Question 9.PS.50: An underground saltmine, 100 000 m³ in volume, contains air ......

An underground saltmine, 100 000 m³ in volume, contains air at 290 K, 100 kPa. The mine is used for energy storage so the local power plant pumps it up to 2.1 MPa using outside air at 290 K, 100 kPa. Assume the pump is ideal and the process is adiabatic. Find the final mass and temperature of the air and the required pump work.

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C.V. The mine volume and the pump
Continuity Eq.6.15:      m _2- m _1= m _{ in }

Energy Eq.6.16:    m _2 u _2- m _1 u _1={ }_1 Q _2-{ }_1 W _2+ m _{ in } h _{ in }

Entropy Eq.9.12:      m _2 s _2- m _1 s _1=\int dQ / T +{ }_1 S _{2 \text { gen }}+ m _{\text {in }} s _{\text {in }}

Process: Adiabatic      { }_1 Q _2=0 \text {, Process ideal } \quad{ }_1 S _{2 \text { gen }}=0, \quad s _1= s _{\text {in }}

\Rightarrow m _2 s _2= m _1 s _1+ m _{ in } s _{ in }=\left( m _1+ m _{ in }\right) s _1= m _2 s _1 \Rightarrow s _2= s _1

Constant s ⇒      P _{ r 2}= P _{ r\,i  }\left( P _2 / P _{ i }\right)=0.9899\left\lgroup \frac{2100}{100} \right\rgroup=20.7879

\text { A.7.2 } \Rightarrow T _2= 6 8 0 \,K , u _2=496.94\,kJ / kg

\begin{aligned}& m _1= P _1 V _1 / RT _1=100 \times 10^5 /(0.287 \times 290)=1.20149 \times 10^5 \,kg \\& m _2= P _2 V _2 / RT _2=100 \times 21 \times 10^5 /(0.287 \times 680)= 1 0 . 7 6 0 \times 1 0^5 \,k g \\& \Rightarrow m _{ in }=9.5585 \times 10^5 \,kg\end{aligned}
\begin{aligned}{ }_1 & W _2= m _{\text {in }} h _{\text {in }}+ m _1 u _1- m _2 u _2 \\& = m _{ in }(290.43)+ m _1(207.19)- m _2(496.94)=- 2 . 3 2 2 \times 1 0^8 \,k J\end{aligned}
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