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

We start by identifying the given quantities and the equations to be used.

- Water velocity v = 0.15 m/s and pipe length L = 11 m. The travel time in the contactor is t = L / v = 11 / 0.15 s = 73.333... s ≈ 1.222 min.
- The pathogen inactivation rate constant for viruses at ToC = 25 C is given by kp = 2.1744 × (1.0726)^T, where T is the temperature in degrees Celsius. For T = 25 C, compute the factor (1.0726)^T.
  - Compute ln(1.0726) ≈ 0.0701.
  - Multiply by T: 25 × 0.0701 ≈ 1.7525.
  - Exponentiate: (1.0726)^25 ≈ e^1.7525 ≈ 5.77.
  - Therefore, kp ≈ 2.1744 × 5.77 ≈ 12.55 mg·min⁻¹.
- The Chick-Watson model for log10 inactivation is log10(No/N) = k_p × c_d × t, where c_d is the disinfectant concentration in mg/L and t is time in minutes.
  - Given c_d = 0.05 mg/L and t ≈ 1.222 min, compute c_d × t = 0.05 × 1.222 ≈ 0.0611 (mg·min)/L.
  - Multiply by kp: k_p × c_d × t ≈ 12.55 × 0.0611 ≈ 0.767.

Putting it together, the Chick–Watson log inactivation credit for viruses at 25 C is approximately 0.77 (to two decimal places). The calculation hinges on consistent unit treatment and correct interpretation of the exponential temperature term, followed by straightforward multiplication of kp, cd, and t.

Note on steps and potential pitfalls:
- Ensure the time t is in minutes when using kp in mg·min⁻¹; convert from seconds to minutes if needed.
- The temperature factor (1.0726)^T grows with T; misplacing the exponent or using a different base would lead to a different kp value.
- If one uses a rounded intermediate value for (1.0726)^25, the final log-inactivation credit can shift slightly; carrying more precise intermediate results reduces rounding error.

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