Question

6. A 500 L tank initially contains 10 g of salt dissolved in 200 L of water. Water containing 1/4 g of salt per liter is poured into the tank at the rate of 4 L/min and the mixture is drained from the tank at the rate of 2 L/min. Find an initial value problem that solve for x(t), the mass of salt in the tank at time t prior to when the tank overflows, then solve for x(t).

Answer 1

lt in th olueet x(t) be amunt tank at minutes. Phe dx vate of ult in - (rate of Aalt out) dt Here 4 9 rate in = 4 L X Aiter min min fist alulate t amount To find rate out ии &Aolution in th tank at ony tims t. Initially th amownt ofolutin s 200 L. After 1 min, th Aolution in th tank s 200 + 4 -2 = 2a 2 L hu after minu tes, th dluton in t tk is = (20D +2t) L. 200 (4-2)t 2 L X ate aut = (2 0+ 24)L min

guuisd differ.tlgutin s Hend du dt 100t ol t Lineer differatl Ltin,hawing Thi's is integrtingator o0t Melthly clt d 00+ (t On integration, t - ii + Ci (l 0 0+ x( = 2

Ainle initially thurefore o0+o)(l0) Phus (200 tt 2 (loo time t, t sohtin in t Wknou thet amy ettbe t tine uhe, tamk 2o02t tank over fow tha at t 5 0 0 200 2t 5 0 tak ouer fows after 150 min tes ie Now amound of Aalt at t = 1 50 s

00 150) (150o 150) lo1 2 100 150 (3 50) (150) 250 5 00 o5 4 X50) J0 9 amount of Dalt whe tmk oreflous Phees (109) grom
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