If the BOD5 of a waste is 220 mg/L and the ultimate BOD is 320 mg/L, what

is the rate constant?

GBG

To calculate the rate constant, we can use the equation:

BOD(t) = BOD(0) * e^(-k*t)

Where:
BOD(t) is the BOD at time t
BOD(0) is the initial BOD
k is the rate constant
t is the time

Given information:
BOD(0) = 220 mg/L
BOD(t) = 320 mg/L

Let's rearrange the equation to solve for k:

BOD(t) / BOD(0) = e^(-k*t)

Taking the natural logarithm of both sides:

ln(BOD(t) / BOD(0)) = -k * t

We have BOD(t) = 320 mg/L and BOD(0) = 220 mg/L, so let's substitute these values:

ln(320 / 220) = -k * t

Now, we can solve for t:

t = ln(320 / 220) / -k

Since we don't have the value of t, we cannot directly calculate the rate constant. However, if you provide the time (t), we can substitute it into the equation above and solve for k.

To find the rate constant, we can use the following formula:

k = (ln(BOD5/BODu))/(t5-tu)

where:
k is the rate constant
BOD5 is the 5-day biochemical oxygen demand
BODu is the ultimate biochemical oxygen demand
t5 is the 5-day measurement time
tu is the ultimate measurement time

Given:
BOD5 = 220 mg/L
BODu = 320 mg/L
t5 = 5 days
tu = unknown

We need to find the value of tu to calculate the rate constant. Let's rearrange the formula and solve for tu:

ln(BOD5/BODu) = k(t5-tu)

First, divide both sides of the equation by k:

ln(BOD5/BODu)/k = t5 - tu

Now, subtract t5 from both sides:

tu = t5 - ln(BOD5/BODu)/k

Substituting the given values:

tu = 5 - ln(220/320)/k

Finally, substitute this value of tu back into the original formula to find k:

k = (ln(BOD5/BODu))/(t5 - tu)

Doing the calculations, we get:

k = (ln(220/320))/(5 - (5 - ln(220/320)/k))

This is a nonlinear equation, so we'll need to iterate to find the value of k. Start by making an initial guess for k, then plug it into the equation to get a new value for k. Repeat this process until you reach a convergence.