Respuesta :
Answer:
Follows are the solution to this question:
Explanation:
In first point:
Ammonia volumetric flow = [tex]{105} \frac{m3}{h} = \frac{7}{240} \ \frac{m^3}{ s}[/tex]
Waste gas mole density, dw[tex]= 0.0407 \ \frac{mol}{L} = {40.7} \ \frac{mol}{m^3}[/tex]
Original ammonia mole proportion in receiving waste gas [tex]= 0.120[/tex]
Mole density for ammonia [tex]= 0.120 \times 40.7[/tex]
[tex]= 4.884 \ \ \frac{ \ mol}{ m^3 \text {of its incoming waste}}[/tex]
Therefore, ammonia rate is extracted from its incoming stream
[tex]= NH_3 \text{molecule density} \times \text {Volume flow rate} \\\\ = 4.884\times (\frac{7}{240})\\\\ = 0.14245 \ \ NH_3 \ \ \text{moles per second}[/tex]
As 90 percent ammonia is removed, that ammonia removal rate:
[tex]= 0.9 \times 0.14245 \\\\ = 0.128 \ \ NH_3 \ \text{moles per second} \\\\ = 460.8 \ \ \frac{mol}{h}[/tex]
In second point:
Ammonia possessing water content [tex]= 0.35\ \ \frac{NH_3}{mole}[/tex]
Consequently, to maintain 0.128 mol ammonia:
water moles necessary [tex]= \frac{1}{0.35} \times 0.128[/tex]
[tex]= 0.365 \ mol \ water[/tex]
Water atomic weight=[tex]18 g[/tex]
Therefore, mass needed of water [tex]= 18 \times 0.365[/tex]
[tex]= 6.57 \ g[/tex]
In this required water range of competencies stream rate:
[tex]= \frac{\text{necessary weight}}{\text{water density}} = \frac{6.57}{0.99} \\\\ = 6.63 \frac{ml}{second} \\\\ = 6.63 \times 10^6 \times 3600 \\\\ = 0.023868 \frac{m^3}{h} \\\\ = 23.868 \frac{L}{h}[/tex]
In third point:
fresh ammonia mol fraction [tex]= \frac{ \text{moles of ammonia from of the outgoing stream} }{\text{maximum moles}}[/tex]
Amino ammonia moles = 10 per cent of the ammonia moles inside an incoming stream = 0.10×initial concentration of the mole:
[tex]= 0.1 \times 4.884\\\\ = 0.484 \ moles[/tex] in an outgoing stream.
Its total outflow moles = the total ammonia moles at the original ammonia:
[tex]\text{stream moles} = 40.7- (90\% \times 4.884)[/tex]
[tex]= 36.3[/tex]
So new ammonina mole fraction:
[tex]= \frac{0.4884}{36.3} \\\\ = 0.01345[/tex]
