Thermodynamics Lecture Series Assoc. Prof. Dr. J.J. Pure substances – Property tables and Property Diagrams & Ideal Gases Applied Sciences Education Research Group (ASERG) Faculty of Applied Sciences Universiti Teknologi MARA email: drjjlanita@hotmail.com http://www5.uitm.edu.my/faculties/fsg/drjj1.html

Properties of Pure Substances- Part 2 CHAPTER 2 Pure substance Properties of Pure Substances- Part 2 Send self-assessments to: Thermopre@salam.uitm.edu.my Thermopost@salam.uitm.edu.my

"What we have to learn to do, we learn by doing." -Aristotle 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." - Oscar Wilde "What we have to learn to do, we learn by doing." -Aristotle

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

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

Example: A steam power cycle. Turbine Mechanical Energy to Generator Heat Exchanger Cooling Water Pump Fuel Air Combustion Products System Boundary for Thermodynamic Analysis Steam Power Plant

Phase Change of Water - Pressure Change 100 P, kPa , m3/kg 1 30 C T = 30 C P = 100 kPa H2O: C. liquid T = 30 C P = 4.246 kPa H2O: Sat. liquid 2 = f @ 30 °C 1 = f@ 30 °C 2 = f@30 °C 4.246 Tsat@100 kPa = 99.63 C Psat@30 C = 4.246 kPa Water when pressure is reduced

Phase Change of Water - Pressure Change T = 30 C P = 4.246 kPa H2O: Sat. liquid 2 = f@ 30 °C , m3/kg 1 100 P, kPa 4.246 30 C 2 = f@30 °C 1 = f@ 30 °C H2O: Sat. Liq. Sat. Vapor 3 = [f + x f g]@ 30 °C 3 Tsat@100 kPa = 99.63 C Psat@30 C = 4.246 kPa Water when pressure is reduced

Phase Change of Water - Pressure Change H2O: Sat. Liq. Sat. Vapor T = 30 C P = 4.246 kPa 1 3 P, kPa , m3/kg 2 = f@ 30 °C 100 4.246 30 C 3 = [f + x f g]@ 30 °C 1 = f@ 30 °C 2 = f@ 30 °C H2O: Sat. Vapor 4 = g@ 30 °C Tsat@100 kPa = 99.63 C Psat@30 C = 4.246 kPa 4 = g@ 30 °C Water when pressure is reduced

Phase Change of Water - Pressure Change 4 = g@ 30 °C 2 = f@100 kPa , m3/kg 1 3 2 = f@ 30 °C 100 P, kPa 4.246 30 C 3 = [f + x f g]@ 30 °C 4 = g@ 30 °C 1 = f@ 30 °C 2 = f@ 30 °C H2O: Sat. Vapor T = 30 C P = 4.246 kPa H2O: Super Vapor T = 30 C P = 2 kPa 5= @2kPa, 30 °C 2 5 Tsat@100 kPa = 99.63 C Psat@30 C = 4.246 kPa Water when pressure is reduced

Phase Change of Water - Pressure Change T = 30 C P = 100 kPa H2O: C. liquid T = 30 C P = 4.246 kPa H2O: Sat. liquid H2O: Sat. Liq. Sat. Vapor T = 30 C P = 4.246 kPa H2O: Sat. Vapor T = 30 C P = 4.246 kPa H2O: Super Vapor T = 30 C P = 2 kPa Tsat@100 kPa = 99.63 C Psat@30 C = 4.246 kPa Water when pressure is reduced

Phase Change of Water- Pressure Change , m3/kg 1 100 P, kPa 30 C Tsat@100 kPa = 99.63 C Psat@30 C = 4.246 kPa 1 = f@ 30 °C 2 = f@ 30 °C 2 = f@ 30 °C 4.246 3 = [f + x f g]@ 30 °C 4 = g@ 30 °C 5= @2kPa, 30 °C 2 5 3 4 = g@ 30 °C Compressed liquid: Good estimation for properties by taking y = yf@T where y can be either , u, h or s.

