Entropy (S)
➢ a measure of how much energy is
not available to do work
➢ a measure of its disorder
nd
2
Law of Thermodynamics
(Kelvin-Planck Statement)
“It is impossible to construct a
device which operates on a
cycle and produces no other
effect than the production of
work and the transfer of heat
from a single body”.
nd
2
Law of Thermodynamics
(Clausius Statement)
“It is impossible to construct
a device which operates on a
cycle and whose sole effect
is the transfer of heat from a
cooler body to a hotter
body”.
nd
2
Law of Thermodynamics
(Entropy & Second Law of Thermodynamics)
states that the state of entropy of the entire
universe, as an isolated system, will always
increase over time. It also states that the
changes in the entropy in the universe can
never be negative.
∆𝐒 ≥ 𝟎
Processes of Ideal Gas
•
•
•
•
•
Isobaric (P=C)
Isometric/Isochoric/Isovolumetric (V=C)
Isothermal (T=C)
Isentropic (S=C)
Polytropic (Actual)
Ideal processes formulas
𝐟
𝐐 = ‫𝐬𝐝𝐓 𝐢׬‬
𝐝𝐔 = 𝐦𝐂𝐕 𝐝𝐓
𝐐 = ∆𝐔 + 𝐖𝐧𝐟 𝐝𝐇 = 𝐦𝐂𝐩 𝐝𝐓
𝐖𝐧𝐟 =
𝐖𝐬𝐟 =
𝐟
‫𝐕𝐝𝐏 𝐢׬‬
𝐟
‫𝐏𝐝𝐕 𝐢׬‬
𝐏𝐕 = 𝐦𝐑𝐓
𝐧
𝐏𝐕
=𝐂
Air @ STP
𝐤 𝐚𝐢𝐫
𝑪𝐏𝒂𝒊𝒓
= 𝟏. 𝟒
=
𝑪𝐕𝒂𝒊𝒓
𝐁𝐓𝐔
𝐂𝐏𝐚𝐢𝐫 = 𝟎.𝟐𝟒
𝐥𝐛𝐦 − 𝐑
𝐤𝐉
𝐂𝐕𝐚𝐢𝐫 = 𝟎.𝟕𝟏𝟖𝟔
𝐤𝐠𝐦 − 𝐊
𝐁𝐓𝐔
𝐂𝐕𝐚𝐢𝐫 = 𝟎.𝟏𝟕𝟏
𝐥𝐛𝐦 − 𝐑
𝐤𝐂𝒂𝒍
𝐂𝐏𝐚𝐢𝐫 = 𝟎.𝟐𝟒
𝐤𝐠𝐦 − 𝐊
𝐤𝐂𝐚𝐥
𝐂𝐕𝐚𝐢𝐫 = 𝟎.𝟏𝟕𝟏
𝐤𝐠𝐦 − 𝐊
𝐤𝐉
𝐂𝐏𝐚𝐢𝐫 = 𝟏.𝟎𝟎𝟔𝟐
𝐤𝐠𝐦 − 𝐊
Isobaric processes (𝐏 = 𝐂)
𝐧=𝟎
Wnf =
Wnf =
f
‫׬‬i PdV
2
P ‫׬‬1 dV
P=C
Wnf = P(V2 − V1 ) P2 = P1
Wnf = P2 V2 − P1 V1 PV = mRT
Wnf = mR(T2 − T1 )
Q = ∆H
@𝐏=𝐂
V1 V2
T2 V2
=
=
1
f
f
T
T
T
V
1
2
1
1
( ‫ ׬‬Tds = mCP ‫׬‬i dT )
i
T
2
‫׬‬1 ds
=
2 dT
mCP ‫׬‬1
T
T2
∆S = mCP ln
T1
V2
∆S = mCP ln
V1
1. A piston – cylinder system contains a gas
which expands under a constant pressure of
2
1200 lbf/ft . If the piston is displaced 12”
during the process, and the piston diameter
is 24”, what is the work done by the gas on
the piston?
