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 email: drjjlanita@hotmail.com http://www.uitm.edu.my/staff/drjj/ Copyrights DR JJ, ASERG, FSG, UiTM Shah Alam, 2005

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

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% Copyrights DR JJ, ASERG, FSG, UiTM Shah Alam, 2005

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 12/4/2018 Copyrights DR JJ, ASERG, FSG, UiTM Shah Alam, 2005

Introduction Objectives: State the conservation of energy principle. 12/4/2018 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. Copyrights DR JJ, ASERG, FSG, UiTM Shah Alam, 2005

Introduction Objectives: 12/4/2018 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

Introduction Objectives: 12/4/2018 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

Introduction Objectives: 12/4/2018 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

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? 12/4/2018 Copyrights DR JJ, ASERG, FSG, UiTM Shah Alam, 2005

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 12/4/2018 Copyrights DR JJ, ASERG, FSG, UiTM Shah Alam, 2005

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 12/4/2018 Copyrights DR JJ, ASERG, FSG, UiTM Shah Alam, 2005

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.

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 12/4/2018 Copyrights DR JJ, ASERG, FSG, UiTM Shah Alam, 2005

First Law – Energy Transfer Movable boundary position gone up System expands E1= U1+KE1+PE1 System System, E1 A change has taken place. 12/4/2018 Copyrights DR JJ, ASERG, FSG, UiTM Shah Alam, 2005

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 12/4/2018 Copyrights DR JJ, ASERG, FSG, UiTM Shah Alam, 2005

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 12/4/2018 Copyrights DR JJ, ASERG, FSG, UiTM Shah Alam, 2005

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 12/4/2018 Copyrights DR JJ, ASERG, FSG, UiTM Shah Alam, 2005

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 12/4/2018 Copyrights DR JJ, ASERG, FSG, UiTM Shah Alam, 2005

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 12/4/2018 Copyrights DR JJ, ASERG, FSG, UiTM Shah Alam, 2005

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 12/4/2018 Copyrights DR JJ, ASERG, FSG, UiTM Shah Alam, 2005

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 12/4/2018 Copyrights DR JJ, ASERG, FSG, UiTM Shah Alam, 2005

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

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 12/4/2018 Copyrights DR JJ, ASERG, FSG, UiTM Shah Alam, 2005

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 12/4/2018 Copyrights DR JJ, ASERG, FSG, UiTM Shah Alam, 2005

First Law of Thermodynamics 12/4/2018 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 12/4/2018 Copyrights DR JJ, ASERG, FSG, UiTM Shah Alam, 2005 Copyrights DR JJ, ASERG, FSG, UiTM Shah Alam, 2005

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. 12/4/2018 Copyrights DR JJ, ASERG, FSG, UiTM Shah Alam, 2005

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 12/4/2018 Copyrights DR JJ, ASERG, FSG, UiTM Shah Alam, 2005

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 12/4/2018 Copyrights DR JJ, ASERG, FSG, UiTM Shah Alam, 2005

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 12/4/2018 Copyrights DR JJ, ASERG, FSG, UiTM Shah Alam, 2005

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 12/4/2018 Copyrights DR JJ, ASERG, FSG, UiTM Shah Alam, 2005

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 12/4/2018 Copyrights DR JJ, ASERG, FSG, UiTM Shah Alam, 2005

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

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 + 0 + 0, kJ/kg 12/4/2018 Copyrights DR JJ, ASERG, FSG, UiTM Shah Alam, 2005

First Law – Closed System Energy Balance –Closed system Qin – Qout+ Win – Wout+ 0 - 0 = DU+ DKE + DPE, kJ qin – qout+ win – wout + 0 – 0 = Du+ Dke + Dpe, kJ/kg 12/4/2018 Copyrights DR JJ, ASERG, FSG, UiTM Shah Alam, 2005

First Law – Stationary & Closed Energy Balance –Stationary Closed system Qin – Qout+ Win – Wout+ 0 - 0 = DU + 0 + 0 qin – qout+ win – wout+ 0 - 0 = Du + 0 + 0, kJ/kg 12/4/2018 Copyrights DR JJ, ASERG, FSG, UiTM Shah Alam, 2005

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) 12/4/2018 Copyrights DR JJ, ASERG, FSG, UiTM Shah Alam, 2005

First Law – Adiabatic Process Energy Balance Stationary Closed system - Special Adiabatic: 0 – 0 + Win – Wout+ 0 - 0 = DU + 0 + 0 0 – 0 + win – wout+ 0 - 0 = Du + 0 + 0, kJ/kg win = welec + wpw + wb,compress and wout = wb,expand kJ/kg 12/4/2018 Copyrights DR JJ, ASERG, FSG, UiTM Shah Alam, 2005

First Law – Adiabatic Process Energy Balance Closed Stationary system - Special System Adiabatic: win – wout = Du + 0 + 0, 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 12/4/2018 Copyrights DR JJ, ASERG, FSG, UiTM Shah Alam, 2005

