Upcoming SlideShare
×

# JJ207 Thermodynamics I Chapter 4

4,319 views
4,104 views

Published on

9 Likes
Statistics
Notes
• Full Name
Comment goes here.

Are you sure you want to Yes No
• salam haikal..leh x sy nak nota termo JJ 207 dan JJ507?sy xleh nak download nota anda

Are you sure you want to  Yes  No

Are you sure you want to  Yes  No
• thnx

Are you sure you want to  Yes  No
Views
Total views
4,319
On SlideShare
0
From Embeds
0
Number of Embeds
406
Actions
Shares
0
0
3
Likes
9
Embeds 0
No embeds

No notes for slide

### JJ207 Thermodynamics I Chapter 4

1. 1. CHAPTER 4 ND 2 LAW OF THERMODYNAMICS4.1.1 The Concept Of The Second Law Of Thermodynamics Satisfying the first law alone does not ensure that the process will actually takeplace. For Example, a hot coffee left in a cooler room eventually cools off. But a reverseprocess, hot coffee getting even hotter in a cooler room as a result of heat transfer fromroom air will never take place. Yet doing so would not violate the first law as long as theamount of energy lost by the air is equal to the amount gained by the coffee. Therefore,processes proceed in a certain direction and in not the reverse direction. The first law sets no limit on the percentage of heat supplied, which can beconverted into work. Nor does it indicate whether the energy conversion process isphysically possible or impossible. We shall see, though, that a limit is imposed by theSecond Law of Thermodynamics, and that the possibility or otherwise of a process can bedetermined through a property of the working fluid called entropy. The use of 2nd Law: 1. Identifying the direction of processes. 2. Provides the necessary means to determine the quality as well as the degradation of energy during a process. 3. Determining the theoretical limits.Thermal energy reservoir: A hypothetical body with a relatively large thermal energycapacity (mass x specific heat) that can supply or absorb finite amounts of heat withoutundergoing any change in temperature.Example: Atmosphere does not warm up as a result of heat losses from residentialbuilding in winter.* any physical body whose thermal energy capacity is large relative to the amount ofenergy it supplies or absorbs can be modeled as reservoir.Heat reservoirs: thermal energy reservoirs.Source: A reservoir that supplies energy in the form of heat.Sink: A reservoir that absorbs energy in the form of heat.Kelvin-Planck Statement:It is impossible for any device that operates on a cycle to receive heat from a singlereservoir and produce a net amount of work. No heat engine can have a thermal efficiency of 100%, or as for a power plant tooperate, the working fluid must exchange heat with the environment as well as thefurnace. The impossibility of having a 100% efficient heat engine is not due to friction orother dissipative effects.*2nd Law limitation: no heat engine can convert all heat it receives to useful work.
2. 2. Clausius Statement:It is impossible to construct a device that operates in a cycle and produces no effect otherthan the transfer from a lower-temperature body to a higher temperature body. This statement does not imply that a cyclic device that transfers heat from a coldmedium to a warmer one is impossible to construct but it simply states that a refrigeratorcannot operates unless its compressor is driven by an external power source, such as anelectric motor.4.1.2 The Principles Of A Heat Engine And A Reverse Heat Engine.Heat Engine We know from experience that work can be converted to heat directly andcompletely, but converting heat to work requires the use of some special devices. Thesedevices are called heat engines. A heat engine is a system operating in a complete cycle and developing net workfrom a supply of heat. The second law implies that a source of heat supply (or hotreservoir) and a sink (or cold reservoir) for the rejection of heat are both necessary, sincesome heat must always be rejected by the system. Heat engines can be characterized by the following: They receive heat from a high-temperature source (for example solar energy, oil furnace, nuclear reactor, steam boiler, etc.) They convert part of this heat to work (usually in the form of a rotating shaft, for example gas turbine, steam turbine, etc.) They reject the remaining waste heat to a low-temperature sink (for example the atmosphere, rivers, condenser, etc.) They operate on a cycle.
