Chapter One 1
Chapter One 1
Chapter One 1
THERMODYNAMICS PROCESS
In this process
V cons tan t
dV 0
2
W PdV 0
1
Q-W= U
Example 1.1
A rigid tank contains air at 500kPa and 150 oC. As a result of
the surrounding, the temperature and pressure in side the tank drop
to 65oC and 400kPa, respectively. Determine the work done during
this process.
Solution: Given T1=150oC and P1=500kPa
T2=65oC and P2=400kPa with no change in
volume because the tank is rigid.
V=constant and dv=0 and so W=0
P cons tan t
P P1 P2
2
W PdV P (V2 V1 )
1
In this process
Fig.hyperbolic process
CHAPTER one FIRST LAW OFTHERMODYNAMIC
A.M.ALHSNAWI CLOSED SYSTEM
c
PV Const. c P
V
2 2 2
c dV V
W pdV dV c c ln 2
1 1
V 1
V V1
V2 V P P
W P1V1 ln P2V2 ln 2 P1V1 ln 1 P2V2 ln 1
V1 V1 P2 P2
this process are called isothermal process for ideal gas(only),
because for ideal gases when PV const. so T const.
where Q=W for ideal gas
Example 1.3
One tenth kg of saturated vapor water is at 2MPa is
compressed in hyperbolic process to a pressure of 4MPa . Find the final
temperature of the water and the work done.
Solution: Given m=0.1kg P1=2MPa sat water vapor
P2=4MPa and the process is PV=constant
At the first state v1 v g at 2 MPa 0.09963m 3 / kg
P1 2MPa
P2 v2 P1v1 OR v2 v1 0.09963m 3 / kg 0.04982m 3 / kg
P2 4MPa
the sat. volume at 4MPa v g 0.04978 m 3 / kg
it is found that v2 v g at 4 MPa so the state is superheated vapor
to find the temperature by using the superheated water table and
interpolation as follows
T oC v m 3 / kg
250.4 0.04978
0.04982
275.0 0.05457
(0.04982 0.04978)
T 250.4 (275 250.4) 250.6 o C
(0.05457 0.04978)
and the work can be calculated by
P1 2
W mP1v1 ln 0.1 2000 0.09963 ln 13.812kJ
P2 4
CHAPTER one FIRST LAW OFTHERMODYNAMIC
A.M.ALHSNAWI CLOSED SYSTEM
This process can be discussed separately for ideal gas and vapor
1- Ideal gas
when the temperature is constant ( T cons tan t ) and from the
ideal gas equation of state, with no change in the mass( PV cons tan t ) .
the process becomes hyperbolic process and
V2 V P P
W P1V1 ln P2V2 ln 2 P1V1 ln 1 P2V2 ln 1
V1 V1 P2 P2
PV mRT
V2 P
W mRT ln mRT ln 1
V1 P2
Example 1.4
One kilogram of air at 500oC is expanded isothermally from
a pressure of 2MPa to a pressure of 0.5MPa, find the work done by the
air.
Solution: Given Air of m=1kg at P1=2MPa P2=0.5MPa T1=T2=T=500oC
It is an ideal gas and isothermal process of expansion
P1
W mRT ln
P2
2
W 1kg 0.287 (500 273.15) ln 307.61kJ
0.5
as the water is still in the saturated region the expansion is also constant
pressure of P=Psat at 120oC =198.53kPa
v f 0.00106m 3 / kg v g 0.8919m 3 / kg v1 v f 0.00106m 3 / kg
V2 mv 2 mf
Vf 2 mf vf v 2 10 v f 101 x v f
10 10 m
9V 9mv 2 10 m g 10
Vg 2 mg v g 2 v2 v g xv g
10 10 9 m 9
vf
v 2 101 x v f xv g x
10 0.00106
0.0106
9 1 0.8919
vg v f 0.00106
9 9
v 2 v f xv g v f 0.00106 .0106(0.8919 .00106) 0.0105m 3 / kg or
10 10
v2 xv g 0.0106 0.8919 0.0105m 3 / kg
9 9
W mPv 2 v1 0.4 198.53 0.0105 0.00106 0.75kJ
e) Polytropic Process
During expansion and compression processes of real gases,
pressure and volume are often related by ( PV n c) where n, and c are
constants. A process of this kind is called a polytropic process.
