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Thermodynamics Lecture Series Applied Sciences Education.

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Presentation on theme: "Thermodynamics Lecture Series Applied Sciences Education."— Presentation transcript:

1 Thermodynamics Lecture Series Applied Sciences Education Research Group (ASERG) Faculty of Applied Sciences Universiti Teknologi MARA Pure substances – Property tables and Property Diagrams & Ideal Gases

2 CHAPTER 2 Properties of Pure Substances- Part 2 Pure substance Send self-assessments to: Send self-assessments to:

3 Quotes "Education is an admirable thing, but it is well to remember from time to time that nothing that is worth knowing can be taught taught." - Oscar Wilde we learn by doing "What we have to learn to do, we learn by doing." -Aristotle

4 Introduction 1.Choose the right property table to read and to determine phase and other properties. 2.Derive and use the mathematical relation to determine values of properties in the wet-mix phase 3.Sketch property diagrams with respect to the saturation lines, representing phases, processes and properties of pure substances. Objectives:

5 Introduction 4.Use an interpolation technique to determine unknown values of properties in the superheated vapor region 5.State conditions for ideal gas behaviour 6.Write the equation of state for an ideal gas in many different ways depending on the units. 7.Use all mathematical relations and skills of reading the property table in problem-solving. Objectives:

6 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

7 Phase Change of Water - Pressure Change Water when pressure is reduced 2 = °C T = 30 C P = 100 kPa T = 30 C P = 100 kPa H 2 O: C. liquid T = 30 C P = kPa T = 30 C P = kPa H 2 O: Sat. liquid 100 P, kPa, m 3 /kg 1 30 C 2 = 30 °C 1 = 30 °C T kPa = C P C = kPa T kPa = C P C = kPa

8 Phase Change of Water - Pressure Change H 2 O: Sat. Liq. Sat. Vapor 3 T = 30 C P = kPa T = 30 C P = kPa H 2 O: Sat. liquid Water when pressure is reduced 2 = 30 °C, m 3 /kg P, kPa C 3 = [ f + x f g 30 °C 2 = °C 1 = 30 °C T kPa = C P C = kPa T kPa = C P C = kPa

9 Phase Change of Water - Pressure Change 4 = 30 °C H 2 O: Sat. Vapor H 2 O: Sat. Vapor H 2 O: Sat. Liq. Sat. Vapor T = 30 C P = kPa T = 30 C P = kPa Water when pressure is reduced 1 3 P, kPa, m 3 /kg 2 = 30 °C C 4 = 30 °C 3 = [ f + x f g 30 °C 1 = 30 °C 2 = 30 °C T kPa = C P C = kPa T kPa = C P C = kPa

10 Phase Change of Water - Pressure Change H 2 O: Sat. Vapor H 2 O: Sat. Vapor T = 30 C P = kPa T = 30 C P = kPa 2 5 H 2 O: Super Vapor H 2 O: Super Vapor T = 30 C P = 2 kPa T = 30 C P = 2 kPa Water when pressure is reduced 4 = 30 °C 2 = kPa, m 3 /kg = 30 °C P, kPa C 3 = [ f + x f g 30 °C 4 = 30 °C 1 = 30 °C 2 = 30 °C 5 30 °C T kPa = C P C = kPa T kPa = C P C = kPa

11 Phase Change of Water - Pressure Change H 2 O: Sat. Vapor H 2 O: Sat. Vapor T = 30 C P = kPa T = 30 C P = 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 = kPa T = 30 C P = kPa T = 30 C P = 100 kPa T = 30 C P = 100 kPa H 2 O: C. liquid T = 30 C P = kPa T = 30 C P = kPa H 2 O: Sat. liquid Water when pressure is reduced T kPa = C P C = kPa T kPa = C P C = kPa

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

13 Phase Change of Water 1, C C C P, C, m 3 /kg C C

14 Phase Change of Water P, C, m 3 /kg C 1, C 10 C 100 C C 22,090 P- diagram with respect to the saturation lines

15 FIGURE 2-19 P-v diagram of a pure substance. Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 2-4

16 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 2-7 FIGURE 2-25 P-T diagram of pure substances.