3. 3. High-temperature HOT RESERVOIR at TH QH WORK OUTPUT HEAT W = QH – QL Note: QH = The heat supplied from the ENGINE source. QL QL = The heat rejected. Low-temperature COLD RESERVOIR W = The net work done. at TL Part of the heat received by the heat engine is converted to work,while the rest is rejected to cold reservoir. Heat engines and other cyclic devices usually involve a fluid that moves to andfro from which heat is transferred while undergoing a cycle. This fluid is called theworking fluid. The work-producing device that best fits into the definition of a heat engine are: The steam power plant The close cycle gas turbineReverse Heat Engine The first and second laws apply equally well to cycles working in the reversedirection to those of heat engine. In general, heat only flows from a high-temperaturesource to a low-temperature sink. However, a reversed heat engine can be utilized topump the heat from a low-temperature region to a high-temperature region. The reversedheat engine is called heat pump. In the case of a reversed cycle, the net work done on the system is equal to the netheat rejected by the system.The work-producing device that best fits into the definition of a heat pump are: The refrigerator. The air-conditioner.
4. 4. High-temperature HOT RESERVOIR at TH QH WORK INPUT HEAT W PUMP QH = W + QL QL Low-temperature COLD RESERVOIR at TL Reverse heat engineQH = magnitude of heat transfer between cyclic device and high-temperature medium at temperature TH.QL = magnitude of heat transfer between cyclic device and low-temperature medium at temperature TL.4.1.3 The Essentials Of Heat Engine According To Working Substance, HeatSource, Mechanical Arrangement And Working Cycle. The steam power plant is an external-combustion engine that is, combustion takesplace outside the engine, and thermal energy released during this process is transferred tothe steam as heat. The schematic of a basic steam power plant is shown below. This is arather simplified diagram, and the discussion of actual steam power plant is given later.The various quantities shown on this figure are as follows: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 atmephere, a ricer, etc.)W out = amount work delivered by steam as it expands in turbine.W in = amount work required to compress water to boiler pressure.
5. 5. 4.1.4 Energy Balance For A Heat Engine (As A Black Box) And Efficiency.Energy Balance The net work output of this power plant is simply the difference between the totalwork output of the plant and the total work input:W net out  W out  W in The net work can also be determined from the heat transfer data alone. The figureabove can be analyzed as a closed system. Recall that for a closed system undergoing acycle, the change in internal energy is zero, therefore:W net out  Q in  Q outThermal Efficiency Note that Qout is never zero. Thus, the net work output of a heat engine is alwaysless than the amount of heat input. The fraction of heat input that is converted to net workoutput is a measure of the performance of a heat engine and is called thermal efficiency,TH.Thermal efficiency = Net work output Total heat input
6. 6. TH = W net, out QH  TH = 1 - QL QH It can be seen that the second law implies that the thermal efficiency of a heatengine must always be less than 100% (Q1 > W ). Thermal efficiency is a measure ofhow efficiently a heat engine converts the heat that it receives to work. The increase inefficiency means the less fuel consumption.Example 1Heat is transferred to a heat engine from a furnace at a rate of 80 MW. If the rate ofwaste heat rejection to a nearby river is 45 MW, determine the net work done and thethermal efficiency for this heat engine. FURNACE QH = 80 MW HEAT W=? ENGINE QL = 45 MW RIVERA schematic of the heat engine is given in the diagram above. The furnace serves as thehigh-temperature reservoir for this heat engine and the river as the low-temperaturereservoir.Assumption: Heat lost through the pipes and other components are negligible.Analysis: The given quantities can be expressed in rate form as; QH = 80 MW QL = 45 MWWnet, out = QH – QL= (80 – 45) MW= 35 MW
7. 7. W 35 MWTH    0.4375 (or 43.75%) Q1 80 MWThat is, the heat engine converts 43.75 percent of the heat it receives to work.4.2.1 Explain the maximum possible efficiency The efficiency of a heat engine cycle greatly depends on how the individualprocesses that make up the cycle are executed. The net work, thus the cycle efficiency,can maximized by using processes that require the least amount of work and deliver themost, that is, by using reversible processes. Therefore the most efficient cycles arereversible cycles, that is, cycles that consist entirely of reversible processes.4.2.2 Explain the reversible and irreversible processesReversible Processes Reversible Process is a process that can be reversed without leaving any trace onthe surroundings. That is allowing system and surroundings to be restored to their initialstates. – no heat transfer – no net work – e.g., adiabatic compression/expansion of a gas in a frictionless piston device: • Reversible processes are considered