Question

A \(40-\mathrm{ft}^{3}\) adiabatic container is initially evacuated. The supply line contains air that is maintained at 150 psia and \(90^{\circ} \mathrm{F}\). The valve is opened until the pressure in the container is the same as the pressure in the supply line. Determine the work potential of the air in this container when it is filled. Take \(T_{0}=80^{\circ} \mathrm{F}\).

Short answer

Based on the given problem, the work potential (exergy) of the air in the 40-ft³ adiabatic container when it is filled is approximately 880.53 Btu. The calculations involved converting temperatures to Rankine, determining the final temperature in the container, and calculating the work potential using the exergy formula.

Step-by-step solution

Convert the given temperatures to the absolute scale (Rankine)

To work with the ideal gas law and thermodynamics equations, it's necessary to use absolute temperature scales. Convert the air supply and ambient temperatures from Fahrenheit to Rankine. $$ T_{1} = 90 + 459.67 = 549.67\,\mathrm{R} $$ $$ T_{0} = 80 + 459.67 = 539.67\,\mathrm{R} $$

Calculate the final temperature in the container

When the air fills the container and reaches equilibrium with the supply line, we'll assume that the filling process is adiabatic. Since \(Q = 0\) in an adiabatic process: $$ W_{12} = -\Delta U_{12} $$ Let's apply this to an ideal gas with constant specific heat at constant volume \(c_v\): $$ W_{12} = -m c_v (T_2 - T_1) $$ To find the mass of air in the container, we can use the ideal gas law and the given conditions: $$ m = \frac{P_1 V}{R T_1} $$ Now, we can use the adiabatic process equation to determine the final temperature: $$ c_p - c_v = R $$ Rearranging the equation and using specific heat values of air (\(c_p = 0.24\,\mathrm{Btu/lbm\cdot R}\) and \(c_v = 0.171\,\mathrm{Btu/lbm\cdot R}\)): $$ R = 0.069\,\mathrm{Btu/lbm\cdot R} $$ Substitute the values and solve for m: $$ m = \frac{150 \times 144 \times 40}{0.069 \times 549.67} = 514.3\,\mathrm{lbm} $$ Now, let's find the final temperature \(T_2\): $$ W_{12} = -m c_v (T_2 - T_1) $$ $$ W_{12} = -514.3 \times 0.171 \times (T_2 - 549.67) $$ A reverse adiabatic process leads to an isentropic process. Hence, we can write: $$ \frac{T_2}{T_1} = \left(\frac{P_2}{P_1}\right)^{(k-1)/k} $$ Where \(k=\frac{c_p}{c_v}=1.4\) Now, in the equilibrium, \(P_2=P_1\), therefore: $$ \frac{T_2}{549.67} = 1 $$ $$ T_2 = 549.67\,\mathrm{R} $$

Calculate the work potential (exergy)

To calculate the work potential (exergy) of the air in the container, we can use the following formula: $$ W_{pot} = m c_v (T_2 - T_0) $$ Substitute the given values and solve for \(W_{pot}\): $$ W_{pot} = 514.3 \times 0.171 \times (549.67 - 539.67) $$ $$ W_{pot} = 880.53\,\mathrm{Btu} $$ The work potential of the air in the container when it is filled is 880.53 Btu.