Module 7 (processes of fluids)

Yuri Melliza
Yuri MellizaMechanical Engineer um Xavier University

THERMODYNAMICS - MODULE 7

MODULE 8
PROCESSES OF FLUIDS
ISOBARIC PROCESS (P = C): An Isobaric Process is an internally reversible Constant Pressure process.
CLOSED SYSTEM
OPEN SYSTEM
2 1
2 1
2 1
For any substance
Q U W 1
W P dV
At P C
W P(V -V ) 2
U m(U -U ) 3
from
h U PV
dh dU PdV VdP
dP 0 at P C
dU PdV dQ
dQ dh
Q h
Q m(h h ) 4
=  + →
= 
=
= →
 = →
= +
= + +
= =
+ =
=
= 
= − →

8
T
T
ln
mC
S
T
dT
mC
T
dh
T
dQ
S
dT
mC
dh
dQ
Gas
Ideal
For
7
S
S
T
dh
T
dQ
S
substance
any
For
CHANGE
ENTROPY
6
)
T
T
(
mC
h
Q
3
)
T
T
(
mC
U
5
)
T
T
(
mR
)
V
V
(
P
W
1
T
V
T
V
Gas
Ideal
For
1
2
p
2
1
p
p
1
2
1
2
p
1
2
v
1
2
1
2
2
2
1
1
→
=

=
=
=

=
=
→
−
=
=
=

→
−
=

=
→
−
=

→
−
=
−
=
→
=





Q h KE PE W 9
W Q h KE PE
W - VdP - KE - PE 10
dP 0 at P C and Q h; - V dP 0
W - KE - PE 11
If KE 0 and PE 0
W 0 12
=  +  +  + →
= −  −  − 
=   →
= = =   =
=   →
 =  =
= →


ISOMETRIC PROCESS (V = C): An Isometric Process is an internally reversible “Constant Volume” process.
CLOSED SYSTEM
OPEN SYSTEM
3
)
U
-
m(U
Q
U
Q
dU
dQ
0
dV
PdV
dU
dQ
2
0
W
0
dV
C
V
At
dV
P
W
1
W
U
Q
substance
any
For
1
2 →
=

=
=
=
+
=
→
=
=
=
•
=
→
+

=

6
T
T
ln
mC
S
T
dT
mC
T
dQ
S
CHANGE
ENTROPY
5
)
T
T
(
mCv
U
Q
4
T
P
T
P
Gas
Ideal
For
1
2
v
v
1
2
2
2
1
1
→
=

=
=

→
−
=

=
→
=
 
1 2
Q h KE PE W 7
W Q h KE PE
W VdP KE PE 8
V dP V(P P ) 9
If KE 0 and PE 0
W V dP 10
=  +  +  + →
= −  −  − 
= − −  −  →
−  = − →
 =  =
= −  →



ISOTHERMAL PROCESS (T = C or PV = C): An Isothermal Process is an internally reversible “Constant Temperature”
Process
CLOSED SYSTEM
4
0
U
T
T
But
)
T
T
(
mC
U
C
V
P
V
P
V
C
P
or
C
PV
Gas
Ideal
For
2
1
1
2
v
2
2
1
1
→
=

=
−
=

=
=
=
=
3
)
U
-
m(U
U
2
dV
P
W
1
W
U
Q
substance
any
For
1
2 →
=

→

=
→
+

=

10
Q
W
e
therefor
0,
U
gas
ideal
For
9
T
Q
S
S
T
Q
C
T
At
Tds
dQ
From
8
S
-
S
S
substance
any
For
CHANGE
ENTROPY
1
2
→
=
=

