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# t6203 - APPLIED FLUID MECHANICS TUTORIAL No.6 DIMENSIONAL...

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D. J. Dunn 1 APPLIED FLUID MECHANICS TUTORIAL No.6 DIMENSIONAL ANALYSIS When you have completed this tutorial you should be able to do the following. Explain the basic system of dimensions. Find the relationship between variables affecting a phenomenon. Define and use dimensionless numbers. Solve problems by the use of model tests. Solve typical exam questions.

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D. J. Dunn 2 1. BASIC DIMENSIONS All quantities used in engineering can be reduced to six basic dimensions. These are the dimensions of Mass M Length L Time T Temperature θ Electric Current I Luminous Intensity J The last two are not used in fluid mechanics and temperature is only used sometimes. All engineering quantities can be defined in terms of the four basic dimensions M,L,T and θ . We could use the S.I. units of kilogrammes, metres, seconds and Kelvins, or any other system of units, but if we stick to M,L,T and θ we free ourselves of any constraints to a particular system of measurements. Let's now explain the above with an example. Consider the quantity density . The S.I. units are kg/m 3 and the imperial units are lb/in 3 . In our system the units would be Mass/Length 3 or M/L 3 . It will be easier in the work ahead if we revert to the inverse indice notation and write it as ML -3 . Other engineering quantities need a little more thought when writing out the basic MLT θ dimensions. The most important of these is the unit of force or the Newton in the S.I. system. Engineers have opted to define force as that which is needed to accelerate a mass such that 1 N is needed to accelerate 1 kg at 1 m/s 2 . From this we find that the Newton is a derived unit equal to 1 kg m/s 2 . In our system the dimensions of force become MLT -2 . This must be considered when writing down the dimensions of anything containing force. Another unit that produces problems is that of angle. Angle is a ratio of two sides of a triangle and so has no units nor dimensions at all. This also applies to revolutions which are angular measurements. Strain is also a ratio and has no units nor dimensions. Angle and strain are in fact examples of dimensionless quantities that will be considered in detail later.
D. J. Dunn 3 WORKED EXAMPLE No. 1 Write down the basic dimensions of pressure p. SOLUTION Pressure is defined as p = Force/Area The S.I. unit of pressure is the Pascal which is the name for 1N/m 2 . Since force is MLT - 2 and area is L 2 then the basic dimensions of pressure are ML -1 T - 2 When solving problems it is useful to use a notation to indicate the MLT dimensions of a quantity and in this case we would write [p] = ML -1 T - 2 WORKED EXAMPLE No.2 Deduce the basic dimensions of dynamic viscosity. SOLUTION Dynamic viscosity was defined in an earlier tutorial from the formula τ = µ du/dy τ is the shear stress, du/dy is the velocity gradient and µ is the dynamic viscosity.

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