ATMOS Lecture 7. The First Law and Its Consequences Pressure-Volume Work Internal Energy Heat Capacity Special Cases of the First Law
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1 TMOS 5130 Lecture 7 The First Law and Its Consequences Pressure-Volume Work Internal Energy Heat Caacity Secial Cases of the First Law
2 Pressure-Volume Work Exanding Volume Pressure δw = f & dx δw = F ds
3 Pressure-Volume Work F = δw = F ds F = δw = ds ds dv = ds δw = dv
4 Pressure-Volume Work δw = dv Intensive Units work or energy er unit mass w W M δw dv M δw dα Where α is the secific volume 5 6 w = 3 δw 3 dα 5 7
5 One-way exansion Closed cycle Closed cycle Positive Work B Positive Net Work B Negative Net Work B α 0 α α 1 α 0 α α 1 α 0 α α 1 The Total work done by (or on) a arcel during a transition from one state to the next deends on how it gets there! 5 6 w = 8 dα 3 9 α dα + 3 ; α dα
6 One-way exansion Closed cycle Closed cycle Positive Work B Positive Net Work B Negative Net Work B α 0 α α 1 α 0 α α 1 α 0 α α 1 The Total work done by (or on) a arcel during a transition from one state to the next deends on how it gets there! 5 6 w = 8 dα 3 9 α dα + 3 ; α dα w = dα = CLOSED CYCLE
7 One-way exansion Closed cycle Closed cycle Positive Work B Positive Net Work B Negative Net Work B α 0 α α 1 α 0 α α 1 α 0 α α 1 The Total work done by (or on) a arcel during a transition from one state to the next deends on how it gets there! 5 6 w = 8 dα 3 9 α dα + 3 ; α dα
8 NY diagram where the total work involved in a rocess is roortional to the area enclosed by the curve reresenting that rocess Skew-T diagram
9 First Law of Thermodynamics Secial Case of the Law of Conservation of Energy Energy stored by the arcel is its internal energy
10 First Law of Thermodynamics Secial Case of the Law of Conservation of Energy Energy stored by the arcel is its internal energy Pressure volume work done by the system = reduction in internal energy + heat sulied by the environment Pressure volume work done on the system = increase in internal energy + heat transferred to the environment
11 First Law of Thermodynamics Secial Case of the Law of Conservation of Energy Energy stored by the arcel is its internal energy Pressure volume work done by the system = reduction in internal energy + heat sulied by the environment Pressure volume work done on the system = increase in internal energy + heat transferred tothe environment δw = du + δq Where δw = increment of work (er unit mass du = change in the internal energy (er unit mass) δq = increment of heat energy (er unit mass) Recall: δw dα δq = du + dα
12 Joule s Law δq = du + dα Ideal gas (kinetic theory of gases) molecules do not exert attractive or reulsive forces From equation above the internal energy (u) is roortional to the temerature of the air arcel!
13 Joule s Law δq = du + dα Ideal gas (kinetic theory of gases) molecules do not exert attractive or reulsive forces From equation above the internal energy (u) is roortional to the temerature of the air arcel!
14 Heat Caacity dd heat (δq) Then temerature will increase (from T to T+dT) δq dt = Heat caacity of the material (or secific heat) Heat Caacity at Constant Volume = Isochoric case c l δq dt 5efghijkhj Heat Caacity at Constant Pressure = Isobaric case c d δq dt defghijkhj
15 Heat Caacity at Constant Volume = Isobaric case V constant V constant
16 Heat Caacity at Constant Volume = Isochoric case c l δq dt 5efghijkhj Recall, Joule s Law: δq = du + dα Since volume is constant, dα=0 (So, heat equals internal energy no work done on system) c l = du dt du = c l dt δq = c l dt + dα c l = 718 J kg 9 K 9
17 Heat Caacity at Constant Pressure = Isobaric case F = F = ds dv = ds P constant P constant
18 Heat Caacity at Constant Pressure = Isobaric case F δw = = dv F = ds dv = ds P constant P constant c d > c l
19 Heat Caacity at Constant Pressure = Isobaric case c d δq c d = c d = Product rule of differentiation d α = dα + α d Substitute the ideal gas law (R is constant) d RT = R dt = dα + α d dα = R dt α d dα = R dt, where is constant c d = du + R dt dt du + R dt dt dt defghijkhj du + dα dt c d = c l + R defghijkhj defghijkhj = du dt + R
20 Review of Heat Caacity c d = c l + R du = c l dt δq = c l dt + dα δq = (c d R) dt + dα Recall R dt = dα + α d δq = c d dt α d c l = 718 J kg 9 K 9 c d = 1005 J kg 9 K 9 = 718 J kg 9 K J kg 9 K 9 htts://
21 Secial Cases of the First Law δq = c l dt + dα δq = c d dt α d Isobaric Process: (d = 0) δq = c d dt Isothermal Process Isochoric Process diabatic Process
22 Secial Cases of the First Law δq = c l dt + dα δq = c d dt α d Isobaric Process: (d = 0) δq = c d dt = c d c l du du = c l dt Isothermal Process Isochoric Process diabatic Process
23 Secial Cases of the First Law δq = c l dt + dα δq = c d dt α d Isobaric Process: (d = 0) δq = c d dt = Isothermal Process: (dt=0) c d c l du δq = α d = dα = δw Isochoric Process diabatic Process
24 Secial Cases of the First Law Isobaric Process: (d = 0) δq = c d dt = Isothermal Process: (dt=0) Isochoric Process: (dα = 0) δq = c l dt + dα δq = c d dt α d c d c l du δq = α d = dα = δw δq = c l dt = du diabatic Process
25 Secial Cases of the First Law Isobaric Process: (d = 0) δq = c d dt = Isothermal Process: (dt=0) Isochoric Process: (dα = 0) δq = c l dt + dα δq = c d dt α d c d c l du δq = α d = dα = δw δq = c l dt = du diabatic Process: (δq = 0) negligible change of heat between the system and environment c l dt = dα c d dt = α d
26 Secial Cases of the First Law δq = c l dt + dα δq = c d dt α d Isobaric Process: (d = 0) δq = c d dt = Isothermal Process: (dt=0) c d c l du δq = α d = dα = δw Secial Significance in Meteorology!! Isochoric Process: (dα = 0) δq = c l dt = du diabatic Process: (δq = 0) negligible change of heat between the system and environment c l dt = dα c d dt = α d
27 Hint for Homework Problem 5.2
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