→
=


=
=
=
→
=

7
P
P
ln
mRT
W
P
P
V
V
6
V
V
ln
mRT
W
5
V
V
ln
V
P
W
V
dV
C
PdV
W
2
1
1
2
1
1
2
1
2
1
1
2
1
1
→
=
=
→
=
→
=
=
=  
OPEN SYSTEM
ISENTROPIC PROCESS (S = C): An Isentropic Process is an internally “reversible adiabatic” process in which the entropy
remains constant where S = C (for any substance) or PVk
= C (for an ideal or perfect gas)
1 1
1 1 1
2 2
2
1
1
2
1 1
1
and applying laws of logarithm
P P
VdP PV ln mRT ln 6
P P
V
VdP mRT ln 7
V
If KE 0 and PE 0
P
W VdP PV ln 8
P
W Q 9
− = = →
− = →
 =  =
= − = − →
= →



2 1
p 2 1
1 2
1 1 2 2
2
1 1
1
Q h KE PE W 1
W= Q h KE PE
W - VdP- KE- PE 2
h m(h h ) 3
For ideal gas
h mC (T -T )
but T T
h 0 4
W= Q KE PE
From
C
PV C or V
P
PV P V C
dP
VdP C
P
P
VdP PV ln 5
P
=  +  +  + →
−  −  − 
=   →
 = − →
 =
=
 = →
−  − 
= =
= =
− = −
− = − →

 

1
V
P
V
P
or
V
V
V
V
P
P
antilog
taking
V
V
ln
V
V
ln
k
V
V
ln
k
P
P
ln
V
dV
k
P
dP
n
integratio
by
V
dV
k
P
dP
PdV
VdP
k
k
2
2
k
1
1
k
2
k
1
k
2
1
1
2
k
2
1
2
1
1
2
1
2
2
1
2
1
→
=
=








=








=
=
−
=
−
=
−
=
−
=


hence
,
k
dU
dh
C
C
but
3
PdV
VdP
dU
dh
2
VdP
dh
0
dQ
VdP
dQ
dh
1
PdV
dU
adiabatic
for
,
0
dQ
PdV
dU
dQ
From
v
p
=
=
→
−
=
→
=
=
+
=
→
−
=
=
+
=
CLOSED SYSTEM
0
S
CHANGE
ENTROPY
5
1
P
P
k
1
V
P
1
P
P
k
1
1
mRT
k
1
)
T
T
(
mR
PdV
W
P
P
T
T
From
4
k
1
)
T
T
(
mR
k
1
V
P
V
P
PdV
W
3
)
T
-
(T
-mC
U
-
W
Gas
Ideal
For
2
U
W
1
0
Q
W
U
Q
substance
any
For
k
1
k
1
2
1
1
k
1
k
1
2
1
2
k
1
k
1
2
1
2
1
2
1
1
2
2
1
2
v
=

→










−








−
=










−








−
=
−
−
=
=








=
→
−
−
=
−
−
=
=
→
=

=
→

−
=
→
=
+

=
−
−
−


OPEN SYSTEM
( )
8
mRT
V
P
7
mRT
V
P
6
PdV
k
VdP
5
k
1
V
P
V
P
k
VdP
4
k
1
V
P
V
P
PdV
egration
int
By
2
2
2
1
1
1
2
1
2
1
1
1
2
2
2
1
1
1
2
2
2
1
→
=
→
=
→
=
−
→
−
−
=
−
→
−
−
=




3
P
C
V
and
V
C
P
C
PV
From
2
V
V
P
P
T
T
T
V
P
T
V
P
and
V
P
V
P
C
T
PV
and
C
PV
g
sin
U
k
1
k
1
k
k
1
k
2
1
k
1
k
1
2
1
2
2
2
2
1
1
1
k
2
2
k
1
1
k
→
=
=
=
→








=








=
=
=
=
=
−
−
p 2 1
2 2 1 1 2 1 2
1
Q h KE PE W
W Q h KE PE
W VdP KE PE
Q 0 1
W h KE PE 2
VdP h 3
For Ideal Gas
h mC (T -T ) 4
If KE 0 and PE 0
W - VdP - h
VdP k PdV
k(P V PV ) kmR(T T ) P
kmRT1
VdP
1 k 1 k 1 k P
=  +  +  +
= −  −  − 
= − −  − 
= →
= − −  −  →
− = − →
 = →
 =  =
= = 
− =
 
− −
− = = = 
− − − 



 

k 1 k 1
k k
1 1 2
1
kPV P
1 1 5
1 k P
− −
   
 
   
