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Use the method of dimensions to derive the form of an equation V in terms of other quantities.

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When liquid flows under pressure P through a tube of length l, the volume flow rate (ie. the volume per unit time), V, depends on the radius,r , of the tube, the pressure gradient (P/l) and the viscocity of the liquid.

Sep 10th, 2014

Using the units cm, g and s

volume flow rate : dimension = cm^3/s

radius                  : dimension = cm

pressure gradient: g.cm/(s^2.cm)= g/s^2

viscosity               (g.cm.s/s^2)/(cm^2) = g.cm/s    

So according to the relationship, the equality in the dimensions become:

cm^3/s = (cm)^x * (g/s^2)^y * (g.cm/s)^z   (x, y z unknown powers)

In order to cancel off the g,  and have s at the bottom, choose y=1 and z=-1

Hence right side simplifies to (cm)^x * 1/s * (cm) ^-1

and for equality of dimensions x= 4

Form of the formula is

V = K (r^4)*(pressure gradient)/ viscosity where k is some constant


Sep 14th, 2014

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