Applied Sciences Education Research Group (ASERG) Faculty of Applied Sciences Universiti Teknologi MARA First Law of Thermodynamics & Energy Balance – Control Mass, Open System Thermodynamics Lecture Series

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CHAPTER 4 The First Law of Thermodynamics

Introduction 1.State the conservation of energy principle. 2.Write an energy balance for a general system undergoing any process. 3.Write the unit-mass basis and unit-time basis (or rate-form basis) energy balance for a general system undergoing any process. 4.Identify the energies causing the system to change. Objectives:

Introduction 5.Identify the energy changes within the system. 6.Write the energy balance in terms of all the energies causing the change and the energy changes within the system. 7.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. Objectives:

Introduction 8.State the conditions for stationary, closed system and rewrite the energy balance and the unit-mass basis energy balance for stationary- closed systems. 9.Apply the energy conservation principle for a stationary, closed system undergoing an adiabatic process and discuss its physical interpretation. Objectives:

Introduction 10.Apply the energy conservation principle for a stationary, closed system undergoing an isochoric, isothermal, cyclic and isobaric process and discuss its physical interpretation. 11.Give the meaning for specific heat and state its significance in determining internal energy and enthalpy change for ideal gases, liquids and solids. 12.Use the energy balance for problem solving. Objectives:

3-1 Energy Transfer -Heat Transfer Lem Oven 200  C Nasi Lemak 20  C Nasi Lemak 20  C q in H 2 O: Sat. Liq. Sat. Vapor P = 100 kPa T = 99.6  C P = 100 kPa T = 99.6  C Q in What happens to the properties of the system after the energy transfer? SODA 5  C SODA 5  C SODA 5  C SODA 5  C q in q out 25  C

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 System Boundary for Thermodynamic Analysis Steam Power Plant Q out W in W out Q in The net work output is Desiredoutput Requiredinput

Energy Transfer – Work Done i i Voltage, V No heat transfer T increases after some time No heat transfer T increases after some time H 2 O: Super Vapor H 2 O: Super Vapor Mechanical work: Piston moves up Boundary work is done by system Mechanical work: Piston moves up Boundary work is done by system Electrical work is done on system H 2 O: Sat. liquid W pw,in,kJ W e,in = Vi  t/100, kJ

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 4-8 FIGURE 4-46 Pipe or duct flow may involve more than one form of work at the same time.

First Law – Energy Transfer System in thermal equilibrium System Total energy E 1 Can it change? How? Why? System’s initial total energy is E 1 = E 1 = U 1 +KE 1 +PE 1 U 1 +KE 1 +PE 1 or e 1 = u 1 +ke 1 +pe 1, kJ/kg

First Law – Energy Transfer A change has taken place. System, E 1 System E 1 = U 1 +KE 1 +PE 1 Movable boundary position gone up System expands

First Law – Energy Transfer A change has taken place System, E 1 System Initial Final System’s final energy is E 2 =U 2 +KE 2 +PE 2 E 1 = U 1 +KE 1 +PE 1 Movable boundary position gone up System expands

First Law – Energy Transfer How to relate changes to the cause Heat as a cause (agent) of change System E 1, P 1, T 1, V 1 To q in, or Q in q out, or, Q out Properties will change indicating change of state E 2, P 2, T 2, V 2

First Law – Energy Transfer Work as a cause (agent) of change System E 1, P 1, T 1, V 1 To Properties will change indicating change of state W in,  in, kJ/kg W out,  in, kJ/kg How to relate changes to the cause E 2, P 2, T 2, V 2

First Law – Energy Transfer How to relate changes to the cause Mass transfer as a cause (agent) of change System E 1, P 1, T 1, V 1 To Properties will change indicating change of state Mass out Mass in E 2, P 2, T 2, V 2

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

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 Known as Conservation of Energy Principle In any process, every bit of energy should be accounte d for!! z =h z =0 z =h/2 E=U+KE+PE = U+PE

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 Known as Conservation of Energy Principle In any process, every bit of energy should be accounte d for!! z =h z =0 z =h/2E=U+KE+PE E=U+KE+PE=U+0+PE

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 Known as Conservation of Energy Principle In any process, every bit of energy should be accounte d for!! z =h z =0 z =h/2E=U+KE+PE E=U+KE+0

First Law Energy Balance 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 system’s energy

First Law of Thermodynamics 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 system’s energy

First Law of Thermodynamics 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

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 4-1 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.

