review ch 3 and ch 4 prior to going on to second...
TRANSCRIPT
ME2121 Engineering Thermodynamics 2004-05
Chapter 5 of Cengel and Boles : Second Law Review Ch 3 and Ch 4 prior to going on to Second Law • First Law requires conservation of energy-
does not place restriction on direction of energy flow
• Second Law says that processes can proceed in
a certain direction. Energy has quality- not just quantity!
• Second Law provides theoretical limits of
performance of engineering systems like refrigerators, heat pumps, reactions etc.
SOME CONCEPTS & DEFINITIONS Thermal energy reservoir: Hypothetical body
with large thermal capacity (or inertia). They can be heat sink or heat source.
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Converting work into other forms of energy is easy but the converse is not true.
Work can be converted into heat completely and
directly. To do the opposite, we need special devices called HEAT ENGINES.
__________________________________________ Heat engines(HE) differ from one another but
must have following features: 1.Receive heat from a higher temperature source
(reservoir) 2.Must convert a part of thermal energy into work
(using a shaft , for example) 3.Reject remaining heat to a sink (e.g. to
atmosphere) 4.Operate in a cycle i.e. return to original state HEs are cyclic devices which usually use a fluid –called a
working fluid, to and form which heat is transferred. Note: thermodynamic cycles not using a working fluid are possible but not commonly used.
Note: IC engines, gas turbines do not operate in a thermodynamic cycle and yet are called heat engines! In HEs only heat and work transfer are permitted at system/surroundings boundary.
===================== By First Law: Wnet = Qin – Qout
since for a closed system undergoing a cycle
there is no change of internal energy over the cycle since the fluid returns to its original state! Note Qout must be finite by Second Law!
Thus, net work output of a HE is equal to net
heat input to the system. Note: Qout must be wasted in order to complete
the cycle-or else we cannot come back to the starting state! Hence it can NEVER be zero! Hence net work output is ALWAYS less than heat input.
Measure of performance of Heat engine: Generally,
PERFORMANCE= Desired output/ Required
Input For Heat Engines: Thermal efficiency: Net work output/ Total
heat input Note: In textbook subscripts H and L refer to
higher and lower temperature thermal reservoirs needed for cyclic devices such as refrigerators, heat pumps etc. They are defined as positive magnitudes.
Efficiency of HE = 1 – ( QL / QH ) See Fig. 5.16 and read text- illustrates why
Qout cannot be made zero! (Simple answer- only with finite waste of heat can the fluid complete a thermodynamic cycle!)
Section 5.5 Refrigerators and Heat Pumps
Refrigerators transfer heat from low to high temperature – by second law external energy must be supplied for this to occur.
Definition: COP: Coefficient of performance for refrigerator COPR :Desired output/required input= QL/ W net,in
where Wnet,in= QH – QL kJ. COPHP = QH/ Wnet,in = 1 / (1 – QL/ QH ) COPHP = 1 + COPR -always greater than unity. At worst heat pump acts as a resistance heater i.e.
supplying as much energy as it consumes. All energy is converted to heat.
Aside: In real HP systems in very cold climates, some of the
QH may be lost to piping etc and COP may fall below 1.0. Under such conditions it is best to switch to simple electric heating.
PV Diagrams Pressure-Volume (PV) diagrams are a primary visualization tool for
the study of heat engines. Since the engines usually involve a gas as a
working substance, the ideal gas law relates the PV diagram to the
temperature so that the three essential state variables for the gas can be
tracked through the engine cycle.
Since work is done only when the volume of the gas changes, the
diagram gives a visual interpretation of work done. Since the internal
energy of an ideal gas depends upon its temperature, the PV diagram
along with the temperatures calculated from the ideal gas law determine
the changes in the internal energy of the gas so that the amount of heat
added can be evaluated from the first law of thermodynamics.
For a cyclic heat engine process, the PV diagram be closed loop. The area inside the loop is a
resentation of the amount of work done during a le. Some idea of the relative efficiency of an engine le can be obtained by comparing its PV diagram with t of a Carnot cycle, the most efficient kind of heat ine cycle.
Heat Engines
A heat engine typically uses energy provided in the
form of heat to do work and then exhausts the heat
which cannot be used to do work. Thermodynamics is
the study of the relationships between heat and work.
The first law and second law of thermodynamics
constrain the operation of a heat engine.
General heat engines can be described by the
reservoir model (left) or by a PV diagram (right)
P-V diagram is generally more useful.
