Applied Sciences Education Research Group (ASERG)

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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 Dynamic Energy Transfer – Heat, Work and Mass Thermodynamics Lecture Series

CHAPTER 2 Properties of Pure Substances- A Review Pure substance

Quotes "Good judgment comes from experience. Experience comes from bad judgment - Anonymous "The roots of education are bitter, but the fruit is sweet." -Aristotle

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

Phase Change of Water Water interacts with thermal energy H 2 O: Sat. Liq. Sat. Vapor P = 100 kPa T = 99.6 C P = 100 kPa T = 99.6 C Q in P = 100 kPa T = 99.6 C P = 100 kPa T = 99.6 C H 2 O: Sat. Vapor H 2 O: Sat. Vapor Q in P = 100 kPa T = 150 C P = 100 kPa T = 150 C H 2 O: Super Vapor H 2 O: Super Vapor Q in P = 100 kPa T = 30 C P = 100 kPa T = 30 C H 2 O: C. liquid Q in P = 100 kPa T = 99.6 C P = 100 kPa T = 99.6 C H 2 O Sat. liquid Q in

Phase Change of Water 99.6 2 = f@100 kPa T, C 30, m 3 /kg 1 4 = g@100 kPa 3 5 = @100 kPa, 150°C 3 = [ f + x f g ] @100 kPa 1 = f@T1 150 100 kPa 5 Compressed liquid: Good estimation for properties by taking y = y f@T where y can be either, u, h or s.

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 2-2 FIGURE 2-16 T-v diagram of constant- pressure phase-change processes of a pure substance at various pressures (numerical values are for water). 99.6 45.8 179.9 T –v diagram: Multiple P

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 2-3 FIGURE 2-18 T-v diagram of a pure substance. T –v diagram: Multiple P

Phase Change of Water - Pressure Change H 2 O: Sat. Vapor H 2 O: Sat. Vapor T = 30 C P = 4.246 kPa T = 30 C P = 4.246 kPa H 2 O: Super Vapor H 2 O: Super Vapor T = 30 C P = 2 kPa T = 30 C P = 2 kPa H 2 O: Sat. Liq. Sat. Vapor T = 30 C P = 4.246 kPa T = 30 C P = 4.246 kPa T = 30 C P = 100 kPa T = 30 C P = 100 kPa H 2 O: C. liquid T = 30 C P = 4.246 kPa T = 30 C P = 4.246 kPa H 2 O: Sat. liquid Water when pressure is reduced T sat@100 kPa = 99.63 C P sat@30 C = 4.246 kPa T sat@100 kPa = 99.63 C P sat@30 C = 4.246 kPa

Phase Change of Water- Pressure Change Compressed liquid: Good estimation for properties by taking y = y f@T where y can be either, u, h or s. 2 = f@ 30 °C 4.246 3 2 5 4 = g@ 30 °C, m 3 /kg 1 100 P, kPa 30 C 3 = [ f + x f g ] @ 30 °C 4 = g@ 30 °C 1 = f@ 30 °C 2 = f@ 30 °C 5 = @2kPa, 30 °C T sat@100 kPa = 99.63 C P sat@30 C = 4.246 kPa T sat@100 kPa = 99.63 C P sat@30 C = 4.246 kPa

Phase Change of Water P, C, m 3 /kg 101.35 g@100 C 1,553.8 1.2276 200 C 10 C 100 C f@100 C 22,090 P- diagram with respect to the saturation lines

Saturated Liquid-Vapor Mixture H 2 O: Sat. Liq. Sat. Vapor Vapor Phase:, V g, m g, g, u g, h g Liquid Phase:, V f, m f, f, u f, h f Mixture:, V, m,, u, h, x Mixtures quality Divide by total mass, m t wherewhere Given the pressure, P, then T = T sat, y f < y <y g Given the pressure, P, then T = T sat, y f < y <y g

Saturated Liquid-Vapor Mixture H 2 O: Sat. Liq. Sat. Vapor Vapor Phase:, V g, m g, g, u g, h g Liquid Phase:, V f, m f, f, u f, h f Mixture:, V, m,, u, h, x Mixtures quality Given the pressure, P, then T = T sat, y f < y <y g Given the pressure, P, then T = T sat, y f < y <y g wherewhere wherewhere If x is known or has been determined, use above relations to find other properties. If either, u, h are known, use it to find quality, x. y can be, u, h

