Question

A 15-ft3 tank contains oxygen initially at 14.7 psia and 80°F. A paddle wheel within the tank is rotated until the pressure inside rises to 20 psia. During the process 20 Btu of heat is lost to the surroundings. Determine the paddle-wheel work done. Neglect the energy stored in the paddle wheel. The gas constant and molar mass of oxygen are R = 0.3353 psia·ft3/lbm·R and M = 32 lbm/lbmol. The specific heat of oxygen at the average temperature of Tavg = (735 + 540)/2 = 638 R is cv ,avg= 0.160 Btu/lbm·R.

Answer 1

Explanation:

Equation for energy balance will be as follows.

         $\Delta E_{system} = E_{in} - E_{out}$

        $\Delta U = W_{in} - Q_{out}$

Hence,    $W_{in} = Q_{out} + mC_{v} (T_{2} - T_{1})$

Therefore, we will calculate the final temperature as follows.

            $\frac{P_{1}V}{T_{1}} = \frac{P_{2}V}{T_{2}}$

       $T_{2} = \frac{20 psia}{14.7}(638 R)$

                   = 868.03 R

Now, we will calculate the mass as follows.

             m = $\frac{P_{1}V}{RT_{1}}$

                 = $\frac{14.7 psia \times 15 ft^{3}}{0.3353 psi ft^{3}/lbm R \times 638 R}$

                 = 1.031 lbm

Hence,

        $W_{in} = Q_{out} + mC_{v} (T_{2} - T_{1})$

Putting the values into the above equation as follows.

            $W_{in} = Q_{out} + mC_{v} (T_{2} - T_{1})$

    $W_{in} = 20 Btu + 1.031 lbm (\frac{0.160 Btu}{lbm R})(735 - 540)R$

            $W_{in}$ = 655.2 Btu

Thus, we can conclude that work done by paddle wheel is 655.2 Btu.