− = − →
  
   
−
  
   
   
17
K
KJ
T
T
ln
mC
S
T
dT
mC
T
dQ
S
CHANGE
ENTROPY
16
1
P
P
n
1
V
P
1
P
P
n
1
1
mRT
n
1
)
T
T
(
mR
PdV
W
P
P
T
T
From
15
n
1
)
T
T
(
mR
n
1
V
P
V
P
PdV
W
14
)
T
-
(T
-mC
U
13
)
T
T
(
mC
Q
12
U
Q
W
U
Q
1
2
n
n
n
1
n
1
2
1
1
n
1
n
1
2
1
2
n
1
n
1
2
1
2
1
2
1
1
2
2
1
2
v
1
2
n
→
=

=
=

→










−








−
=










−








−
=
−
−
=
=








=
→
−
−
=
−
−
=
=
→
=

→
−
=
→

=
+

=
 


−
−
−
POLYTROPIC PROCESS (PVn
= C): A Polytropic Process is an internally reversible process of an ideal or perfect gas in
which PVn
= C, where n stands for any constants.
CLOSED SYSTEM
( )
8
mRT
V
P
7
mRT
V
P
6
PdV
n
VdP
5
n
1
V
P
V
P
n
VdP
4
n
1
V
P
V
P
PdV
egration
int
By
2
2
2
1
1
1
2
1
2
1
1
1
2
2
2
1
1
1
2
2
2
1
→
=
→
=
→
=
−
→
−
−
=
−
→
−
−
=




3
P
C
V
and
V
C
P
C
PV
From
2
V
V
P
P
T
T
1
T
V
P
T
V
P
and
V
P
V
P
C
T
PV
and
C
PV
g
sin
U
n
1
n
1
n
n
1
n
2
1
n
1
n
1
2
1
2
2
2
2
1
1
1
n
2
2
n
1
1
n
→
=
=
=
→








=








=
→
=
=
=
=
−
−
heat
specific
Polytropic
n
1
n
k
C
C
11
)
T
-
(T
mC
Q
m
g
Considerin
10
)
T
T
(
C
Q
dT
C
dQ
n
1
n
k
C
C
:
let
dT
n
1
n
k
C
n
1
n
k
dT
C
dQ
n
1
1
k
n
1
dT
C
n
1
1
k
1
dT
C
dQ
v
n
1
2
n
1
2
n
n
v
n
v
v
v
v
→






−
−
=
→
=
→
−
=
=






−
−
=






−
−
=






−
−
=






−
−
+
−
=






−
−
+
=






−
−
+
=
−
−
+
=
=
−
=
−
+
=
+
=
→
−
=
−
−
=
−

−

=

=
→
+

=

n
1
1
k
CvdT
CvdT
dQ
n
1
dT
C
dT
kC
dT
C
dQ
kC
C
C
C
R
n
1
RdT
dT
C
dQ
dW
dU
dQ
10
n
1
RdT
dW
n
1
)
T
T
(
R
n
1
P
P
Pd
W
9
W
U
Q
From
v
v
v
v
p
v
p
v
1
2
1
1
2
2
OPEN SYSTEM
ISOENTHALPIC PROCESS or THROTTLING PROCESS (h = C): An Iso-enthalpic Process is a steady state, steady flow,
process in which W = 0, KE = 0, PE = 0, and Q = 0, where the enthalpy h remains constant.
h1 = h2 or h = C
IRREVERSIBLE OR PADDLE WORK
m
U
Q
W
WP



=
=

=











−








−
=










−








−
=
−
−
=
−
−
=
−
→
−
=
→
=


−

=


+

=

+
=
→

−

−

−
=
→

−

−
−
=
→
+

+

+

=
−
−
VdP
-
W
0
PE
and
0
KE
If
1
P
P
n
1
V
nP
1
P
P
n
1
nmRT
n
1
)
T
T
(
nmR
n
1
)
V
P
V
P
(
n
VdP
22
)
T
T
(
mC
Q
1
2
)
T
-
(T
mC
h
U
h
)
PV
(
)
PV
(
U
h
PV
U
h
20
PE
KE
h
Q
W
19
PE
KE
VdP
W
18
W
PE
KE
h
Q
n
1
n
1
2
1
1
n
1
n
1
2
1
1
2
1
1
2
2
1
2
n
1
2
p
work
Paddle
or
le
Irreversib
Wp
:
Where
W
W
U
Q P
−
−
+

