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In: Civil Engineering

An unvented kerosene heater is operated for 1 hour in an apartment building having a volume...

An unvented kerosene heater is operated for 1 hour in an apartment building having a volume of 300 m3. The heater emits SO2 at a rate of 50 ug/s. The ambient air concentration and the indoor air concentration of SO2 are 80 ug/m3. If the rate of ventilation is70 liters/s and the apartment is assumed to be well-mixed, what is the indoor air concentration of SO2 at the end of 1 hour? Assume a first order decay rate using Table 7.14 in the textbook.

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Expert Solution

Given:- • Volume of apartment building , V = 300 m^3

• Emission rate of SO2 pollutant into apartment from unvented kerosene heater, E = 50 μg/s

• Ambient air concentration of SO2, Ca = 80 μg/m^3

• Initial indoor air concentration of SO2, Co = 80 μg/m^3

• Rate of ventilation, Q = 70 L/s = 0.07 m^3/s

• Apartment is assumed to be well mixed.

• Assume a first order decay rate.

To find:- The indoor air concentration (Ct) of SO2 at the end of 1 hour.

Solution:- The whole solution is shown in below images. The decay rate for SO2 pollutant from table is K = ( 6.39 × 10^-5 ) sec^-1. The indoor air concentration (Ct) is determined using the general solution form of the indoor air quality model which is shown in below images. Substituting all the respective parameter values in the general solution form equation of the indoor air quality model and find the value of the indoor air concentration of SO2 at the end of time 1 hour.

The required indoor air concentration of SO2 obtained at the end of 1 hour is 437.098 μg/m^3.

Ally Wells answered 11 months ago
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