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Thermodynamics Lecture Series Applied Sciences Education.

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Presentation on theme: "Thermodynamics Lecture Series Applied Sciences Education."— Presentation transcript:

1 Thermodynamics Lecture Series email: drjjlanita@hotmail.com http://www5.uitm.edu.my/faculties/fsg/drjj1.htmldrjjlanita@hotmail.com Applied Sciences Education Research Group (ASERG) Faculty of Applied Sciences Universiti Teknologi MARA Second Law – Quality of Energy-Part1

2 Quotes It does not matter how slowly you go, so long as you do not stop. --Confucius To be wronged is nothing unless you continue to remember it. --Confucius

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4 Symbols Q q W V E

5 Introduction - Objectives 1.Explain the need for the second law of thermodynamics on real processes. 2.State the general and the specific statements of the second law of thermodynamics. 3.State the meaning of reservoirs and working fluids. 4.List down the characteristics of heat engines. Objectives:

6 Introduction - Objectives 5.State the difference between thermodynamic heat engines and mechanical heat engines. 6.Sketch an energy-flow diagram indicating the flow of energy and label all the energies and all the reservoirs for a steam power plant. 7.Sketch a schematic diagram for a steam power plant and label all the energies, flow of energies and all the reservoirs. Objectives:

7 Introduction - Objectives 8.State the desired output and required input for a steam power plant. 9.State the meaning of engines performance and obtain the performance of a steam power plant in terms of the heat exchange. 10.State the Kelvin-Planck statement on steam power plant. Objectives:

8 Introduction - Objectives 11.Sketch an energy-flow diagram indicating the flow of energy and label all the energies and all the reservoirs for a refrigerator. 12.Sketch a schematic diagram for a refrigerator and label all the energies, flow of energies and all the reservoirs. 13.State the desired output and required input for a refrigerator. Objectives:

9 Introduction - Objectives 14.Obtain the performance of a refrigerator in terms of the heat exchange. 15.Sketch an energy-flow diagram indicating the flow of energy and label all the energies and all the reservoirs for a heat pump. 16.Sketch a schematic diagram for a a heat pump and label all the energies, flow of energies and all the reservoirs. Objectives:

10 Introduction - Objectives 17.State the desired output and required input for a a heat pump. 18.Obtain the performance of a a heat pump in terms of the heat exchange. 19.State the Clausius statement for refrigerators and heat pumps. 20.Solve problems related to steam power plants, refrigerators and heat pumps. Objectives:

11 Review - First Law All processes must obey energy conservation do not obey energy conservation cannot happen.Processes which do not obey energy conservation cannot happen. do not obey mass conservation cannot happenProcesses which do not obey mass conservation cannot happen Piston-cylinders, rigid tanks Turbines, compressors, Nozzles, heat exchangers

12 How to relate changes to the cause Dynamic Energies as causes (agents) of change System E 1, P 1, T 1, V 1 To E 2, P 2, T 2, V 2 Properties will change indicating change of state Mass out Mass in W in W out Q in Q out Review - First Law

13 Energy Balance Amount of energy causing change change must be equal to amount of energy change change of system Energy Entering a system - Energy Leaving a system = Change of systems energy Review - First Law

14 Energy Balance E in – E out = E sys, kJ or e in – e out = e sys, kJ/kg or Energy Entering a system - Energy Leaving a system = Change of systems energy Review - First Law

15 Mass Balance m in – m out = m sys, kg or Mass Entering a system - Mass Leaving a system = Change of systems mass Review - First Law

16 E sys = 0, kJ; e sys = 0, kJ/kg, V sys = 0, m3;m3; m sys = 0 or m in = m out, kg Energy Balance – Control Volume Steady-Flow Steady-flow Steady-flow is a flow where all properties properties within boundary of the system remains constant with time Review - First Law

17 Mass & Energy Balance–Steady- Flow CV q in + in + = q out + out + out, kJ/kg Energy balance Mass balance Review - First Law

18 Mass & Energy Balance–Steady-Flow: Single Stream q in – q out + in – out = – in, kJ/kg Energy balance Mass balance Review - First Law = h out - h in + ke out – ke in + pe out - pe in, kJ/kg

