A tank contains 11,00o L of brine with 22 kg of dissolved selt. Pure waber enter

Question

A tank contains 11,00o L of brine with 22 kg of dissolved selt. Pure waber enters the tank at a rate of 110 min, The solution is kept thoroughly mixed and drains from the tark at the same rate. How much sat is in the tank after t inutes? Step 1 Let y(t) be the amount of salt in the tenk after t minutes Then y(o)-22 8223 Step 2 There are 11000 L in the tank at all times, and so the concentration of salt per liter of water is given by Step 3 Since pure water flows into the tank at 110 /min and the misture drains at the same rate, then the rate of change of salt in the tank is given by Y temi .. @) 11000 Stap 4 The dnerential equation for this stuation -1 Separating gives us ly dy /1007t Stap 5 dy- Ignoring the constant of iteretion, integrating the len side gves Step 6 ../aa- eving the constant of integratien, integrating the Ignoring the constant of integration, Integrating the right side gives Snce y is .card no-22. Since y is elways postive, we now have In)- 100 c and to)-22 and solving fr caves usc-E01+ . Substtuting and soling for C gives us c-Ihol + How much salt is in the tank after 20 minutes?

Explanation / Answer

7. At t=0. y=22

So,

ln(22) = -0/100 + C

Thus, C = ln(22) = 3.0910

Part b.

When t=20,

ln(y) = -20/100 + 3.0910 = 2.8910

So, y = e2.8910 = 18.0121

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