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- Lec_15_S07.ppt

Sree Ram M.

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ENGR 107: Engineering Fundamentals Lecture 15 Chap 6: Dimensions, Units, and Conversions Department of Electrical and Computer Engineering George Mason University References http://go.hrw.com/resources/go_sc/hst/HP1PE729.PDF http://www.dot.state.ak.us/stwddes/research/assets/documents/rfps/examples/metric.doc http://www.hazelwood.k12.mo.us/~grichert/sciweb/convert.htm http://www.epa.gov/ORD/NRMRL/pubs/600r01106/600R01106appA.pdf SI Unit Conversion Tables Paper & on-line Physical Quantities Fundamental Dimensions: For example: Length (L) = 20 meters Units: magnitude of physical quantities only be understood when compared with pre-determined reference amounts called units Time Force Length Temperature Mass These are abstract qualities of measurement without scale, (i.e. length) Dimensions Units are used to describe physical quantities can be conveniently and usefully manipulated when expressing all physical quantities of a particular field of science or engineering e.g. length (meter) Fundamental Dimensions: Precision today: Inch: Cubit: tip of middle finger to the elbow distance covered by three barley corns, round and dry, laid end to end Length of a meter is determined by the distance traveled by light in a vacuum during a specified amount of time Derived Demensions a combination of fundamental dimensions e.g. velocity: length/time (mps) i.e., a combination of two fundamental dimensions e.g. In this case, length and time Derived Dimensions: Units Two fundamental systems of units are used in the world today: The US has adopted this standard and is in the very slow process of implementing this system The metric system,modified over the years now called the System International (SI). SI is now considered the new international standard system of units 2. Engineering System used in the United States ? is a system based on the foot, foot-pound, and second 1. Metric System used in almost every industrial country in the world today ? it?s a decimal, absolute, system based on the meter, kilogram, and second (MKS) International System Divided into three classes of units Supplementary units Base units there are seven base units in the SI system Derived units ? combination of units formed by combining base, supplementary, and other derived units The International System of Units (SI) was developed and maintained by the General Conference on Weights and Measures and intended as a basis for worldwide standardization of measurements. The key element is that it?s a decimal system Base Units Quantity Name Symbol Length meter m Mass kilogram kg Time time s Electric current ampere A Thermodynamic temp kelvin K Amount of substance mole mol Luminous intensity candela cd Supplementary Units Quantity Name Symbol Plane angle radian rad Solid angle steradian sr Plane angle: a rad is the plane angle between two radii of a circle that cuts off a circumference of an arc equal in length to the radius. A circle subtends two pi radians about the origin. Steradian: measure of the angular ?area? subtended by a two dimensional surface about the origin in three dimensional space. A sphere subtends 4 pi steradians about the origin. Derived Units Frequency Hz hertz s-1 Force N newton kg*m*s-2 Pressure stress Pa pascal kg*m-1*s-2 Energy or work J joule kg*m2**s-2 Quantity of heat J joule kg*m2*s-2 Power radiant flux W watt kg*m2*s-3 Electric charge C coulomb A*s Electric Potential V volt kg*m2*s-3*A-1 Potential difference V volt kg*m2*s-3*-1 Electromotive force V volt kg*m2*s-3*A-1 Capacitance F farad A2*s4*kg-1*m -2 Quantity SI Unit Name Base Units Symbol Derived Units Electric resistance ? ohm kg*m2*s-3*A-2 Conductance S siemens kg-1*m2*s3*A2 Magnetic flux Wb weber kg-1*m*s-2*A-1 Magnetic flux density T tesla kg*s2*A-1 Inductance H henry kg*m2*s-2*A-2 Luminous flux lm lumen cd*sn Illuminance lx lux cd*sn*m-2 Celsius temperature C degree Celsius K Activity (radionuclides) Bq becqueret s -1 Adsorbed dose Gy gray m2*s -2 Dose equivalent S sievert m 2*s -2 Multiplier SI Unit Name Base Units Symbol Decimal Multiples 1018 exa E 1015 peta P 1012 tera T 109 giga G 106 mega M 103 kilo k 102 hecto h 101 deka da 10-1 deci d 10-2 centi c 10-3 milli m 10-6 micro U 10-9 nano N 10-12 pico p Quantity Prefix name Symbol Rarely used Engineering Notation Engineering Notation is expressed in powers that are multiples of 103 For example: 12 345 678* Ordinary Notation Scientific Notation Engineering Notation 1.234 567 8 x 107 12.345 678 x 106 * An exact number 0.039 A 3.9 x 10-2 A 0.39 x 10-3 A 12 M? = ? 