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faltersainse [42]
4 years ago
12

A piston-cylinder device containing a fluid is fitted with a paddle wheel stirring device operated by the fall of an external we

ight 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.
Physics
1 answer:
tangare [24]4 years ago
5 0

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

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