Question

A rigid-walled cubical container is completely filled with water at 40 °F and sealed. The water is then heated to 100 °F. Determine the pressure that develops in the container when the water reaches this higher temperature. Assume that the volume of the container remains constant and the value of the bulk modulus of the water remains constant and equal to 300,000 psi.

Answer 1

Explanation:

The given data is as follows.

For the constant volume condition,

       $T_{1}$ = 40 F      

      $V_{1} = 0.01602 ft^{3}/lb$    

      $V_{2}$ = 0.01613

Therefore, expression for the compressibility is as follows.

          $\frac{dP}{\frac{dV}{V}}$ = K  

   $\frac{dP}{\frac{(0.01602 - 0.01613)}{0.01602}} = 30 \times 10^{4}$

           dP = $30 \times 10^{4} (\frac{0.01602 - 0.01613}{0.01602})$

                 = 2059.925 psi

                 = $2.059 \times 10^{3}$ psi

therefore, we can conclude that pressure that develops in the container when the water reaches this higher temperature is $2.059 \times 10^{3}$ psi.