Energy Reservoir Model
One of the general ways to illustrate a heat engine is the energy reservoir model. The engine takes energy from a hot reservoir and uses part of it to do work, but is constrained by the second law of
thermodynamics to exhaust part of the energy to a cold reservoir. In the case of the automobile engine, the hot reservoir is the burning fuel and the cold reservoir is the environment to which the combustion products are exhausted.
The efficiency expression given isgeneral one, but the maximum ficiency is limited to that of the rnot cycle. This limitation is often lled the thermal bottleneck.
Second Law of Thermodynamics: It is not
possible for heat to flow from a colder body to a
warmer body without any work having been done to
accomplish this flow.
Energy will not flow spontaneously from a low
temperature object to a higher temperature object.
This precludes a perfect refrigerator. The statements
about refrigerators apply to air conditioners and heat
pumps, which embody the same principles.
This is the "second form" or Clausius statement of the second law.
Note: Perfect refrigerator is not permitted by
Second Law!
The Thermal Bottleneck
If the first law of thermodynamics says you can't win, then the Second Law of Thermodynamics says you can't even break even.
The First Law is essentially a statement of conservation of energy and asserts that you can't get more energy out of a heat engine than you put in. But the Second Law says that no heat engine can use all the heat produced by a fuel to do work. The Carnot cycle sets the ideal efficiency which can be obtained if there is no friction, mechanical losses, leakage, etc., but real machine efficiencies are much less.
Thermal conversion into electricity has low
efficiency. Hydro power generation does not go thru the
THERMAL BOTTLENECK!
Some Qualitative Statements: Second Law of Thermodynamics
The second law of thermodynamics is a profound principle of nature which affects the way energy can be used. There are several approaches to stating this
principle qualitatively. Here are some approaches to giving the basic sense of the principle.
1. Heat will not flow spontaneously from a cold bject to a hot object.
2. Any system which is free of external influences ecomes more disordered with time. This disorder n be expressed in terms of the quantity called
ntropy.
3. You cannot create a heat engine which extractseat and converts it all to useful work.
4. There is a thermal bottleneck which constrains evices which convert stored energy to heat and thene the heat to accomplish work. For a given echanical efficiency of the devices, a machine hich includes the conversion to heat as one of the eps will be inherently less efficient than one which purely mechanical.
Is Second Law valid all the time?
The Second Law of Thermodynamics, Evolution, and Probability
Frank Steiger
Creationists believe that the second law of thermodynamics does not permit order to arise from disorder, and therefore the macro evolution of complex living things from single-celled ancestors could not have occurred. The creationist argument is based on their interpretation of the relationship between probability and a thermodynamic property called "entropy."
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A quote re the Second Law of Thermodynamics
"Nothing in life is certain except death, taxes and the second law of thermodynamics. All three are processes in which useful or accessible forms of some quantity, such as energy or money, are transformed into useless, inaccessible forms of the same quantity. That is not to say that these three processes don't have fringe benefits: taxes pay for roads and schools; the second law of thermodynamics drives cars, computers and metabolism;
Seth Lloyd, writing in Nature 430, 971 (26 August 2004);
Second Law: Kelvin Planck Statement
It is impossible for any cyclic device to receive heat forma single reservoir and produce a net amount of work.
Thus no heat engine can be 100% efficient!
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Clausius Statement of Second Law:
It is impossible to construt a cyclic device that produces no effect than transfer of heat from a lower temperature body to a higher temperature body.
This implies that a device like a refrigerator can only operate if we provide net work input to it.
Note: Both are negative statements- cannot be proven. Both are equivalent. No experiment contradicts it!
A violation of one statement amounts to vilation of the other. Proof is only by logic. It can be shown that violation leads to absurd conclusions. See text section 5.5.
Setion 5.6: Perpetual motion machines
Banned by US patent office since 1918!
A device that vilates First Law is called a PMM1- PMM of first kind. Similarly PMM2.
Power plant that outputs work and heat without input of energy is impossible as PMM1. One that receives heat form one rservoir only violates second law and hence is PMM2
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Section 5.7 Reversible and Irreversible Porcesses
What is highest thermodynamoic efficiency of an ideal engine? Must define reversible processes for this!
Reversible process:
A process that can be reversed without leaving any trace on surroundings. Possible ONLY if the net heat and work exhange between system and suroundings is zero for the COMBINED original and reverse processes.ALL other proceses are IRREVERSIBLE. Examples?
Irreversibilties can be due to:
• Friction • Mixing • Unrestrained expansion of a gas • Heat transfer thru a finite temperature
difference. Howver, as dT becomes vanishingly small process approaches reversibility!
