Question

Consider the air over a city to be a box 100 km on two sides that reaches up to an altitude of 1.0km. Clean air is blowing into the box along one of its sides with a speed of 4 m/s. Suppose an air pollutant with a degradation reaction rate k = 0.2 hr-1 is emitted into the box at a total rate of 10.0 kg/s.

If the windspeed in this above problem suddenly drops to 1 m/s, estimate the concentration of the pollutant two hours later.

Answer 1

Step-by-step explanation:

Consider the following image(attached)

input Rate=

`10(kg/s) * 10^9(\mu/kg)\\=10*10^9\mu g/s\\k=0.2/hr`

To find the concentration,

Input rate= Output rate + Decay rate ...(1)

`Volume=(100* 10^3)* (100* 10^3)* (1* 10^3)\\1000* 10^(10)m^3`

Output rate = Area x Velocity x concentration

`=(100* 10^3(m)* 1* 10^3 (m))* 4(m/s)*C(\mu g/m^3)\\=4* 10^8 C \mu g/s`

Decay rate = reaction rate x Volume x C

`=(0.2/hr)/(3600s/hr)* 1000* 10^(10)(m^3)* C(\mu g/m^3)\\=5.556* 10^8 \mu g/s`

Substitute the values in equation(1)

`10* 10^9(\mu g/s)=4* 10^8 C(\mu g/s)+5.556* 10^8 (\mu g/s)\\\\C=(10* 10^9)/(4* 10^8 +5.556* 10^8)\\\\=10.46\mu g/m^3`

If wind speed is suddenly drops to 1m/s concentration of the pollutant two hours later:

Concentration:

Output rate = Area x Velocity x concentration

`=(100* 10^3(m)* 1* 10^3 (m))* 1(m/s)*C(\mu g/m^3)\\=1* 10^8 C \mu g/s`

Decay rate=
`5.556* 10^8\mu g/s`

`10* 10^9(\mu g/s)=1* 10^8 C(\mu g/s)+5.556* 10^8 (\mu g/s)\\\\C=(10* 10^9)/(1* 10^8 +5.556* 10^8)\\\\=15.25\mu g/m^3`

Concentration after two hours

`C(t)=[C_0-C_(2hr)]exp[-(K+Q/V)t]+C_(2hr)\\\\C(2hr)=[10.46-15.25]exp[-((0.2)/(hr)+(1* 10^8(m^3/s)* 3600(s/hr))/(1000* 10^(10)(m^3)))2hr]+15.25\\\\\\[10.446-15.25]* 0.624 +15.25\\\\12.26\mu g/m^3`