Answer:
T = -22/r + 39
(a) y = [130 − 130e^(-0.015t)] / 3
(b) 22.9
Explanation:
d²T/dr² + 2/r dT/dr = 0
If S = dT/dr, then:
dS/dr + 2/r S = 0
dS/dr = -2/r S
dS / S = -2/r dr
ln S = -2 ln r + C
S = e^(-2 ln r + C)
S = Ce^(-2 ln r)
S = C [e^(ln r)]^-2
S = C / r²
dT/dr = C / r²
dT = C dr / r²
T = -C / r + D
When r=1, T=17.
17 = -C + D
When r=2, T=28.
28 = -C/2 + D
Solving:
11 = C/2
C = 22
D = 39
T = -22/r + 39
We can check our answer by finding dT/dr and d²T/dr².
dT/dr = 22/r²
d²T/dr² = -44/r³
Plug into the original equation:
d²T/dr² + 2/r dT/dr = 0
-44/r³ + 2/r (22/r²) = 0
0 = 0
The total flow into the tank is 15 L/min, and the total flow out is 15 L/min, so the volume is constant at 1000 L. The amount of salt added to the tank is:
(0.05 kg/L) (5 L/min) + (0.04 kg/L) (10 L/min) = 0.65 kg/min
If y is the amount of salt in the tank at time t, and y/1000 is the concentration in the tank at time t, then the amount of salt removed from the tank is:
(y/1000 kg/L) (15 L/min) = 0.015y kg/min
The change in salt over time equals the amount of salt in minus the amount of salt out.
dy/dt = 0.65 − 0.015y
-200 dy/dt = 3y − 130
dy / (3y − 130) = -0.005 dt
3 dy / (3y − 130) = -0.015 dt
ln(3y − 130) = -0.015t + C
3y − 130 = Ce^(-0.015t)
3y = 130 + Ce^(-0.015t)
y = [130 + Ce^(-0.015t)] / 3
At t=0, y=0:
0 = (130 + C) / 3
C = -130
y = [130 − 130e^(-0.015t)] / 3
At t=50:
y = [130 − 130e^(-0.75)] / 3
y ≈ 22.9