2. A cylinder/piston arrangement contains water at 105ºC, 85% quality, with a volume of 1 L. The system is heated, causing the piston to rise and encounter a linear spring. At this point the volume is 1.5 L, the piston diameter is 150 mm, and the spring constant is 100 N/mm. The heating continues, so the piston compresses the spring. What is the cylinder temperature when the pressure reaches 200 kPa?

Explanation:

The given data is as follows.

      ,     x = 0.85

        = 1 L = 0.001   (as 1 L = 0.1 )  

      = 1.5 L = 0.0015 ,

      = 150 mm = 0.15 m,     K = 100 N/mm = 100 kN/m

      = 200 kPa

According to the table,

      = 120.8 kPa,     = 0.001047 m/kg

So, we have

                    = 0.001047 + 0.85 \times (1.41831 m^{3}/kg)" alt="m^{3}/kg" />

So, we have

                    = 0.001047 + 0.85 \times (1.41831 m^{3}/kg)" />    

                    = 1.2066

Now, we will calculate as follows.

                   =

                   = 0.000828 kg

Now,  Area =

                   =

                   = 0.0176

As we have,

                       =

                       = 1.8099

From table, for = 120.8 kPa and .

    200 =

Therefore, from the table

               T = ,      v = 2.01297

               T = ,     v = 2.2443

                =

Thus, we can conclude that the cylinder temperature when the pressure reaches 200 kPa is .