First Law – Interaction Energies Energy Balance – The Agent E in = Q in +W in +E mass,in,kJ For Closed system: E mass,in = 0, kJ,  in = 0, kJ/kg e in = q in +  in +  in, kJ/kg

First Law - Interaction Energies Energy Balance – The Agent E out = Q out +W out +E mass,out,kJ For Closed system: E mass,out = 0, kJ,  out = 0, kJ/kg e out = q out +  out +  out, kJ/kg

First Law - System’s Energy Energy Balance – The Change WIthin Energy change within the system,  E sys = E 2 -E 1 Internal energy change,  U = U 2 – U 1 kinetic energy change,  KE = KE 2 – KE 1 potential energy change,  PE = PE 2 – PE 1 is the sum of

First Law – Energy Change Energy Balance – The Change WIthin  E sys =  U+  KE+  PE, kJ For Stationary system:  KE=  PE = 0, kJ  e sys =  u+  ke+  pe, kJ/kg  ke=  pe=0, kJ/kg

First Law – General Energy Balance 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 system’s energy

First Law – General Energy Balance Energy Balance – General system Q in + W in + E mass,in – Q out – W out - E mass,out q in +  in +  in – q out –  out –  out =  U+  KE  +  PE, kJ =  u+  ke  +  pe, kJ/kg

First Law – Stationary System Energy Balance – Stationary system Q in – Q out + W in – W out + E mass,in - E mass,out q in – q out +  in –  out +  in -  out =  U+  +  =  u +  + , kJ/kg

First Law – Closed System Energy Balance – Closed system =  U+  KE  +  PE, kJ =  u+  ke  +  pe, kJ/kg Q in – Q out + W in – W out q in – q out +  in –  out +  – 

First Law – Stationary & Closed Energy Balance – Stationary Closed system Q in – Q out + W in – W out =  U +  +  q in – q out +  in –  out +  -  =  u +  + , kJ/kg

First Law – Historical Perspective Energy Balance – Stationary Closed system-History = Q net,in – W net,out q –  q net,in –  net,out = (q in –q out) – (  out –  in ) (Q in – Q out ) - (W out - W in ) (Q in – Q out ) + (W in - W out ) = = Q – W

First Law – Adiabatic Process Energy Balance Stationary Closed system - Special Adiabatic Adiabatic: 0 – 0 + W in – W out =  U +  +  0 – 0 +  in –  out +  -  =  u +  + , kJ/kg  in =  elec +  pw +  b,compress and  out =  b,expand kJ/kg

First Law – Adiabatic Process Energy Balance Closed Stationary system - Special Adiabatic Adiabatic:  in –  out =  u +  + , kJ/kg Work is expansion work: 0 -  out = -  b,expand =  u < 0  in =  and  out =  b,expand kJ/kg u 2 < u 1. Final u is smaller than initial u, T drops Spontaneous Expansion: piston- cylinder device System

First Law – Adiabatic Process Energy Balance Closed Stationary system - Special Adiabatic Adiabatic:  in –  out -  =  u+  + , kJ/kg Work is compression work:  in =  b,compress =  u > 0  out =  and  in =  b,compress kJ/kg u 2 > u 1. Final u is bigger than initial u; T increases Compression: piston-cylinder device System

First Law – Cyclic Process Energy Balance Closed Stationary system - Special Cyclic Cyclic: Q in – Q out + W in – W out =  E sys = 0 q in – q out +  in –  out +  -  =  e sys = 0, kJ/kg q in - q out = Expansion: q in - 0 =  out =  b,expand All heat absorbed is used to do expansion work or q net,in =  net,out  out –  in

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 3-4 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 Total work is area of A minus area of B. Total work is shaded area Input power Output power Q in Q out

First Law – Cyclic Process Energy Balance Closed Stationary system - Special Compression: 0 -  in = -  b,compress = q in – q out All compression work is removed as heat q in - q out =  out –  in Cyclic Cyclic: Q in – Q out + W in – W out =  E sys = 0 q in – q out +  in –  out +  -  =  e sys = 0, kJ/kg = - q out

First Law – Isochoric Process Energy Balance Closed Stationary system - Special : Q in – Q out + W in – W out =  U q in – q out +  in –  out +  -  =  u, kJ/kg Since,  in -  out =  others + 0 =  elec +  pw Then, q in - q out +  elec +  pw =  u Isochoric Rigid Tank

First Law - Isobaric Process Energy Balance Closed Stationary system - Special : Q in – Q out + W in – W out =  U q in – q out +  in –  out +  -  =  u, kJ/kg For expansion,  in -  out =  elec +  pw -  b,expand q in – q out +  elec +  pw +  -  =  b,expand +  u, kJ/kg IsobaricPiston-Cyl

First Law - Isobaric Process Energy Balance Closed Stationary system - Special Isobaric expansion q in – q out +  in –  out +  -  =  u, kJ/kg q in – q out +  elec +  pw +  -  =  b,expand +  u, kJ/kg

First Law - Isobaric Process Energy Balance Closed Stationary system - Special Isobaric expansion

First Law - Closed System Energy Balance Closed Stationary system For pure substances, use property table and mathematical manipulations to determine u 2 and u 1. Then For pure substances, use property table and mathematical manipulations to determine u 2 and u 1. Then  u = u 2 – u 1. q in – q out +  in –  out +  -  =  u, kJ/kg And h 2 and h 1,. Then And h 2 and h 1,. Then  h = h 2 – h 1.