Wnf = P∆V
𝐖𝐧𝐟 = 𝟑𝟕𝟔𝟗. 𝟗 𝐟𝐭 − 𝐥𝐛𝐟
2. Ammonia weighing 22 kg is confined inside a
cylinder equipped with a piston has an initial
pressure of 413 kPa at 380C. if 2900 kJ of heat
is added to the ammonia until its pressure and
0
the temperature are 413 kPa and 100 C
respectively. What is the amount of work done
by the fluid in kJ? NH3 (MW = 17)
ഥ
R
(t 2 −t1 )
Wnf = mR(T2 − T1 ) = m
MW
𝐖𝐧𝐟 = 𝟔𝟔𝟕 𝐤𝐉
3. Determine the change in enthalpy per
lbm of nitrogen gas as its temperature
changes from 500 °F to 200 °F.
∆h = Cp ∆t
kR
Cp =
k−1
ഥ
R
R=
MW
ഥ∆t
kR
𝐁𝐓𝐔
∆h =
∆h
=
−𝟕𝟒.
𝟒𝟕
MW(k − 1)
𝐥𝐛𝐦
4. Calculate the change in enthalpy as 1 kg
of nitrogen is heated from 1000K to 1500K,
assuming the nitrogen is an ideal gas at a
constant pressure. The temperature
dependent specific heat of nitrogen is CP =
−1.5
−2
39.06 − 512.79T
+ 1072.7T −
−3
820.4T where CP is in kJ/kg-mol and T is in
K.
dH = mCp dT
2
2
න dH = m න Cp dT
1500
1
1
kJ
∆H = 1kgන(39.06 − 512.79T−1.5 + 1072.7T−2 − 820.4T−3)dT
kgmol
kJ 1mol N2
∆H = 19 524.41
28
mol
1000
∆𝐇 = 𝟔𝟗𝟕. 𝟑 𝐤𝐉
5. A closed gaseous system undergoes a
reversible process during which 25 BTU are
rejected, the volume changing from 5 ft3 to 2
ft3, and the pressure remains constant at 50
psia. Find the change of internal energy.
Q = ∆U + Wnf
Wnf = P(V2 − V1 )
∆U = 𝟐. 𝟕𝟔 𝐁𝐓𝐔
Isometric processes (𝐕 = 𝐂)
𝐧 =∞
Wnf =
f
‫׬‬i PdV
V = C dV = 0
Wnf = 0
Q = ∆U + Wnf
@𝐕=𝐂
P1 P2
T2 P2
=
=
1
f
f
T
T
T
P
1
2
1
1
( ‫ ׬‬Tds = mCV ‫׬‬i dT )
i
T
2
‫׬‬1 ds
=
2 dT
mCV ‫׬‬1
T
T2
∆S = mCV ln
T1
P2
∆S = mCV ln
P1
6. An amount 1.63 lb of gas is contained in a
3
tank, the volume of which is 2 ft . If the
internal energy is decreases by 27.4 Btu as
temperature changes from 200 °F to 100 °F,
compute the specific heat at constant volume
conditions.
∆U = mC ∆t
V
−27.4BTU = 1.63 CV 100 − 200 R
𝐁𝐓𝐔
𝐂𝐕 = 𝟎. 𝟏𝟔𝟖
𝐥𝐛 − 𝐑
3
ft
7. Two lb of air 50 psia & occupying 10
undergo a process at constant volume
o
until the temperature has risen 500 R.
Calculate the initial & final pressures &
temperatures, the heat added, and the
heat gain of internal energy and enthalpy.
8. In the p-V diagram shown, that addition
occurs between pts 1 & 2. Given that CV =
𝐵𝑇𝑈
0.336
, what is the entropy produced
𝑙𝑏𝑚−𝑅
during this step?
P
o
2
461 psia, 2762 R
1
181 psia, 1080oR
V
8. In the p-V diagram shown, that addition
occurs between pts 1 & 2. Given that CV =
𝐵𝑇𝑈
0.336
, what is the entropy produced
𝑙𝑏𝑚−𝑅
during this step?
