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Units conversion by factor-label

Many, if not most, parameters and measurements in the physical sciences and engineering are expressed as a numerical quantity and a corresponding dimensional unit; for example: 1000 kg/m³, 100 kPa/bar, 50 miles per hour, 1000 Btu/lb. Converting from one dimensional unit to another is often somewhat complex and being able to perform such conversions is an important skill to acquire. The factor-label method, also known as the unit-factor method or dimensional analysis, is a widely used approach for performing such conversions. It is also used for determining whether the two sides of a mathematical equation involving dimensions have the same dimensional units.

The factor-label method for converting units

The factor-label method is the sequential application of conversion factors expressed as fractions and arranged so that any dimensional unit appearing in both the numerator and denominator of any of the fractions can be cancelled out until only the desired set of dimensional units is obtained. For example, 10 miles per hour can be converted to meters per second by using a sequence of conversion factors as shown below:

10 mile 1609 meter 1 hour meter -- ---- × ---- ----- × ---- ------ = 4.47 ------ 1 hour 1 mile 3600 second second

It can be seen that each conversion factor is equivalent to the value of one. For example, starting with 1 mile = 1609 meters and dividing both sides of the equation by 1 mile yields 1 mile / 1 mile = 1609 meters / 1 mile, which when simplified yields 1 = 1609 meters / 1 mile.

So, when the units mile and hour are cancelled out and the arithmetic is done, 10 miles per hour converts to 4.47 meters per second.

As a more complex example, the concentration of nitrogen oxides (i.e., NOx) in the flue gas from an industrial furnace can be converted to a mass flow rate expressed in grams per hour (i.e., g/h) of NOx by using the following information as shown below:

NOx concentration :
= 10 parts per million by volume = 10 ppmv = 10 volumes/106 volumes
NOx molar mass :
= 46 kg/kgmol (sometimes also expressed as 46 kg/kmol)
Flow rate of flue gas :
= 20 cubic meters per minute = 20 m³/min
The flue gas exits the furnace at 0 °C temperature and 101.325 kPa absolute pressure.
The molar volume of a gas at 0 °C temperature and 101.325 kPa is 22.414 m³/kgmol.

10 m³ NOx 20 m³ gas 60 minute 1 kgmol NOx 46 kg NOx 1000 g g NOx --- ------ × -- ------ × -- ------ × ------ --------- × -- --------- × ---- -- = 24.63 ----- 106m³ gas 1 minute 1 hour 22.414 m³ NOx 1 kgmol NOx 1 kg hour

After cancelling out any dimensional units that appear both in the numerators and denominators of the fractions in the above equation, the NOx concentration of 10 ppmv converts to mass flow rate of 24.63 grams per hour.

Checking equations that involve dimensions

The factor-label method can also be used on any mathematical equation to check whether or not the dimensional units on the left hand side of the equation are the same as the dimensional units on the right hand side of the equation. Having the same units on both sides of an equation does not guarantee that the equation is correct, but having different units on the two sides of an equation does guarantee that the equation is wrong.

For example, check the Universal Gas Law equation of P·V = n·R·T, when:

  • the pressure P is in pascals (Pa)
  • the volume V is in cubic meters (m³)
  • the amount of substance n is in moles (mol)
  • the universal gas law constant R is 8.3145 Pa·m³/(mol·K)
  • the temperature T is in kelvins (K)

mol (Pa)(m³) K (Pa)(m³) = ----- × ---------- × --- 1 (mol)(K) 1

As can be seen, when the dimensional units appearing in the numerator and denominator of the equation's right hand side are cancelled out, both sides of the equation have the same dimensional units.

Limitations

The factor-label method can convert only unit quantities for which the units are in a linear relationship intersecting at 0. Most units fit this paradigm. An example for which it cannot be used is the conversion between degrees Celsius and kelvins (or Fahrenheit). Between degrees Celsius and kelvins, there is a constant difference rather than a constant ratio, while between Celsius and Fahrenheit there is both a constant difference and a constant ratio. Instead of multiplying the given quantity by a single conversion factor to obtain the converted quantity, it is more logical to think of the original quantity being divided by its unit, being added or subtracted by the constant difference, and the entire operation being multiplied by the new unit. Mathematically, this is an affine transform (ax+b), not a linear transform (ax). Formally, one starts with a displacement (in some units) from one point, and ends with a displacement (in some other units) from some other point.

