Q. 0.2 kG of air with P =1.5 bar and T=300 K is compressed to a
pressure of 15 bar. According to law Pv
1.25=constant.
Determine air work in kJ.
Q. An
i
nsulated tank of 100 Lt capacity contain an ideal gas (1.4) at 10
bar and 400K. A valve situated at the top of tank is opened allowing the
gas to escape till the pressure is reduce to 2 bar. What will be the
amount of gas (in kg) left in tank.
Q. A gas in a piston- cylinder assembly undergoes an expansion
process for which the relationship between pressure and volume is
given as PV
n=constant and the initial pressure 0.3MPa, the initial
volume is 0.1 m
3and final volume is 0.2 m
3.Calculate the work
for the process in kJ if
(a)n=1.5
Heat Engines and Heat Pumps
A
heat engine – any device that is capable of
converting thermal energy (
heating
) into mechanical
energy (
work
).
Heat Pumps:
Perpetual Motion Machines are Impossible
Perpetual Motion Machines of the first type – these designs seek to create the energy required for their operation out of nothing.
Perpetual Motion Machines of the second type - these designs extract the energy required for their operation in a manner that decreases the entropy of an isolated system.
violation of the First Law (energy conservation)
violation of the Second Law
hot reservoir
TH
work
h
ea
t
Word of caution: for non-cyclic processes,
100% of heat can be transformed into work without violating the Second Law.
Example: an ideal gas expands isothermally
being in thermal contact with a hot reservoir. Since U = const at T = const, all heat has been transformed into work.
Fundamental Difference between Heating and Work
- is the difference in the entropy transfer!
Transferring purely mechanical energy to or from a system does not (necessarily) change its entropy: S = 0 for reversible processes. For this reason, all forms of work are thermodynamically equivalent to each other - they are freely convertible into each other and, in particular, into mechanical work.
Work can be completely converted into heat, but the inverse is not true. The transfer of energy by heating is accompanied with the entropy transfer
Thus, entropy enters the system with heating, but does not leave the system with the work. On the other hand, for a continuous operation of a heat engine, the net entropy change during a cycle must be zero!
Q
dS
T
The Price Should be Paid...
en
tr
o
p
y heat
work
h
ea
t
hot reservoir, TH
cold reservoir, TC
Thus, the only way to get rid of the accumulating entropy is to absorb more internal energy in the heating process than the amount converted to work, and to “flush” the entropy with the flow of the waste heat out of the system.
An engine can get rid of all the entropy received from the hot reservoir by transferring only part
of the input thermal energy to the cold reservoir.
“Working substance” – the system that absorbs heat, expels waste energy, and does work (e.g., water vapor in the steam engine)
T
Q
S
d
Essential parts of a heat engine (any continuously operating reversible device generating work from “heat”)
Perfect Engines (no extra
S
generated)
H H H Q S T
C C C Q S T
H Q
C Q
W en tr o p y heat work h ea thot reservoir, TH
cold reservoir, TC
The condition of continuous operation:
C H S
S
H C
H C Q Q T T
C C H H T Q Q T
The work generated during one cycle of a reversible process:
H C
H C H
H
T T
W Q Q Q
T
Carnot efficiency:
the highest possible value of the energy conversion efficiency
max 1 C 1
Consequences
Any difference TH–TC 0 can be exploited to generate mechanical energy.
The greater the TH –TC difference, the more efficient the engine. Energy waste is inevitable.
Example: In a typical nuclear power plant, TH = 3000C (~570K), T
C =
400C (~310K), and the maximum efficiency e
max=0.45. If the plant
generates 1000 MW of “work”, its waste heat production is at a rate
- more fuel is needed to get rid of the entropy then to generate useful power.
1
1 1220 MW
C H
Q Q W W
e
REFRIGERATORS
AND HEAT PUMPS
The objective of a refrigerator is to remove heat (Q
L) from the cold
medium; the objective of a heat pump is to supply heat (Q
H) to a warm
medium.
The transfer of heat from a
low-temperature region to a
high-temperature one requires special
devices called
refrigerators
.
Refrigerators and heat pumps are
essentially the same devices; they
differ in their objectives only.
THE REVERSED
CARNOT CYCLE
Schematic of a
Carnot
refrigerator and
T-s diagram of the
reversed Carnot
cycle.
