Bth_AssignmentI_Units2

Bth_AssignmentI_Units2 - ME 3533: Thermodynamics Part I:...

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ME 3533: Thermodynamics Part I : Introduction, Concepts, and Units 2. Thermodynamic Quantities & Units Physical quantities of interest in thermodynamics Dimensions and Units Conversion of Units Basic and derived quantities and units Assignment#2.1 : Basic Quantities & Units Mechanical quantities and units. Work interaction Assignment#2.2 : Mechanical Quantities & Units Thermal quantities of interest. Heat interaction Thermodynamic properties. Assignment#2.3 : Thermal quantities & Units Objectives: (The Student) 1. Understands the use of dimensions and units to characterize a measurable physical quantity. 2. Understands mechanical and work-related concepts in physics and thermodynamics 3. Understands thermal and heat-related concepts 4. Knows the key thermodynamic properties Significant figures for solutions You should not give your solution to a problem to an unrealistic level of precision. As a general rule, the precision level for your result cannot exceed the lowest level of precision for the variables used in the equation for your calculation. Level of precision is expressed by the number of digits or significant figures used. The number of significant figures is the number of digits that are known with some degree of confidence. Thus, a density given as 2320 kg/m 3 has three significant figures (2, 3, and 2). Rules for counting significant figures are straightforward. Zeros within a number are always significant (for example, 203 and 2003). Zeros that do nothing else than set the decimal point do not count as significant. For example, 0.00043 can be re-written in scientific notation as 4.3 × 10 −4 . It has just two significant figures. The three leading zeros after the decimal point are not significant. On the other hand, trailing zeros that are not needed to hold the decimal point are significant. For example, 4.500 has four significant figures. A general rule followed in this class is to retain high levels of precision in intermediate calculations but give the final result with three significant figures. Abridged Units and Conversion information: 0.3048 m = 1 ft; 1 mi = 1609.3 m 1 kg = 2.205 lbm 1 kW = 3412 Btu/h = 1.341 hp; 1 kWh = 3412 Btu; 1 Btu = 1055.1 J = 778.16 ft·lbf 1 kPa = 0.145 lbf/in 2 ; 1 atm = 14.7 lbf/in 2 = 101.325 kPa g = 9.81 m/s 2 ; g c = 32.178 lbm ft/lbf s 2 T (K) = 273.15 + T ( ° C); T ( ° R) = 460 + T ( ° F); (C ) (F ) 32 100 180 TT °° = Dr. George Adebiyi Page 1 1/28/2008
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TABLE 2.4 Conversion Factors Physical Quantity Conversion Factor length 1 ft = 0.3048 m 1 in. = 2.54 cm 1 mi = 1.6093 km velocity 1 ft/s = 0.3048 m/s 1 mi/h = 0.44703 m/s area 1 ft 2 = 0.0929 m 2 1 in. 2 = 6.452 × 10 -4 m 2 mass 1 lbm = 0.4536 kg 1 kg 2.205 = 1 slug = 14.594 kg mass flow rate 1 lbm/h = 0.000126 kg/s 1 lbm/s = 0.4536 kg/s volume 1 ft 3 = 7.48052 US gal = 0.02832 m 3 = 28.32 l (liter) 1 in. 3 = 1.6387 × 10 -5 m 3 volumetric flow rate 1 ft 3 /min = 0.000472 m 3 /s density 1 lbm/ft 3 = 16.018 kg/m 3 force 1 lbf = 4.448 N pressure 1 lbf/in. 2 = 6894.8 Pa (N/m 2 ) 1 bar 14.50 = 1 lbf/ft 2 = 47.88 Pa (N/m 2 ) 1 atm = 14.696 lbf/in.
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This note was uploaded on 08/21/2008 for the course ME 3513 taught by Professor Felicelli during the Spring '04 term at Mississippi State.

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Bth_AssignmentI_Units2 - ME 3533: Thermodynamics Part I:...

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