An ozone disinfection system uses a section of pipe as the disinfection contactor. The concentration of ozone in the water is 0.09 mg/L, the velocity in the pipe is 0.21 m/s, the pipe length is 15 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 0.09 mg/L, the velocity in the pipe is 0.21 m/s, the pipe length is 15 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

We start by restating the problem and identifying the given quantities: a disinfection contactor with water ozone concentration cd = 0.09 mg/L, flow velocity v = 0.21 m/s, pipe length L = 15 m, and the pathogen inactivation rate constant kp for viruses at ToC (assumed to be 25°C). We are asked to calculate the log inactivation credit for viruses at 25°C using the Chick–Watson model: log10(N0/N) = kp * cd * t, where t is the residence time in standard time units (minutes in this formulation).

First, determine the residence time t in the pipe. With a uniform velocity, the residence time t is the length divided by velocity: t = L / v = 15 m / 0.21 m/s ≈ 71.4286 s. If the model’s time unit is minutes, convert seconds to minutes: t ≈ 71.4286 s × (1 min / 60 s) ≈ 1.19048 min. This t will be used in the log inactivation calculation.

Next, interpret the kp expression. The problem provides a specific form for kp: kp = 2.1744 × (1.0726)^{T_L mg. min}. The formatting is unusual, and it is not entirely clear how to parse the exponent. The most common interpretation in this context would be that kp is a function of temperature, often written kp(T) = A × (B)^{T}, with T in °C. If we take T = ToC = 25°C, a plausible interpretation is kp ≈ 2.1744 × (1.0726)^{25}. Calculating this would yield a substantial kp value (since (1.0726)^{25} ≈ e^{25 × ln(1.0726)} with ln(1.0726) ≈ 0.07, giving roughly e^{1.75} ≈ 5.75; kp ≈ 2.1744 × 5.75 ≈ 12.5).

However, there is a critical caveat: the units in the given kp expression are not clearly specified in the text (the line shows “kp = 2.1744 × (1.0726)^{T L mg. min}” which is ambiguous). The Chick–Watson model requires that kp be in units compatible with cd (mg/L) and t (minutes) so that log10(N0/N) is dimensionless. If kp is indeed in min^{-1} (or in a unit that, when multiplied by cd in mg/L and t in minutes, yields a dimensionless log value), the multiplication kp × cd × t should be performed directly. So, with cd = 0.09 mg/L and t ≈ 1.1905 min, the product kp × cd × t would yield the log inactivation.

Assuming the interpretable form kp ≈ 2.1744 × (1.0726)^{25} ≈ 12.5 (as a plausible estimate), we can compute log10(N0/N) ≈ kp × cd × t ≈ 12.5 × 0.09 × 1.1905 ≈ 12.5 × 0.1072 ≈ 1.34. This would give a log inactivation of about 1.34 log units. If, instead, a different interpretation of the kp expression or a different time unit were intended (e.g., if t were in seconds or if kp were in a different unit scale), the resulting log value would change accordingly. The key point is that the calculation hinges on a clear, consistent interpretation of the kp expression and the time unit, both of which are not unambiguously specified in the statement.

Another possible pitfall is treating the exponent base (1.0726) as a temperature multiplier raised to temperature in degrees Celsius. If someone mistakenly uses T = 25 in a different formulation, or misreads the exponent (for example treating the exponent as 25 in a unitless fashion without converting the time to minutes), the resulting kp value would change, leading to a markedly different log inactivation credit. In practice, small misinterpretations of kp’s temperature dependence or the time unit can yield noticeably different results, so it is essential to confirm the exact units and the intended form of the kp expression from the source.

In summary, the calculation path requires (1) obtaining the residence time t from length and velocity (t ≈ 1.1905 min), (2) evaluating kp at 25°C from the given expression (which is ambiguous in its formatting, but a common approach yields a kp on the order of several to a few tens per minute, e.g., ≈12.5 min^{-1} under one interpretation), and (3) computing log10(N0/N) = kp × cd × t. With the numbers above, the result is sensitive to the exact kp interpretation, and this sensitivity explains why different readings of the same formula can produce noticeably different log inactivation credits. The core lesson is that unit consistency and precise interpretation of kp's temperature dependence are essential to obtain a reliable result.

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