I n l e t a i r r e q u i r e m e n t = 3 , 125 m g m i n ÷ 1000 2 ÷ 0.2769 k g m 3 × 1000 L m 3 = 11.29 L m i n {\displaystyle Inlet\ air\ requirement=3,125\ {\frac {mg}{min}}\div 1000^{2}\div 0.2769\ {\frac {kg}{m^{3}}}\times 1000\ {\frac {L}{m^{3}}}=11.29\ {\frac {L}{min}}}