17 Property Table Saturated water – Temperature table Temp T sat, C Specific internal energy, kJ/kg u f, kJ/kgu fg, kJ/kgu g, kJ/kg Specific volume, m 3 /kg f, m 3 /kg g, m 3 /kg Sat. P. P, kPa P, MPa

18 Property Table Saturated water – Temperature table Temp T sat, C Specific enthalpy, kJ/kg h f, kJ/kgh fg, kJ/kgh g, kJ/kg Enthalpy H = U+PV, kJ. or h=u+P, kJ/kg: Sum of internal energy and flow energy. Latent heat of vaporization. Specific internal energy, kJ/kg u f, kJ/kgu fg, kJ/kgu g, kJ/kg

19 T, C, m 3 /kg T – v diagram - Example 70 = C = kPa P, kPa T, C 5070 Phase, Y? Compressed Liquid, T P sat, m 3 /kg C T sat, C P sat, kPa

20 P, kPa, m 3 /kg P – v diagram - Example P, kPa T, C 5070 Phase, Y? Compressed Liquid, P > P sat or T < T sat, m 3 /kg C 70 C T sat, C P sat, kPa = C =

21 400 C P – v diagram - Example P, kPa T, C P- diagram with respect to the saturation lines Phase, Why? Sup. Vap., T >T sat P sat, kPa T sat, C NA120.2 P, kPa, m 3 /kg 22,090.0 = , m 3 /kg kPa = kPa = C

22 T – v diagram - Example T, C, m 3 /kg 1,000 kPa P, kPau, kJ/kg 1,0002,000 T- diagram with respect to the saturation lines Phase, Why? Wet Mix., u f < u < u g P sat, kPa T sat, C kPa = kPa = T, C179.9 = [ f + x f g kPa

23 Saturated Liquid-Vapor Mixture H 2 O: Sat. Liq. Sat. Vapor Given the pressure, P, then T = T sat, y f < y { "@context": "http://schema.org", "@type": "ImageObject", "contentUrl": "http://images.slideplayer.com/1650779/7/slides/slide_22.jpg", "name": "Saturated Liquid-Vapor Mixture H 2 O: Sat.Liq. Sat.", "description": "Vapor Given the pressure, P, then T = T sat, y f < y

24 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 { "@context": "http://schema.org", "@type": "ImageObject", "contentUrl": "http://images.slideplayer.com/1650779/7/slides/slide_23.jpg", "name": "Saturated Liquid-Vapor Mixture H 2 O: Sat.Liq. Sat.", "description": "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

25 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 { "@context": "http://schema.org", "@type": "ImageObject", "contentUrl": "http://images.slideplayer.com/1650779/7/slides/slide_24.jpg", "name": "Saturated Liquid-Vapor Mixture H 2 O: Sat.Liq. Sat.", "description": "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

26 Interpolation: Example – Refrigerant-134a T, C, m 3 /kg THTH L TLTL H T = ?? m1m1 m2m2 P, kPa, m 3 /kg Phase, Why? Sup. Vap., > g P sat, kPa T sat, C T, C ?? Assume properties are linearly dependent. Perform interpolation in superheated vapor phase. Assume properties are linearly dependent. Perform interpolation in superheated vapor phase.

27 Interpolation: Example – Refrigerant-134a P, kPa, m 3 /kg Assume properties are linearly dependent. Perform interpolation in superheated vapor phase. Phase, Why? Sup. Vap., > g P sat, kPa T sat, C T, C ?? T, C, m 3 /kg u, kJ/kg T L = 0 L = u L = < T L < T H = < u L < u H T H = 10 H = u H =

28 Interpolation: Example – Refrigerant-134a P, kPa, m 3 /kg P- diagram with respect to the saturation lines Phase, Why? Sup. Vap., > g P sat, kPa T sat, C P, kPa, m 3 /kg 22,090.0 T, C3.35 kPa = kPa = C u, kJ/kg = T = 3.35 C P sat ??