For instance, the freezing point of water is 0 in Celsius and 32 in Fahrenheit, and a 5 degrees change in Celsius correspond to a 9 degrees change in Fahrenheit. Thus to convert from Fahrenheit to Celsius one subtracts 32 (displacement from one point), divides by 9 and multiplies by 5 (scales by the ratio of units), and adds 0 (displacement from new point). Reversing this yields the formula for Celsius; one could have started with the equivalence between 100 Celsius and 212 Fahrenheit, though this would yield the same formula at the end.


Dimensional analysis

In physics and science, dimensional analysis is a tool to find or check relations among physical quantities by using their dimensions. The dimension of a physical quantity is the combination of the basic physical dimensions (usually mass, length, time, electric charge, and temperature) which describe it; for example, speed has the dimension length / time, and may be measured in meters per second, miles per hour, or other units. Dimensional analysis is based on the fact that a physical law must be independent of the units used to measure the physical variables. A straightforward practical consequence is that any meaningful equation (and any inequality and inequation) must have the same dimensions in the left and right sides. Checking this is the basic way of performing dimensional analysis.

Dimensional analysis is routinely used to check the plausibility of derived equations and computations. It is also used to form reasonable hypotheses about complex physical situations that can be tested by experiment or by more developed theories of the phenomena, and to categorize types of physical quantities and units based on their relations to or dependence on other units, or their dimensions if any.

The basic principle of dimensional analysis was known to Isaac Newton (1686) who referred to it as the "Great Principle of Similitude". James Clerk Maxwell played a major role in establishing modern use of dimensional analysis by distinguishing mass, length, and time as fundamental units, while referring to other units as derived. The 19th-century French mathematician Joseph Fourier made important contributions based on the idea that physical laws like should be independent of the units employed to measure the physical variables. This led to the conclusion that meaningful laws must be homogeneous equations in their various units of measurement, a result which was eventually formalized in the Buckingham π theorem. This theorem describes how every physically meaningful equation involving n variables can be equivalently rewritten as an equation of dimensionless parameters, where m is the number of fundamental dimensions used. Furthermore, and most importantly, it provides a method for computing these dimensionless parameters from the given variables.

A dimensional equation can have the dimensions reduced or eliminated through nondimensionalization, which begins with dimensional analysis, and involves scaling quantities by characteristic units of a system or natural units of nature. This gives insight into the fundamental properties of the system, as illustrated in the examples below.

Introduction

Definition

The dimensions of a physical quantity are associated with combinations of mass, length, time, electric charge, and temperature, represented by sans-serif symbols M, L, T, Q, and Θ, respectively, each raised to rational powers.

The term dimension is more abstract than scaleunit: mass is a dimension, while kilograms are a scale unit (choice of standard) in the mass dimension.

As examples, the dimension of the physical quantity speed is distance/time (L/T or LT−1), and the dimension of the physical quantity force is "mass Ã— acceleration" or "mass×(distance/time)/time" (ML/T2 or MLT−2). In principle, other dimensions of physical quantity could be defined as "fundamental" (such as momentum or energy or electric current) in lieu of some of those shown above. Most physicists do not recognize temperature, Θ, as a fundamental dimension of physical quantity since it essentially expresses the energy per particle per degree of freedom, which can be expressed in terms of energy (or mass, length, and time). Still others do not recognize electric charge, Q, as a separate fundamental dimension of physical quantity, since it has been expressed in terms of mass, length, and time in unit systems such as the cgs system. There are also physicists that have cast doubt on the very existence of incompatible fundamental dimensions of physical quantity.

The unit of a physical quantity and its dimension are related, but not identical concepts. The units of a physical quantity are defined by convention and related to some standard; e.g., length may have units of meters, feet, inches, miles or micrometres; but any length always has a dimension of L, independent of what units are arbitrarily chosen to measure it. Two different units of the same physical quantity have conversion factors that relate them. For example: 1 in = 2.54 cm; then (2.54 cm/in) is the conversion factor, and is itself dimensionless and equal to one. Therefore multiplying by that conversion factor does not change a quantity. Dimensional symbols do not have conversion factors.