2
W PdV
1
c
PV n c P n
cV n
V
2 2
cV n 1 PV nV 1 n
2 2
PV
W cV dV
n
1
1 n 1
1 n 1
1 n 1
P2V2 P1V1
W
1 n
Fig. polytropic process
n
P2V2 P1V1 P2 V1
W
1 n
or P1 V2
mRT2 T1
1
W V2 P1 n
1 n
V1 P2
for ideal gas in polytropic process we can drive the following
relation:
P1V1 mRT1 P2V2 mRT2
and
PV C
1 1
n
P2V2n C
P1V1n P2V 2n
n
n
P2 V1 T2 n 1
V
T
P1 2 1
1 1
V2 P1 n T1 n 1 for ideal gas only
T
V1 P2 2
n 1
n 1
T2 P2 n V1
P
V
T1 1 2
Example 1.6
Carbon dioxide with mass of 5kg at 100kPa pressure and
300K temperature is compressed polyropically according to the law
PV1.32=C until the pressure of 500kPa. Find (a) initial and final volume
(b) the final temperature (c) the work done
Solution: Given CO2 gas m 5kg P1 100kPa T1 300 o C
P2 500 kPa for CO2 the gas constant R=0.2968kJ/kg.K
mRT1 5 0.1889 300
V1 2.8335m 3
P1 100
1 1
P n 100 1.32
V2 V1 1 2.8335 0.8371m 3
2
P 500
n 1 1.32 1
P n 500 1.32
T2 T1 2 300 443.2 K
P1 100
mR (T2 T1 ) 5 0.1889443.2 300
W 422.5kJ or
1 n 1 1.32
P V P1V1 500 0.8371 100 2.8335
W 2 2 422.5
1 n 1 1.32
CHAPTER one FIRST LAW OFTHERMODYNAMIC
A.M.ALHSNAWI CLOSED SYSTEM
for a cyclic process, the initial and final state are identical, and therefore
U U 2 U 1 0 . then the first law relation for a cycle simplifies to
Q W 0
Example 1.8
A rigid vessel of .1m3 volume contains refrigerant-12 5%
liquid and 95% vapor by volume at 24 oC. the vessel is heated until the
refrigerant exit as saturated vapor. Find (a) the initial pressure in the
vessel (b) the mass of Refrigerant-12, (c) the final pressure and
temperature, and (d) heat transfer occurs during the process:
Solution: Given V=0.1m3 10% liquid 90% vapor T1=24oC second state
is sat. vapor.
V f 0.05V 0.05 0.1 0.005m 3 , V g 0.95V 0.95 0.1 0.095m 3
and from the sat. R-12 table we find that the following properties at 24 oC
Psat 634.05kPa, v f 0.0007607m 3 / kg , v g 0.02759m 3 / kg ,
u f 58.25kJ / kg , u g 179.85kJ / kg
(a) The initial pressure: because the initial state is saturated mixture at
24oC, then P1=Psat at 24 oC =634.05 kPa
(b) The mass of R-12 in the vessel
Vf 0.005 Vg 0.095
mf 6.573kg , m g 3.443kg
vf 0.0007607 v g 0.02759
m m f m g 6.573 3.443 10.016kg
(c )the final pressure and temperature of the R-12 in the vessel: the
V 0.1
final state is saturated vapor with v2 .01m 3 / kg
m 10.016
and at the second state v g v 2 0.01m 3 / kg
from the pressure table we find the following data and using
extrapolation we can find the data at the vg =0.01m3/kg
P kPa T oC vf m3/kg vg m3/kg uf kJ/kg ug kJ/kg
1400 56.09 0.0008448 0.01222 90.28 191.11
1600 62.19 0.0008660 0.01054 96.80 192.95
1664.4 64.15 0.0008728 0.01 98.90 193.54
o
So P2=1664.4kPa, T2 =64.15 C
(d)The heat transfer: because there is no change in volume so W=0
Q m(u 2 u1 )
u1 u f x(u g u f ) 58.25 0.344(179.85 58.25) 100.08kJ / kg
u 2 u g at sec ond state
193.54kJ / kg
Example 1.9
A rigid insulated tank of 0.5m3, contains 5 kg of water at
100oC. An electric heater is passing through the tank with a voltage of
200V and a current of 5A for 30 minutes. Find the final state of water.
