Respuesta :
Explanation:
It is known that equation for steady state concentration is as follows.
[tex]C_{a} = \frac{QC}{Q + kV}[/tex]
where, Q = flow rate
k = rate constant
V = volume
C = concentration of the entering air
Formula for volume of the box is as follows.
V = [tex]a^{2}h[/tex]
= [tex]100 \times 100 \times 1[/tex]
= [tex]10000 km^{3}[/tex]
Now, expression to determine the discharge is as follows.
Q = Av
= [tex]100 \times 1 \times \frac{4 m}{s} \times \frac{km}{1000 m}[/tex]
= 0.4 [tex]km^{3}/s[/tex]
And, m (loading) = 10kg/s,
k = 0.20/hr
as 1 [tex]km^3 = 10^{12}[/tex] L (if u want kg/L as concentration)
Now, calculate the concentration present inside as follows.
[tex]C_{in} = \frac{10kg/s}{0.4 km^3/s}[/tex]
= 25 [tex]kg/km^3[/tex]
Now, we will calculate the concentration present outwards as follows.
[tex]C_{out} = {C_{in}}{(1 + k \times t)}[/tex],
and, t = [tex]\frac{V}{Q}[/tex]
= 25000 s or 6.94 hr
Hence, [tex]C_{out} = \frac{25}{(1 + 0.20 \times 6.94)}[/tex]
= 10.47 [tex]kg/km^3[/tex]
Thus, we can conclude that the the steady-state concentration if the air is assumed to be completely mixed is [tex]C_{out} = 10.47 kg/km^3[/tex] and [tex]C_{in} = 25 kg/km^3[/tex] .
The steady-state concentration if the air is assumed to be completely mixed is;
C_in = 25 kg/km³
C_out = 10.47 kg/km³
We are given;
Side lengths of box; x = 100 km
Height of box; h = 1 km
speed of clean air; v = 4 m/s = 0.004 km/s
Reaction rate; k = 0.2 /hr
Emission rate; E' = 10 kg/s
Volume of box is; V = length * width * height
Thus;
V = 100 * 100 * 1
V = 10000 km³
Formula for the discharge of air;
Q = Av
where A is area and v is speed. Thus;
Q = 100 * 1 * 0.004
Q = 0.4 km³/s
Formula for concentration present inside;
C_in = E'/Q
C_in = 10/0.04
C_in = 25 kg/km³
Now, time is;
t = V/Q
t = 10000/0.4
t = 25000 s = 6.944 hours
Formula for concentration present outside;
C_out = C_in/(1 + kt)
Thus;
C_out = 25/(1 + (0.2 * 6.944))
C_out = 10.47 kg/km³
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