Internally vs externally reversible processes
No internal (to system) or external reversiblities (in surroundings).
Totally reversible or reversible processe are ones where no reversibilities exist internal to system or in surroundings.
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Section 5.8 THE CARNOT CYCLE
Proposed by Sadi Carnot (1824)-best known ideal reversible cycle for Carnot Heat engine.
Cycle composed of two ISOTHERMAL and two ADIABATIC processes- 4 reversible processes. This can be executed in a closed or a steady flow system.
See figures 5.43 and 5.44 as you read below.
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STEPS: Gas in a cylinder-piston device such that the cylinder head can be insulated or brought in contact with hot or cold reservoirs for heat transfer as required.Frictionless piston!
1. Reversible isothermal expansion (1-2 TH= const) Energy source at higher temperature –allow cylinder head to exapnd SLOWLY doing WORK on SURROUNDINGS keeping temperature constant at TH by heat input with small dT (reversibility requirement)Total heat input QH.
2.Reversible Adiabatic expansion(2-3)
No heat input as gas expands..temperature drops to TL. Work continues to be done on surroundings.
3. Reversible isothermal compression(3-4):
Remove insulation, bring cylinder head in contact with reservoir at TL. Piston pushed forward by EXTERNAL force doing work on tha gas. Heat rejected in this step is QL.
4. Adiabatic reversible compression(4-1)
Back to original state 1 ! Completes cycle.
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Compress gas reversibly to return to intial state at TH. Step 3 essential to return to originla state i.e. energy wastage cannot be avoided by second law!
On PV diagram area under process curve represents boundary work. Area enclosed by paths of cycle represents NET WORK done during cycle.
Carnot cycle can be executed in steady flow systems as well- part of power cycles you will study later.
Note : Carnot cycle can be reversed –called Carnot Refrigeration cycle.
CARNOT PRINCIPLES
1. Efficinecy of an irreversible engine is always less than that of a reversible heat engine operating between same two reservoirs.
2. Efficiencies of all REVERSIBLE heat engines operating between same two reservoirs are the same.
Violation of either amounts to violation of the Second Law.
Note: See simple logical proof in Section 5.9 of textbook.
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Reading Assignment: Chapter 5
Study examples given in this chapter of textbook..
September 7, 2004
Absolute Temperature Scale: Kelvin Scale See Figure 5.49 of textbook for the following. Thermodynamic absolute temperature scale must be independent of substance used to measure temperature. Lord Kelvin used Carnot engine concept to derive such a scale. Consider three reservoirs and three Carnot engines A,B and C. T1 and T3 are the high
and low temperatures and T2 is the intermediate one. Q1 is same for A and C and Q3 same for B and C. Note: A operates between T1 and T2, B between T2 and T3 while C operates between T1 and T3. We know Carnot efficiency depends on reservoir temperatures alone. Hence, = 1- F(TL- TH ) F denotes function of. Q1/Q2= F(T1,T2) ; Q2/Q3= F(T2/T3); and Q1/Q3=F(T1,T3) But, Q1/Q3= Q1Q2/Q2Q3. Hence, F(T1,T3) = F(T1,T2) x F(T2,T3) Hence, F must be such that F(T1,T2)=F1(T1)/F1(T2) etc. Kelvin chose simplest form : Q1/Q2=T1/T2 Thus, Carnot efficiency can be expressed in terms of reservoir temperatures alone as
Efficiency = 1 – (TL / TH ) = 1 – (QL/ QH ) Form of F gives only ratio of absolute temperatures and not absolute value. Objective: Find a suitable Function F! Assume a Carnot engine that receives heat at steam point and rejects it at ice point. Efficiency of such an ideal reversible engine would be 26.80%. Since T(ice pt)/T(steam pt) = 0.7320. This is one eqn relating TL and TH. Other equation can be obtained by requiring the magnitude of the degree on absolute scale to correspond to one on Celsius scale i.e. the temperature difference between TH and TL to be 100. Solving, we get T (Celsius) = 273.15 = T(Kelvin)
If the temperature difference is chosen to be 212 we get T(F) + 459.67 = T (Rankine) Example Problem/Solution Problem: A heat engine receives heat at 1MW from 550 C source rejects at 300K. Work produced by this engine is 450kW. Find engine efficiency. Compare with Carnot efficiency. Solution:
Energy eqn. yields QL = 1000- 450 = 550 kW rejected heat to lower temperature sink. Real efficiency of engine: 450/1000 = 45% Carnot efficiency = 1 – 300/(550 +273) = 0.635 Real efficiency is lower than ideal Carnot efficiency due to irreversibilties.