Solution: Given V=0.5m3 , rigid , insulated Q=0, m=5kg water, T=100 oC,
electric heater V=200Volt , I=5A, time=30minutes=1800sec.
The energy equation can be written as:
Q We Wb U
where Q=0 for insulated tank, Wb=boundary work=0 rigid tank
We= electric work= V I time / 1000 200 5 1800 / 1000 1800 kJ
This work is negative because it is done in the system.
We mu
(1800) 5u
u 360kJ
V 0.5
from the first state T=100oC, and v1 .1m 3 / sec
m 5
and it is shown that the state is saturated mixture because v f v1 v g
v1 v f 0.1 .001044
x1 .06
vg v f 1.6729 .001044
u1 u f x1u fg 418.94 0.06 2087.6 542.5kJ / kg
u 2 u1 u 542.5 360 902.5kJ / kg
v 2 v1 0.1m 3 / kg
it is shown from the values of the internal energy and specific volume
that the water is still saturated mixture. And by trial and error we can get
the temperature or pressure.
T2 134.9 o C , P2 312.3kPa, x .1696 16.96%
Example 1.11
A piston cylinder device contains water at 300kPa, and
250 C with a volume of 0.4m3. If the weight of the piston is required a
o
pressure of 300kPa to rise it. The heat is transfer until the water become
saturated mixture with quality of 80%. (a) prove that the heat transfer in a
constant pressure process equal to the change in enthalpy.(b) the work
done (c) heat transfer during the process.
Solution: Given P1 300kPa, V1 0.4m 3 , T1 250 o C constant pressure
process. x 2 0.8
(a) for constant pressure process W P(V2 V1 ) PV2 PV1
the energy equation for closed system Q W U
Q PV2 PV1 U 2 U 1 ( P2V2 U 2 ) ( P1V1 U 1 )
Q H 2 H 1 H
CHAPTER one FIRST LAW OFTHERMODYNAMIC
A.M.ALHSNAWI CLOSED SYSTEM
Example 1.12
Air is in a rigid tank of volume 1m3 at initial pressure of
500kPa and temperature of 300K. It is heated to a final temperature of
700K.(a) find the final pressure in the tank, (b) the change in internal
energy and enthalpy , (c) heat transfer to the system.
Solution: Given P1=500kPa, T1=300K, The tank is rigid with volume of
V=1m3 , T2 =700K, from table of ideal gas properties.
R=0.287kJ/kg.K, Cp=1.005kJ/kg.K, Cv=0.718kJ/kg.K
(a)
T2 700
P2 P1 500 1166.67kPa
T1 300
P1V 500 1
(b) m 5.807kg
RT1 0.287 300
U mCv(T2 T1 ) 5.807 0.718 (700 300) 1667.77kJ
H mCp(T2 T1 ) 5.807 1.005 (700 300) 2334.414kJ
( c) because the tank is rigid, W=0
Q U 1667 .77 kJ
Example 1.13
Nitrogen gas is heated in a piston-cylinder device from 30oC
to 120oC at constant pressure of 200kPa. The mass of nitrogen in the
system is 0.2 kg. Find the work done and heat transfer during the process.