=
3
3
1
1
3
2
2
2 2
1 1 2 2
2
2
1 1
2 2
2
2 2 1 1
m
V 6 L x 0.006 m
1000L
P 100 KPa
V 2 L 0.002 m
PV C
PV P V C
C
P
V
PV
P 900 KPa
V
P V PV
W PdV
1 n
W 1.2 KJ
W 1.2 KJ work is done on the system
= =
=
= =
=
= =
=
= =
−
= =
−
= −
= →

SAMPLE PROBLEMS PURE SUBSTANCE & PROCESSES
1. If 6 L of a gas at a pressure of 100 KPa are compressed reversibly according to PV2
= C until the volume becomes 2 L,
Find the final pressure and the work.
P
V
dV
2
1
 
−
= dP
V
Area
C
PV2
=
2. An ideal gas with R = 2.077 KJ/kg-K and a constant k= 1.659 undergoes a constant pressure process during which 527.5
KJ are added to 2.27 kg of the gas. The initial temperature is 38C. Find the S in KJ/K.
Given:
R = 2.077 KJ/kg-K; k = 1.659
Q = 527.5 KJ; m = 2.27 kg
T1 = 38 + 273 = 311 K
Process: P = C
Q = mCp(T2 – T1) ;
p
Rk
C 5.72KJ / kg K
k 1
= = −
−
K
352
T
mCp
Q
T 1
2 
=
+
=
K
/
KJ
6
.
1
T
T
ln
mCp
S
1
2
=
=

3. A perfect gas has a molecular weight of 26 kg/kgm and a value of k = 1.26. Calculate the heat rejected when 1 kg of the
gas is contained in a rigid vessel at 300 KPa and 315C, and is then cooled until the pressure falls to 150 KPa. (- 361 KJ)
KJ
-91.5
61.5
-
-30
W
-
Q
U
KJ
5
.
61
)
14
.
0
55
.
0
(
150
W
)
V
-
P(V
dV
P
W
C
P
at
PdV
W
W
U
Q
1
2
=
=
=

=
−
=
=
=
=
=
+

=


(rejected)
KJ
2
.
361
Q
)
588
294
(
23
.
1
(
1
)
T
T
(
mC
Q
294
300
)
588
(
150
T
T
P
T
P
C
V
At
588
273
315
T
23
.
1
1
k
R
C
32
.
0
26
3143
.
8
R
1
2
v
2
2
2
1
1
1
v
=
−
=
−
=
=
=
=
=
=
+
=
=
−
=
=
=
4. A closed gaseous system undergoes a reversible process in which 30 KJ of heat are rejected and the volume changes
from 0.14 m3
to 0.55 m3
. The pressure is constant at 150 KPa. Determine the change in internal energy of the system and
the work done.
5. An ideal gas has a mass of 1.5 kg and occupies 2.5 m3 while at a temperature of 300K and a pressure of 200 KPa.
Determine the ideal gas constant for the gas.
Given:
m = 1.5 kg
V = 2.5 m3
T = 300K
P = 200 KPa
6. A cylinder fitted with a frictionless piston contains 5 kg of superheated water vapor at 1000 KPa and 250C. The system
is now cooled at constant pressure until the water reaches a quality of 50%. Calculate the work done and the heat
transferred.
From
h = u + PV
dh = du + PdV + VdP
but
dQ = du + PdV
dh = dQ + VdP
K
kg
KJ
11
.
1
)
300
(
5
.
1
)
5
.
2
(
200
mT
PV
R
mRT
PV