Phase Change of Water g@100 C P, C , m3/kg 101.35 100 C f@100 C 1,553.8 200 C 1.2276 10 C

P-  diagram with respect to the saturation lines Phase Change of Water P, C , m3/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

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

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

Property Table Saturated water – Temperature table Tsat, C 10 50 100 200 300 374.14 Sat. P. P, kPa 1.2276 12.349 P, MPa 0.10235 1.5538 8.581 22.09 Specific volume, m3/kg f, m3/kg g, m3/kg 0.001000 106.38 0.001012 12.03 0.001044 1.6729 0.001157 0.13736 0.001404 0.02167 0.003155 Specific internal energy, kJ/kg uf, kJ/kg ufg, kJ/kg ug, kJ/kg 42.00 2347.2 2389.2 209.32 2234.2 2443.5 418.94 2087.6 2506.5 850.65 1744.7 2593.3 1332.0 1231.0 2563.0 2029.6

Property Table Saturated water – Temperature table Tsat, C 10 50 100 300 374.14 Specific internal energy, kJ/kg uf, kJ/kg ufg, kJ/kg ug, kJ/kg 42.00 2347.2 2389.2 209.32 2234.2 2443.5 418.94 2087.6 2506.5 1332.0 1231.0 2563.0 2029.6 Specific enthalpy, kJ/kg hf, kJ/kg hfg, kJ/kg hg, kJ/kg 42.01 2377.7 2519.8 209.33 2382.7 2592.1 419.04 2257.0 2676.1 1332.0 1940.7 2793.2 2099.3 Enthalpy H = U+PV, kJ. or h=u+P, kJ/kg: Sum of internal energy and flow energy. Latent heat of vaporization.

T < Tsat or P > Psat T – v diagram - Example P, kPa T,  C 50 70 Tsat, C 81.33 Psat, kPa 31.19 , m3/kg f@70 C Phase, Y? Compressed Liquid, T < Tsat or P > Psat T, C , m3/kg 50 kPa 81.3 3.240 0.001030 70 =f@70 C = 0.001023

P > Psat or T < Tsat P – v diagram - Example P, kPa T,  C 50 70 Psat, kPa 31.19 Tsat, C 81.33 , m3/kg f@70 C Phase, Y? Compressed Liquid, P > Psat or T < Tsat P, kPa , m3/kg 70 C =f@70 C = 0.001023 50 31.19 5.042 0.001023

P-  diagram with respect to the saturation lines P – v diagram - Example P, kPa T,  C 200 400 Phase, Why? Sup. Vap., T >Tsat Psat, kPa Tsat, C NA 120.2 , m3/kg 1.5493 P, kPa , m3/kg 22,090.0 400 C P-  diagram with respect to the saturation lines 200 f@200 kPa = 0.001061 g@200 kPa = 0.8857 120.2 C  = 1.5493

T-  diagram with respect to the saturation lines T – v diagram - Example P, kPa u, kJ/kg 1,000 2,000 Psat, kPa Tsat, C 179.9 T,  C 179.9 Phase, Why? Wet Mix., uf < u < ug T, C , m3/kg 1,000 kPa T-  diagram with respect to the saturation lines 374.1 f@1,000 kPa = 0.001127 179.9 g@1,000 kPa = 0.19444   = [f + x f g]@1,000 kPa

Saturated Liquid-Vapor Mixture Given the pressure, P, then T = Tsat, yf < y <yg H2O: Sat. Liq. Sat. Vapor Vapor Phase:, Vg, mg, g, ug, hg Liquid Phase:, Vf, mf, f, uf, hf Mixture:, V, m, , u, h, x Mixture’s quality More vapor, higher quality x = 0 for saturated liquid x = 1 for saturated vapor Specific volume of mixture?? Since V=m 

Saturated Liquid-Vapor Mixture Given the pressure, P, then T = Tsat, yf < y <yg H2O: Sat. Liq. Sat. Vapor Vapor Phase:, Vg, mg, g, ug, hg Liquid Phase:, Vf, mf, f, uf, hf Mixture:, V, m, , u, h, x Mixture’s quality Divide by total mass, mt where

Saturated Liquid-Vapor Mixture Given the pressure, P, then T = Tsat, yf < y <yg H2O: Sat. Liq. Sat. Vapor Vapor Phase:, Vg, mg, g, ug, hg Liquid Phase:, Vf, mf, f, uf, hf Mixture:, V, m, , u, h, x Mixture’s quality where y can be , u, h where 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.