Solution: Given N2, m=0.2 kg, T1=30oC=303K, T2 =120oC=393K at
P=200kPa, from table R=0.2968kJ/kg.K, Cp=1.039kJ/kg.K
W mRT 0.2 0.2968 (120 30) 5.3424kJ
Q mCpT 0.2 1.039 (120 30) 18.702kJ
CHAPTER one FIRST LAW OFTHERMODYNAMIC
A.M.ALHSNAWI CLOSED SYSTEM
Example 1.14
A rigid tank of volume 0.5m3 contains saturated water mixture with
quality of 50% at 120oC is heated until its temperature becomes 200 oC.
Find (a) final state of the water and pressure, (b) heat transfer (c) change
of entropy.
Solution: The given rigid tank V=0.5m3 sat water T1=120oC, x1=0.5
The final temperature is T2=200oC
At T1= 120oC and from sat. water table it is found that
v f 1 0.001060 m 3 / kg v g1 0.8919 m 3 / kg
u f 1 503.5kJ / kg u fg1 2025.8kJ / kg
Example 1.15
Piston cylinder contains 2kg of steam at a pressure of 200kPa and
quality of 75% is heated with constant pressure until it becomes dry
saturated vapor. Find (a) the work done (b) the heat transfer and (c) the
change in entropy.
Solution: the given sat. water mixture at P1=200kPa and x1=0.75 m=2kg,
the final state is sat. vapor at the same pressure of 200kPa.
CHAPTER one FIRST LAW OFTHERMODYNAMIC
A.M.ALHSNAWI CLOSED SYSTEM
Example 1.16
One kg of steam at a pressure of 700kPa and quality of 80% is
expanded hyperbolically to a pressure of 150kPa. Determine (a) the final
state of the vapor and (b) change in entropy
Solution: the given m=1kg , P1=700kPa, x1=0.8 , P2=150kPa
The Process is hyperbolically P1V1=P2V2
At P1=700kPa, x1=0.8
v1 v f x1 v g v f 0.001108 0.8 0.0.2729 0.001108 0.21854 m 3 / kg
s1 s f x1 s fg 1.9922 0.8 4.7158 5.7648 kJ / kg .K
P1v1 700 0.21854
v2 1.01985m 3 / kg
P2 150
v 2 v g at P2 150kPa
v2 v f 1.0198 0.001053
x2 0.8796 87.96%
vg v f 1.1593 0.001053
s 2 s f x 2 s fg 1.4336 0.8796 5.7897 6.526 kJ / kg .K
S ms 2 s1 1 6.526 5.7648 0.7612 kJ / K
Example 1.17
Piston cylinder device contains 0.5kg of refrigerant-12 at 20oC and
quality of 40%. The refrigerant is expanded isothermally until its final
pressure becomes 240kPa. Find (1) the change in entropy (2) heat transfer
(3) the work done.
Solution: the given R-12 T1=T2=20oC, x1=0.4 , P2=240kPa
At the initial state
u1 u f x1u fg 54.44 0.4 178.32 54.44 103.992kJ / kg
s1 s f x1 s fg 0.2078 .4 0.6884 0.2078 0.4kJ / kg.K
at the final state
u 2 182.53kJ / kg s 2 0.7624kJ / kg.K
CHAPTER one FIRST LAW OFTHERMODYNAMIC
A.M.ALHSNAWI CLOSED SYSTEM
Example 1.18
Q 0
Q TdS
The temperature is in absolute value so it is not equal zero.
Then
dS 0
S cons tan t
therefore, the adiabatic process is called a constant entropy or (isentropic
process). T
For ideal gas from the third equation in which
P V
S m CvLn 2 CpLn 2
P1 V1
for adiabatic process S 0
P V
0 CvLn 2 CpLn 2
P1 V1
k
P V V
Ln 2 kLn 2 Ln 1 Q
P1 V1 V2
by exponential the equation we get s
k
P2 V1
Fig.