P
o
T2
∆S = mCV ln
T1
𝐁𝐓𝐔
∆𝐒 = 𝟎. 𝟑𝟏𝟔
𝐥𝐛𝐦 − 𝐑
2
461 psia, 2762 R
1
181 psia, 1080oR
V
9. A rigid tank contains air at 500 kPa
and 150 degree C. As a result of heat
transfer to the surroundings, the
temperature and pressure inside the
tank drop to 65 degree C and 400 kPa
respectively. Determine the boundary
work done during the process.
𝐖𝐧𝐟 = 𝟎
10. There are 1.36 kg of gas for which R=377
J/kgK and k=1.25 that undergo a non-flow
constant volume process from 551.6 kPa and 60
degree C to 1655 kPa. During the process the gas
is internally stirred and there are also added 105.5
kJ of heat. Determine the change of entropy.
P2
∆S = mCV ln
P1
R
CV =
k−1
mR
P2
∆S =
ln
k − 1 P1
∆𝐒 = 𝟐. 𝟐𝟓 𝐤𝐉/𝐊
11. A 1 – Hp stirring motor is applied to a tank
containing 22.7 kg of water. The stirring action is
applied for 1 hour and the tank loses 850 kJ/hr
of heat. Calculate the rise in the temperature of
the tank after 1 hr., assuming that the process
occurs at constant volume process.
Q + Win = ∆U + Wnf ∆U = mCV ∆T
∆𝐓 = 𝟏𝟗. 𝟑𝟏 𝐊
Isothermal processes(𝐓 = 𝐂)
(𝐏𝐕 = 𝐂) 𝐧 = 𝟏
Wnf =
Wnf =
Wnf =
thus,
f
‫׬‬i PdV
V2 P1
=
V1 P2
C
P=
V
2 dV
C ‫׬‬1
V
V2
C ln
V1
P2 V2 = P1 V1
Wnf =
P1
C ln
P2
where,
𝐂 = P1 V1 = P2 V2 = mRT
Q = ∆U + Wnf
f
( ‫׬‬i Tds
=
f
‫׬‬i PdV )
2P
2
‫׬‬1 ds= ‫׬‬1 T dV
𝐏𝐕 = 𝐦𝐑𝐓
𝐏 𝐦𝐑
=
𝐓
𝐕
1
T
2 dV
2
‫׬‬1 ds= mR ‫׬‬1 V
V2
∆S = mR ln
V1
@ 𝐓 = 𝐂 P2 V2 = P1 V1
P1
∆S = mR ln
P2
V2 P1
=
V1 P2
12. Determine the theoretical horsepower
required for the isothermal compression of
3
800 ft /min of air from 14.7 to 120 psia.
Wnf =
P1
P1 V1 ln
P2
𝐖𝐧𝐟 = −𝟏𝟎𝟖 𝐇𝐩
13. Helium (R = 0.4968 BTU/lbm – R) is
compressed isothermally from 14.7 psia and
68 °F. The compression ratio is 4. Calculate
the work done by the gas.
wnf = RT
V2
ln
V1
rK =
V1
V2
𝐰𝐧𝐟 = −𝟑𝟔𝟒 𝐁𝐓𝐔/𝐥𝐛𝐦
14.
A piston cylinder device initially
3
contains 0.4 m of air at 100 kPa and 80
degree C. the air is now compressed to 0.1
3
m in such a way that the temperature inside
the cylinder remains constant. Determine the
work done during the process.
Wnf =
V2
P1 V1 ln
V1
𝐖𝐧𝐟 = −𝟓𝟓. 𝟒𝟓 𝐤𝐉
15. Air flows steadily through an engine at
constant temperature of 400 K. Find the work
per kg if the exit pressure is one-third of the
inlet pressure and the inlet pressure is 207 kPa.
wnf = RT
P1
ln
P2
𝐰𝐧𝐟 = 𝟏𝟐𝟔. 𝟏𝟐 𝐤𝐉/𝐤𝐠
16. Work done by a substance in reversible
nonflow manner in accordance with V=100/P
3
ft , where P is in psia. Evaluate the work done
on or by the substance as the pressure
increases from 10 psia to 100 psia.