ideal processes – no energy is “wasted”, i.e., all energy can be recovered or restored – they can produce the maximum amount of work (e.g., in a turbine) – they can consume the least amount of work (e.g., in a compressor or pump) – they can produce the maximum KE increase (e.g., in a nozzle) – when configured as a cycle, they produce the maximum performance (i.e., the highest th or COP)Irreversible Processes • Irreversible Process - process that does not allow system and surroundings to be restored to initial state – such a process contains “irreversibilities” – all real processes have irreversibilities – examples: • heat transfer through a temperature difference • unrestrained expansion of a fluid • spontaneous chemical reaction • spontaneous mixing of different fluids • sliding friction or viscous fluid flow
8. 8. • electric current through a resistance • magnetization with hysteresis • inelastic deformationInternally Reversible Processes • A process is called internally reversible if no irreversibilities occur within the boundary of the system – the system can be restored to its initial state but not the surroundings – comparable to concept of a point mass, frictionless pulley, rigid beam, etc. – allows one to determine best theoretical performance of a system, then apply efficiencies or correction factors to obtain actual performanceExternally Reversible Processes • A process is called externally reversible if no irreversibilities occur outside the boundary of the system – heat transfer between a reservoir and a system is an externally reversible process if the outer surface of the system is at the reservoir temperature4.2.4 The Carnot cycle, Carnot principles, Carnot heat engines, Carnot refrigeratorheat pump.The Carnot Cycle • The Carnot cycle is the best-known reversible cycle, consisting of four reversible processes: – adiabatic compression from temperature TL to TH – isothermal expansion with heat input QH from reservoir at TH – adiabatic expansion from temperature TH to TL – isothermal compression with heat rejection QL to reservoir at TL • Note: – the heat transfers (QH , QL) can only be reversible if no temperature difference exists between the reservoir and system (working fluid) – the processes described constitute a power cycle; it produces net work and operates clockwise on a P-v diagram – The Carnot heat engine can be reversed (operating counter-clockwise on a P-v diagram) to become a Carnot refrigerator or heat pump – the thermal efficiency and coefficients of performance of Carnot cycles correspond to maximum performanceThe Carnot Principles • Several corollaries (the Carnot principles) can be deduced from the Kelvin-Planck statement: – the thermal efficiency of any heat engine must be less than 100% Wnet,out QL th   1 QH QH
9. 9. – th of an irreversible heat engine is always less than that of a reversible heat engine – all reversible heat engines operating between the same two thermal reservoirs must have the same thCarnot Heat Engine • Carnot Heat Engine: TL th ,rev  1  THCarnot Refrigerator and Heat Pump • Carnot Refrigerator: • Carnot Heat Pump:
10. 10. Tutorial1. Study the statements in table below and decide if the statements are TRUE (T) or FALSE (F). STATEMENTS TRUE or FALSE i. The Second Law of Thermodynamics is represented by the equation Q1 – Q2 = W. ii. The heat engine receives heat from a high-temperature source. iii. The heat engine convert part of the heat to internal energy. iv. The work-producing device of a heat engine are the steam power plant and a close cycle gas turbine. v. A reversed heat engine is called a heat pump. vi. The work producing device for a heat pump is the refrigerator. vii. In heat engines, the net work done must be greater than the gross heat supplied, i.e W > Q1 .2. The work done by heat engine is 20 kW. If the rate of heat that enters into the hot reservoir is 3000 kJ/min, determine the thermal efficiency and the rate of heat rejection to the cold reservoir.3. Heat is transferred to a heat engine from a hot reservoir at a rate of 120 MW. If the net work done is 45 MW, determine the rate of waste heat rejection to a cold reservoir and the thermal efficiency of this heat engine.4. 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.5. The food compartment of a refrigerator is maintained at 4°C by removing heat 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. b. The rate of heat rejection to the room that houses the refrigerator.6. 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 -2ºC, the house is estimated to lose heat at a rate of 20,000 kJ/h. If the heat pump under this conditions has a COP of 2.5, determine: a. The power consumed by the heat pump b. The rate at which heat is absorbed from the cold outdoor air