−
=
=
=
=
2 1
for a cons tan t pressure process, P C
dP 0; therefore
dh dQ; and by int egration
dh h and dQ Q
Q h m(h - h ) 5(1768.57 - 2942)
Q -5867.2 KJ
Q 5867.2 KJ (Heat is rejected)
=
=
=
=  =
=  = =
=
=
 
2
2 1
1
2 1
Q = ΔU + W
KJ
W = PdV at P = C; W = P(υ - υ ) in KJ
kg
W = m P(υ - υ ) = -676.43 KJ
W = 676.43 KJ (Work is done on the system)

From table or software at 1000 KPa and 250C
h1 = 2942 KJ/kg: 1 = 0.233 m3
/kg
At P = 1000 KPa and quality x = 0.50
h2 = 1768.57 KJ/kg; 2 = 0.097714 m3
/kg
7. A throttling calorimeter is connected to the de-superheated steam line supplying steam to the auxiliary feed pump of a
ship. The line pressure measures 2.5 MPa (2500 KPa). The calorimeter pressure is 110 KPa and the temperature is
150C. Determine the line steam quality.
From Superheated table, at 110 KPa and 150C, h2 = 2775.6 KJ/kg
From Saturated liquid and saturated vapor table
hf1 = 962.11 KJ/kg; hfg = 1841.0 KJ/kg
h1 = hf1 + x1(hfg1)
h1 = h2
1 f1
1
fg1
1
h -h 2775.6-962.11
x 0.985
h 1841.0
x 98.5 %
= = =
=
Thank You

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Module 7 (processes of fluids)