19 First Law – Heat Exchanger Heat Exchanger Boundary has 2 inlets and 2 exits

20 First Law – Heat Exchanger Heat Exchanger Boundary has 1 inlet and 1 exit

21 First Law of Thermodynamics Heat Exchanger -no mixing -2 inlets and 2 exits In, 1Exit, 2 In, 3 Exit, 4

22 First Law of Thermodynamics Heat Exchanger -no mixing -1 inlet and 1 exit In, 1Exit, 2 In, 3 Exit, 4

23 First Law of Thermodynamics Heat Exchanger Case 1 Energy balance Mass balance

24 First Law of Thermodynamics Heat Exchanger Case 2 Q out Energy balance Mass balance

25 First Law of Thermodynamics Energy balance: Case 1 Mass balance: Heat Exchanger Purpose: Purpose: Remove or add heat where

26 First Law of Thermodynamics Energy balance: Case 2 Mass balance: Heat Exchanger Purpose: Remove or add heat where Q out

27 First Law of Thermodynamics Energy balance: Case 2 Mass balance: Heat Exchanger Purpose: Remove or add heat where Q in

28 Second Law First Law quantity or amount of energyFirst Law involves quantity or amount of energy to be conserved in processes Q out T sys,initial =40 C Q out T surr =25 C This is a natural process!!! Q flows from high T to low T medium until thermal equilibrium is reached 0 -q out + 0 - 0 = - u = u 1 - u2, u2, kJ/kg T sys,final =25 C OK for this cup

29 Second Law T surr =25 C Q in T sys,initial =25 C This is NOT a natural process process!!! Q does not flow from low T to high T medium. Never will the coffee return to its initial state. q in – 0 + 0 - 0 = u = u 2 - u1, u1, kJ/kg T sys,final =40 C

30 Second Law T surr =25 C Q in T sys,initial =25 C NOT a natural process This is NOT a natural process!!! does not flow Q does not flow from low T to high T medium. Never will the coffee return to its initial state. q in – 0 + 0 - 0 = u = u 2 - u 1, kJ/kg T sys,final =40 C

31 Second Law First Law quantity or amount of energyFirst Law involves quantity or amount of energy to be conserved in processes T surr =25 C Q in T sys,initial =25 C NOT a natural process This is a NOT a natural process!!! does not flow Q does not flow from low T to high T medium. Never will equilibrium be reached q in – 0 + 0 - 0 = u =u 2 -u 1 kJ/kg T sys,final =40 C But is the process in this cup possible??

32 Second Law First Law is not sufficientFirst Law is not sufficient to determine if a process can or cannot proceed

33 Second Law First Law is not sufficientFirst Law is not sufficient to determine if a process can or cannot proceed Heat Heat (thermal energy) flows from high temperature medium to low temperature medium. Introduce the second law of thermodynamics thermodynamics – processes occur in its natural direction. Energy Energy has quality & quality is higher with higher temperature. More work can be done.

34 Second Law Considerations: Requires Requires a special device – heat engine. WorkWork can can be converted to heat directly & totally. HeatHeat cannot cannot be converted to work directly & totally.

35 Second Law Heat Engine Characteristics: ReceiveReceive heat from a high T source. ConvertConvert part of the heat into work. RejectReject excess heat into a low T sink. OperatesOperates in a cycle.

36 Second Law Heat Engines ThermodynamicsThermodynamics heat engines engines – external combustion: steam power plants MechanicalMechanical heat engines engines – internal combustion: jets, cars, motorcycles Combustion inside system Performance = Desired output / Required input Combustion outside system

37 Second Law Steam Power Plant High T Res., T H Furnace q in = q H net,out Low T Res., T L Water from river An Energy-Flow diagram for a SPP q out = q L Working fluid: Water q in - q out = out - in net,out = q in - q out Purpose: Produce work, W out, out q in = net,out + q out

38 Second Law Pump Boiler Turbin e Condenser High T Res., T H Furnace q in = q H in out Low T Res., T L Water from river A Schematic diagram for a Steam Power Plant q out = q L Working fluid: Water q in - q out = out - in q in - q out = net,out

39 Second Law Thermal Efficiency for steam power plants

40 Second Law Thermal Efficiency for steam power plants

41 Second Law Kelvin Planck Statement for steam power plants It is impossible for engines operating in a cycle to receive heat from a single reservoir and convert all of the heat into work. Heat engines cannot be 100% efficient.

