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:

kp=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:

log10(
No

N
)=kPcdt

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:

kp=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:

log10(
No

N
)=kPcdt

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

BlackTom AI · Accepted Answer

Let’s parse the problem: we have an ozone contactor with water concentration cd = 0.05 mg/L, flow velocity v = 0.15 m/s, and pipe length L = 11 m. The inactivation rate constant for viruses at ToC is given by kp = 2.1744 × (1.0726)^T, where T = 25°C. The Chick-Watson model is log10(No/N) = kp × cd × t, with t being the residence time in minutes.

Step 1 — Compute the temperature-dependent kp: First determine the multiplicative term (1.0726)^T at T = 25. Since 1.0726 is the base and 25 is the exponent, we evaluate (1.0726)^25. Taking natural logs for a rough check: ln(1.0726) ≈ 0.0701, so 25 × ln(1.0726) ≈ 1.7525, and exp(1.7525) ≈ 5.77. Multiplying by 2.1744 gives kp ≈ 2.1744 × 5.77 ≈ 12.6 mg·min⁻¹. This is the inactivation rate constant at 25°C.

Step 2 — Compute the residence time t in minutes: The residence time t = pipe length / velocity = 11 m / 0.15 m/s ≈ 73.333... seconds. Convert to minutes: t ≈ 73.333... / 60 ≈ 1.2222 minutes.

Step 3 — Assemble the log inactivation credit: Using log10(No/N) = kp × cd × t, substitute kd, cd, and t: kp × cd × t ≈ 12.6 × 0.05 × 1.2222. Multiply 12.6 by 0.05 to get 0.63, then 0.63 × 1.2222 ≈ 0.77 (keeping two decimal places as requested).

Step 4 — Interpret the result: The calculation combines the temperature-adjusted inactivation rate, the ozone dose in water, and the residence time to yield the log reduction credit for viruses. The arithmetic shows the product is on the order of a few tenths, specifically about 0.77 when rounded to two decimals.

Note on method and units: kp has units of mg·min⁻¹, cd is mg/L, and t is minutes, so the product kp × cd × t yields a dimensionless log reduction credit in base-10 (log10). The key checks are ensuring T is properly substituted into kp, converting velocity and length to residence time in minutes, and multiplying in the correct order to obtain log10(No/N).

ENGR10004_2025_SM1 Homework Assessment 6 - Water Treatment Quiz

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