  • 1. MODULE 8 PROCESSES OF FLUIDS ISOBARIC PROCESS (P = C): An Isobaric Process is an internally reversible Constant Pressure process. CLOSED SYSTEM OPEN SYSTEM 2 1 2 1 2 1 For any substance Q U W 1 W P dV At P C W P(V -V ) 2 U m(U -U ) 3 from h U PV dh dU PdV VdP dP 0 at P C dU PdV dQ dQ dh Q h Q m(h h ) 4 =  + → =  = = →  = → = + = + + = = + = = =  = − →  8 T T ln mC S T dT mC T dh T dQ S dT mC dh dQ Gas Ideal For 7 S S T dh T dQ S substance any For CHANGE ENTROPY 6 ) T T ( mC h Q 3 ) T T ( mC U 5 ) T T ( mR ) V V ( P W 1 T V T V Gas Ideal For 1 2 p 2 1 p p 1 2 1 2 p 1 2 v 1 2 1 2 2 2 1 1 → =  = = =  = = → − = = =  → − =  = → − =  → − = − = → =      Q h KE PE W 9 W Q h KE PE W - VdP - KE - PE 10 dP 0 at P C and Q h; - V dP 0 W - KE - PE 11 If KE 0 and PE 0 W 0 12 =  +  +  + → = −  −  −  =   → = = =   = =   →  =  = = →  
  • 2. ISOMETRIC PROCESS (V = C): An Isometric Process is an internally reversible “Constant Volume” process. CLOSED SYSTEM OPEN SYSTEM 3 ) U - m(U Q U Q dU dQ 0 dV PdV dU dQ 2 0 W 0 dV C V At dV P W 1 W U Q substance any For 1 2 → =  = = = + = → = = = • = → +  =  6 T T ln mC S T dT mC T dQ S CHANGE ENTROPY 5 ) T T ( mCv U Q 4 T P T P Gas Ideal For 1 2 v v 1 2 2 2 1 1 → =  = =  → − =  = → =   1 2 Q h KE PE W 7 W Q h KE PE W VdP KE PE 8 V dP V(P P ) 9 If KE 0 and PE 0 W V dP 10 =  +  +  + → = −  −  −  = − −  −  → −  = − →  =  = = −  →   
  • 3. ISOTHERMAL PROCESS (T = C or PV = C): An Isothermal Process is an internally reversible “Constant Temperature” Process CLOSED SYSTEM 4 0 U T T But ) T T ( mC U C V P V P V C P or C PV Gas Ideal For 2 1 1 2 v 2 2 1 1 → =  = − =  = = = = 3 ) U - m(U U 2 dV P W 1 W U Q substance any For 1 2 → =  →  = → +  =  10 Q W e therefor 0, U gas ideal For 9 T Q S S T Q C T At Tds dQ From 8 S - S S substance any For CHANGE ENTROPY 1 2 → = =  → =   = = = → =  7 P P ln mRT W P P V V 6 V V ln mRT W 5 V V ln V P W V dV C PdV W 2 1 1 2 1 1 2 1 2 1 1 2 1 1 → = = → = → = = =  
  • 4. OPEN SYSTEM ISENTROPIC PROCESS (S = C): An Isentropic Process is an internally “reversible adiabatic” process in which the entropy remains constant where S = C (for any substance) or PVk = C (for an ideal or perfect gas) 1 1 1 1 1 2 2 2 1 1 2 1 1 1 and applying laws of logarithm P P VdP PV ln mRT ln 6 P P V VdP mRT ln 7 V If KE 0 and PE 0 P W VdP PV ln 8 P W Q 9 − = = → − = →  =  = = − = − → = →    2 1 p 2 1 1 2 1 1 2 2 2 1 1 1 Q h KE PE W 1 W= Q h KE PE W - VdP- KE- PE 2 h m(h h ) 3 For ideal gas h mC (T -T ) but T T h 0 4 W= Q KE PE From C PV C or V P PV P V C dP VdP C P P VdP PV ln 5 P =  +  +  + → −  −  −  =   →  = − →  = =  = → −  −  = = = = − = − − = − →     1 V P V P or V V V V P P antilog taking V V ln V V ln k V V ln k P P ln V dV k P dP n integratio by V dV k P dP PdV VdP k k 2 2 k 1 1 k 2 k 1 k 2 1 1 2 k 2 1 2 1 1 2 1 2 2 1 2 1 → = =         =         = = − = − = − = − =   hence , k dU dh C C but 3 PdV VdP dU dh 2 VdP dh 0 dQ VdP dQ dh 1 PdV dU adiabatic for , 0 dQ PdV dU dQ From v p = = → − = → = = + = → − = = + =
