1. Thermodynamics Lecture Series
12/4/2018
Thermodynamics Lecture Series
Assoc. Prof. Dr. J.J.
First Law of Thermodynamics & Energy Balance – Control Mass, Open System
Applied Sciences Education Research Group (ASERG)
Faculty of Applied Sciences
Universiti Teknologi MARA
Copyrights DR JJ, ASERG, FSG, UiTM Shah Alam, 2005

2. 12/4/2018
Quotes
"One who learns by finding out has sevenfold the skill of the one who learned by being told.“ - Arthur Gutterman
"The roots of education are bitter, but the fruit is sweet." -Aristotle
Copyrights DR JJ, ASERG, FSG, UiTM Shah Alam, 2005

3. 12/4/2018
Research Findings
Research Findings – Retention % of Learning After 3 days period
Read only – 10%
See only 30%
Hear only – 20%
See + hear – 50%
Say only – 70%
Say & do simultaneously - 90%
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4. Copyrights DR JJ, ASERG, FSG, UiTM Shah Alam, 2005
CHAPTER 4
The First Law of Thermodynamics
Goal: Identifying sources of energy interactions and write energy balances thermodynamic processes
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5. Introduction Objectives: State the conservation of energy principle.
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Introduction
Objectives:
State the conservation of energy principle.
Write an energy balance for a general system undergoing any process.
Write the unit-mass basis and unit-time basis (or rate-form basis) energy balance for a general system undergoing any process.
Identify the energies causing the system to change.
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6. Introduction Objectives:
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Introduction
Objectives:
Identify the energy changes within the system.
Write the energy balance in terms of all the energies causing the change and the energy changes within the system.
Write a unit-mass basis and unit-time basis (or rate-form basis) energy balance in terms of all the energies causing the change and the energy changes within the system.
Copyrights DR JJ, ASERG, FSG, UiTM Shah Alam, 2005

7. Introduction Objectives:
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Introduction
Objectives:
State the conditions for stationary, closed system and rewrite the energy balance and the unit-mass basis energy balance for stationary-closed systems.
Apply the energy conservation principle for a stationary, closed system undergoing an adiabatic process and discuss its physical interpretation.
Copyrights DR JJ, ASERG, FSG, UiTM Shah Alam, 2005

8. Introduction Objectives:
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Introduction
Objectives:
Apply the energy conservation principle for a stationary, closed system undergoing an isochoric, isothermal, cyclic and isobaric process and discuss its physical interpretation.
Give the meaning for specific heat and state its significance in determining internal energy and enthalpy change for ideal gases, liquids and solids.
Use the energy balance for problem solving.
Copyrights DR JJ, ASERG, FSG, UiTM Shah Alam, 2005

9. Energy Transfer -Heat Transfer
H2O:
Sat. Liq.
Sat. Vapor
P = 100 kPa
T = 99.6 C
Qin
Lem
Oven
200C
Nasi
Lemak 20C
qin
SODA
5C
qout
25C
qin
What happens to
the properties of
the system after the
energy transfer?
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10. Copyrights DR JJ, ASERG, FSG, UiTM Shah Alam, 2005
Example: A steam power cycle.
Steam
Turbine
Mechanical Energy
to Generator
Heat
Exchanger
Cooling Water
Pump
Fuel
Air
Combustion
Products
System Boundary
for Thermodynamic
Analysis
Steam Power Plant
Qin
Required
input
Wout
Desired
output
Qout
Win
The net work output is
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11. Energy Transfer – Work Done
Mechanical work:
Piston moves up
Boundary work is
done by system
Electrical work is done on system
H2O:
Super
Vapor
No heat transfer
T increases
after some time
Wpw,in ,kJ
H2O:
Sat. liquid i
Voltage, V
We,in = Vit/100, kJ
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Copyrights DR JJ, ASERG, FSG, UiTM Shah Alam, 2005

12. Copyright © The McGraw-Hill Companies, Inc
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
FIGURE 4-46 Pipe or duct flow may involve more than one form of work at the same time.