29 Property Table Saturated water – Temperature table Temp T sat, C Specific internal energy, kJ/kg u f, kJ/kgu fg, kJ/kgu g, kJ/kg Specific volume, m 3 /kg f, m 3 /kg g, m 3 /kg Sat. P. P, kPa P, MPa

30 Property Table Saturated water – Temperature table Temp T sat, C Specific enthalpy, kJ/kg h f, kJ/kgh fg, kJ/kgh g, kJ/kg Enthalpy H = U+PV, kJ. or h=u+P, kJ/kg: Sum of internal energy and flow energy. Latent heat of vaporization. Specific internal energy, kJ/kg u f, kJ/kgu fg, kJ/kgu g, kJ/kg

31 Sample Questions (a)Briefly, give a description that best fit each of the following: i)processii)cyclic process iii) saturated vaporiv) kinetic energy (a)Briefly, give a description that best fit each of the following: i)processii)cyclic process iii) saturated vaporiv) kinetic energy (i) A property change or a change of state of a system (ii) A process where the initial and the final state is the same. (i) A property change or a change of state of a system (ii) A process where the initial and the final state is the same. (iii) Vapor ready to condense. (iv) Energy associated with the movement(speed) of a system. (iii) Vapor ready to condense. (iv) Energy associated with the movement(speed) of a system. ANSWER:(2 marks each)

32 Sample Questions Using the property table, read the saturation temperatures and the saturated volumes for water at pressures of 30 kPa and 300 kPa, respectively. (12 marks) System P, kPa T sat, C f, m 3 /kg g, m 3 /kg Water 30 Water 300 ANSWER:

33 Sample Questions Using the property table, read the saturation temperatures and the saturated volumes for water at pressures of 30 kPa and 300 kPa, respectively. (12 marks) System P, kPa T sat, C f, m 3 /kg g, m 3 /kg Water Water ANSWER:

34 Sample Questions Using the property table, determine the missing properties in the table below. [Note: The symbols used are the following: P for pressure, T for temperature, and u for the specific internal energy. When indicating the quality, use NA for the compressed liquid and the superheated vapor phase and use 0 < x < 1 for the wet mix phase. When indicating properties such as, u, h or s in the wet mix phase, just write down an expression on how to get it. NO CALCULATIONS ARE REQUIRED.]

35 Sample Questions (5 marks) Substance P kPa P kPa P kPa P kPa T C T C T P C T P C Quali ty x Quali ty x u kJ/kg u kJ/kg Phase & Reaso n Phase & Reaso n Water

36 Sample Questions (5 marks) ANSWER: Substance P kPa P kPa P kPa P kPa T C T C T P C T P C Quali ty x Quali ty x u kJ/kg u kJ/kg Phase & Reaso n Phase & Reaso n Water u=u f +x u fg Wet mix 0 < x <1 Wet mix 0 < x <1

37 Sample Questions (6 marks) iii) Refrigerant 134a P kPa P kPa P kPa P kPa T C T C T C T C Quality x Quality x u kJ/kg u kJ/kg Phase & Reason Phase & Reason

38 Sample Questions (6 marks) ANSWER: NA u C = u C = Comp. Liquid T P sat Comp. Liquid T P sat P kPa P kPa P kPa P kPa T C T C T C T C Quality x Quality x u kJ/kg u kJ/kg Phase & Reason Phase & Reason

39 Sample Questions In the question that follows, you need to use the data in the table above along with the property table to sketch property diagrams with respect to the saturation lines to indicate the state of the system. In your diagram, draw and label the temperature or the pressure lines. Place an X mark on the graph to indicate the state. Insert values for T, T sat, P, P sat,, f and g. Use the space below to draw a T - diagram for the system in part (iii) of the table.