Mathematical properties

Dimensional symbols, such as L, form a group: The identity is defined as L0 = 1, and the inverse to L is 1/L or L−1. L rais

Image analysis

Image analysis is the extraction of meaningful information from images; mainly from digital images by means of digital image processing techniques. Image analysis tasks can be as simple as reading bar coded tags or as sophisticated as identifying a person from their face.

Computers are indispensable for the analysis of large amounts of data, for tasks that require complex computation, or for the extraction of quantitative information. On the other hand, the human visual cortex is an excellent image analysis apparatus, especially for extracting higher-level information, and for many applications — including medicine, security, and remote sensing — human analysts still cannot be replaced by computers. For this reason, many important image analysis tools such as edge detectors and neural networks are inspired by human visual perception models.

Computer image analysis

Computer image analysis largely contains the fields of computer or machine vision, and medical imaging, and makes heavy use of pattern recognition, digital geometry, and signal processing. This field of computer science developed in the 1950s at academic institutions such as the MIT A.I. Lab, originally as a branch of artificial intelligence and robotics.

It is the quantitative or qualitative characterization of two-dimensional (2D) or three-dimensional (3D) digital images. 2D images are, for example, to be analyzed in computer vision, and 3D images in medical imaging. The field was established in the 1950s—1970s, for example with pioneering contributions by Azriel Rosenfeld, Herbert Freeman, Jack E. Bresenham, or King-Sun Fu.

Techniques

There are many different techniques used in automatically analysing images. Each technique may be useful for a small range of tasks, however there still aren't any known methods of image analysis that are generic enough for wide ranges of tasks, compared to the abilities of a human's image analysing capabilities. Examples of image analysis techniques in different fields include:

Digital image analysis

Digital Image Analysis is when a computer or electrical device automatically studies an image to obtain useful information from it. Note that the device is often a computer but may also be an electrical circuit, a digital camera or a mobile phone. The applications of digital image analysis are continuously expanding through all areas of science and industry, including:

Object-based image analysis

Object-Based Image Analysis (OBIA) is a sub-discipline of From Yahoo Answers

Question:i'm having a lot of trouble with this problem on my homework, and she didnt explain how to do it very well: Calculate the volume of the cube in mL as measured in the graduated cylinder. Convert to cubic centimeters knowing that one cubic centimeter = 1 mL the volume in mL is 4.5 thanks!

Answers:it is the same since 1mL is 1 cubic centimeter

Question:An average man requires 2.00 mg of riboflavin per day. How many pounds of cheese would a man have to eat per day if this were his only source of riboflavin and if the cheese were to contain 5.5 micrograms of riboflavin per gram? This is how I was trying to solve the problem. I know that 200 mg is equal to 200,000 micrograms. I divided 200,00 micrograms by 5.5 micrograms and got 36363.63636. Then I converted 36363.63636 grams to pounds and got 80.1680953 pounds. The answer is 0.80 pounds/day. 200 mg = 200,000 micrograms 1 gram of cheese= 5.5 micrograms of riboflavin 200,000 micrograms/5.5 micrograms = 36363.63636 grams of cheese 36363.63636 grams = 80.1680953 pounds/day Please tell me what I'm doing wrong!

Answers:How did you go from 2.0 mg of riboflavin/day, as stated by the problem, to 200 mg, the number you used in your calculation? Fix that and the answer will come out right.