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 12/4/2018 Copyrights DR JJ, ASERG, FSG, UiTM Shah Alam, 2005

First Law – Cyclic Process Energy Balance Closed Stationary system - Special Cyclic: Qin – Qout+ Win – Wout+ 0 - 0 = DEsys = 0 qin – qout+ win – wout+ 0 - 0 = 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 12/4/2018 Copyrights DR JJ, ASERG, FSG, UiTM Shah Alam, 2005

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 12/4/2018 Copyrights DR JJ, ASERG, FSG, UiTM Shah Alam, 2005 3-4

First Law – Cyclic Process Energy Balance Closed Stationary system - Special Cyclic: Qin – Qout+ Win – Wout+ 0 - 0 = DEsys = 0 qin – qout+ win – wout+ 0 - 0 = Desys = 0, kJ/kg qin - qout = wout – win Compression: 0 - win = -wb,compress = qin – qout = - qout All compression work is removed as heat 12/4/2018 Copyrights DR JJ, ASERG, FSG, UiTM Shah Alam, 2005

First Law – Isochoric Process Energy Balance Closed Stationary system - Special Isochoric Rigid Tank : Qin – Qout+ Win – Wout+ 0 - 0 = DU qin – qout+ win – wout+ 0 - 0 = Du, kJ/kg Since, win - wout = wothers + 0 = welec + wpw+ 0 + 0 Then, qin - qout + welec + wpw+ 0 + 0 = Du 12/4/2018 Copyrights DR JJ, ASERG, FSG, UiTM Shah Alam, 2005

First Law - Isobaric Process Energy Balance Closed Stationary system - Special Isobaric Piston-Cyl : Qin – Qout+ Win – Wout+ 0 - 0 = DU qin – qout+ win – wout+ 0 - 0 = Du, kJ/kg For expansion, win - wout = welec + wpw - wb,expand qin – qout+ welec + wpw + 0 - 0 = wb,expand + Du, kJ/kg 12/4/2018 Copyrights DR JJ, ASERG, FSG, UiTM Shah Alam, 2005

First Law - Isobaric Process Energy Balance Closed Stationary system - Special Isobaric expansion qin – qout+ win – wout+ 0 - 0 = Du, kJ/kg qin – qout+ welec + wpw + 0 - 0 = wb,expand + Du, kJ/kg 12/4/2018 Copyrights DR JJ, ASERG, FSG, UiTM Shah Alam, 2005

First Law - Isobaric Process Energy Balance Closed Stationary system - Special Isobaric expansion 12/4/2018 Copyrights DR JJ, ASERG, FSG, UiTM Shah Alam, 2005

First Law - Closed System Energy Balance Closed Stationary system qin – qout+ win – wout+ 0 - 0 = 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. 12/4/2018 Copyrights DR JJ, ASERG, FSG, UiTM Shah Alam, 2005

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 12/4/2018 Copyrights DR JJ, ASERG, FSG, UiTM Shah Alam, 2005

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) 12/4/2018 Copyrights DR JJ, ASERG, FSG, UiTM Shah Alam, 2005

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 12/4/2018 Copyrights DR JJ, ASERG, FSG, UiTM Shah Alam, 2005

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 12/4/2018 Copyrights DR JJ, ASERG, FSG, UiTM Shah Alam, 2005

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 12/4/2018 Copyrights DR JJ, ASERG, FSG, UiTM Shah Alam, 2005

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) 12/4/2018 Copyrights DR JJ, ASERG, FSG, UiTM Shah Alam, 2005

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. 12/4/2018 Copyrights DR JJ, ASERG, FSG, UiTM Shah Alam, 2005

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= f@150 kPa, = 0.0010528 m3/kg h1= hf@150 kPa, = 467.1 kJ/kg 2= [f + x2vfg]@150 kPa, = 0.579938 m3/kg h2= [hf + xhfg]@150 kPa =1580 kJ/kg 12/4/2018 Copyrights DR JJ, ASERG, FSG, UiTM Shah Alam, 2005

First Law –Example – Prob 4-20 Energy balance, Ein - Eout = DEsys 0 + Win + 0 – 0 – Wout - 0 = DU+ 0 + 0, kJ Wpw,in + We,in= DU + Wout where Then 12/4/2018 Copyrights DR JJ, ASERG, FSG, UiTM Shah Alam, 2005

First Law –Example – Prob 4-20 Energy balance, Ein - Eout = DEsys Since Electrical work done is Voltage source is 12/4/2018 Copyrights DR JJ, ASERG, FSG, UiTM Shah Alam, 2005

First Law –Example – Prob 4-20 Energy balance, Ein - Eout = DEsys Voltage source is Note that the unit kJ/s = Volts-Ampere or VA 12/4/2018 Copyrights DR JJ, ASERG, FSG, UiTM Shah Alam, 2005