A tank contains 10000 liters of Brine of 200 kg dissolve salt. Fresh water flows into the tank at the rate of 100 liters/minute and mixed with uniform continuity and the same amount runs out with the same rate. If Q is total amount of salt present at time t in the brine, we have
Answers
Answer:
Let s(t) = amount, in kg of salt at time t. Then we have
ds
dt
= (rate of salt into tank) − (rate of salt out of tank)
= (0.05 kg/L · 5 L/min) + (0.04 kg/L · 10 L/min) −
s kg
1000 L
· 15 L/min
= 0.25 kg/min + 0.4 kg/min −
15s
1000
kg/min
So we get the differential equation
ds
dt
= 0.65 −
15s
1000
=
65
100
−
15s
1000
ds
dt
=
130 − 3s
200
We separate s and t to get
1
130 − 3s
ds =
1
200
dt
Integrate
Z
1
130 − 3s
ds =
Z
1
200
dt
−
1
3
· ln |130 − 3s| =
1
200
t + C1
ln |130 − 3s| = −
3
200
t + C2
|130 − 3s| = e− 3
200 t+C2
|130 − 3s| = C3e
−3t/200
130 − 3s = C4e
−3t/200
−3s = −130 + C4e
−3t/200
s =
130 − C4e
−3t/200
3
Since we begin with pure water, we have s(0) = 0. Substituting,
0 =
130 − C4e
−3·0/200
3
0 = 130 − C4
C4 = 130
1Stewart, Calculus, Early Transcendentals, p. 607, #48.
So our function is
s(t) = 130 − 130e−3t/200
3
After one hour (60 min), we have
s(60) = 130 − 130e−3·60/200
3
s(60) ≈ 25.7153
Thus, after one hour there is about 25.72 kg of salt in the tank.
Step-by-step explanation:
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