Question:I really do not understand dimensional analysis in chemistry. My teacher doesn't understand that nobody in the class understands it. i have homework due tomorrow and im going to get in trouble if i do not do it. Please do not think that i am lazy and just dont want to do it. if you want just tell me how to work out each problem and i will get the answer myself. any kind of help is appreciated. thanks. 1. A chemical plant process requires that a cylindrical reaction tank be filled with a certain liquid in 238 s. the tank is 1.2 m in diameter and 4.6 m high. what flow rate in liters per minute is required to fill the reaction tank in the specified time? 2.The radioactive decay of 2.8 g of plutonium-238 generates 1.0 joule of heat every second. Plutonium has a density of 19.86 g/cm^3. How many calories (1 cal=4.184 J) of heat will a ractangular piece of plutonium that is 1.5 cm x 3.05 cm x 15 cm generate per hour? 3. the mass of the earth is 5.974x10^24 kg. Assume that earth is a sphere of diameter 1.28x10^4 km and calculate the average density of earth in grams per cubic centimeter. 4.An automobile can travel 38 km on 4.0 L of gasoline if the automobile is driven 75% of the days in a year and the average distance traveled each day is 86 km, how many liters of gasoline will be consumed in one year (assume the year has 365 days)? 5. Automobile batteries are filled with a solution of sulfuric acid, which has a density of 1.285 g/cm^3. the solution used to fill the battery is 38% (by mass) sulfuric acid. How many grams of sulfuric acid are present in 500 mL of battery acid?

Answers:"dimensional analysis" here is just telling you that you need to pay close attention to the units in the problem. I have read through all the questions here. None of them is even slightly difficult. If you do not know much chemistry, please consider them to be math problems. 1. The volume of the tank is: 3.14*(1.2/2)^2*4.6 = 5.2 m^3 = 5.2x10^3 L To fill this tank in 238s, the rate must be: (5.2x10^3 L)/(238s) = 21.8 (L/s) = 1.3x10^3 (L/min) *** Do you see I do all the unit change? 2. A rectangular piece of plutonium that is 1.5 cm x 3.05 cm x 15 cm is of volume: 1.5 cm x 3.05 cm x 15 cm = 68.6 cm^3 and thus have a mass of: (68.6 cm^3)*(19.86 g/cm^3) = 1.36x10^3 g That can generate this much heat per second: (1.36x10^3 g)*(1.0 joule/s)/(2.8 g) = 0.487x10^3 joule/s Now convert this to calories per hour: (0.487x10^3 J/s)*(cal/4.184J)*(3600s/hr) = 4.2x10^5 cal/hr ***You also need to pay attention to the significant figures. Get it?? Please solve the rest.

Question:If i wanted to convert 1Joule/(mol K) to cal/(g C), how would I go about doing it? For chlorine. I know that Kelvin = Celsius +273.16, but I'm not sure how to work that into dimensional analysis joule/mol K * 1 cal/ 4.186 J * 1 mol Cl/ 35.453 g * ? = cal/g C thanks. Also, is it possible to have a unit with two different kinds of temperatures? ie can I have a unit of Rankine Kelvin? RK?

Answers:joule/mol K * 1 cal/ 4.186 J * 1 mol Cl/ 35.453 g * ? = cal/g C 0.0067 cal/gC or 0.0067 cal/gK ANS teddy boy

From Youtube

Dimensional Analysis :Edited by Dan Rosenthal, Everything else by Mark Matthews

Chemistry: Dimensional Analysis :www.mindbites.com In this lesson, you will learn how to convert measurements from one unit to another. You will learn how to create a conversion factor, which is always a factor of one that is constructed from a known quality. When creating a conversion factor, you will want to make sure to choose one that allows you to cancel out unwanted units. Prof. Yee also discusses significant figures when using conversion factors. Since conversion factors are considered exact, they have an infinite number of significant figures that are not limited. The data will limit the number of significant figures that you will use in an answer. Objects that are counted are also considered exact with unlimited significant figures. You will learn how to link multiple conversion factors, which is sometimes necessary when converting from one unit to another. Prof. Yee reminds you to make sure you always ""take the units along for the ride"" to ensure that you are finding your answer in the proper units. Finally, you will learn the proper conversion factors for converting both from degrees Farenheit to degrees Celsius and from degrees Celsius to degrees Kelvin. Taught by Professor Yee, this lesson was selected from a broader, comprehensive course, Chemistry. This course and others are available from Thinkwell, Inc. The full course can be found at www.thinkwell.com The full course covers atoms, molecules and ions, stoichiometry, reactions in aqueous solutions, gases, thermochemistry, Modern Atomic ...