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LECTURE NOTES 04/07  STKM3212: FOOD PROCESSING TECHNOLOGY   THERMODYNAMICS  (Termodinamik)  SAIFUL IRWAN ZUBAIRI   PMIFT, Grad B.E.M.   B. Eng. (Chemical-Bioprocess) (Hons.), UTM M. Eng. (Bioprocess), UTM ROOM NO.: 2166, CHEMISTRY BUILDING, TEL. (OFF.): 03-89215828, FOOD SCIENCE PROGRAMME, CENTRE OF CHEMICAL SCIENCES AND FOOD TECHNOLOGY,  UKM BANGI, SELANGOR
1.1 OUTLINES ,[object Object],[object Object],[object Object],[object Object],[object Object]
1.2 DEFINITION OF THERMODYNAMICS ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
1.3 INTERNAL ENERGY ,[object Object],[object Object],[object Object],[object Object],[object Object]
CONTINUE: ,[object Object]
CONTINUE: ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
CONTINUE: ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
CONTINUE: WHY specific heat (C p ) of H 2 O is much larger that copper?     When the sample of water and copper are  both heated by 1°C , the addition to the  kinetic energy (KE) is the same (because of the same temperature given) . But to achieve this increase of 1  0 C for water, a much larger proportional energy must be added to the  potential energy (PE)  portion of the  internal energy (U)  in water  (molekul air adalah berselerak, jadi wujud tenaga yg berkaitan dgn tarikan graviti iaitu [PE]) . So the total energy required to increase the temperature of the water is much larger.  Much heat (Q) must be supplied to the H 2 O molecules than the copper to increase to 1 0 C - REASON: the H 2 O has a lot of scattered molecules as compared to copper. Scattered molecules take time to transfer the energy to the other molecules. Copper has an uniform, tight & static molecules. When the heat is supplied, the molecules besides will vibrate/transfer the energy to the other molecules at instantly.
1.4  1 st  LAW OF THERMODYNAMICS ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
CONTINUE: ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
CONTINUE: ,[object Object],[object Object],[object Object],[object Object],[object Object], U (Joules) = Q (Joules) + W (Joules)
CONTINUE: ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
CONTINUE: ,[object Object],[object Object],[object Object]
CONTINUE: ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
1.5 IMPORTANT TERMS IN THERMODYNAMICS ,[object Object],[object Object],[object Object]
CONTINUE: ,[object Object],[object Object]
CONTINUE: ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
CONTINUE: ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
CONTINUE: ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],Ice cube  Liq H 2 O  Vapor   LH, T=ambient LH, T=100 0 C
EXAMPLE 1: ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],0
CONTINUE: ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],0 Conversion factor: 1 J = 1 kg.m 2 .s -2  = 1 N.m 0 0
CONTINUE: ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],0
CONTINUE: ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
CONTINUE: ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
CONTINUE: ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],Conversion factor:  1 J = 1 Pa.m 3  = 10 -5  Bar.m 3 1 bar = 10 5  Pa   U (Joules) = Q (Joules) + W (Joules) ---  (non-flow process & constant mass)
EXAMPLE 2: ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],0.001 m 3  kg -1  --- given basis 1 kg, so to get the volume only --- 0.001 m 3  kg -1  × 1 kg = 0.001 m 3
CONTINUE: ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
CONTINUE:
1.6 2 nd  LAW OF THERMODYNAMICS RECALL: 1 st  LAW
CONTINUE: ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
CONTINUE: ,[object Object],[object Object],[object Object]
CONTINUE: A generic heat pump
CONTINUE: ,[object Object],[object Object],[object Object],[object Object]
CONTINUE: ,[object Object],[object Object],[object Object]
CONTINUE: A generic heat engine
CONTINUE: ,[object Object],[object Object],[object Object],[object Object],[object Object]
CONTINUE: A simplified internal combustion engine
CONTINUE: ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
CONTINUE: ,[object Object],[object Object],[object Object],[object Object]
CONTINUE: ,[object Object],[object Object],[object Object],[object Object],[object Object]

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Lecture 04: STKM3212

  • 1. LECTURE NOTES 04/07 STKM3212: FOOD PROCESSING TECHNOLOGY THERMODYNAMICS (Termodinamik) SAIFUL IRWAN ZUBAIRI PMIFT, Grad B.E.M. B. Eng. (Chemical-Bioprocess) (Hons.), UTM M. Eng. (Bioprocess), UTM ROOM NO.: 2166, CHEMISTRY BUILDING, TEL. (OFF.): 03-89215828, FOOD SCIENCE PROGRAMME, CENTRE OF CHEMICAL SCIENCES AND FOOD TECHNOLOGY, UKM BANGI, SELANGOR
  • 2.
  • 3.
  • 4.
  • 5.
  • 6.
  • 7.
  • 8. CONTINUE: WHY specific heat (C p ) of H 2 O is much larger that copper?  When the sample of water and copper are both heated by 1°C , the addition to the kinetic energy (KE) is the same (because of the same temperature given) . But to achieve this increase of 1 0 C for water, a much larger proportional energy must be added to the potential energy (PE) portion of the internal energy (U) in water (molekul air adalah berselerak, jadi wujud tenaga yg berkaitan dgn tarikan graviti iaitu [PE]) . So the total energy required to increase the temperature of the water is much larger. Much heat (Q) must be supplied to the H 2 O molecules than the copper to increase to 1 0 C - REASON: the H 2 O has a lot of scattered molecules as compared to copper. Scattered molecules take time to transfer the energy to the other molecules. Copper has an uniform, tight & static molecules. When the heat is supplied, the molecules besides will vibrate/transfer the energy to the other molecules at instantly.
  • 9.
  • 10.
  • 11.
  • 12.
  • 13.
  • 14.
  • 15.
  • 16.
  • 17.
  • 18.
  • 19.
  • 20.
  • 21.
  • 22.
  • 23.
  • 24.
  • 25.
  • 26.
  • 27.
  • 29. 1.6 2 nd LAW OF THERMODYNAMICS RECALL: 1 st LAW
  • 30.
  • 31.
  • 32. CONTINUE: A generic heat pump
  • 33.
  • 34.
  • 35. CONTINUE: A generic heat engine
  • 36.
  • 37. CONTINUE: A simplified internal combustion engine
  • 38.
  • 39.
  • 40.