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

Answer:

The value is $W_N = 11849 \ J$

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  

        $A = \pi * \frac{d^2}{4}$

        $A = 3.142 * \frac{0.51^2}{4}$

So

       $W_p =101*10^{3} * 3.142 * \frac{0.51^2}{4} * 0.71$

=>       $W_p = 14651$

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

       $W_N = -2799+14651$

      $W_N = 11849 \ J$