13. First Law – Energy Transfer
Can it change? How? Why?
System’s initial total energy is
E1= U1+KE1+PE1 or
System
Total energy E1
e1= u1+ke1+pe1, kJ/kg
System in thermal equilibrium
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14. First Law – Energy Transfer
Movable boundary position gone up
System expands
E1= U1+KE1+PE1
System
System, E1
A change has taken place.
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Copyrights DR JJ, ASERG, FSG, UiTM Shah Alam, 2005

15. First Law – Energy Transfer
Movable boundary position gone up
System
expands
System
E1= U1+KE1+PE1
System
System, E1
Final
Initial
System’s final energy is E2=U2+KE2+PE2
A change has taken place
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Copyrights DR JJ, ASERG, FSG, UiTM Shah Alam, 2005

16. First Law – Energy Transfer
Properties will change indicating change of state
How to relate changes to the cause
System
E1, P1, T1, V1 To
qin, or Qin
qout, or, Qout
E2, P2, T2, V2
Heat as a cause (agent) of change
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17. First Law – Energy Transfer
Properties will change indicating change of state
How to relate changes to the cause
System
E1, P1, T1, V1 To
Win, in, kJ/kg
Wout, in, kJ/kg
E2, P2, T2, V2
Work as a cause (agent) of change
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Copyrights DR JJ, ASERG, FSG, UiTM Shah Alam, 2005

18. First Law – Energy Transfer
Properties will change indicating change of state
How to relate changes to the cause
System
E1, P1, T1, V1 To
Mass in
Mass out
E2, P2, T2, V2
Mass transfer as a cause (agent) of change
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19. First Law – Energy Transfer
Properties will change indicating change of state
How to relate changes to the cause
System
E1, P1, T1, V1 To
Win
Wout
Mass in
Mass out
E2, P2, T2, V2
Qin
Qout
Dynamic Energies as causes (agents) of change
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Copyrights DR JJ, ASERG, FSG, UiTM Shah Alam, 2005

20. First Law-Conservation of Energy Principle
Energy must be conserved in any process. Energy cannot be created nor destroyed. It can only change forms. Total Energy before a process must equal total energy after process
In any process, every bit of energy should be accounted for!!
E=U+KE+PE = U+PE
z =h
z =h/2
z =0
Known as Conservation of Energy Principle
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Copyrights DR JJ, ASERG, FSG, UiTM Shah Alam, 2005

21. First Law-Conservation of Energy Principle
Energy must be conserved in any process. Energy cannot be created nor destroyed. It can only change forms. Total Energy before a process must equal total energy after process
In any process, every bit of energy should be accounted for!!
z =h
E=U+KE+PE=U+0+PE
z =h/2
E=U+KE+PE
z =0
Known as Conservation of Energy Principle
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Copyrights DR JJ, ASERG, FSG, UiTM Shah Alam, 2005

22. First Law-Conservation of Energy Principle
Energy must be conserved in any process. Energy cannot be created nor destroyed. It can only change forms. Total Energy before a process must equal total energy after process
In any process, every bit of energy should be accounted for!!
z =h
z =h/2
E=U+KE+PE
E=U+KE+0
z =0
Known as Conservation of Energy Principle
12/4/2018
Copyrights DR JJ, ASERG, FSG, UiTM Shah Alam, 2005

23. 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
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24. First Law of Thermodynamics
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
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Copyrights DR JJ, ASERG, FSG, UiTM Shah Alam, 2005

25. First Law of Thermodynamics
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First Law of Thermodynamics
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
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Copyrights DR JJ, ASERG, FSG, UiTM Shah Alam, 2005

26. Copyrights DR JJ, ASERG, FSG, UiTM Shah Alam, 2005
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
FIGURE 4–7 The energy change of a system during a process is equal to the net work and heat transfer between the system and its surroundings.
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Copyrights DR JJ, ASERG, FSG, UiTM Shah Alam, 2005

27. First Law – Interaction Energies
Energy Balance – The Agent
Ein = Qin+Win+Emass,in ,kJ
ein = qin+ in+ qin, kJ/kg
For Closed system: Emass,in = 0, kJ, qin= 0, kJ/kg
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Copyrights DR JJ, ASERG, FSG, UiTM Shah Alam, 2005

28. First Law - Interaction Energies
Energy Balance – The Agent
E out = Q out +W out +Emass,out ,kJ
eout = qout+ out+ qout, kJ/kg
For Closed system: Emass,out = 0, kJ, qout= 0, kJ/kg
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Copyrights DR JJ, ASERG, FSG, UiTM Shah Alam, 2005

29. First Law - System’s Energy
Energy Balance – The Change WIthin
Energy change within the system, DEsys = E2-E1
is the sum of
Internal energy change, DU = U2 – U1
kinetic energy change, DKE = KE2 – KE1
potential energy change, DPE = PE2 – PE1
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Copyrights DR JJ, ASERG, FSG, UiTM Shah Alam, 2005

30. First Law – Energy Change
Energy Balance – The Change WIthin
DEsys = DU+DKE+DPE, kJ
Desys = Du+Dke+Dpe, kJ/kg
For Stationary system: DKE=DPE = 0, kJ
Dke=Dpe=0, kJ/kg
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Copyrights DR JJ, ASERG, FSG, UiTM Shah Alam, 2005