P1 V2 T-s diagram
or
P1V1k P2V2k
then for adiabatic process of an ideal gas the relation between pressure
and volume is
PVk=Const.
For this process we can show that
P2 Pr1
P1 Pr 2
This process is look like the polytropic process except n=k the specific
heat ratio (also called the adiabatic index). Then the work for this process
is
P2V2 P1V1 mR(T2 T1 )
W mCvT2 T1 U 2 U 1 U
1 k 1 k
The heat transfer from a system during a process is defined by the
relation
2 2
Q TdS or q Tds
1 1
This relation is look like the work done during the process relation
with a pressure and differential form of volume. So the area under the
curve on a T-s diagram represents the heat transfer during the process as
shown in fig.6.3. On this diagram the constant temperature process is a
horizontal line, and the constant entropy process as a vertical line as
CHAPTER one FIRST LAW OFTHERMODYNAMIC
A.M.ALHSNAWI CLOSED SYSTEM
shown on the fig.6.4. From this we denoted that at constant entropy line
the area under the curve is equal to zero. So through this process there is
no heat transfer.
The Carnot cycle of heat engine is shown on the fig.. the analysis
of this cycle is as follows;
s1 s 4 , s 2 s3 and T1 T2 TH , T3 T4 TL
q H TH s 2 s1
q L TL s3 s 4 TL s 2 s1
w q H q L TH TL s 2 s1
the efficiency of the cycle
qL T s s T
th 1 1 L 2 1 1 L
qH TH s 2 s1 TH
also the work can be represented by the area rounded by the cycle .
Example 1.20
Saturated vapor refrigerant-12 at –10oC, is compressed
adiabatically in piston and cylinder device to a pressure of 600kPa,
calculate the final temperature and the work done per unit mass.
Solution: given R-12 saturated vapor at T1=-10oC, adiabatically s1=s2,
Q=0.0, P2=600kPa
s1 s g 0.7019 kJ / kg .K , u1 u g 166 .39 kJ / kg
s 2 s1 0.7019 kJ / kg .K at P2 600 kPa
by using interpolation
0.7019 0.6878
T2 22 30 22 27.94 o C
0.7068 0.6878
0.7019 0.6878
u 2 179.09 184.01 179.09 182.74kJ / kg
0.7068 0.6878
w u 182 .74 166 .39 16 .35 kJ / kg
Example 1.21
5kg of Air at 1000K and 2MPa is expanded adiabatically in a
closed system to the temperature of 600K, find the final pressure and
volume of the air, and the work done.
Solution: given air m 5kg , T1 1000K , P2 2MPa 2000kPa,
T2 600 K , expansion is adiabatically, (s2=s1),or ( P1V1K P2V2K )
mRT1 5 0.287 1000
V1 0.7175m 3
P1 2000
1 / k 1 1 / 1.4 1
T 1000
V2 V1 1 0.7175 2.573m 3
T2 600
CHAPTER one FIRST LAW OFTHERMODYNAMIC
A.M.ALHSNAWI CLOSED SYSTEM
2
1
3
4
s
q H TH s 2 s1
q L TL s 2 s1
s 3 s f x 3 s fg , s 4 s f x 4 s fg , at T L , or PL
CHAPTER one FIRST LAW OFTHERMODYNAMIC
A.M.ALHSNAWI CLOSED SYSTEM
PROBLEM
1.1 Calculate the work done and heat transfer of 2kg of air,
when it is heated at constant volume from 100kPa to 400kPa
.
1.2 Air occupies 0.084m3 at 1.25MPa and 537oC. It is expanded
at a constant temperature to a final volume of 0.336m3.
Calculate :the pressure at the end of expansion, (ii) work
done during expansion (iii) heat transfer to the air
1.8A rigid tank contains air at 500 kPa and 150°C. As a result of
heat transfer to the surroundings,
the temperature and pressure inside the tank drop to 65°C and
400 kPa,
respectively. Determine the boundary work done during this
process.