Thermodynamics II
LECTURE 1
INTRODUCTION
Energy
• Kinetic Energy is found in movement
• GRAVITATIONAL
• CHEMICAL
• NUCLEAR
• ELASTIC
• MOTION
• THERMAL ENERGY AND TEMPERATURE
• Electric
Energy in thermodynamics
• Heat, Work and Internal energy (1st Law of thermodynamics)
• Close (heat and work) and open system (mass, heat and work)
For open system, add flow
work, h = u + pv
Two Phase System
Steam Tables
• Subcooled, saturated and
superheated.
• At T = 220 oC, specific volume of
liquid = 0.001194 m3/Kg and
specific volume of sat vapour =
0.086094 m3/Kg.
• At T = 220 oC and x (quality) =
0.325, specific volume:
v = vf + x vfg = 0.001190 + 0.325 *
0.086094 = 0.02917
• Other variables : v, u, h, s
Interpolation for Intermediate Properties
• You may need double interpolation
Slope of line
Super Heated Table
• At 75 oC, find entropy s. From table
Temperature Entropy S
50 (X1) 8.1488 (Y1)
75 (X) Y = ?
100 (X2) 8.1741 (Y2)
𝑆𝑙𝑜𝑝𝑒 =
𝑌2 − 𝑌1
𝑋2 − 𝑋1
=
8.1741 − 8.1488
100 − 50
= 0.000506
Y= 𝑌1 + 𝑆𝑙𝑜𝑝𝑒 × 𝑋 − 𝑋1
Y= 8.1488 + 0.000506 × 75 − 50 =8.16145
𝑠𝑜 𝑎𝑡 𝑡𝑒𝑚𝑝𝑟𝑎𝑡𝑢𝑟𝑒, 𝑇 = 75 𝑜 𝐶, 𝐸𝑛𝑡𝑜𝑝𝑦 𝑆 = 8.16145
𝑘𝐽
𝑘𝑔. 𝐾
Heat Engine and Second Law of Thermodynamics
Second Law of Thermodynamics
• Kelvin Planck Statement
Second Law of Thermodynamics
• Clausius statement
Entropy
• Disorder of system
• Friction, heat transfer and non equilibrium
process
PV and TS diagrams?
Work done during process or cycle
Polytypic process
What about the adiabatic process?
Work done during process or cycle
Isothermal process
Some Important Formulae
PV n= constant
M= 28.97 g/mol,
therefore Ra =
0.287 kJ/kg.K
𝑄 = 𝑚𝐶 𝑝∆𝑇
𝑄 = 𝑚𝐶𝑣∆𝑇
Extra

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Thermodynamics Lecture I

  • 3. Energy • Kinetic Energy is found in movement • GRAVITATIONAL • CHEMICAL • NUCLEAR • ELASTIC • MOTION • THERMAL ENERGY AND TEMPERATURE • Electric
  • 4. Energy in thermodynamics • Heat, Work and Internal energy (1st Law of thermodynamics) • Close (heat and work) and open system (mass, heat and work) For open system, add flow work, h = u + pv
  • 6. Steam Tables • Subcooled, saturated and superheated. • At T = 220 oC, specific volume of liquid = 0.001194 m3/Kg and specific volume of sat vapour = 0.086094 m3/Kg. • At T = 220 oC and x (quality) = 0.325, specific volume: v = vf + x vfg = 0.001190 + 0.325 * 0.086094 = 0.02917 • Other variables : v, u, h, s
  • 7. Interpolation for Intermediate Properties • You may need double interpolation Slope of line
  • 8. Super Heated Table • At 75 oC, find entropy s. From table Temperature Entropy S 50 (X1) 8.1488 (Y1) 75 (X) Y = ? 100 (X2) 8.1741 (Y2) 𝑆𝑙𝑜𝑝𝑒 = 𝑌2 − 𝑌1 𝑋2 − 𝑋1 = 8.1741 − 8.1488 100 − 50 = 0.000506 Y= 𝑌1 + 𝑆𝑙𝑜𝑝𝑒 × 𝑋 − 𝑋1 Y= 8.1488 + 0.000506 × 75 − 50 =8.16145 𝑠𝑜 𝑎𝑡 𝑡𝑒𝑚𝑝𝑟𝑎𝑡𝑢𝑟𝑒, 𝑇 = 75 𝑜 𝐶, 𝐸𝑛𝑡𝑜𝑝𝑦 𝑆 = 8.16145 𝑘𝐽 𝑘𝑔. 𝐾
  • 9. Heat Engine and Second Law of Thermodynamics
  • 10. Second Law of Thermodynamics • Kelvin Planck Statement
  • 11. Second Law of Thermodynamics • Clausius statement
  • 12. Entropy • Disorder of system • Friction, heat transfer and non equilibrium process PV and TS diagrams?
  • 13. Work done during process or cycle Polytypic process What about the adiabatic process?
  • 14. Work done during process or cycle Isothermal process
  • 15. Some Important Formulae PV n= constant M= 28.97 g/mol, therefore Ra = 0.287 kJ/kg.K 𝑄 = 𝑚𝐶 𝑝∆𝑇 𝑄 = 𝑚𝐶𝑣∆𝑇
  • 16. Extra