Interpolation: Example – Refrigerant-134a P, kPa , m3/kg 200 0.10600 Phase, Why? Sup. Vap.,  >  g Psat, kPa Tsat, C - -10.09 T,  C ?? Assume properties are linearly dependent. Perform interpolation in superheated vapor phase. T, C , m3/kg TH L TL  H T = ?? m2  m1

Interpolation: Example – Refrigerant-134a P, kPa , m3/kg 200 0.10600 Phase, Why? Sup. Vap.,  >  g Psat, kPa Tsat, C - -10.09 T,  C ?? T,  C , m3/kg u, kJ/kg TL= 0 L = 0.10438 uL = 229.23 0 < TL < TH  = 0.10600 0 < uL < uH TH= 10 H = 0.10922 uH = 237.05 Assume properties are linearly dependent. Perform interpolation in superheated vapor phase.

Interpolation: Example – Refrigerant-134a P, kPa , m3/kg 200 0.10600 Phase, Why? Sup. Vap.,  >  g Psat, kPa Tsat, C - -10.09 T,  C 3.35 u, kJ/kg 231.85 P, kPa , m3/kg 22,090.0  = 0.10600 T = 3.35 C Psat ?? f@200 kPa = 0.0007532 g@200 kPa = 0.0993 200 -10.09 C P-  diagram with respect to the saturation lines

Property Table Saturated water – Temperature table Tsat, C 10 50 100 200 300 374.14 Sat. P. P, kPa 1.2276 12.349 P, MPa 0.10235 1.5538 8.581 22.09 Specific volume, m3/kg f, m3/kg g, m3/kg 0.001000 106.38 0.001012 12.03 0.001044 1.6729 0.001157 0.13736 0.001404 0.02167 0.003155 Specific internal energy, kJ/kg uf, kJ/kg ufg, kJ/kg ug, kJ/kg 42.00 2347.2 2389.2 209.32 2234.2 2443.5 418.94 2087.6 2506.5 850.65 1744.7 2593.3 1332.0 1231.0 2563.0 2029.6

Property Table Saturated water – Temperature table Tsat, C 10 50 100 300 374.14 Specific internal energy, kJ/kg uf, kJ/kg ufg, kJ/kg ug, kJ/kg 42.00 2347.2 2389.2 209.32 2234.2 2443.5 418.94 2087.6 2506.5 1332.0 1231.0 2563.0 2029.6 Specific enthalpy, kJ/kg hf, kJ/kg hfg, kJ/kg hg, kJ/kg 42.01 2377.7 2519.8 209.33 2382.7 2592.1 419.04 2257.0 2676.1 1332.0 1940.7 2793.2 2099.3 Enthalpy H = U+PV, kJ. or h=u+P, kJ/kg: Sum of internal energy and flow energy. Latent heat of vaporization.

Sample Questions (a)Briefly, give a description that best fit each of the following: i)process ii)cyclic process iii) saturated vapor iv) kinetic energy ANSWER: (2 marks each) (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.

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.   ANSWER: System P, kPa Tsat, C f, m3/kg g,m3/kg Water 30 300 (12 marks)

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.   ANSWER: System P, kPa Tsat, C f, m3/kg g,m3/kg Water 30 69.1 0.001022 5.229 300 133.55 0.001073 0.6058 (12 marks)

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.]  