CHAPTER 3 Energy Transfer by Heat, Work, and Mass Goal: Identify forms of energy interactions and ways of representing it in thermodynamics processes

Introduction 1.Identify the types of dynamic energies interacting with a system. 2.Distinguish the difference and relate between heat transfer and thermal energy. 3.Write the different symbols and the conventions used to represent heat transfer. 4.Differentiate between heat transfer and work done. Objectives:

Introduction 5.Write the symbols and convention used for work done. 6.Obtain a mathematical relation representing mechanical work done for any system. 7.Obtain the amount of work done from a P – V or P - graph. 8.Write down the relationship between mass and volume flow rate. Objectives:

Introduction 9.Obtain a mathematical relation representing mass flow rate in terms of the mass velocities and the systems inlet or exit area. 10.Write the specific energy carried by a flowing mass. 11.Use all mathematical relations and graphing skills to solve problems involving interaction energies. Objectives:

3-1 Energy Transfer -Heat Transfer SODA 5 C SODA 5 C q in 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 q out

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 3-1 FIGURE 3-9 Specifying the directions of heat and work. Isochoric Process – Closed System

Isobaric Process – Closed System H 2 O: Sat. Liq. Sat. Vapor P = 100 kPa T = 99.6 C P = 100 kPa T = 99.6 C Q in P = 100 kPa T = 99.6 C P = 100 kPa T = 99.6 C H 2 O: Sat. Vapor H 2 O: Sat. Vapor Q in P = 100 kPa T = 150 C P = 100 kPa T = 150 C H 2 O: Super Vapor H 2 O: Super Vapor Q in P = 100 kPa T = 30 C P = 100 kPa T = 30 C H 2 O: C. liquid Q in P = 100 kPa T = 99.6 C P = 100 kPa T = 99.6 C H 2 O Sat. liquid Q in Source of thermal energy System expands, volume increases

Isobaric Process – Closed System System expands, volume increases H 2 O: Sat. Liq. Sat. Vapor P = 100 kPa T = 99.6 C P = 100 kPa T = 99.6 C P = 100 kPa T = 99.6 C P = 100 kPa T = 99.6 C H 2 O: Sat. Vapor H 2 O: Sat. Vapor P = 100 kPa T = 150 C P = 100 kPa T = 150 C H 2 O: Super Vapor H 2 O: Super Vapor P = 100 kPa T = 30 C P = 100 kPa T = 30 C H 2 O: C. liquid P = 100 kPa T = 99.6 C P = 100 kPa T = 99.6 C H 2 O Sat. liquid Source of thermal energy Q in, kJ

Symbols and Convention – Heat transfer Total heat entering, Q in = 100 kJ, m = 5 kg Specific heat Rate of heat transfer if heat is allowed to interact for 50 seconds

Symbols and Convention – Heat transfer Total heat leaving, Q out = 20 kJ, m = 5 kg Specific heat Rate of heat transfer if heat is allowed to interact for 50 seconds

Symbols and Convention – Heat transfer Net heat transfer Net specific heat transfer Net rate of heat transfer if heat is allowed to interact for 50 seconds Net rate of heat transfer if heat is allowed to interact for 50 seconds H 2 O: Super Vapor H 2 O: Super Vapor Q out Q in

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

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,kJ W e = Vi t, kJ

Symbols and conventions for Work – Isothermal Process H 2 O: Sat. Vapor H 2 O: Sat. Vapor T = 30 C P = 4.246 kPa T = 30 C P = 4.246 kPa H 2 O: Super Vapor H 2 O: Super Vapor T = 30 C P = 2 kPa T = 30 C P = 2 kPa H 2 O: Sat. Liq. Sat. Vapor T = 30 C P = 4.246 kPa T = 30 C P = 4.246 kPa T = 30 C P = 100 kPa T = 30 C P = 100 kPa H 2 O: C. liquid T = 30 C P = 4.246 kPa T = 30 C P = 4.246 kPa H 2 O: Sat. liquid no thermal energy or heat Water when pressure is reduced – no thermal energy or heat T sat@100 kPa = 99.63 C P sat@30 C = 4.246 kPa T sat@100 kPa = 99.63 C P sat@30 C = 4.246 kPa System expands, volume increases-boundary work is done