31. First Law – General Energy Balance
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
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Copyrights DR JJ, ASERG, FSG, UiTM Shah Alam, 2005

32. First Law – General Energy Balance
Energy Balance –General system
Qin + Win + Emass,in – Qout – Wout - Emass,out
= DU+ DKE + DPE, kJ
qin + win + qin – qout – wout – qout
= Du+ Dke + Dpe, kJ/kg
12/4/2018
Copyrights DR JJ, ASERG, FSG, UiTM Shah Alam, 2005

33. First Law – Stationary System
Energy Balance –Stationary system
Qin – Qout+ Win – Wout+ Emass,in - Emass,out
= DU+0+0
qin – qout+ win – wout+ qin - qout
= Du , kJ/kg
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Copyrights DR JJ, ASERG, FSG, UiTM Shah Alam, 2005

34. First Law – Closed System
Energy Balance –Closed system
Qin – Qout+ Win – Wout
= DU+ DKE + DPE, kJ
qin – qout+ win – wout + 0 – 0
= Du+ Dke + Dpe, kJ/kg
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35. First Law – Stationary & Closed
Energy Balance –Stationary Closed system
Qin – Qout+ Win – Wout = DU
qin – qout+ win – wout = Du , kJ/kg
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36. First Law – Historical Perspective
Energy Balance –Stationary Closed system-History
(Qin – Qout) + (Win - Wout) =
(Qin – Qout) - (Wout - Win)
= Qnet,in – Wnet,out
= Q – W
q – w = qnet,in – wnet,out = (qin –qout) – (wout – win)
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37. First Law – Adiabatic Process
Energy Balance Stationary Closed system - Special
Adiabatic: 0 – 0 + Win – Wout = DU
0 – 0 + win – wout = Du , kJ/kg
win = welec + wpw + wb,compress
and wout = wb,expand kJ/kg
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Copyrights DR JJ, ASERG, FSG, UiTM Shah Alam, 2005

38. First Law – Adiabatic Process
Energy Balance Closed Stationary system - Special
System
Adiabatic: win – wout = Du , kJ/kg
System
Spontaneous Expansion: piston-cylinder device
win = 0, and wout = wb,expand kJ/kg
Work is expansion work: 0 - wout = -wb,expand = Du < 0
u2 < u1. Final u is smaller than initial u, T drops
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Copyrights DR JJ, ASERG, FSG, UiTM Shah Alam, 2005

39. First Law – Adiabatic Process
Energy Balance Closed Stationary system - Special
System
Adiabatic: win – wout - 0 = Du+0+0, kJ/kg
System
Compression: piston-cylinder device
wout = 0, and win = wb,compress kJ/kg
Work is compression work: win = wb,compress = Du > 0
u2 > u1. Final u is bigger than initial u; T increases
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Copyrights DR JJ, ASERG, FSG, UiTM Shah Alam, 2005

40. First Law – Cyclic Process
Energy Balance Closed Stationary system - Special
Cyclic: Qin – Qout+ Win – Wout = DEsys = 0
qin – qout+ win – wout = Desys = 0, kJ/kg
qin - qout =
wout – win
or qnet,in = wnet,out
Expansion: qin - 0 =wout = wb,expand
All heat absorbed is used to do expansion work
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41. Work done - Cyclic process
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
FIGURE 3-22 The net work done during a cycle is the difference between the work done by the system and the work done on the system.
Work done - Cyclic process
Total work is area of A minus area
of B. Total work is shaded area
Input power
Qin
Qout
Output power
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3-4

42. First Law – Cyclic Process
Energy Balance Closed Stationary system - Special
Cyclic: Qin – Qout+ Win – Wout = DEsys = 0
qin – qout+ win – wout = Desys = 0, kJ/kg
qin - qout =
wout – win
Compression: 0 - win = -wb,compress = qin – qout
= - qout
All compression work is removed as heat
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Copyrights DR JJ, ASERG, FSG, UiTM Shah Alam, 2005

