An ozone disinfection system uses a section of pipe as the disinfection contactor. The concentration of ozone in the water is 0.05 mg/L, the velocity in the pipe is 0.15 m/s, the pipe length is 11 m.

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

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

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 0.05 mg/L, the velocity in the pipe is 0.15 m/s, the pipe length is 11 m.

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

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

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

BlackTom AI · Accepted Answer

First, restate the problem elements to ensure we have the full setup in view: an ozone contactor with water ozone concentration cd = 0.05 mg/L, flow velocity v = 0.15 m/s, pipe length L = 11 m, and we are to compute the log inactivation credit for viruses at 25°C using the Chick-Watson model. The Chick-Watson model for log inactivation is typically expressed as: log10(N0/N) = k * t, where k is the first-order rate constant (often represented as kp in some formulations) and t is the contact time in appropriate units. In many process calculations for UV or disinfection, t would be the residence time, which in a plug-flow (or well-mixed) assumption is t = L / v, provided the flow is fully developed and the fluid behaves as a continuous flow with negligible back-mixing.

Step 1: Calculate the residence time t in the pipe. Using t = L / v, we get t = 11 m / 0.15 m/s = 73.333... seconds (approximately 73.33 s). This determines the exposure duration of viruses to the disinfectant under the given flow and geometry.

Step 2: Identify the inactivation rate constant kp at ToC (25°C). The prompt mentions kp is given for viruses at ToC, but the numerical value for kp is not provided in the problem statement. Without kp, we cannot compute the Chick-Watson log inactivation credit because the product kp * t cannot be evaluated. If kp were provided (for example, in units of 1/s or 1/min depending on the convention), we would compute log10(inactivation) as log10(N0/N) = kp * t, with t in seconds if kp is in 1/s, or convert t to minutes if kp is in 1/min, ensuring unit consistency.

Step 3: Check unit consistency. The Chick-Watson equation requires consistent time units with kp. If kp is given in 1/s, then t in seconds is fine (73.33 s). If kp is given in 1/min, convert t to minutes: 73.33 s ≈ 1.2222 min, and compute log inactivation as kp * 1.2222 min. If kp is given in a different unit, we must convert accordingly.

Step 4: Compare to the provided answer. The given answer is 0.77 log. To assess plausibility, we would need the specific kp value used by the problem setter. For example, if kp were approximately 0.0105 1/s, then kp * t = 0.0105 * 73.33 ≈ 0.77, yielding log10(N0/N) ≈ 0.77. However, this is contingent on kp being 0.0105 1/s. Without the kp value, the calculation cannot be uniquely verified.

Step 5: Summary of what’s missing and how an exact result would be obtained. The missing piece is the numerical kp value at ToC for viruses. Once kp is provided, compute t = 11 / 0.15 ≈ 73.33 s, then evaluate log10(N0/N) = kp * t (with consistent units). If the resulting product equals 0.77, the stated answer is correct under those units and conditions. Absence of kp prevents a definitive verification. Additionally, ensure the Chick-Watson model form used in the course materials matches the one applied here (some texts use natural log ne = k*t with log base e; others use log10). If your course uses natural log, convert accordingly: log10(N0/N) = (1/ln(10)) * (kp * t).

Conclusion (for this analysis): The numerical result shown (0.77) is plausible only if kp at 25°C is approximately 0.0105 s^-1 (assuming t in seconds and log base 10). Since kp is not provided in the prompt, we cannot confirm the exact calculation; please supply the kp value to verify the 0.77 log credit.

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