39 mA Rules for Using SI Units No periods after symbols except at the end of a sentence Unit symbols written in lowercase unless the symbols are derived from a proper name Lowercase Uppercase m, kg, s, mol, cd A, K, Hz, Pa, C Symbols rather than self-styled abbreviations should be used to represent units Correct Not Correct A amp s sec SI Rules (continued) ?s? never added to the symbol (plural) Space between numerical values and the unit symbol Correct Not Correct 43.7 km 43.7km 0.25 Pa 0.25Pa Exception: no space between the numerical values and the symbols for: degree, minute, and second of angles and for degree Celsius No space between the prefix and the unit symbols Correct Not Correct mm, Mohm k m , ? F SI Rules (continued) Unit names: lowercase all letters except at the beginning of a sentence, even if the unit is derived from a proper name Plurals: required when writing unit names. Example Singular Plural henry henries exceptions are noted: Singular Plural lux lux hertz hertz siemens siemens No hyphen or space between a prefix and the unit name. In three cases the final vowel is omitted: megohm, kilohm, and hectare SI Rules (continued) Symbols used in preference to unit names; symbols are standardized An exception is made when a number is written, in that case the unit name should be used. For example: ten meters not ten m and 10 m not 10 meters Multiplication and Division When writing unit names as a product, always use a space (preferred) or a hyphen Correct Usage Newton meter or newton-meter When expressing a quotient using unit names, always use the word per and not a solidus (/). The solidus or slash is reserved for use with symbols. Correct Not Correct meter per second meter / second mps or m/s Multiplication and Division When writing a unit name that requires a power, use a modifier, such as squared or cubed, after the unit name. For an area or volume the modifier can be placed before the unit name. Correct Not Correct degrees squared square degrees When expressing products using unit symbols, the center dot is preferred. Preferred N?m for newton meter or newton-meter Multiplication and Division SI Rules (Cont) Example: m/s or m?s-1 or m s In more complicated cases, use negative powers or parentheses, e.g. m/s2 or m·s-2 but not m/s/s for acceleration When denoting a quotient by unit symbols, any of the following methods are accepted form. Numbers To denote a decimal point: period When expressing numbers less than one: zero before the decimal Examples: 15.6 or 0.93 Comma is used in many countries to denote a decimal point Avoid using commas to separate data into groups of three, instead, counting from the decimal to the right or left insert a space Correct and Recommended Procedure 6.513 824 76 851 7 434 296 0.187 62 $5649733250 Try reading these numbers: $5 649 733 250 or Calculating with SI Units One unit is used to represent each physical quantity, e.g. meter for length. SI unit are coherent; that is, each new derived unit is a product or quotient of the fundamental and supplemental units without any numerical factors. A newton is the force required to impart an acceleration of one meter per second squared (1.0 m/s2) to a mass of one kilogram (1.0 kg) 1 N = (1.0 kg) (1.0 m/s2) Applying Newton?s second law F = ma/gc, where gc = ma/F or gc = 1.0 kg · 1.0 m N · s2 Calculations Using SI Fundamental relationships are simple and easier to use because of coherence. Recognize how to manipulate units. e.g. Since Watt = J/s = N·m/s you should realize that N·m/s = (n/m2)(m3/s) = (pressure)(volume flow rate) Understand the advantage of adjusting all variables to base units such as, replacing N with kg·m/s2 or Pa with kg·m-1·s-2 and so on. Develop a proficiency with exponential notation to be used with unit prefixes. (1 mm)3 = (10-3 m)3 = 10-9 m3 1 ns-1 = (10-9 s)-1 = s-1 109 Non-SI Units Accepted for Use in the US Time minute hour day min h d 60 s 3 600 s 86 400 s Plane angle degree minute second º ' " ?/180 rad ?/10 800 rad ?/648 000 rad Volume Liter L or l 10-3 m3 Mass metric ton unified atomic mass unit t m 103 kg 1.660 57x10-27 kg L is recommended Non-SI Units Accepted for Use in the US Land area Hectare ha 104 m2 Energy electronvolt eV 1.602x10-19 J (approx) SI Rules Use the letter ?L? not ?l? to represent Liters. What does this mean? 16.007 l m Does this mean 16.007 liter meters? Does it mean (16.007 l) meters ? There is no way you can tell without using L for liters. 