  • 5. CLOSED SYSTEM 0 S CHANGE ENTROPY 5 1 P P k 1 V P 1 P P k 1 1 mRT k 1 ) T T ( mR PdV W P P T T From 4 k 1 ) T T ( mR k 1 V P V P PdV W 3 ) T - (T -mC U - W Gas Ideal For 2 U W 1 0 Q W U Q substance any For k 1 k 1 2 1 1 k 1 k 1 2 1 2 k 1 k 1 2 1 2 1 2 1 1 2 2 1 2 v =  →           −         − =           −         − = − − = =         = → − − = − − = = → =  = →  − = → = +  = − − −   OPEN SYSTEM ( ) 8 mRT V P 7 mRT V P 6 PdV k VdP 5 k 1 V P V P k VdP 4 k 1 V P V P PdV egration int By 2 2 2 1 1 1 2 1 2 1 1 1 2 2 2 1 1 1 2 2 2 1 → = → = → = − → − − = − → − − =     3 P C V and V C P C PV From 2 V V P P T T T V P T V P and V P V P C T PV and C PV g sin U k 1 k 1 k k 1 k 2 1 k 1 k 1 2 1 2 2 2 2 1 1 1 k 2 2 k 1 1 k → = = = →         =         = = = = = − − p 2 1 2 2 1 1 2 1 2 1 Q h KE PE W W Q h KE PE W VdP KE PE Q 0 1 W h KE PE 2 VdP h 3 For Ideal Gas h mC (T -T ) 4 If KE 0 and PE 0 W - VdP - h VdP k PdV k(P V PV ) kmR(T T ) P kmRT1 VdP 1 k 1 k 1 k P =  +  +  + = −  −  −  = − −  −  = → = − −  −  → − = − →  = →  =  = = =  − =   − − − = = =  − − −        k 1 k 1 k k 1 1 2 1 kPV P 1 1 5 1 k P − −           − = − →        −           
  • 6. 17 K KJ T T ln mC S T dT mC T dQ S CHANGE ENTROPY 16 1 P P n 1 V P 1 P P n 1 1 mRT n 1 ) T T ( mR PdV W P P T T From 15 n 1 ) T T ( mR n 1 V P V P PdV W 14 ) T - (T -mC U 13 ) T T ( mC Q 12 U Q W U Q 1 2 n n n 1 n 1 2 1 1 n 1 n 1 2 1 2 n 1 n 1 2 1 2 1 2 1 1 2 2 1 2 v 1 2 n → =  = =  →           −         − =           −         − = − − = =         = → − − = − − = = → =  → − = →  = +  =     − − − POLYTROPIC PROCESS (PVn = C): A Polytropic Process is an internally reversible process of an ideal or perfect gas in which PVn = C, where n stands for any constants. CLOSED SYSTEM ( ) 8 mRT V P 7 mRT V P 6 PdV n VdP 5 n 1 V P V P n VdP 4 n 1 V P V P PdV egration int By 2 2 2 1 1 1 2 1 2 1 1 1 2 2 2 1 1 1 2 2 2 1 → = → = → = − → − − = − → − − =     3 P C V and V C P C PV From 2 V V P P T T 1 T V P T V P and V P V P C T PV and C PV g sin U n 1 n 1 n n 1 n 2 1 n 1 n 1 2 1 2 2 2 2 1 1 1 n 2 2 n 1 1 n → = = = →         =         = → = = = = − − heat specific Polytropic n 1 n k C C 11 ) T - (T mC Q m g Considerin 10 ) T T ( C Q dT C dQ n 1 n k C C : let dT n 1 n k C n 1 n k dT C dQ n 1 1 k n 1 dT C n 1 1 k 1 dT C dQ v n 1 2 n 1 2 n n v n v v v v →       − − = → = → − = =       − − =       − − =       − − =       − − + − =       − − + =       − − + = − − + = = − = − + = + = → − = − − = −  −  =  = → +  =  n 1 1 k CvdT CvdT dQ n 1 dT C dT kC dT C dQ kC C C C R n 1 RdT dT C dQ dW dU dQ 10 n 1 RdT dW n 1 ) T T ( R n 1 P P Pd W 9 W U Q From v v v v p v p v 1 2 1 1 2 2
  • 7. OPEN SYSTEM ISOENTHALPIC PROCESS or THROTTLING PROCESS (h = C): An Iso-enthalpic Process is a steady state, steady flow, process in which W = 0, KE = 0, PE = 0, and Q = 0, where the enthalpy h remains constant. h1 = h2 or h = C IRREVERSIBLE OR PADDLE WORK m U Q W WP    = =  =            −         − =           −         − = − − = − − = − → − = → =   −  =   +  =  + = →  −  −  − = →  −  − − = → +  +  +  = − − VdP - W 0 PE and 0 KE If 1 P P n 1 V nP 1 P P n 1 nmRT n 1 ) T T ( nmR n 1 ) V P V P ( n VdP 22 ) T T ( mC Q 1 2 ) T - (T mC h U h ) PV ( ) PV ( U h PV U h 20 PE KE h Q W 19 PE KE VdP W 18 W PE KE h Q n 1 n 1 2 1 1 n 1 n 1 2 1 1 2 1 1 2 2 1 2 n 1 2 p work Paddle or le Irreversib Wp : Where W W U Q P − − +  =