43. First Law – Isochoric Process
Energy Balance Closed Stationary system - Special
Isochoric
Rigid Tank
: Qin – Qout+ Win – Wout = DU
qin – qout+ win – wout = Du, kJ/kg
Since, win - wout = wothers + 0 = welec + wpw
Then, qin - qout + welec + wpw = Du
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44. First Law - Isobaric Process
Energy Balance Closed Stationary system - Special
Isobaric
Piston-Cyl
: Qin – Qout+ Win – Wout = DU
qin – qout+ win – wout = Du, kJ/kg
For expansion, win - wout = welec + wpw - wb,expand
qin – qout+ welec + wpw = wb,expand + Du, kJ/kg
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45. First Law - Isobaric Process
Energy Balance Closed Stationary system - Special
Isobaric expansion
qin – qout+ win – wout = Du, kJ/kg
qin – qout+ welec + wpw = wb,expand + Du, kJ/kg
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46. First Law - Isobaric Process
Energy Balance Closed Stationary system - Special
Isobaric expansion
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47. First Law - Closed System
Energy Balance Closed Stationary system
qin – qout+ win – wout = Du, kJ/kg
For pure substances, use property table and mathematical manipulations to determine u2 and u1. Then Du = u2 – u1.
And h2 and h1, . Then Dh = h2 – h1.
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48. First Law –Specific Heat
Energy Balance Closed Stationary system C  , Cp,
Specific heats to find DU and DH
Specific Heat Capacity
C  at constant volume, Cp, at constant pressure
Amount of heat necessary to increase temperature of a unit mass by 1K or 1degree Celcius
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49. First Law –Specific Heat
Energy Balance Closed Stationary system C  , Cp,
For Ideal Gases: Estimate internal energy change and enthalpy change
Assume smooth change of C with T, & approximate to be linear over small DT (approx. a few hundred degrees)
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50. First Law –Specific Heat
Energy Balance Closed Stationary system C  , Cp,
For Ideal Gases : Estimate internal energy change and enthalpy change
Assume smooth change of C with T, & approximate to be linear over small DT (approx. a few hundred degrees)
C , avg is determined using interpolation technique & use Table A-2b
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51. First Law –Specific Heat
Energy Balance Closed Stationary system C  , Cp,
For solids & Liquids: May consider as incompressible or constant volume
Cv = Cp = C , kJ/kg K ,
Du = Cav (T2 - T1), kJ/kg
Enthalpy h = u + P,
So, dh =du + dP +Pd, kJ/kg
Dh = Du + DP + PD, kJ/kg
Hence the enthalpy change,
Dh = Cav DT + DP + PD, kJ/kg
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Copyrights DR JJ, ASERG, FSG, UiTM Shah Alam, 2005

52. First Law –Specific Heat
Energy Balance Closed Stationary system C  , Cp,
For solids & liquids: Du = Cav (T2 - T1), kJ/kg
Cv = Cp = C , kJ/kg K ,
For solids: Enthalpy, Dh = Cav DT + DP + 0, kJ/kg
Since, DP = 0, Then, Dh  Cav DT, kJ/kg
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Copyrights DR JJ, ASERG, FSG, UiTM Shah Alam, 2005

53. First Law –Specific Heat
Energy Balance Closed Stationary system C  , Cp,
For Liquids: Du = Cav (T2 - T1), kJ/kg
Enthalpy, Dh = Cav DT + DP + 0, kJ/kg
Heaters where DP = 0,
Dh = Du  Cav DT, kJ/kg
Pumps where DT = 0,
Dh = DP, kJ/kg
Or written as h2 - h1= (P2 - P1)
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54. First Law –Example – Prob 4-20
H2O
Initial: V = 5 L
phase sat. liq. P = 150 kPa,
Wpw,in ,kJ
Wpw,in = 300 kJ, i
Voltage, V
Current, i = 8 A,
Dt = 45 min x 60s/min
Dt = 45 x 60s
P1= P2
Final phase is sat. liq.-vapor mix at 150 kPa. Quality of steam is 0.5. Since mg = m/2, hence x = mg/m = m/2m =0.5.
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55. First Law –Example – Prob 4-20
H2O
Initial: V = 5 L
phase sat. liq. P = 150 kPa,
Wpw,in ,kJ
Wpw,in = 300 kJ, i
Voltage, V
Current, i = 8 A,
Dt = 45 min x 60s/min
Dt = 45 x 60s
P1= P2
x2 = 0.5
1= kPa, = m3/kg
h1= kPa, = kJ/kg
2= [f + kPa, = m3/kg
h2= [hf + kPa =1580 kJ/kg
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56. First Law –Example – Prob 4-20
Energy balance,
Ein - Eout = DEsys
0 + Win + 0 – 0 – Wout - 0 = DU , kJ
Wpw,in + We,in= DU + Wout
where
Then
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57. First Law –Example – Prob 4-20
Energy balance,
Ein - Eout = DEsys
Since
Electrical work done is
Voltage source is
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58. First Law –Example – Prob 4-20
Energy balance,
Ein - Eout = DEsys
Voltage source is
Note that the unit kJ/s = Volts-Ampere or VA
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