Question

Air in the amount of 2 lbm is contained in a well-insulated, rigid vessel equipped with a stirring paddle wheel. The initial state of this air is 30 psia and 60 F. How much work, in Btu, must be transferred to the air with the paddle wheel to raise the air pressure to 40 psia? Also, what is the nal temperature of air?

Answer 1

Explanation:

It is known that energy balance relation is as follows.

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

Also,   $W_{in} = \Delta U$

so,       $W_{in} = mC_{v}(T_{2} - T_{1})$  

According to the ideal gas equation,

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

Putting the values into the above equation as follows.

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

                         = $(520R) \frac{40psia}{30psia}$

                         = 693.3 R

Now, we will convert the temperature into degree Fahrenheit as follows.

            693.3 - 458.67

          = $234.63^{o}F$

From table A-$2E_{a}$

 $C_{p}$ = 0.240 Btu/lbm R  and   $C_{v}$ = 0.171 Btu/lbm

Now, we will substitute the energy balance as follows.

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

                         = $2 lbm \times 0.171 Btu/lbm R (693.3 - 520)$

                         = 59.3 Btu

Thus, we can conclude that final temperature of air is 59.3 Btu.