If we move through a pool of water we experience a resistance to our motion. This shows that there is a frictional force in liquids. We say this is due to the viscosity of the liquid. If the frictional force is comparatively low, as in water, the viscosity of the liquid is low; if the frictional force is large, as in glue or glycerine, the viscosity of two liquids by filling two measuring cylinders with each of them, and allowing identical small steel ball-bearings to fall through each liquid. The sphere falls more slowly through the liquid of higher viscosity.

As we shall see later, the viscosity of a lubricating oil is one of the factors which decide whether it is suitable for use in an engine. The Ministry of Aircraft Production, for example, listed viscosity values to which lubricating oils for aero-engines must conform. The subject of viscosity has thus considerable practical importance.

** Newton’s Formula and Coefficient of Viscosity.**

When water flows slowly and steadily through a pipe, the layer A of the liquid in contact with the pipe is practically stationary, but the central part C of the water is moving relatively fast. At other layers between A and C, such as B, the water has a velocity less than at C, the magnitude of the velocities being represented by the length of the arrowed lines now as in the case of two solid surfaces.

Moving over each other, a frictional force is exerted between two liquid layers when they move over each other. Thus because the velocities of neighboring layers are different, a frictional force occurs between the various layers of a liquid when flowing through a pipe.

The basic formula for the frictional force, F, in a liquid was first suggested by Newton. He saw that the larger the area of the surface of liquid considered, the greater was the frictional force F. He also stated that F was directly proportional to the velocity gradient at the part of the liquid considered. This is the case for most common liquids, called Newtonian liquids. If v^{1}, v^{2} are the velocities of C, and h is their distances apart, the velocity gradient between the liquids is defined as (v^{1} – v^{2})/h. The velocity gradient can thus be expressed in (m/s)/m, or as ‘s^{-1}’

Thus if A is the area of the liquid surface considered, the frictional force F on the surface is given by

F α A x velocity gradient,

or F = *ŋA *x velocity gradient, .. … .. .(1)

where *ŋ* is a constant of the liquid known as the coefficient of viscosity.

The expression for the frictional force in a liquid should be contrasted with the case of solid friction, in which the frictional force is independent of the area of contact and of the relative velocity between the solid surfaces concerned.

**Definition, Units, and Dimensions of Coefficient of Viscosity**

The magnitude of *ŋ* is given by

The unit of F is a newton, the unit of A is m^{2}, and the unit velocity gradient is 1 m/s per m. thus, n may be defined as the frictional force per unit area of a liquid when it is in a region of unit velocity gradient.

The ‘unit velocity gradient’ = 1 ms^{-1} change per m. since the ‘m’ cancels, the ‘unit velocity gradient’ = 1 per second. From n = F/(A x velocity gradient), it follows that n may be expressed in units of newton sm^{-2} (N s m^{-2}), or ‘dekapoise’.

The coefficient of viscosity of water at 10^{o}C is 1.3 x 10^{-3} N s m^{-2}.

Since F = *ŋA* x velocity gradient, the frictional force on an area of 10cm^{2} in water at 10^{o}C between two layers of water 0.1 cm apart which move with a relative velocity of 2 cm s^{-1} is found as follows:

Coefficient of viscosity *ŋ *= 1.3 x 10^{-3} newton m^{-2}, A = 10 x 10^{-4} m^{2}, velocity gradient = 2 x 10^{-2} m s^{-1} ÷ 0.1 x 10^{-2} m = 2/0.1 s^{-1}.

** Steady Flow of Liquid Through Pipe: The Poiseuille’s Formula**

The steady flow of liquid through a pipe was first investigated thoroughly by Poiseuille in 1844, who derived an expression for the volume of liquid issuing per second from the pipe. The proof of the formula is given on page****. But we can derive most of the formula by the method of dimensions.

The volume of liquid issuing per second from the pipe depends on (i) the coefficient of viscosity, *ŋ*, (ii) the radius, a, of the pipe, (iii), the pressure gradient, g, set up along the pipe. The pressure gradient = p/l, l is its length. Thus x, y, z being indices which require to be found, suppose

Volume per second= k* ŋ*^{x}a^{y}g^{z} …`…`…(1)

Now the dimensions of volume per second are L^{3}T^{-1}; the dimensions of *ŋ* are ML^{-1}T^{–}, the dimension of a is L; and the dimensions of g are

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