Wnf =
P1
P1 V1 ln
P2
𝐖𝐧𝐟 = −𝟑𝟑𝟏𝟓𝟕. 𝟐 𝐟𝐭 − 𝐥𝐛𝐟
(𝐒 = 𝐂)
Isentropic
processes
𝐤
PV
(𝐏𝐕 = 𝐂) 𝐧 = 𝐤
f
‫׬‬i PdV
−k
CV
Wnf =
2
1−k 1
2 −k
P2 V2 − P1 V1
Wnf = C ‫׬‬1 V dV
Wnf =
1−k
V−k+1 2
k
Wnf = C
C = PV
PV
=
mRT
−k+1 1
mR(T
−
T
)
1−k
2
2
1
V
Wnf
=
k
Wnf = PV
1
−
k
1−k
Wnf =
P=
1
Isentropic processes
Relationship of T, P, and V
k
k
P2 V2 = P1 V1
k−1
k−1
P2 k
V1
T2
=
T1
P1
=
V2
17. Nitrogen is expands isentropically. Its
temperature change from 620 °F to 60 °F.
Find the pressure ratio (p1/p2).
P1
T1
=
P2
T2
k−1
k
620 + 460
=
60 + 460
𝐫𝐏 = 𝟏𝟐. 𝟗
1.399−1
rP 1.399
18. Nitrogen is expands isentropically. Its
temperature change from 620 °F to 60 °F. The
volumetric ratio is V2/V1 = 6.22, and the value R
for nitrogen is 0.0707 BTU/ lbm – R. What is the
work done by the gas?
R(t 2 − t1 ) 0.0707(60 − 620)
wnf =
=
1−k
1 − 1.399
𝐰𝐧𝐟 = 𝟗𝟗. 𝟐𝟑 𝐁𝐓𝐔/𝐥𝐛𝐦
19.
How much work is necessary to
compress air in an insulated cylinder from 0.20
3
3
o
m to 0.01 m . Use T1=20 C and P1= 100kPa.
P2 V2 − P1 V1
Wnf =
1−k
k
P2 V2
=
k
P1 V1
𝐖𝐧𝐟 = −𝟏𝟏𝟓. 𝟕 𝐤𝐉
20. Air undergoes an isentropic
compression from 14.7 psia to 180.6 psia.
o
If the initial temperature is 68 F and the
o
final temperature is 621.5 F , Calculate
the work done by the gas.
R(t 2 − t1 )
wnf =
1−k
𝐰𝐧𝐟 = −𝟗𝟒. 𝟖𝟕 𝐁𝐓𝐔/𝐥𝐛𝐦
21. Air having an initial pressure of 6,516 kPa
3
and an initial volume of 0.113 m is compressed
3
adiabatically to a final volume of 0.057 m .
Calculate the pressure at the end of
compression.
k
P2 V2
1.4
P2 0.113
k
P1 V1
=
= 6516 0.057
𝐏𝟐 = 𝟏𝟔 𝟗𝟖𝟓 𝐤𝐏𝐚𝐚
1.4
22. A group of 50 persons attend a secret meeting
in room which is 12 m wide by 10 m long and a
ceiling height of 3 m. The room is completely
sealed off and insulated. Each person gives off 150
kCal per hour of heat and occupies a volume of
0.20 m3. The room has an initial pressure of 101.3
kPa and temperature of 16 ℃. Calculate the room
temperature after 10 minutes. Use R=0.287 kJ/kg-K
and CV=0.171 kCal/kg-K.
𝐭 𝟐 = 𝟑𝟑. 𝟏 ℃
23. During an isentropic non-flow process
with air, the pressure drops from 60 psia
to 20 psia. The air has mass of 0.10 lb and
o
the initial temperature was 280 F. Find:
A.
B.
C.
D.
