An isobaric process is a constant pressure process where pressure (P) remains constant. Work (W) done is equal to pressure (P) times the change in volume (ΔV). For an ideal gas, the change in internal energy (ΔU) is equal to heat (Q) added. An isometric process is a constant volume process where volume (V) remains constant. Work (W) done is zero since there is no change in volume. For any substance, the change in internal energy (ΔU) is equal to the heat (Q) added. An isothermal process is a constant temperature process where temperature (T) remains constant. For an ideal gas, the ratio

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