It is found that ozone reaction in water nearly follows first-order kinetics in ozone concentration decreases by 50% in 12 minutes (t1/2 = 12 min). A water supplier wants to inject ozone into a pipe bringing water to thee water treatment plant to pre-disinfect the influent. The 3 ft diameter pipe is 3400 feet long with a steady flow rate of 10000 gal/min. What concentration of ozone in mg/L should be injected at the head of the pipe so that there will be an ozone concentration of 1 mg/L at the pipe's exit into the plant?

Compute the average velocity of the water flowing in the pipe.

Get this from the relation
Q = A * V
where Q is the volume flow rate and A is the cross sectional area. I recommend that you convert Q from gallons/min to ft^3/min, to get a flow velocity in ft/s.

The time it takes the water to reach the plant is
T = L/V

Compute that time.

The ozone density will decrease by a factor
(1/2)^(T/t1/2)

Use that factor to determine the ozone density requirement at the source

To determine the concentration of ozone in mg/L that should be injected at the head of the pipe, we need to calculate the rate of ozone decay in the pipe.

First, we need to convert the flow rate from gallons per minute (gal/min) to liters per minute (L/min) since the concentration is measured in mg/L and the flow rate needs to be in the same unit.

1 gallon = 3.78541 liters
1 minute = 60 seconds

Flow rate in L/min = 10000 gal/min * 3.78541 L/gal * (1/60) min/sec
Flow rate in L/min = 631.569 L/min

Next, we need to convert the length of the pipe from feet to meters since it will be used to calculate the residence time.

1 meter = 3.28084 feet

Pipe length in meters = 3400 ft * (1/3.28084) m/ft
Pipe length in meters = 1036.9694 m

Now we can calculate the residence time of the water in the pipe, which is the time it takes for water to travel from the head of the pipe to the exit.

Residence time = Pipe length / Flow rate
Residence time = 1036.9694 m / 631.569 L/min
Residence time = 1.641 min

Since the ozone reaction follows first-order kinetics with a half-life of 12 minutes, we can calculate the rate constant (k).

t1/2 = 0.693 / k
12 min = 0.693 / k

k = 0.693 / 12 min
k = 0.05775 min^-1

Next, we can use the first-order rate equation to calculate the concentration of ozone at the exit of the pipe.

ln(Ct / C0) = -k * t

where:
Ct = concentration of ozone at time t (1 mg/L)
C0 = initial concentration of ozone at the head of the pipe (unknown)
t = residence time (1.641 min)
k = rate constant (0.05775 min^-1)

Rearranging the equation, we get:

C0 = Ct * e^(k * t)
C0 = 1 mg/L * e^(0.05775 min^-1 * 1.641 min)
C0 ≈ 1.274 mg/L

Therefore, the concentration of ozone that should be injected at the head of the pipe is approximately 1.274 mg/L to achieve an ozone concentration of 1 mg/L at the pipe's exit into the water treatment plant.

To determine the concentration of ozone that should be injected at the head of the pipe, we'll need to use the first-order reaction kinetics equation. The equation is as follows:

C = C0 * e^(-kt)

Where:
- C is the concentration of ozone at a given time (mg/L)
- C0 is the initial concentration of ozone at t = 0 (mg/L)
- k is the rate constant of the reaction
- t is time (min)
- e is the base of the natural logarithm (approximately 2.71828)

Given that the half-life (t1/2) is 12 minutes, we can use this information to determine the rate constant (k) as follows:

t1/2 = 0.693 / k

Solving for k:
k = 0.693 / t1/2
k = 0.693 / 12 min
k = 0.05775 min^(-1)

Now, we can use the rate constant (k) to determine the initial concentration of ozone (C0) at the head of the pipe. We can refer to the following equation:

C0 = C * e^(kt)

Given that we want the ozone concentration (C) at the pipe's exit to be 1 mg/L, we can rearrange the equation as follows:

C * e^(kt) = C0
1 mg/L * e^(0.05775 min^(-1) * 3400 ft / (10000 gal/min * 7.48 gal/L)) = C0

Note: We need to convert the units of the pipe length and flow rate to ensure consistency.

Now, we can calculate C0.