Can someone explain this problem to me, please A tank is filled with 1000 liters

Question

Can someone explain this problem to me, please

A tank is filled with 1000 liters of pure water. Brine containing 0.08 kg of salt per liter enters the tank at 5 liters per minute. Another brine solution containing 0.06 kg of salt per liter enters the tank at 9 liters per minute. The contents of the tank are kept thoroughly mixed and the drains from the tank at 14 liters per minute. Determine the differential equation which describes this system. Let S(t) denote the number of kg of salt in the tank after t minutes. Then dS/dt Solve the differential equation for S(t). S(t) =

Explanation / Answer

A)rate of change of concentration of salt = rate of salt input -rate of salt output

rate of change of concentration of salt = flowrate*concentration of salt input -flowrate*concentration of salt output

ds/dt =5*0.08 +0.06*9 - 14*s/(1000)

ds/dt =0.94 -0.014s , s(0)=0

B)ds/dt =0.94 -0.014s, s(0)=0

ds =0.94dt -0.014s dt

ds +0.014s dt =0.94 dt ,

integrating factor =e 0.014dt

integrating factor =e0.014t

multiply on both sides of equation by integrating factor

ds e0.014t + s0.014e0.014t dt =0.94 e0.014tdt ,

u'v+uv'=(uv)'

(se0.014t)' =0.94 e0.014tdt ,

integrate on both sides

(se0.014t)' = 0.94 e0.014tdt ,

se0.014t= (0.94/0.014)e0.014t+c

we have s(0)=0

0e0= (0.94/0.014)e0+c

0=(0.94/0.014)+c

c=(0.94/0.014)

se0.014t= (0.94/0.014)e0.014t+(0.94/0.014)

s= [(0.94/0.014)e0.014t+(0.94/0.014)]/e0.014t

s= (0.94/0.014)+(0.94/0.014)e-0.014t

s= (0.94/0.014)(1+e-0.014t)

s(t)= 67.143(1+e-0.014t)

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