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lorasvet [3.4K]
3 years ago
11

Many households in developing countries prepare food over indoor cook stoves with no ducting system to exhaust the combustion pr

oducts outside. A woman in a household lights a fire in her cookstove at 4 pm. The fire emits benzo(a)pyrene, a carcinogenic compound, at a rate of 0.01 ng/min. The house has a size of 40 m3 and a ventilation rate of 1 m3/h. The ambient indoor air concentration of benzo(a)pyrene was 0.2 ng/m3 when she started the fire, but no significant concentration of benzo(a)pyrene is found in the outdoor air. What is the indoor benzo(a)pyrene concentration in the household at 6 pm?
Engineering
1 answer:
lorasvet [3.4K]3 years ago
7 0

Answer:

   C = 0.22857 ng / m³

Explanation:

Let's solve this problem for part the total time in the kitchen is

          t = 2h (60 min / 1h) = 120 min

The concentration (C) quantity of benzol pyrene is the initial quantity plus the quantity generated per area minus the quantity eliminated by the air flow. The amount removed can be calculated assuming that an amount of extra air that must be filled with the pollutant

amount generated

         C = co + time_generation rate / (area_house + area_flow)

         C = 0.2 + 0.01 120 / (40+ 2)

         C = 0.22857 ng / m³

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Explanation:

Data given

For this case we have the following data given:

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p = 2*0.05m +2*0.001m= 0.102m

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A= wh= 0.05m *0.001m = 0.00005 m^2

Solution to the problem

For this case we assume that we have steady conditions, the temperature of the fins varies just in one direction, the heat transfer coefficient not changes with the time and the thermal properties of the fin not change.

We can determine the temperature if the fin at x=5 cm=0.05 m from the base with the following formula:

\frac{T-T_{\infty}}{T_b -T_{\infty}} = e^{-mx}

Where m is a coefficient given by:

m = \sqrt{\frac{hp}{kA}}=\sqrt{\frac{20 W/m^2 C 0.102 m}{200 W/ mC 0.00005 m^2}}= 14.28 m^{-1}

The value of x for this case represent the distance x =5 cm =0.05m

T_b =130 C represent the base temperature

T_{\infty}= 20 represent the temperature of the sorroundings or the ambient.

If we replace we have this:

\frac{T-20}{130-20}= e^{-14.28*0.05}

And if we solve for T we got:

T= 20 + 110e^{-14.28*0.05} = 73.86 C

The answer for this case would be T = 73.86 C at 5cm from the base of the fin.

3 0
3 years ago
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