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

Answer:

Explanation:

Consider the following image(attached)

input Rate=

$10(kg/s) \times 10^9(\mu/kg)\\=10\times10^9\mu g/s\\k=0.2/hr$

To find the concentration,

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

$Volume=(100\times 10^3)\times (100\times 10^3)\times (1\times 10^3)\\1000\times 10^{10}m^3$

Output rate = Area x Velocity x concentration

$=(100\times 10^3(m)\times 1\times 10^3 (m))\times 4(m/s)\timesC(\mu g/m^3)\\=4\times 10^8 C \mu g/s$

Decay rate = reaction rate x Volume x C

$=\frac{0.2/hr}{3600s/hr}\times 1000\times 10^{10}(m^3)\times C(\mu g/m^3)\\=5.556\times 10^8 \mu g/s$

Substitute the values in equation(1)

$10\times 10^9(\mu g/s)=4\times 10^8 C(\mu g/s)+5.556\times 10^8 (\mu g/s)\\\\C=\frac{10\times 10^9}{4\times 10^8 +5.556\times 10^8}\\\\=10.46\mu g/m^3$

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

Concentration:

Output rate = Area x Velocity x concentration

$=(100\times 10^3(m)\times 1\times 10^3 (m))\times 1(m/s)\timesC(\mu g/m^3)\\=1\times 10^8 C \mu g/s$

Decay rate=$5.556\times 10^8\mu g/s$

$10\times 10^9(\mu g/s)=1\times 10^8 C(\mu g/s)+5.556\times 10^8 (\mu g/s)\\\\C=\frac{10\times 10^9}{1\times 10^8 +5.556\times 10^8}\\\\=15.25\mu g/m^3$

Concentration after two hours

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