Question

A piston-cylinder device containing a fluid is fitted with a paddle wheel stirring device operated by the fall of an external weight of mass 51kg. As the mass drops by a height of 5.6m, the paddle wheel makes 10100 revolutions. Meanwhile the free moving piston (frictionless and weightless) of 0.51m diameter moves out by a distance of 0.71m. Determine the net work for the system if atmospheric pressure is 101 kPa.

Answer 1

The value is
`W_N = 11849 \ J`

Step-by-step explanation:

From the question we are told that

The mass of the external weight is
`m = 51 \ kg`

The height through which the mass drops is
`h = 5.6 \ m`

The number of revolution made is
`N  = 10100 \  kg`

The diameter of the free moving piston is
`d  = 0.51 \  m \  kg`

The distance moved by the free moving piston is
`s  = 0.71 \  m \  kg`

The atmospheric pressure is
`P  = 101 \ kPa = 101*10^(3)\ Pa`

Generally the workdone by the external weight is mathematically represented as

`W_w  =m *  g   *  h`

=
`51 *  9.8  *  5.6`

=
`2799 N`

Generally the workdone by the free moving piston is mathematically represented as

`W_p  =P *  A  *  s`

Here A is the cross-sectional area with value

`A  =  \pi *  (d^2)/(4)`

`A  =  3.142 *  (0.51^2)/(4)`

So

`W_p  =101*10^(3) * 3.142 *  (0.51^2)/(4)  *  0.71`

=>
`W_p  = 14651`

So

The net workdone is mathematically evaluated as

`W_N   = -W_w+W_p`

The negative sign shows that it is acting in opposite direction to
`W_N`

So

`W_N   = -2799+14651`

`W_N   = 11849 \ J`