  • 8. 3 3 1 1 3 2 2 2 2 1 1 2 2 2 2 1 1 2 2 2 2 2 1 1 m V 6 L x 0.006 m 1000L P 100 KPa V 2 L 0.002 m PV C PV P V C C P V PV P 900 KPa V P V PV W PdV 1 n W 1.2 KJ W 1.2 KJ work is done on the system = = = = = = = = = = = − = = − = − = →  SAMPLE PROBLEMS PURE SUBSTANCE & PROCESSES 1. If 6 L of a gas at a pressure of 100 KPa are compressed reversibly according to PV2 = C until the volume becomes 2 L, Find the final pressure and the work. P V dV 2 1   − = dP V Area C PV2 = 2. An ideal gas with R = 2.077 KJ/kg-K and a constant k= 1.659 undergoes a constant pressure process during which 527.5 KJ are added to 2.27 kg of the gas. The initial temperature is 38C. Find the S in KJ/K. Given: R = 2.077 KJ/kg-K; k = 1.659 Q = 527.5 KJ; m = 2.27 kg T1 = 38 + 273 = 311 K Process: P = C Q = mCp(T2 – T1) ; p Rk C 5.72KJ / kg K k 1 = = − − K 352 T mCp Q T 1 2  = + = K / KJ 6 . 1 T T ln mCp S 1 2 = =  3. A perfect gas has a molecular weight of 26 kg/kgm and a value of k = 1.26. Calculate the heat rejected when 1 kg of the gas is contained in a rigid vessel at 300 KPa and 315C, and is then cooled until the pressure falls to 150 KPa. (- 361 KJ)
  • 9. KJ -91.5 61.5 - -30 W - Q U KJ 5 . 61 ) 14 . 0 55 . 0 ( 150 W ) V - P(V dV P W C P at PdV W W U Q 1 2 = = =  = − = = = = = +  =   (rejected) KJ 2 . 361 Q ) 588 294 ( 23 . 1 ( 1 ) T T ( mC Q 294 300 ) 588 ( 150 T T P T P C V At 588 273 315 T 23 . 1 1 k R C 32 . 0 26 3143 . 8 R 1 2 v 2 2 2 1 1 1 v = − = − = = = = = = + = = − = = = 4. A closed gaseous system undergoes a reversible process in which 30 KJ of heat are rejected and the volume changes from 0.14 m3 to 0.55 m3 . The pressure is constant at 150 KPa. Determine the change in internal energy of the system and the work done. 5. An ideal gas has a mass of 1.5 kg and occupies 2.5 m3 while at a temperature of 300K and a pressure of 200 KPa. Determine the ideal gas constant for the gas. Given: m = 1.5 kg V = 2.5 m3 T = 300K P = 200 KPa 6. A cylinder fitted with a frictionless piston contains 5 kg of superheated water vapor at 1000 KPa and 250C. The system is now cooled at constant pressure until the water reaches a quality of 50%. Calculate the work done and the heat transferred. From h = u + PV dh = du + PdV + VdP but dQ = du + PdV dh = dQ + VdP K kg KJ 11 . 1 ) 300 ( 5 . 1 ) 5 . 2 ( 200 mT PV R mRT PV  − = = = = 2 1 for a cons tan t pressure process, P C dP 0; therefore dh dQ; and by int egration dh h and dQ Q Q h m(h - h ) 5(1768.57 - 2942) Q -5867.2 KJ Q 5867.2 KJ (Heat is rejected) = = = =  = =  = = = =  
  • 10. 2 2 1 1 2 1 Q = ΔU + W KJ W = PdV at P = C; W = P(υ - υ ) in KJ kg W = m P(υ - υ ) = -676.43 KJ W = 676.43 KJ (Work is done on the system)  From table or software at 1000 KPa and 250C h1 = 2942 KJ/kg: 1 = 0.233 m3 /kg At P = 1000 KPa and quality x = 0.50 h2 = 1768.57 KJ/kg; 2 = 0.097714 m3 /kg 7. A throttling calorimeter is connected to the de-superheated steam line supplying steam to the auxiliary feed pump of a ship. The line pressure measures 2.5 MPa (2500 KPa). The calorimeter pressure is 110 KPa and the temperature is 150C. Determine the line steam quality. From Superheated table, at 110 KPa and 150C, h2 = 2775.6 KJ/kg From Saturated liquid and saturated vapor table hf1 = 962.11 KJ/kg; hfg = 1841.0 KJ/kg h1 = hf1 + x1(hfg1) h1 = h2 1 f1 1 fg1 1 h -h 2775.6-962.11 x 0.985 h 1841.0 x 98.5 % = = = = Thank You