1. Thermodynamics Lecture Series
Assoc. Prof. Dr. J.J.
Second Law – Quality of Energy-Part1
Applied Sciences Education Research Group (ASERG)
Faculty of Applied Sciences
Universiti Teknologi MARA

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

4. Symbols Q q W  V  E 

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

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

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

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

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

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

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

12. Review - First Law Properties will change indicating change of state
How to relate changes to the cause
System
E1, P1, T1, V1 To
E2, P2, T2, V2
Win
Wout
Mass in
Mass out
Qin
Qout
Dynamic Energies as causes (agents) of change

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

14. Review - First Law Energy Balance Ein – Eout = Esys, kJ or
Energy Entering a system -
Energy Leaving a system =
Change of system’s energy
Energy Balance
Ein – Eout = Esys, kJ or
ein – eout = esys, kJ/kg or

15. Change of system’s mass
Review - First Law
Mass Entering a system -
Mass Leaving a system =
Change of system’s mass
Mass Balance
min – mout = msys, kg or

16. Energy Balance – Control Volume Steady-Flow
Review - First Law
Energy Balance – Control Volume Steady-Flow
Steady-flow is a flow where all properties within boundary of the system remains constant with time
DEsys= 0, kJ; Desys= 0 , kJ/kg, DVsys= 0, m3; Dmsys= 0 or min = mout , kg

17. Mass & Energy Balance–Steady-Flow CV
Review - First Law
Mass & Energy Balance–Steady-Flow CV
Mass balance
Energy balance
qin + win + qin = qout+ wout+ qout, kJ/kg

18. Mass & Energy Balance–Steady-Flow: Single Stream
Review - First Law
Mass & Energy Balance–Steady-Flow: Single Stream
Mass balance
Energy balance
qin – qout+ win – wout = qout – qin, kJ/kg
= hout - hin + keout– kein + peout - pein, kJ/kg

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

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

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

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

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

24. First Law of Thermodynamics
Heat Exchanger
Qout
Case 2
Mass balance
Energy balance

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

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

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

28. Second Law 0 -qout+ 0 - 0 = -Du = u1 - u2, kJ/kg
First Law involves quantity or amount of energy to be conserved in processes
Qout
Tsys,initial=40C
OK for this cup
Tsurr=25C
Tsys,final=25C
This is a natural process!!!
Q flows from high T to low T
medium until thermal
equilibrium is reached

29. Second Law qin – 0 + 0 - 0 = Du = u2 - u1, kJ/kg Tsys,initial=25C
Tsurr=25C
Tsys,final=40C
This is NOT a natural process!!!
Q does not flow from low T to
high T medium. Never will the
coffee return to its initial state.

30. Second Law qin – 0 + 0 - 0 = Du = u2 - u1, kJ/kg Tsys,initial=25C
Tsurr=25C
Tsys,final=40C
This is NOT a natural process!!!
Q does not flow from low T to
high T medium. Never will the
coffee return to its initial state.

31. Second Law qin – 0 + 0 - 0 = Du =u2-u1 kJ/kg
First Law involves quantity or amount of energy to be conserved in processes
Qin
Tsys,initial=25C
Tsurr=25C
Tsys,final=40C
This is a NOT a natural process!!!
Q does not flow from low T to
high T medium. Never will
equilibrium be reached
But is the process in this cup possible??

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

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

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

35. Second Law Heat Engine Characteristics:
Receive heat from a high T source.
Convert part of the heat into work.
Reject excess heat into a low T sink.
Operates in a cycle.

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

37. An Energy-Flow diagram for a SPP
Second Law
Working fluid:
Water
High T Res., TH
Furnace
Purpose:
Produce work,
Wout, out
qin = qH
qin - qout = out - in
qin = net,out + qout
Steam Power Plant
net,out
net,out = qin - qout
qout = qL
Low T Res., TL
Water from river
An Energy-Flow diagram for a SPP

38. A Schematic diagram for a Steam Power Plant
Second Law
Working fluid:
Water
High T Res., TH
Furnace
qin = qH
Boiler
Turbine
Pump
out
in
Condenser
qout = qL
qin - qout = out - in
Low T Res., TL
Water from river
qin - qout = net,out
A Schematic diagram for a Steam Power Plant

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
Working fluid:
Ref-134a
High T Res., TH, Kitchen room / Outside house
qout – qin = in - out
qout = qH
Refrigerator/ Air Cond
net,in
qin = qL
net,in = qout - qin
Purpose:
Maintain space
at low T by
Removing qL
Low Temperature
Res., TL, Inside fridge or house
net,in = qH - qL
An Energy-Flow diagram for a Refrigerator/Air Cond.

43. A Schematic diagram for a Refrigerator/Air Cond.
Second Law
Working fluid:
Refrigerant-134a
High T Res., TH
Kitchen/Outside house
qout = qH
pressor
Com
Condenser
in
Throttle Valve
Evaporator
qin = qL
Low T Res., TL
Ref. Space/Room
A Schematic diagram for a Refrigerator/Air Cond.

44. Second Law Coefficient of Performance for a Refrigerator
Divide top and
bottom by qin

45. Second Law
Coefficient of Performance for a Refrigerator

46. An Energy-Flow diagram for a Heat Pump
Second Law
Working fluid:
Ref-134a
High Temperature
Res., TH, Inside house
Purpose:
Maintain space
at high T by
supplying qH
qout = net,in + qin
qout = qH
Heat Pump
net,in
net,in = qout - qin
qin = qL
net,in = qH - qL
Low Temperature
Res., TL, Outside house
An Energy-Flow diagram for a Heat Pump

47. A Schematic diagram for a Heat Pump
Second Law
Working fluid:
Refrigerant-134a
High T Res., TH
Inside house
qout = qH
pressor
Com
Condenser
in
Throttle Valve
Evaporator
qin = qL
Low T Res., TL
Outside the house
A Schematic diagram for a Heat Pump

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

51. Second Law – Dream Engine
Carnot Cycle
Isothermal expansion
Slow adding of Q resulting in work done by system (system expand)
Qin – Wout = U = 0. So, Qin = Wout . Pressure drops.
Adiabatic expansion
0 – Wout = 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
- Qout + Win = U = 0. So, Qout = Win .
Pressure increases.
Adiabatic compression
0 + Win = 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
, m3/kg
qout
qin 2 3 4 1

54. Second Law – Dream Engine
Reverse Carnot Cycle
P, kPa
qout
qin 3 4 2 1
, m3/kg
P -  diagram for a Carnot (ideal) refrigerator

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. An Energy-Flow diagram for a Carnot SPPs
Second Law
Working fluid:
Not a factor
High T Res., TH
Furnace
qin = qH
P1: 1 = 2 = 3
Steam Power Plants
net,out
P2: real < rev
qout = qL
Low T Res., TL
Water from river
An Energy-Flow diagram for a Carnot SPPs

57. Second Law Rev. Fridge/ Heat Pump
Working fluid:
Not a factor
High T Res., TH, Kitchen room / Outside house
qout = qH
Rev. Fridge/ Heat Pump
net,in
qin = qL
Low Temperature
Res., TL, Inside fridge or house
An Energy-Flow diagram for Carnot Fridge/Heat Pump