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Saturday, 17 March 2012

1.10 Rounding Off Number


For numerical calculations, the accuracy obtained from solution of problem generally can never be better than the accuracy of the problem data. This is what is to be expected, but often handheld calculators or computer involve more figures in the answer than the number of significant figures used for the data. For this reason, a calculated result should always be "rounded off" to an appropriate number of significant figures.
To convey appropriate accuracy, the following rules for rounding off a number to n significant figures apply:
  • If the n + 4 digit is less than 5, n + 1 digit and other following it are dropped. For example, 2.326 and 0.451 rounded to n = 2, significant figures would be 2.3 and 4.5.
  • If n + 1 digit is equal to 5 with zero following it, then round off the nth digit to an even number. For example, 1.245 and 0.8655 rounded to n = 3 significant figures become 1.24 and 0.866.
  • If the n + 1 digit is greater than 5 or equal to 5 with any nonzero digits following it, increase the nth digit by 1 and drop n + 1 digits and other following it. For example, 0.723 87 and 565.500 3 rounded off to n = 3 significant figures become 0.724 and 566. 



Friday, 16 March 2012

1.9 A Remark on Significant Digits

In engineering calculations, the information given is not known to more than a certain number of significant digits, usually three digits. Consequently, the results obtained cannot possibly be accurate to more significant digits. Reporting results in more significant digits implies greater accuracy than exists, and it should be avoided. For example, consider a 3.75-L container filled with gasoline whose density is 0.845 kg/L, and try to determine its mass. Probably the first thought that comes to your mind is to multiply the volume and density to obtain 3.16875 kg for the mass, which falsely implies that the mass determined is accurate to six significant digits. In reality, however, the mass cannot be more accurate than three significant digits since both the volume and the density are accurate to three significant digits only. Therefore, the result should be rounded to three significant digits, and the mass should be reported to be 3.17 kg instead of what appears in the screen of the calculator. The result 3.16875 kg would be correct only if the volume and density were given to be 3.75000 L and 0.845000 kg/L, respectively. The value 3.75 L implies that we are fairly confident that the volume is accurate within ±0.01 L, and it cannot be 3.74 or 3.76 L. However, the volume can be 3.746, 3.750, 3.753, etc., since they all round to 3.75 L (Fig. 1–62). It is more appropriate to retain all the digits during intermediate calculations, and to do the rounding in the final step since this is what a computer will normally do. When solving problems, we will assume the given information to be accurate to at least three significant digits. Therefore, if the length of a pipe is given to be 40 m, we will assume it to be 40.0 m in order to justify using three significant digits in the final results. You should also keep in mind that all experimentally determined values are subject to measurement errors, and such errors will reflect in the results obtained. For example, if the density of a substance has an uncertainty of 2 percent, then the mass determined using this density value will also have an uncertainty of 2 percent. You should also be aware that we sometimes knowingly introduce small errors in order to avoid the trouble of searching for more accurate data. For example, when dealing with liquid water, we just use the value of 1000 kg/m3 for density, which is the density value of pure water at 0°C. Using this value at 75°C will result in an error of 2.5 percent since the density at this temperature is 975 kg/m3. The minerals and impurities in the water will introduce additional error. This being the case, you should have no reservation in rounding the final results to a reasonable number of significant digits. Besides, having a few percent uncertainty in the results of engineering analysis is usually the norm, not the exception.

1.8 Prefixes And Rules For Their Use


When a numerical quantity is either very large or very small, the units used to define its size may be modified using a prefix. Some of the prefixes used in the SI system are shown in table.  Each represent a multiple or submultiples which, if applied successively, moves the decimal point to every third place. Except for some volume and area measurement, the use of these prefixes is to be avoided in sciences and engineering.
Multiple
Exponential Form
Prefix
SI Symbol
1 000 000 000
109
giga
G
1 000 000
106
mega
M
1 000
103
kilo
K

Submultiple



0.001
10-3
milli
M
0.000 001
10-6
micro
ยต
0.000 000 001
10-9
nano
N

Rules for Use
The following rules are given for the proper use of various SI symbols:
  • A symbol is never written with a plural “s”, since it may be confused with the unit for second (s).
  • Symbols are always written in lowercase letters, with the following exceptions: symbols for the two largest prefixes, giga and mega, are capitalized as G and M, respectively, and symbols named after an individual are also capitalized.
  • Quantities defined by several units which are multiple of one another are separated by a dot to avoid confusion with prefix notation, as indicated by N = 1 kg.m.s-2. Also, m.s (meter-second), whereas ms (milli-second).
  • The exponential power represented for a unit having a prefix refers to both the unit and its prefix.
  • With the exception of the base unit the kilogram, in general avoid the use of a prefix in the denominator of composite function.




1.7 The International System of Units


The International System of Units, abbreviated SI, is accepted in the United States and throughout the world, and is modern version of metric system. By international agreement, SI units will in time replace other systems. In SI, the kilogram (kg) for mass, meter (m) for length, and second (s) for time are selected as the base units, and the unit Newton (N) for force is derived from the preceding three by F = ma. Thus force (N) = mass (Kg) × acceleration (ms-2) or
N = kg. ms-2
Thus, 1 newton is the force reqiured to give a mass of 1 kilogram an acceleration of 1 meter per second per second.