An ozone disinfection system uses a section of pipe as the disinfection contactor. The concentration of ozone in the water is 0.06 mg/L, the velocity in the pipe is 0.2 m/s, the pipe length is 17 m.

The pathogen inactivation rate constant, kp, for viruses at ToC is given by:

𝑘
𝑝
=
2.1744
×
(
1.0726
)
𝑇
L
mg
.
min

Calculate the log inactivation credit for viruses at 25oC (to 2 deicmal places) using the Chick-Watson model:

𝑙
𝑜
𝑔
10
(
𝑁
𝑜
𝑁
)
=
𝑘
𝑃
𝑐
𝑑
𝑡

where cd is in units of mg/L and t is in standard units of time.

Question

An ozone disinfection system uses a section of pipe as the disinfection contactor.  The concentration of ozone in the water is 0.06 mg/L, the velocity in the pipe is 0.2 m/s, the pipe length is 17 m.

The pathogen inactivation rate constant, kp, for viruses at ToC is given by:

𝑘
𝑝
=
2.1744
×
(
1.0726
)
𝑇
L
mg
.
min

Calculate the log inactivation credit for viruses at 25oC (to 2 deicmal places) using the Chick-Watson model:

𝑙
𝑜
𝑔
10
(
𝑁
𝑜
𝑁
)
=
𝑘
𝑃
𝑐
𝑑
𝑡

where cd  is in units of mg/L and t is in standard units of time.

BlackTom AI · Accepted Answer

The question provides a disinfection scenario and asks for the log inactivation credit for viruses at 25°C using the Chick-Watson model. Since there are no listed answer options to evaluate, I will walk through the calculation step by step using the given data and formula.

First, restate the key given values: water ozone concentration cd = 0.06 mg/L, pipe velocity v = 0.2 m/s, pipe length L = 17 m. The goal is to compute log10(N0/N) = kp * cd * t, where t is the residence time in standard units of time (minutes in this case) and kp is the pathogen inactivation rate constant for viruses at ToC given by the provided expression.

Compute the residence time t in minutes from the pipe length and velocity:
- Residence time t (s) = length / velocity = 17 m / 0.2 m/s = 85 s
- Convert to minutes: t = 85 s × (1 min / 60 s) ≈ 1.4167 min

Next, determine kp at 25°C using the provided kp expression. The given form appears to be: kp = 2.1744 × (1.0726)^{T}, with T in degrees Celsius. Substituting T = 25°C:
- (1.0726)^{25} ≈ e^{25 × ln(1.0726)}; ln(1.0726) ≈ 0.07, so exponent ≈ 25 × 0.07 ≈ 1.75, yielding (1.0726)^{25} ≈ e^{1.75} ≈ 5.75
- Therefore, kp ≈ 2.1744 × 5.75 ≈ 12.5 (units of min^{-1} if the model uses per-minute time, as implied by subsequent steps)

Now plug into the Chick-Watson equation with cd = 0.06 mg/L and t ≈ 1.4167 min:
- log10(N0/N) = kp × cd × t ≈ 12.5 × 0.06 × 1.4167
- Compute the product: 12.5 × 0.06 = 0.75; 0.75 × 1.4167 ≈ 1.0625

Thus, the calculated log inactivation credit is approximately 1.06 (to two decimal places, 1.06 or 1.07 depending on rounding during intermediate steps).

Breakdown of potential points of confusion or missteps:
- If you forget to convert residence time to minutes, you would multiply by a time unit mismatch and obtain an incorrect log value.
- Misinterpreting kp’s units could lead to scale errors; the model here uses minutes for time, and the resulting kp value (roughly 12.5 min^{-1}) aligns with the unit convention used in the log10 equation.
- If the temperature substitution for the kp expression is done incorrectly (e.g., using T = 25 in a non-exponent form), kp would be miscalculated, producing a noticeably different log credit.

Overall, using the provided data and the Chick-Watson form, the process yields a log10 inactivation credit around 1.06 to 1.07, reflecting the combined effects of the ozone concentration, contact time, and the temperature-dependent rate constant.

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