40 Sample Questions = f = g = C, m 3 /kg P, kPa C Accepted pairs of saturated values and must be marked correctly on the graph in m 3 /kg are: kPa = kPa = C = C = (8 marks)

41 Sample Questions Heat is isothermally transferred into a piston-cylinder device containing water at 100 o C, 400 kPa until the pressure drops to 50 kPa. Use the space below to draw a T - diagram with respect to the saturation lines for this process. Indicate the initial and final states and the direction of the process. Draw and label the temperature lines and insert values for T 1, T 2, T sat, P 1, P 2, 1, 2, f, and g.

42 Sample Questions T, C = kPa 400 kPa, m 3 /kg = kPa = C = Accepted pairs of saturated values and must be marked correctly on the graph in m 3 /kg are: C = C = kPa = kPa = kPa = kPa = (12 marks)

43 Properties of Pure Substances- Ideal Gases Equation of State Ideal Gases

44 Equation of StateEquation of State –An equation relating pressure, temperature and specific volume of a substance. –Predicts P- -T behaviour quite accurately –Any properties relating to other properties –Simplest EQOS of substance in gas phase is ideal-gas (imaginary gas) equation of state

45 Ideal Gases Equation of State for ideal gasEquation of State for ideal gas –Boyles Law: Pressure of gas is inversely proportional to its specific volume P Equation of State for ideal gasEquation of State for ideal gas –Charless Law: At low pressure, volume is proportional to temperature

46 Ideal Gases Equation of State for ideal gasEquation of State for ideal gas –Combining Boyles and Charles laws: and where R u is universal gas constant R u = kJ/kmol.K and where M is molar mass So,So, EQOS: since the mass m = MN where N is number of moles: So,So, where gas constant R is EQOS: Since the total volume is V = m, so : = V/m

47 Ideal Gases Equation of State for ideal gasEquation of State for ideal gas –Real gases with low densities behaves like an ideal gas P > T cr where,where, Hence real gases satisfying conditions R u = kJ/kmol.K, V = m and m = MN Obeys EQOS

48 Real Gases Gas Mixtures – Ideal Gases Low density (mass in 1 m 3 ) gases Molecules are further apart Real gases satisfying condition P gas > T crit P gas > T crit, have low density and can be treated as ideal gases High density Low density Molecules far apart

49 Real Gases Gas Mixtures – Ideal Gases Equation of State Equation of State - P- -T behaviour P = RT R T P = RT (energy contained by 1 kg mass) where is the specific volume in m 3 /kg, R is gas constant, kJ/kg K, T is absolute temp in Kelvin. High density Low density Molecules far apart

50 Real Gases Gas Mixtures – Ideal Gases Equation of State Equation of State - P- -T behaviour P =RT P =RT, since = V/m then, P(V/m) =RT. So, PV = mRT PV = mRT, in kPa m 3 =kJ. Total energy of a system. Low density High density

51 Real Gases Gas Mixtures – Ideal Gases Equation of State Equation of State - P- -T behaviour PV =mRT PV =mRT = NMRT = N(MR)T PV = NR u T Hence, can also write PV = NR u T 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 = kJ/kmol K Low density High density

52 Other EQOS Equations of States Van Der Waals Equation of State Van Der Waals Equation of State a/ 2, P replaced by P+a/ 2 Considers intermolecular interactions, a/ 2, which then increases the pressure: P replaced by P+a/ 2 Low density High density wherewhere EQOS is: b,Replace by -b, then Considers volume occupied by molecules, b, at high pressures: Replace by -b, then

53 T – v diagram - Example T, C, m 3 /kg 1,000 kPa P, kPau, kJ/kg 1,0002,000 T- diagram with respect to the saturation lines Phase, Why? Wet Mix., u f < u < u g P sat, kPa T sat, C kPa = kPa = T, C179.9 = [ f + x f g kPa


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