First Law –Specific Heat C, C p, Energy Balance Closed Stationary system C, C p, Specific heats to find  U and  H Specific Heat Capacity C at constant volume, C p, at constant pressure C at constant volume, C p, at constant pressure Amount of heat necessary to increase temperature of a unit mass by 1K or 1degree Celcius

First Law –Specific Heat For Ideal Gases: Estimate internal energy change and enthalpy change Assume smooth change of C with T, & approximate to be linear over small  T  T (approx. a few hundred degrees) C, C p, Energy Balance Closed Stationary system C, C p,

First Law –Specific Heat For Ideal Gases : Estimate internal energy change and enthalpy change C, avg is determined using interpolation technique & use Table A-2b Assume smooth change of C with T, & approximate to be linear over small  T (approx. a few hundred degrees) C, C p, Energy Balance Closed Stationary system C, C p,

First Law –Specific Heat For solids & Liquids: May consider as incompressible or constant volume Cv Cv Cv Cv = C p C p = C, kJ/kg  K  K, Hence the enthalpy change, C, C p, Energy Balance Closed Stationary system C, C p, Enthalpy h = u + P, So, dh =du + dP dP +Pd, +Pd, kJ/kg  u  u = C av C av (T 2 (T 2 - T 1 ), T 1 ), kJ/kg  h  h = C av C av  T  T +  P P + P , P , kJ/kg  h  h =  u  u +  P P + P , P , kJ/kg

First Law –Specific Heat For solids & liquids:  u = C av (T 2 - T 1 ), kJ/kg Since,  P P = 0, 0, Then,  h  h  C av C av  T,  T, kJ/kg C, C p, Energy Balance Closed Stationary system C, C p, C v = C p = C, kJ/kg  K, For solids: solids: Enthalpy,  h  h = C av C av  T  T +  P P + 0, 0, kJ/kg

First Law –Specific Heat For Liquids:  u = C av (T 2 - T 1 ), kJ/kg Enthalpy,  h = C av  T + P + 0, kJ/kg Heaters Heaters where  P  P = 0,0,0,0, C, C p, Energy Balance Closed Stationary system C, C p, Pumps Pumps where  T  T = 0,  h  h =  u  u  C av C av  T,  T, kJ/kg  h  h =  P, P, kJ/kg Or written as h2 h2 h2 h2 - h 1 = h 1 = (P 2 (P 2 - P1)P1)P1)P1)

First Law –Example – Prob 4-20 Initial: V = 5 L phase sat. liq. P = 150 kPa, Final phase is sat. liq.-vapor mix at 150 kPa. kPa. Quality of steam is 0.5. Since mg mg mg mg = m/2, hence x = m g /m m g /m = m/2m =0.5. P 1 = P 2 H2OH2OH2OH2O W pw,in,kJ W pw,in = 300 kJ, i i Voltage, V Current, i = 8 A,  t = 45 min x 60s/min  t = 45 x 60s

First Law –Example – Prob 4-20 Initial: V = 5 L phase sat. liq. P = 150 kPa, 1 = 1 = kPa, kPa, = m 3 /kg P 1 = P 2 H2OH2OH2OH2O W pw,in,kJ W pw,in = 300 kJ, i i Voltage, V Current, i = 8 A,  t = 45 min x 60s/min  t = 45 x 60s 2 = 2 = [ f [ f + x 2 v fg x 2 v fg kPa, kPa, = m 3 /kg h 2 = [h f + xh fg kPa =1580 kJ/kg h 1 = h kPa, = kJ/kg h 1 = h kPa, = kJ/kg x 2 = 0.5

First Law –Example – Prob 4-20 E in - E out =  E sys Energy balance, 0 + W in + 0 – 0 – W out - 0 =  U+  + , kJ W pw,in + W e,in =  U + W out where Then

First Law –Example – Prob 4-20 E in - E out =  E sys Energy balance, Electrical work done is Since Voltage source is

First Law –Example – Prob 4-20 E in - E out =  E sys Energy balance, Voltage source is Note that the unit kJ/s = Volts-Ampere or VA