show all steps A water system consists of a 10,000 L tank of well mixed water. W
ID: 2968685 • Letter: S
Question
show all steps
A water system consists of a 10,000 L tank of well mixed water. Water is pumped into the
tank at 1000 L/day and is distributed to the town at the same rate. Usually the water is
pollution free. But after a storm, the runo from surrounding area causes the concentration
of pollution in the incoming water to jump from 0 to 1% for 2 days. Safety regulations specify
that the pollution concentration in the drinking water must remain below 0.1%
(a) Set up a DE and solve it to show that the water will be unsafe to drink at the end of the
two days.
(b) Suppose the water system consists of two well mixed 5,000 L tanks. Water is pumped
into the rst tank, ows into the second tank and is distributed to the town all at 1000
L/day. Set up a DE and solve it to determine if this system will make the water safe to
drink at the end of the two days?
Explanation / Answer
(a)
Let Q be amount of pollutant either in grams or kg or pounds or whatever...
-amount of Q divided by volume will give concentration of Q (that is amount of Q per unit VOLUME)
-concentration divided by time will give the rate of concentration
(dQ/dt = rate in - rate out ) means how the concentration of Q in the tank changes with time
rate in = (inflow volume rate) x (inflow concentration) = (1000L/day)(0.01 /L) = 10 /day
rate out = (outflow volume rate) x (tank concentration )
NB: the outflow concentration wil be the same as the tank concentration!!!
rate out = (1000L/day) x [ Q / (10000L) ] = (Q/10) /day
ignore the units for now
dQ/dt = rate in - rate out = 10 - (Q/10)
dQ/dt = 10 - Q/10
-==============================
10dQ/dt = 100 - Q
dQ/(100 - Q) = 0.1 dt
Q - 100 = Ae^(-0.1t)
Q = 100 + Ae^(-0.1t)
initial conditions: t = 0, Q = 0 that is water was pollutant free initially
0 = 100 + Ae^(-0.1x 0)
A = -100
Q = 100 - 100e^(-0.1t) = 100[1 - e^(-0.1t) ]
Q = 100[1 - e^(-0.1t) ]
after two days: Q = 100[1 - e^(-0.1t) ] = 100[1 - e^(-0.1x2) ] = 18.13
Q/10000L = 18.13/10000 = 0.001813 = 0.18% which is greater than the required 0.1%
====================================================================================
(b)
1st tank:
dQ?/dt = rate in - rate out = (1000L/day)(0.01 /L) - (1000L/day)(Q?/5000L) = 10 - Q?/5
dQ?/ ( Q? - 50) = - 0.2dt
ln( Q? - 50) = -0.2t + C
( Q? - 50) = e^(-0.2t) e^C
initial values: Q? = 0, t = 0
(Q? - 50) = -50e^(-0.2t)
Q? = 50 - 50e^(-0.2t)
2nd tank:
dQ?/dt = rate in - rate out
dQ?/dt = (1000L/day)[ ( 50 - 50e^(-0.2t) ) / 5000L] - (1000L/day)(Q?/5000L)
dQ?/dt = ( 10 - 10e^(-0.2t) ) - Q?/5
Q? = -10[t + 5 - 5e^(0.2t)] e^(-0.2t)
after 2 days
Q? = -10[t + 5 - 5e^(0.2t)] e^(-0.2t) = 3.07759677750526
(Q?/5000L x 100%) = 0.062%
this is less than the required 0.1% so this system will make water safe after two days
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