principles and applications to tribology (bhushan/introduction) || appendix: a units, conversions,...

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A Units, Conversions, and Useful Relations A.1 Fundamental Constants Constant Symbol SI units Avogadro’s constant AN 6.022 × 10 23 mol 1 Boltzmann’s constant k 1.381 × 10 23 J/ K Molar gas constant R = AN k 8.315 J/(K mol) (or Pa m 3 /(K mol)) Electronic charge e 1.602 × 10 19 C (A s) Permittivity of free space ε 0 8.854 × 10 12 C 2 /(J m) Mass of 1 12 of 12 C atom (atomic mass unit) amu 1.661 × 10 27 kg Mass of electron m e 9.109 × 10 31 kg Gravitational constant G 6.670 × 10 11 Nm 2 /kg 2 Gravitational acceleration (New York) g 9.807 m/s 2 Speed of light in vacuum c 2.998 × 108 m/s A.2 Conversion of Units 1 nm = 10 A (angstrom) 1 liter () = 10 3 m 3 = 1000 cm 3 (cc) 1 gallon (US) = (1/7.4805) ft 3 = 3.78 1N = 10 5 dyne = (1/9.807) kgf = (1/4.448) 1b 1 m N/m = 1 dyne/cm = 1 erg/cm 2 = 1 mJ/m 2 (unit of surface tension) 1 Pa (N/m 2 ) = 10 dyne/cm 2 = 10 5 bar = (1/6894) psi 1 atm = 760 mm Hg = 1.013 × 10 5 Pa (N/m 2 ) = 1.013 × 10 6 dyne/cm 2 = 1.013 bar 1 torr = 1 mm Hg = 1.316 × 10 3 atm = 133.3 Pa (N/m 2 ) 1J = 1Nm = 1Ws = 10 7 dyne cm =10 7 ergs Principles and Applications of Tribology, Second Edition. Bharat Bhushan. © 2013 John Wiley & Sons, Ltd. Published 2013 by John Wiley & Sons, Ltd.

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Page 1: Principles and Applications to Tribology (Bhushan/Introduction) || Appendix: A Units, Conversions, and Useful Relations

AUnits, Conversions,and Useful Relations

A.1 Fundamental Constants

Constant Symbol SI units

Avogadro’s constant AN 6.022 × 1023 mol−1

Boltzmann’s constant k 1.381 × 10−23 J/ KMolar gas constant R = AN k 8.315 J/(K mol) (or Pa m3/(K mol))Electronic charge −e 1.602 × 10−19 C (A s)Permittivity of free space ε0 8.854 × 10−12 C2/(J m)Mass of 1

12 of 12C atom (atomic mass unit) amu 1.661 × 10−27kgMass of electron me 9.109 × 10−31kgGravitational constant G 6.670 × 10−11 N m2/kg2

Gravitational acceleration (New York) g 9.807 m/s2

Speed of light in vacuum c 2.998 × 108 m/s

A.2 Conversion of Units

1 nm = 10 A (angstrom)1 liter (�) = 10−3 m3 = 1000 cm3 (cc)1 gallon (US) = (1/7.4805) ft3 = 3.78 �

1 N = 105 dyne = (1/9.807) kgf = (1/4.448) 1b1 m N/m = 1 dyne/cm = 1 erg/cm2 = 1 mJ/m2 (unit of surface tension)1 Pa (N/m2) = 10 dyne/cm2 = 10−5 bar = (1/6894) psi1 atm = 760 mm Hg = 1.013 × 105 Pa (N/m2) = 1.013 × 106 dyne/cm2 = 1.013 bar1 torr = 1 mm Hg = 1.316 × 10−3 atm = 133.3 Pa (N/m2)1 J = 1 N m = 1 W s = 107 dyne cm =107 ergs

Principles and Applications of Tribology, Second Edition. Bharat Bhushan.© 2013 John Wiley & Sons, Ltd. Published 2013 by John Wiley & Sons, Ltd.

Page 2: Principles and Applications to Tribology (Bhushan/Introduction) || Appendix: A Units, Conversions, and Useful Relations

964 Appendix A: Units, Conversions, and Useful Relations

1 cal = 4.187 J (Mechanical equivalent of heat)1 BTU = 778.2 ft 1b = 252 cal1 W (N m/s) = 1 J/s = 107 dyne cm/s1 hp = 550 ft 1b/s = 0.746 kW1 kT = 4.114 × 10−14 erg = 4.114 × 10−21 J at 298 K (∼25◦C)1 kT per molecule = 0.592 kcal/mol = 2.478 kJ/mol at 298 K1 eV = 1.602 × 10−12 erg = 1.602 × 10−19 J1 eV per molecule = 23.06 kcal/mol = 96.48 kJ/mol1 poise (P) = 1 dynes s/cm2 = 10−1 kg/(m s) = 10−1 Pa s(unit of absolute or dynamic viscosity)1 Reyn = 1 1b s/in2 = 68,750 Poise1 Stoke (St) = 102 mm2/s (unit of kinematic viscosity = absolute viscosity/density)1 kg/m3 = 62.43 × 10−3 1b/ft3 (density)◦C = (◦F – 32)/1.80◦C = 273.15 K (triple point of water)1 W/(m K) = (1/4.187 × 102) cal/(s cm ◦C) = 57.79 BTU/(h ft ◦F) (thermal conductivity)

A.3 Useful Relations

The energy equivalent, mc2, of one atomic mass unit = 1.492 × 10−10 JThe mass of an atom or molecule = (molecular weight) (1.661 × 10−27 kg)The kgf (in metric units) is the force required to support a standard kilogram (kg) body

against gravity in a vacuum; or the force applied to give a body the standard acceleration. Theword kilogram is used for the unit of mass,

1 kgf = (1 kg)(9.807 m/s2)

The Newton is the force (in SI units) which, if applied to a standard kilogram body, wouldgive that body an acceleration of 1 m/s2,

1 N = 1 kg m/s2 = 1/9.807 kgf

The dyne is the force which if applied to a standard gram body, would give that body anacceleration of 1 cm/s2, i.e.,

1 dyne = 1 g cm/s2 = 10−5 N = (1/980.7)gf

The specific gravity of a solid or liquid is the ratio of the mass of the body to the mass of anequal volume of water at some standard temperature (typically 4◦C in physics and 15.6◦C inengineering); density of water at 4◦C = 103 kg/m3. The specific gravity of gases is usuallyexpressed in terms of that of hydrogen or air.