16.007 L leaves no doubt. Example Problem 6.1 Problem: A weight of 100 kg (the unit itself indicates mass) is suspended by a rope. Calculate the tension in the rope in newtons to hold the mass stationary when the local gravitational acceleration is 9.087 m/s2 1.63 m/s2 Theory: Tension in the rope or the force required to hold the object when the mass is at rest or moving at constant velocity is: F= mgL/gc Example Problem 6.1 Continued F= mgL/gc Where gL replaces a and is the local acceleration of gravity, gc is the proportionality constant, and m is the mass of the object. Due to coherence, gc = 1.0 kg?m N ?s2 Assumptions: Neglect the mass of the rope Example Problem 6.1 Continued 1.000 x 102 kg Solution a) For gL = 9.807 m/s2 F = mgL/gC = (100 kg )(9.807m/s2)/ 1.0 kg ? m/N ? s2 = 980.7 N F = (100 kg) (9.807 m) x 1 (N ? s2) = 980.7 N s2 kg ? m Example Problem 6.1 Continued 1.00 x 102 kg F = (100 kg) (1.63 m) x 1 (N ? s2) = 163 ? s2 kg ? m b) For gL = 1.63 m/s2 F = mgL/gC = (100 kg )(1.63 m/s2)/ 1.0 kg ? m/N ? s2 = .163 kN Common Conversion Factors C x 9/5+32 = F (F- 32) x 5/9 = C C x 9/5 +32 = F C +273 = K F +460 = K See conversion tables, Eide Text beginning page 501 References at the beginning of the lesson, most complete is ref. 4 No parentheses required, apply PEMDAS Parentheses required, apply PEMDAS No parentheses required, apply PEMDAS Conversion of Units Remember: a conversion factor relates two equal physical quantities Example: 1 m = 3.208 8 ft Convert 2.97 ft to m 2.97 ft 1m 1 3.208 8 ft Pretty straight forward, but what if I said the conversion factor between feet and meters is 0.3048? Convert 4.8 feet to meters. 4.8 ft 1 m ft Are there 0.3048 ft in a m? Are there 0.3048 m in a ft? 0.3048 1 = 1.463 m = 0.926 m = 1.5 m The conversion relates two equal physical quantities. Conversion of Units Using the tables of Unit Conversions beginning on page 501 of the Eide text, convert: Convert 4.80 ft to m Convert 4.80 ft Multiply: By: To Obtain 3.048 x 10-1 m From page 502 = 1.463 m 4.80 ft 3.048 x 10-1 m 1 ft = 1.46 m Conversion of Temperatures Convert 60.0° F to degrees C (F- 32) x 5/9 = C (60.0 ? 32.0)(5/9) = Convert 15.6° C to degrees K C +273 = K = 15.6 ° C 15.5556 15.6 ° C +273 = 288.6 K 15.6 ° C +273 = = 289 K Note: no degree symbol used with Kelvin Engineering System Units Quality Unit Symbol Mass pound-mass lbm Length foot ft Time second s Force pound-force lbf Conversion of Units Example problem 6.2 Convert 6.7 in to millimeters Solution: Write the identity - 6.7 in = 6.7 in 1 Then multiply by the appropriate conversion factor. 6.7 in = 6.7 in * 25.4 mm = 1.7 x 102 mm 1 1 in Conversion of Units Example Problem. 6.3 Convert 85.0 lbm/ft3 to kilograms per cubic meter. Solution: 85.0 lbm/ft3 = 85.0 lbm (1 ft)3 0.453 6 kg 1 ft3 (0.3048m)3 lbm = 1.36 x 103 kg/m3 Note errors in text, p219: 85.0 lbm not 850 lbm and 0.304 8 not 6.304 8 m /ft Conversion of Units Example Problem 6.4 Determine the gravitational force (in newtons) on an automobile with a mass of 3 645 lbm. The acceleration of gravity is known to be 32.2 ft/s2. Theory: F = mgL/gC Solution: m = 3 645 lbm 1 kg = 1 653 kg 1 2.204 6 lbm Conversion of Units Example Problem 6.4 continued gL = 32.2 ft 0.304 8 m = 9.814 6m/s2 1 s 2 1 ft = 1.62 x104 N F = (1 653 kg)(9.814 6 m/s2) N ?s2 1 kg?m F = (1 653.36 kg)(9.814 6 m/s2) / (1.0 kg?m/N?s2) F = mgL/gC where m = 1.653 kg Alternate Solution Example Problem 6.4 Solution: F = mgL/ gC = 3 645 lbm 32.2 ft 1 kg 0.304 8 m 1N?s2 1 1 s2 2.204 6 lbm 1 ft 1.0 kg?m F = 16.2 kN Conversion of Units Example Problem 6.5 Convert a mass flow rate of 195 kg/s (typical of the airflow through a turbofan) to slugs per minute. 782 slugs min 1 slug = 14.594 kg 60 s 1 min Solution: 195 kg/s = 195 kg 1 s From conversion tables of Eide text Page 505: Multiply by To Obtain Slugs 1.459 4 x 101 kg Conversion of Units Example 6.6 Compute the power output of a 225 hp engine in a) Btu and b) kilowatts Solution 225 hp = 225 hp 2.546 1 x 103 Btu 1 h = 9.55x103 Btu/min 1 1hp?h 60 min b) 225hp = 225 hp 0.745 70 kW = 168 kW 1 1 hp Homework Assignment Study: Data Analysis and Statistics (Chapter 8, Eide, et al; pages 251 ? 268 Problems: 5.1a,d,g; 5.2 b, i; 5.4b; 6.1b,f; 6.2 d, f; 6.3b,c; 6.7b (Reminder)

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