42 Second Law Refrigerator/ Air Cond High T Res., T H, Kitchen room / Outside house q out = q H net,in Low Temperature Res., T L, Inside fridge or house An Energy-Flow diagram for a Refrigerator/Air Cond. q in = q L Working fluid: Ref-134a q out – q in = in - out net,in = q out - q in Purpose: Maintain space at low T by Removing qLqL net,in = q H - q L

43 Second Law Condenser Evaporator High T Res., T H Kitchen/Outside house q out = q H in Low T Res., T L Ref. Space/Room A Schematic diagram for a Refrigerator/Air Cond. q in = q L Working fluid: Refrigerant-134a Com pressor Throttle Valve

44 Second Law Coefficient of Performance for a Refrigerator Divide top and bottom by q in

45 Second Law Coefficient of Performance for a Refrigerator

46 Second Law Heat Pump High Temperature Res., T H, Inside house q out = q H net,in Low Temperature Res., T L, Outside house An Energy-Flow diagram for a Heat Pump q in = q L Working fluid: Ref-134a q out = net,in + q in net,in = q out - q in Purpose: Maintain space at high T by supplying qHqH net,in = q H - q L

47 Second Law Condenser Evaporator High T Res., T H Inside house q out = q H in Low T Res., T L Outside the house A Schematic diagram for a Heat Pump q in = q L Working fluid: Refrigerant-134a Com pressor Throttle Valve

48 Second Law Coefficient of Performance for a Heat Pump

49 Second Law Clausius Statement on Refrigerators/Heat Pump It is impossible to construct a device operating in a cycle and produces no effect other than the transfer of heat from a low T to a high T medium. Must do external work to the device to make it function. Hence more energy removed to the surrounding.

50 Second Law – Energy Degrade Factors of irreversibilities less heat can be converted to work –Friction between 2 moving surfaces –Processes happen too fast –Non-isothermal heat transfer What is the maximum performance of real engines if it can never achieve 100%??

51 Second Law – Dream Engine Carnot Cycle Isothermal expansionIsothermal expansion Slow adding of Q resulting in work done by system (system expand) Q in – W out = U = 0. So, Q in = W out. Pressure drops. Adiabatic expansionAdiabatic expansion 0 – W out = U. Final U smaller than initial U. T & P drops.

52 Second Law – Dream Engine Carnot Cycle Isothermal compression Work done on the system Slow rejection of Q - Q out + W in = U = 0. So, Q out = W in. Pressure increases. Adiabatic compression 0 + W in = U. Final U higher than initial U. T & P increases.

53 Second Law – Dream Engine Carnot Cycle P - diagram for a Carnot (ideal) power plant P, kPa, m 3 /kg q out q in 2 34 1

54 Second Law – Dream Engine Reverse Carnot Cycle P - diagram for a Carnot (ideal) refrigerator P, kPa q out qinqin 3 4 2 1, m 3 /kg

55 Second Law – Dream Engine Carnot Principles For heat engines in contact with the same hot and cold reservoir All reversible engines have the same performance. Real engines will have lower performance than the ideal engines.

56 Second Law Steam Power Plants High T Res., T H Furnace q in = q H net,out Low T Res., T L Water from river An Energy-Flow diagram for a Carnot SPPs q out = q L Working fluid: Not a factor P1: 1 = 2 = 3 P2: real < rev

57 Second Law Rev. Fridge/ Heat Pump High T Res., T H, Kitchen room / Outside house net,in Low Temperature Res., T L, Inside fridge or house An Energy-Flow diagram for Carnot Fridge/Heat Pump Working fluid: Not a factor q in = q L q out = q H


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