𝐨𝐅
𝐭 𝟐 = 𝟖𝟎. 𝟔𝟒
The final temperature,
The heat transfer, BTU 𝐐 = 𝟎
The change of internal energy, BTU
−𝟑. 𝟒𝟐 𝐁𝐓𝐔
The work done, BTU 𝟑. 𝟒𝟐 𝐁𝐓𝐔
oF
𝐧
polytropic processes(𝐏𝐕 = 𝐂)
2
PV
𝐧=𝐧 f
Wnf =
Wnf = ‫׬‬i PdV P =
−n
CV
1−n 1
2 −n
P2 V2 − P1 V1
Wnf = C ‫׬‬1 V dV
Wnf =
−n+1
2
1−n
V
n
Wnf = C
C = PV
PV
=
mRT
−n+1 1
1−n
mR(T
−
T
)
2
2
1
V
n
Wnf
=
Wnf = PV
1
−
n
1−n
1
polytropic processes
Relationship of T, P, and V
n
n
P2 V2 = P1 V1
n−1
n−1
n
P2
V1
T2
=
P1
T1
=
V2
polytropic processes
𝐐 = 𝐦𝐂𝐧 ∆𝐓
𝐂𝐧 = 𝐂𝐕
𝐤−𝐧
𝟏−𝐧
Q = mCn dT
f
( ‫ ׬‬Tds =
i
2
‫׬‬1 ds =
f
mCn ‫׬‬i dT
2 dT
mCn ‫׬‬1
T
T2
∆S = mCn ln
T1
1
)
T
24. Determine the value of “n” in the
n
process PV = constant if this process has
the following end states: 100 psia, 1
cu.ft.; 20 psia, 6 cu.ft.
n
P2 V2
n
20 6
=
n
P1 V1
= 100 1
𝐧 = 𝟎. 𝟗
n
25. Work of a polytropic (n=1.21)
compression of air (Cp/Cv = 1.40) in a
system with moving boundary from P1 =
3
15 psia, V1 = 1.0 ft to P2 = 150 psia, V2 =
3
0.15 ft .
P2 V2 − P1 V1
Wnf =
1−n
𝐖𝐧𝐟 = −𝟓𝟏𝟒𝟐. 𝟗 𝐟𝐭 − 𝐥𝐛𝐟
26. Three cubic ft of air is compressed
slowly in a cylinder from 15 psia to 100
o
psia. The initial temperature is 71 F ,
n=1.2. Calculate the final temperature,
heat added and work done.
T2
P2
=
T1
P1
n−1
n
Q = mCn ∆T Cn = CV
mR(t 2 − t1 )
Wnf =
1−n
k−n
1−n
27. 1 gas at 1 atmosphere pressure originally
3
occupies a volume of 20 ft . The gas is
polytropically compressed with n=1.15 until the
pressure is 48.51 psia. What work is required
for this compression?
n
P2 V2
=
n
P1 V1
P2 V2 − P1 V1
Wnf =
1−n
𝐖𝐧𝐟 = −𝟒𝟕 𝟓𝟔𝟓. 𝟗 𝐟𝐭 − 𝐥𝐛𝐟
28. One pound of air (consider here a perfect
gas) with an initial temperature of 200F is
allowed to expand without flow between
pressures of 90 and 15 psia. Which of the
three processes, PV=C, PVk=C, PV1.5=C will
produce the maximum work with minimum
heat transfer.
if max work PV = C
k
if max work with min heat PV = C
29. Air is compressed in a cylinder during a nonflow reversible polytropic process from an initial
o
temperature and pressure of 80 F and 15psia to
o
a final temperature and pressure of 285 F and
75psia. The cylinder contains 0.01lb of air, and
2
the area of the piston is 0.20 ft . Find:
A.
B.
C.
D.
E.
The value of polytropic exponent, n 𝐧 = 𝟏. 𝟐𝟓
The work done in ft-lb −𝟒𝟑𝟕. 𝟓 𝐟𝐭 − 𝐥𝐛𝐟
The change of internal energy, BTU 𝟎. 𝟑𝟓𝟏𝐁𝐓𝐔
The heat transferred, BTU −𝟎. 𝟐𝟏𝐁𝐓𝐔
The distance the piston moves during the
process, inches 𝟓. 𝟕𝟖 𝐢𝐧.