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Q 1218010, 0, 00 (-12-23) 1 - (* —- (1 - V₁ × *) *| (H-V₂ 20 -exp 27 exp Oy 20 C(x, y, z)= C(x, 0, 0) = w(x, 0, 0) = Q 2 mu Oy Oz -exp QV₂ επισ, σε - exp (H–V¸ × ²)²] 202² ×11) ²* H-V₁ × 20²2/2 Wind velocity = 4 m/s, settling velocity = 0.02 m/s, stack height = 80 m, PM emission rate = 25 g/s, oy = 100 m, o₂ = 55 m. The deposition rate at 1000 m downwind is g/m²-s. (Note: Report in scientific notation if needed).

Q 1218010, 0, 00 (-12-23) 1 - (* —- (1 - V₁ × *) *| (H-V₂ 20 -exp 27 exp Oy 20 C(x, y, z)= C(x, 0, 0) = w(x, 0, 0) = Q 2 mu Oy Oz -exp QV₂ επισ, σε - exp (H–V¸ × ²)²] 202² ×11) ²* H-V₁ × 20²2/2 Wind velocity = 4 m/s, settling velocity = 0.02 m/s, stack height = 80 m, PM emission rate = 25 g/s, oy = 100 m, o₂ = 55 m. The deposition rate at 1000 m downwind is g/m²-s. (Note: Report in scientific notation if needed).

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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C(x, y, z) = C(x, 0, 0) = w(x, 0, 0) = Q 2 mu Oy Oz Q -exp • (-21723) 0² 2 Mu Oy Oz Q Vt 2 muo₂ o₂ ·(H-V₁₂ × 2- |- (1-1/2 x + ²) exp 20²/2 exp -exp - (z − (H - V₁ × 202/2 (H-V₁ × 1² 20 Wind velocity = 4 m/s, settling velocity = 0.02 m/s, stack height = 80 m, PM emission rate = 25 g/s, oy = 100 m, oz = 55 m. The deposition rate at 1000 m downwind is g/m²-s. (Note: Report in scientific notation if needed).
Transcribed Image Text:C(x, y, z) = C(x, 0, 0) = w(x, 0, 0) = Q 2 mu Oy Oz Q -exp • (-21723) 0² 2 Mu Oy Oz Q Vt 2 muo₂ o₂ ·(H-V₁₂ × 2- |- (1-1/2 x + ²) exp 20²/2 exp -exp - (z − (H - V₁ × 202/2 (H-V₁ × 1² 20 Wind velocity = 4 m/s, settling velocity = 0.02 m/s, stack height = 80 m, PM emission rate = 25 g/s, oy = 100 m, oz = 55 m. The deposition rate at 1000 m downwind is g/m²-s. (Note: Report in scientific notation if needed).
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