Aissa thermo1 chapter 06kau.edu.sa/Files/0056587/Subjects/Aissa_thermo1_chapter_06.pdf · Title:...
Transcript of Aissa thermo1 chapter 06kau.edu.sa/Files/0056587/Subjects/Aissa_thermo1_chapter_06.pdf · Title:...
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Thermodynamics I
Spring 1432/1433H (2011/2012H)
Saturday, Wednesday 8:00am -
10:00am & Monday 8:00am - 9:00am
MEP 261 Class ZA
Dr. Walid A. AissaAssociate Professor, Mech. Engg. Dept.
Faculty of Engineering at Rabigh, KAU, KSA
Chapter #6October XX, 2011
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Announcements:Dr. Walid’s e-mail and Office Hours
Office hours for Thermo 01 will be every
Sunday and Tuesday from 9:00 – 12:00 am
Dr. Walid’s office (Room 5-213)in Dr. Walid’s office (Room 5-213).Text book:
Thermodynamics An Engineering Approach
Yunus A. Cengel & Michael A. Boles7th Edition, McGraw-Hill Companies,
ISBN-978-0-07-352932-5, 2008
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Chapter 6
THE SECOND LAW OF THERMODYNAMICSTHERMODYNAMICS
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Objectives of CH6: To• Introduce 2nd law of thermodynamics.
• Identify valid processes that satisfy both
1st and 2nd laws of thermodynamics.
• Discuss thermal energy reservoirs,
reversible and irreversible processes, reversible and irreversible processes,
heat engines, refrigerators, and heat
pumps.
• Describe the Kelvin–Planck and Clausius
statements of 2nd law of thermodynamics.
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• Discuss the concepts of perpetual-motion
M/Cs.
* Apply the 2nd law of thermodynamics to
cycles and cyclic devices.
* Apply the 2nd law to develop the absolute
thermodynamic temperature scale.thermodynamic temperature scale.
• Describe the Carnot cycle.
•Examine the Carnot principles, idealized
Carnot heat engines, refrigerators, and
heat pumps.
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• Determine the expressions for the
thermal efficiencies and coefficients of
performance for reversible heat engines, heat pumps, and refrigerators.
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Chapter 6THE SECOND LAW OF THERMODYNAMICS
6–1 ■ INTRODUCTION TO THE 2nd LAW
Processes occur in a
certain direction, and
not in the reverse
direction.
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6–2 ■ THERMAL ENERGY RESERVOIRS
Bodies with relatively large thermal
masses can be modeled as thermal
energy reservoirs.
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A source supplies energy in the form
of heat, and a sink absorbs it
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6–3 ■ HEAT ENGINES
Heat engines can be characterized by the
following :
1. They receive heat from a high-1. They receive heat from a high-
temperature source (solar energy, oil
furnace, nuclear reactor, etc.).
2. They convert part of this heat to work
(usually in the form of a rotating shaft).
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Part of the heat Part of the heat
received
by a heat engine is
converted to work,
while the rest is
rejected to a sink.
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3. They reject the remaining waste heat
to a low-temperature sink (the
atmosphere, rivers, etc.).
4. They operate on a cycle.
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The work-producing device that best fits
into the definition of a heat engine is the
steam power plant
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Q in = amount of heat supplied to steam
in boiler from a high-temperature source
(furnace)
Q out = amount of heat rejected from
steam in condenser to a low temperaturesteam in condenser to a low temperature
sink (the atmosphere, a river, etc.)
Wout = amount of work delivered by
steam as it expands in turbine
W in = amount of work required to
compress water to boiler pressure
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The net work output of this power plant is
simply the difference between the total work
output of the plant and the total work input
(6-1)
The net work output of the system is also
equal to the net heat transfer to the system:
(6-2)
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Thermal Efficiency
Thermal efficiency of a heat engine can
be expressed as
(6-3)(6-3)
or
(6-4)
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By substitution by Wnet, out from Eq. (6-2) in
Eq. (6-4) to get
(6-2)
(6-5)(6-5)
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(6-5)
(6-6)
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EXAMPLE 6–1 Net Power Production of a Heat Engine.
Heat is transferred to a heat engine from
a furnace at a rate of 80 MW. If
the rate of waste heat rejection to a
nearby river is 50 MW, determine the
net power output and the thermal
efficiency for this heat engine.