Sample Questions (5 marks) Substance P kPa Psat@T T C Tsat@P Quality   Substance P kPa Psat@T T C Tsat@P Quality x u kJ/kg Phase & Reason Water 200 - 0.5 (5 marks)

Sample Questions ANSWER: (5 marks) Substance P kPa Psat@T T C Tsat@P   Substance P kPa Psat@T T C Tsat@P Quality x u kJ/kg Phase & Reason Water 200 - 120.23 0.5 u=uf+xufg Wet mix 0 < x <1 (5 marks)

Sample Questions iii) Refrigerant 134a (6 marks) P kPa Psat@T T C   P kPa Psat@T T C Tsat@P Quality x u kJ/kg Phase & Reason 200 -16 (6 marks)

T<Tsat or P>Psat Sample Questions ANSWER:   P kPa Psat@T T C Tsat@P Quality x u kJ/kg Phase & Reason 200 157.48 -16 -10.09 NA uf@-16C = 29.18 Comp. Liquid T<Tsat or P>Psat (6 marks)

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, Tsat, P, Psat, , f and g.   Use the space below to draw a T -  diagram for the system in part (iii) of the table.

Sample Questions (8 marks) P, kPa 200 -10.09 C 157.48 -16 C , m3/kg Accepted pairs of saturated values and must be marked correctly on the graph in m3/kg are: f@200 kPa = 0.0007532 g @200 kPa = 0.0993 f@-16 C = 0.0007428 g @-16 C = 0.1247   200 157.48  = f =0.0007428 g = 0.1247 -16 C , m3/kg P, kPa -10.09 C (8 marks)

Sample Questions Heat is isothermally transferred into a piston-cylinder device containing water at 100 oC, 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 T1, T2, Tsat, P1, P2, 1, 2, f, and g.  

Sample Questions (12 marks) T, C 400 kPa 101.35 kPa 50 kPa 143.63 1 Accepted pairs of saturated values and must be marked correctly on the graph in m3/kg are: f@100 C = 0.001044 g @100 C = 1.6729 f@400 kPa = 0.001084 g @400 kPa = 0.4625 f@50 kPa = 0.001030 g @50 kPa = 3.240   T, C 143.63 81.33 1 = 0.001084 50 kPa 400 kPa , m3/kg 0.4625 2 = 3.418 100 101.35 kPa 2 1 1 = f@100 C = 0.001044 (12 marks)

Properties of Pure Substances- Ideal Gases Equation of State

Ideal Gases Equation 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

Equation of State for ideal gas Ideal Gases Equation of State for ideal gas Boyle’s Law: Pressure of gas is inversely proportional to its specific volume P Equation of State for ideal gas Charles’s Law: At low pressure, volume is proportional to temperature

and where M is molar mass Ideal Gases Equation of State for ideal gas Combining Boyles and Charles laws: where gas constant R is and where M is molar mass and where Ru is universal gas constant Ru = 8.314 kJ/kmol.K EQOS: Since the total volume is V = m, so :  = V/m So, EQOS: since the mass m = MN where N is number of moles: So,

Hence real gases satisfying conditions P << Pcr, T >> Tcr Ideal Gases Equation of State for ideal gas Real gases with low densities behaves like an ideal gas Hence real gases satisfying conditions P << Pcr, T >> Tcr Obeys EQOS where, Ru = 8.314 kJ/kmol.K, m = MN V = m and

Gas Mixtures – Ideal Gases High density Low density Molecules far apart Real Gases Low density (mass in 1 m3) gases Molecules are further apart Real gases satisfying condition Pgas << Pcrit; Tgas >> Tcrit , have low density and can be treated as ideal gases

Gas Mixtures – Ideal Gases Real Gases Equation of State - P--T behaviour P = RT (energy contained by 1 kg mass) where  is the specific volume in m3/kg, R is gas constant, kJ/kgK, T is absolute temp in Kelvin. High density Low density Molecules far apart

Gas Mixtures – Ideal Gases Real Gases Equation of State - P--T behaviour P=RT , since = V/m then, P(V/m) =RT. So, PV = mRT, in kPam3=kJ. Total energy of a system. High density Low density

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

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

T-  diagram with respect to the saturation lines T – v diagram - Example P, kPa u, kJ/kg 1,000 2,000 Psat, kPa Tsat, C 179.9 T,  C 179.9 Phase, Why? Wet Mix., uf < u < ug T, C , m3/kg 1,000 kPa T-  diagram with respect to the saturation lines 374.1 f@1,000 kPa = 0.001127 179.9 g@1,000 kPa = 0.19444   = [f + x f g]@1,000 kPa