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
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
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
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
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
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
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
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
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
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 =− &&
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
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
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
14
DESIGNED AND PREPARED BY : MOHD KAMAL ARIFFIN/2003
HEAT TRANSFER DURING THERMODYANAMICS PROCESSES
( ) ( )1-n
T - TmR + T - TmcQ 2112v12 =
1Rcv −
=γ
( ) ( ) 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
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
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
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