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Boundary Work, W b, out Final, n FIGURE 3-19 A gas does a differential amount of work W b as it forces the piston to move by a differential amount ds. InitialInitial Total work done by system to expand from initial to final state Total work done by system to expand from initial to final state Work done involves force moving an object in the direction of movement Work done involves force moving an object in the direction of movement 11 22

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Boundary Work, W b, out Pressure exerted on a surface is the ratio of force applied w.r.t. the area of surface Pressure exerted on a surface is the ratio of force applied w.r.t. the area of surface As ds approaches zero Final, n FIGURE 3-19 A gas does a differential amount of work W b as it forces the piston to move by a differential amount ds. InitialInitial 11 22

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Boundary Work, W b, out Final, n FIGURE 3-19 A gas does a differential amount of work W b as it forces the piston to move by a differential amount ds. InitialInitial 11 22 Pressure exerted on a surface is the ratio of force applied w.r.t. the area of surface Pressure exerted on a surface is the ratio of force applied w.r.t. the area of surface ThenThen SoSo ButButHenceHence

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Boundary Work, W b, out Final, n FIGURE 3-19 A gas does a differential amount of work W b as it forces the piston to move by a differential amount ds. InitialInitial 11 22 When the pressure is kept constant, Isobaric process, When the pressure is kept constant, Isobaric process, ThenThen OrOr wherewhere

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Specific Boundary Work, b, out Final, n FIGURE 3-19 A gas does a differential amount of work W b as it forces the piston to move by a differential amount ds. InitialInitial 11 22 When the pressure is kept constant, Isobaric process, When the pressure is kept constant, Isobaric process, ThenThen OrOr wherewhere

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 3-3 FIGURE 3-20 The area under the process curve on a P-V diagram represents the boundary work. Boundary work on a P – V graph

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

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 3-5 FIGURE 3-48 Schematic for flow work. Flow work is energy required to overcome resistance at the Boundary both while entering and leaving the system Flow work is energy required to overcome resistance at the Boundary both while entering and leaving the system Work done – Flow work InletInlet ExitExit

Energy Transfer – Mass Conservation Consider mass flowing through cylinder of length and area A Mass must be conserved in any process. Inlet Exit A A Length, oror For steady flow flow ThenThen

Energy Transfer – Energy of Moving Mass Consider mass flowing through cylinder of length and area A Let the mass flow through for time interval of t, such as 10 seconds Let the mass flow through for time interval of t, such as 10 seconds Let the mass flow with a velocity Let the mass flow with a velocity Since volume is Then volume flow rate is Then volume flow rate is Inlet Exit A A Length, oror Then mass Flow rate Then mass Flow rate

Energy Transfer – Energy of Moving Mass Consider mass flowing through cylinder of length and area A Let the mass flow through for time interval of t, such as 10 seconds Let the mass flow through for time interval of t, such as 10 seconds Let the mass flow with a velocity Let the mass flow with a velocity Since volume is Then volume flow rate is Then volume flow rate is Inlet Exit A A Length, oror Then mass Flow rate Then mass Flow rate

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. FIGURE 3-51 The total energy consists of three parts for a nonflowing fluid and four parts for a flowing fluid. Energy Transfer – Energy of Moving Mass Kinetic energy, ke Potential energy, pe Internal energy, u

Gas Mixtures – Ideal Gases Equation of State Equation of State - P- -T behaviour Low density High density Hence, can also write where N N is no of kilomoles, kmol, M M is molar mass in kg/kmole and R u R u =MR R u is universal gas constant; R u =MR. R u = 8.314 kJ/kmol K where N N is no of kilomoles, kmol, M M is molar mass in kg/kmole and R u R u =MR R u is universal gas constant; R u =MR. R u = 8.314 kJ/kmol K PV =mRT PV = NR u T

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