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Solution:
Hence, ηth = 1- (50/80)
i.e., ηth = 0.375*100%= 37.5%
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6–4 ■ REFRIGERATORS AND HEAT PUMPS
REFRIGERATOR
Basic components
of a refrigeration
system and typical
operating
conditions.
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The objective
of a refrigeratorof a refrigerator
is to remove Q L
from the cooled
space.
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Coefficient of Performance
The efficiency of a refrigerator is
expressed in terms of the coefficient of
performance (COP),
(6-7)
Wnet,in = QH - QL (kJ) (6-8)
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Substituting by Wnet,in from Eq. (6-8) in
Eq. (6-7) to get,
(6-9)
Notice that the value of COPR can be
greater than unity.
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Heat Pumps
The objective
of a heat pump of a heat pump
is to supply Q H
into the warmer
space.
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(6-10)
which can also be expressed as
(6-11)(6-11)
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EXAMPLE 6–3 Heat Rejection by a Refrigerator.
The food compartment of a
refrigerator, shown in the figure, is
maintained at 4°C by removing heat
from it at a rate of 360 kJ/min. If the from it at a rate of 360 kJ/min. If the
required power input to the refrigerator
is 2 kW, determine (a) the coefficient
of performance of the refrigerator and
(b) the rate of heat rejection to the room that houses the refrigerator.
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Solution:
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i.e.
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From Eq. (6-7)
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EXAMPLE 6–4 Heating a House by a Heat Pump.
A heat pump is used to meet the heating
requirements of a house and maintain it at
20°C. On a day when the outdoor air
temperature drops to -2°C, the house is temperature drops to -2°C, the house is
estimated to lose heat at a rate of 80,000
kJ/h. If the heat pump under these
conditions has a COP of 2.5, determine
(a) the power consumed by the heat pump
and
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(b) the rate at which heat is absorbed
from the cold outdoor air.
Solution:
But, from Eq. (6-10)
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Hence,
i.e.,
But, from Eq. (6-8)
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Hence,
i.e.,
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6–10 ■ THE CARNOT HEAT ENGINE
(6-6)
Thermal efficiency of any heat engine
is given by Eq. 6–6 as
(6-6)
But by definition, for Carnot engine, or
any reversible heat engine
(6-16)
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Then the efficiency of a Carnot engine, or
any reversible heat engine, becomes
(6-18)
Carnot efficiency represents the limit of
efficiency of any thermal heat engine, i.e.
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EXAMPLE 6–5 Analysis of a Carnot Heat Engine.
A Carnot heat engine, shown in the
figure receives 500 kJ of heat per cycle
from a high-temperature source at
652°C and rejects heat to a low-652°C and rejects heat to a low-
temperature sink at 30°C. Determine
(a) the thermal efficiency of this Carnot
engine and (b) the amount of heat
rejected to the sink per cycle.
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Solution:
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Q H = 500 kJ
T H = 652 ° C TL = 30 ° C
Hence,
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Hence, for this reversible heat engine
Hence,
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6–11 ■ THE CARNOT REFRIGERATORAND HEAT PUMP
A refrigerator or a heat pump that
operates on the reversed Carnot cycle
is called a Carnot refrigerator, or a is called a Carnot refrigerator, or a
Carnot heat pump. The coefficient of
performance of any refrigerator or heat pump, reversible or irreversible, is.
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But by definition, for Carnot engine, or
any reversible heat engine
(6-16)
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Hence,
(6-20)
(6-21)
and
Carnot coefficient of performance
represents the limit of coefficient of
performance of any refrigerator or heat
pump, i.e.
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and
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EXAMPLE 6–7 Heating a House by a Carnot Heat Pump
A heat pump is to be used to heat a house
during the winter, as shown in the figure.
The house is to be maintained at 21°C at
all times. The house is estimated to be all times. The house is estimated to be
losing heat at a rate of 135,000 kJ/h when
the outside temperature drops to 5°C.
Determine the minimum power required
to drive this heat pump.
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From Eq. (6-21)
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Hence,Hence,
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Homework
6–17, 6–18, 6–20, 6–21, 6–29C, 6–39,
6–40, 6–46, 6–47, 6–50, 6–51, 6–71,
6–72, 6–77, 6–78, 6–86, 6–87, 6–88, 6–90, 6–91, 6–94, 6–95, 6–96.6–90, 6–91, 6–94, 6–95, 6–96.