2nd law of thermodynamics -...
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2nd Law of Thermodynamics
A process must satisfy both the first
and the second laws of
thermodynamics to proceed
1st law 2nd law
2nd Law of Thermodynamics
1st law : is concerned with the conversion of
energy from one form to another.
2nd law : controls the process direction.
The first law places no restriction on the direction of a process, but satisfying the first law does not ensure that that process will actually occur
The second law of thermodynamics states that processes
occur in a certain direction, not in just any direction.
•Water flows down a waterfall.
•Gases expand from a high pressure to a low pressure.
•Heat flows from a high temperature to a low temperature.
Once it has taken place, a spontaneous process can be
reversed, but it will not reverse itself spontaneously. Some
external inputs, energy, must be expanded to reverse the
process.
2nd Law of Thermodynamics
A cup of hot cappuccino in a cooler room
eventually cools off Processes proceed in a certain direction
and not in the reverse direction
2nd Law of Thermodynamics
A cup of cappuccino does not get hotter in a cooler room
A process must satisfy both the first
and the second laws of
thermodynamics to proceed
1st law 2nd law
2nd Law of Thermodynamics
Heat Engine
Heat can be converted to work directly and
completely by a device called heat engine
Heat engine is characterized by the following;
• It receives heat from a high-temperature source
• It converts part of this energy (heat to work)
• It rejects the remaining waste heat to a low-
temperature sink
Heat Engines
High-temperature HEAT SOURCE
Low-temperature HEAT SINK
Heat Engine
Qout
Wnet,out
Qin
Schematic of heat engine. Part of the heat
received by heat engine is converted to
work, whereas the rest is rejected to a sink
Heat engine
Steam power plant is the best example
for heat engine
The main devices in steam power plant;
1) Pump
2) Boiler
3) Turbine
4) Condenser
11
Heat Engines
Energy source (such as furnace)
Energy sink (such as the ocean)
Condenser
Boiler
Pump Turbine
System boundary
Wout
QOUT
Win
QIN
Schematic of a steam power plant
Qin - amount of heat supplied to steam in boiler from a high-
temperature source (furnace)
Qout - amount of heat rejected from steam in condenser to a low-
temperature sink (the atmosphere, a river, etc)
Wout - amount of work delivered by steam as it expands in turbine
Win - amount of work required to compress water to boiler pressure
The net work output
Wnet,out = Wout - Win
Heat Engines
Now apply the first law to the cyclic heat engine.
Q W U
W Q
W Q Q
net in net out
net out net in
net out in out
, ,
, ,
,
0 (Cyclic)
The cycle thermal efficiency may be written as
th
net out
in
in out
in
out
in
W
Q
Q Q
Q
Q
Q
,
1
Cyclic devices such as heat engines, refrigerators, and heat
pumps often operate between a high-temperature reservoir
at temperature TH and a low-temperature reservoir at
temperature TL.
The thermal efficiency of the
device becomes
thL
H
Q
Q 1
Condenser
Boiler
Pump Turbine Wout
QOUT
Win
QIN
LETS DO
EXERCISE 1
.
Exercise 1
A steam power plant produces 50 MW of net work while burning fuel
to produce 150 MW of heat energy at the high temperature.
Determine the cycle thermal efficiency and the heat rejected by the
cycle to the surroundings.
th
net out
H
W
Q
MW
MW
,
.50
1500 333 or 33.3%
W Q Q
Q Q W
MW MW
MW
net out H L
L H net out
,
,
150 50
100
Heat Pump
A heat pump is a thermodynamic system operating in a
thermodynamic cycle that removes heat from a low-temperature
body and delivers heat to a high-temperature body. To
accomplish this energy transfer, the heat pump receives external
energy in the form of work or heat from the surroundings.
While the name “heat pump” is the thermodynamic term used to
describe a cyclic device that allows the transfer of heat energy
from a low temperature to a higher temperature, we use the terms
“refrigerator” and “heat pump” to apply to particular devices.
Reverse heat engines is a
system that operates in
thermodynamics cycle
that transfer heat from
low-temperature to high-
temperature media
Refrigerators
Refrigerator
Involve four main components; an evaporator, a
compressor, a condenser and an expansion
valve
Heat is absorbed from the-low temperature
reservoir (cold enviroment) by evaporator and
then the heat is rejected to the high-
temperature reservoir (hot environment) by
condenser
Condenser
Compressor
Evaporator
Expansion valve
Refrigerated space(SINK)
Surrounding (SOURCE) System boundary
Wnet,in
QL
QH
Main components of a refrigeration system
Refrigerators and Heat Pumps
Coefficient of Performance, COP
The index of performance of a refrigerator or heat pump is
expressed in terms of the coefficient of performance, COP,
the ratio of desired result to input. This measure of
performance may be larger than 1, and we want the COP
to be as large as possible.
COP Desired Result
Required Input
For the refrigerator the desired result is the heat supplied
at the low temperature and the input is the net work into
the device to make the cycle operate.
COPQ
WR
L
net in
,
Now apply the first law to the cyclic refrigerator.
( ) ( )
,
Q Q W U
W W Q Q
L H in cycle
in net in H L
0 0
and the coefficient of performance becomes
COPQ
Q QR
L
H L
Refrigerant efficiency
Overall energy balance,
QH – QL = WC
Efficiency is expressed in terms of
coefficient of performance (COP);
1
1
NeededEnergy
Achieved Cooling
L
HLH
L
C
L
QQQQ
Q
W
Q
For the device acting like a “heat pump,” the primary
function of the device is the transfer of heat from low-
temperature to the high-temperature system.
