q23 = m[(u2 - u1) + (u3 - u2)] + W23

Wtotal= W12 + W23

1

2 3350

υ2υ1

150

= m(u3 - u1) + W23= mcv (T3 - T1) + W23= 0.6969(0.718)(1400 - 300) + 140 = 690.49 kJ υ

pkPa

0= p(V3 - V2)

3

33

1

11TVp

TVp

=

Solution

kg 0.69690.287x300150x0.4

RTVpm

1

11 ===

Wtotal = mR(T3 – T2) = 0.6969 x 0.287(1400 – 700) = 140 kJ

( ) K 70015035027327

ppTT

1

212 =⎟

⎠⎞

⎜⎝⎛+=⎟⎟

⎞⎜⎜⎝

⎛=

2

2

1

1Tp

Tp

=2

22

1

11TVp

TVp

=

CHAPTER 4 : ENERGY ANALYSIS OF CLOSED SYSTEMS

Page 16: Chap4 Closed System

16

MY EXAMPLE 4-10

DESIGNED AND PREPARED BY : MOHD KAMAL ARIFFIN/2003

Q12 = m(u2 – u1) = 0.5(1636.9 – 2583.6 )= -473.35 kJ

0.5 kg of saturated water vapor is contained in a closed system at 10 bar. The vapor is then cooled at constant volume until the pressure is reduced to 5 bar. Determine the heat transfer during the process and sketch the process on T- υ dan p-υ diagrams.

From Table A5 at 10 bar, (sat vapor), υ1 = υg = 0.1944 m3/kg = υ2u1 = ug = 2583.6 kJ/kg

The work done during constant volume process is zero (W12 = 0)

neglected) is ( 0.5190.37490.1944x f

g

22 υ

υυ

===

u2 = uf + x2ufg= 639.68 + 0.519(2561.2 – 639.68) = 1636.9 kJ/kg

TS2= 151.8 oC

10

5

TS1= 179.9 oC

1

2

p(bar)

υ (m3/kg)

p1= 10 bar

179.9

151.8

1

2

p2= 5 bar

T(oC)

υ (m3/kg)

υg at 5 bar is 0.3749 m3/kgυf < υ2 < υg mixture

Q12 = m(u2 – u1) + W12

0

CHAPTER 4 : ENERGY ANALYSIS OF CLOSED SYSTEMS

Solution

MY EXAMPLE 4-11

DESIGNED AND PREPARED BY : MOHD KAMAL ARIFFIN/2003

A closed system contains 1.5 kg of water, initially at 7 bar and quality of 0.7. The water undergoes an isothermal expansion process until the pressure reduces to 1.5 bar. During the process 1000 kJ heat is supplied to the system. Determine the final temperature of the water and the work done during the process in kJ. Sketch the process on p-υ and T-υ diagram.

T2 = T1 = Ts at 7 bar = 164.97 oC

At state 1 the water is mixture, thusu1 = uf + xufg = 696.44 + 0.7(2572.5 – 696.44)

= 2009.7 kJ/kg

At 1.5 bar, Ts = 111.4 oC. T2 > Ts, shs

W12 = Q12 – m(u2 – u1) = 1000 – 1.5(2602.8 –2009.9) = 110.65 kJ

Interpolation at 1.5 bar (shs)

( )

kJ/kg 2602.8

2579.92579.92656.3150200150165u2

=

+−⎥⎦⎤

⎢⎣⎡

−−

=

p1 = 7 bar

165

111.4

1 2

p2 = 1.5 bar

Ts2= 111.4oC

7

1.5

TS1 = 165 oC

1

2

p(bar)

υ (m3/kg)

p(bar)

υ (m3/kg)

• •

Solution

CHAPTER 4 : ENERGY ANALYSIS OF CLOSED SYSTEMS

Page 17: Chap4 Closed System

17

MY EXAMPLE 4-12

DESIGNED AND PREPARED BY : MOHD KAMAL ARIFFIN/2003

5 kg of R134a at 10 bar and 0.06 m3 undergoes a polytropic expansion process according to the law of pυ1.3 = constant until the pressure drops to 5 bar. Determine the heat transfer during the process and sketch the process on p-υ and T- υ.

kg/m 0120.0506.0

mV 31

1 ===υ

0.5940.02020.0120x

g

11 ==

υυ

=

υg at 10 bar = 0.0202 m3/kg mixture

u1 = uf + x1ufg= 104.42 + 0.594(247.77 – 104.42)= 189.57 kJ/kg

kg/m 0205.05

10 0120.0pp 3

3.113.1

1

2

112 =⎟

⎠⎞

⎜⎝⎛=⎟⎟

⎞⎜⎜⎝

⎛υ=υ

At 5 bar υg = 0.0409 m3/kg υ2 < υg, mixture

u2 = uf + x2ufg = 70.93 + 0.501(235.64 – 70.93) = 153.45 kJ/kg

0.5010.04090.0205x

g

22 ==

υυ

=

⎥⎦⎤

⎢⎣⎡

−υ−υ

+−=1npp)uu(mQ 2211

1212

( )kJ 303.42

13.1)0205.0(10x5)0120.0(10x1045.15330.2085

22

=

⎥⎦

⎤⎢⎣

⎡−−

+−=

10

5

1

2

pυ1.3 = C

p1 = 10 bar

39.39

15.74

1

2

p2 = 5 bar

pυ1.3 = C

p(bar)

υ (m3/kg)T

(oC)

υ (m3/kg)

Ts=39.39oC

Ts=15.74oC

n

2

1

1

2

pp

⎟⎟⎠

⎞⎜⎜⎝

⎛υυ

=

CHAPTER 4 : ENERGY ANALYSIS OF CLOSED SYSTEMS

THE ENDDD


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