Both COPs increase as the difference between the two temperatures decreases, that is, as TL
rises or TH falls. The reversed Carnot cycle is the most efficient refrigeration
cycle operating between TL and TH.
However, it is not a suitable model for refrigeration cycles since processes 2-3 and 4-1 are not practical because
Definition of efficiency
benefit
General definition: efficiency =
cost
benefit
cost
efficiency
heat engine W
Q
hW/Q
hrefrigerator Q
cW
Q
c/W
Real Engines
W en tr o p y heat work h ea thot reservoir, TH
cold reservoir, TC
Real heat engines have lower efficiencies because the processes within the devices are not perfectly reversible – the entropy will be generated by irreversible processes:
max
1
C H HT
W
e
e
Q
T
e = emax only in the limit of reversible
operation.
Some sources of irreversibility:
heat may flow directly between reservoirs; not all temperature difference TH – TC may be available (temperature drop across thermal resistances in the path of the heat flow);
part of the work generated may be converted to heat by friction;
gas may expand irreversibly without doing work (as gas flow into vacuum).
Problem
The temperature inside the engine of a helicopter is 20000C, the temperature of the
exhaust gases is 9000C. The mass of the helicopter is M = 2103 kg, the heat of
combustion of gasoline is Qcomb = 47103 kJ/kg, and the density of gasoline is = 0.8
g/cm3. What is the maximum height that the helicopter can reach by burning V = 1
liter of gasoline?
The work done on lifting the helicopter:
W
MgH
H C H
T
T
Q
W
e
1
For the ideal Carnot cycle (the maximum efficiency):
V
q
Q
H
comb
The heat released in the gasoline combustion:
m
K
K
m/s
kg
liter
kg/liter
0.8
kJ/kg
2
2273
928
Carnot Cycle
1 – 2 isothermal expansion (in contact with TH) 2 – 3 isentropic expansion to TC
3 – 4 isothermal compression (in contact with TC) 4 – 1 isentropic compression to TH
(isentropic adiabatic+quasistatic)
Efficiency of Carnot cycle for an ideal gas:
(Pr. 4.5) H
C
T T emax 1
T S TH TC 1 2 3 4
On the S -T diagram, the work done is the area of the loop:
en tr o py co nt ai ne d in g a s
dU 0
TdS
PdVThe heat consumed at TH (1 – 2) is the area surrounded by the broken line:
H C
H
H T S S
Q S - entropy
contained in gas
- is not very practical (too slow), but operates at the maximum efficiency allowed by the Second Law.
TC P V TH 1 2 3 4 absorbs heat rejects heat
Refrigerators
H H H T Q S C C C T Q S H Q C Q W heat work h ea thot reservoir, TH
cold reservoir, TC
The purpose of a refrigerator is to make thermal energy flow from cold to hot. The coefficient of
performance for a fridge:
en tr o p y 1 / 1 C H C H C C Q Q Q Q Q W Q COP C H C H T T Q Q C H C T T T COP COP max
COP is the largest when TH and TC are close to each other!
For a typical kitchen fridge TH ~300K, TC~ 250K COP ~ 6 (for each J of el. energy, the coolant can suck as much as 6 J of heat from the inside of the freezer).
A fridge that cools something from RT to LHe
More on Refrigerators
H H H
T Q S
C C C
T Q S
H
Q
C
Q
W
heat
work
h
ea
t
hot reservoir, TH
cold reservoir, TC
We can create a refrigerator by
running
a
Carnot
engine
backwards: the gas extracts heat
from the cold reservoir and deposit
it in the cold reservoir.
en
tr
o
p
y
TC
P
V
TH
1
2
3 4
rejects heat
Example:
A “perfect” heat engine with e = 0.4 is used as a refrigerator (the heat reservoirs remain the same). How much heat QC can be transferred in one cycle from the cold reservoir to the hot one if the supplied in one cycle work is W =10 kJ?
H C H H Q Q Q Q W
e
H C H C C H c c Q Q Q Q Q Q Q W Q COP 1 e Q Q H
C 1
5 . 1 4 . 0 6 . 0 1 1 1
1
e e e e COP kJ 15 cycle) one
(in W COP
Clasius Inequality:-
The Cyclic integral of is less than or
equal to zero.
0
dQT0
dQT
0T dQ