The coefficient of performance for a heat pump is
COPQ
W
Q
Q QHP
H
net in
H
H L
,
HEAT PUMP
Heat pump efficiency
The heat pump efficiency is expressed as
coefficient of performance (β’)
H
LLH
H
C
H
QQQQ
Q
W
Q
1
1
NeededEnergy
Sought Heating
LETS DO
EXERCISE 2 & 3
.
LETS DO
QUIZ 1
QUIZ 2
QUIZ 3
.
Kelvin-Planck statement (HE)
‘ …it is impossible for any device that operates on a cycle to
receive heat from a single reservoir and produce a net amount
of work… ’
That is, a heat engine must exchange heat with low-
temperature sink as well as a high-temperature source to keep
operating
‘ …no heat engine can have a thermal efficiency of 100 percent
, or as for a power plant to operate, the working fluid must
exchange heat with the environment as well as the furnace… ‘
Impossible to have 100 percent efficient heat engine. This is
due to limitation that applies to both the idealized and the
actual heat engines
It is impossible for
any device that
operates on a cycle
to receive heat from
a single reservoir
and produce a net
amount of work.
Kelvin-Planck statement (HE)
Heat Engines
Clausius Statement
Heat Pump
It is impossible to construct a device that operates in a
cycle and produces no effect other than the transfer of
heat from a cooler body to a hotter body.
This statement simply said that a refrigerator cannot
operate without a net work input from an external
source (compressor)
Clausius Statement
It is impossible to construct a
device that operates in a cycle
and produces no effect other
than the transfer of heat from a
lower-temperature body to a
higher-temperature body.
Clausius Statement
Carnot cycle
Carnot Cycle is a reversible heat engine.
It has the highest efficiency of a heat engine
operating between the two thermal energy
reservoirs at temperature TL and TH
The thermal efficiency of actual heat engines
can be maximized by be a Carnot engine
The Carnot cycle is composed of four reversible
processes
Carnot cycle
Thermodynamic Temperature Scale
It is proven that Carnot engine operates
between two identical thermal reservoirs
will have the same efficiency.
It means that, the Carnot engine
efficiency depends to the temperature.
H
L
H
L
CarnotT
T
Q
Q 11
)(
)(
H
L
H
L
T
T
Q
Q
Reversed Carnot Device Coefficient of Performance
If the Carnot device is caused to operate in the reversed
cycle, the reversible heat pump is created. The COP of
reversible refrigerators and heat pumps are given in a
similar manner to that of the Carnot heat engine as
COPQ
Q Q Q
Q
T
T T T
T
RL
H L H
L
L
H L H
L
1
1
1
1
COPQ
Q Q
Q
Q
Q
Q
T
T T
T
T
T
T
HPH
H L
H
L
H
L
H
H L
H
L
H
L
1
1
Again, these are the maximum possible COPs for a
refrigerator or a heat pump operating between the
temperature limits of TH and TL.
LETS DO
EXERCISE 5
EXERCISE 6
.
Exercise 5 (Q1)
A Carnot heat engine receives 500 kJ of heat per
cycle from a high-temperature heat reservoir at
652oC and rejects heat to a low-temperature heat
reservoir at 30oC. Determine
(a) The thermal efficiency of this Carnot engine.
(b) The amount of heat rejected to the low-
temperature heat reservoir.
a.
th revL
H
T
T
K
K
or
,
( )
( )
. .
1
130 273
652 273
0 672 67 2%
Q
Q
T
T
K
K
Q kJ
kJ
L
H
L
H
L
( )
( ).
( . )
30 273
652 2730 328
500 0 328
164
b.
QL
WOUT
QH
TH = 652oC
TL = 30oC
HE
Exercise 5 (Q2)
An inventor claims to have developed a refrigerator that
maintains the refrigerated space at 2oC while operating in a
room where the temperature is 25oC and has a COP of 13.5. Is
there any truth to his claim?
QL
Win
QH
TH = 25oC
TL = 2oC
R
COPQ
Q Q
T
T T
K
K
RL
H L
L
H L
( )
( )
.
2 273
25 2
1196
The claim is false since no refrigerator may have a COP larger
than the COP for the reversed Carnot device.
Exercise 6
A heat pump is to be used to heat a building during the winter.
The building is to be maintained at 21oC at all times. The building
is estimated to be losing heat at a rate of 135,000 kJ/h when the
outside temperature drops to -5oC. Determine the minimum
power required to drive the heat pump unit for this outside
temperature.
21 oC
HP -5 oC
QLost
Win
QL
QH
Q QkJ
hH Lost 135000
COPQ
Q Q
T
T T
K
K
HPH
H L
H
H L
( )
( ( ))
.
21 273
21 5
1131
Using the basic definition of the COP
COPQ
W
WQ
COP
kJ h h
s
kW
kJ s
kW
HPH
net in
net inH
HP
, /
. /
.
,
,
135 000
1131
1
3600
1
3316
The heat lost by the building has to be supplied by the heat pump.
LETS DO
QUIZ 4
QUIZ 6
QUIZ 8
QUIZ 9
.