Chptr7PrblmSolns

# Chptr7PrblmSolns - SOLUTIONS FOR CHAPTER 7 7.1 From(1.9 mg...

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Pg. 7.1 SOLUTIONS FOR CHAPTER 7 7.1 From (1.9), mg/ m 3 = ppm x mol wt 24.465 (at 1 atm and 25 o C) a. CO 2 3 = 5000ppm x 12+ 2x16 ( ) 24.465 =8992mg/m 3 9000mg/ m 3 b. HCHO ppm = 24.465 x 3.6 mg/m 3 2x1 + 12 + 16 ( ) = 2.94ppm c. NO mg/m 3 = 25ppm x 14+16 ( ) 24.465 =30.7mg/m 3 7.2 70% efficient scrubber, find S emission rate: 600 MWe η = 0.38 600/0.38=1579 MWt 9000 Btu/lb coal 1% S Input = 600,000 kWe 0.38 x 3412 Btu kWhr x lb coal 9000Btu x 0.01 lb S lb coal = 5986 lb S/hr 70% efficient, says release 0.3 x 5986 lbS/hr = 1796 lb S/hr 1800 lbS/hr 7.3 If all S converted to SO 2 and now using a 90% efficient scrubber: SO 2 = 0.1 x 5986 lbS hr x (32 + 2x16) lb SO 2 32 lb S = 1197 lb SO 2 / hr 1200 lb SO 2 /hr 7.4 70% scrubber, 0.6 lb SO 2 /10 6 Btu in, find % S allowable: a. X lbs S lbs coal x 0.3 lbs S out 1 lb S in x 2 lbs SO 2 lb S x lb coal 15,000Btu = 0.6 lb SO 2 10 6 Btu X = 15,000x0.6 0.3x2x10 6 = 0.015 = 1.5% S fuel b. X lbs S lbs coal x 0.3 lbs S out 1 lb S in x 2 lbs SO 2 lb S x lb coal 9,000Btu = 0.6 lb SO 2 10 6 Btu X = 9,000x0.6 0.3x2x10 6 = 0.009 = 0.9% S fuel

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Pg. 7.2 7.5 Compliance coal: 1.2 lbs SO 2 10 6 Btu = lb coal 12,000 Btu x X lb S lb coal x 2 lb SO 2 lb S X = 1.2x12,000 2x10 6 = 0.0072 = 0.7%S 7.6 Air Quality Index: _________________________________________________________ Pollutant Day 1 Day 2 Day 3 _________________________________________________________ O 3 , 1-hr (ppm) 0.15 0.22 0.12 CO, 8-hr (ppm) 12 15 8 PM 2.5 , 24-hr ( μ g/m 3 ) 130 150 10 10 , 24-hr ( μ g/m 3 ) 180 300 100 SO 2 , 24-hr (ppm) 0.12 0.20 0.05 NO 2 , 1-hr (ppm) 0.4 0.7 0.1 ___________________________________________________________ Using Table 7.3: a. Day 1: Unhealthy, AQI 151-200 triggered by PM . b. Day 2: Very Unhealthy, AQI 201-300, triggered by both O 3 and NO 2 c. Day 3: Moderate, AQI 51-100, triggered by CO, PM and SO 2 7.7 8 hrs of CO at 50 ppm, from (7.6): %COHb = 0.15%1 e 0.402/ hr x 8hr ( ) x50 = 7.2% 7.8 Tractor pull at 436 ppm CO: a. 1 hr exposure: %COHb = e 0.402t ( ) ppm ( )= 0.15% 1 e 0.402x1 ( ) x436 = 21.6% b. To reach 10% COHb, 10 = 0.15 1 e 0.402t ( ) = 65.4 65.4e 0.402t e 0.402t = 55.4 65.4 = 0.871 so t =- 1 0.402 ln 0.871 ( ) = 0.41 hr 7.9 RH to produce HCHO: RO • + O 2 HO 2 • + R'CHO (7.19) for R' CHO to be HCHO, R' must be H so that +O 2 2 • + HCHO for the reaction to balance , R = CH 3 which says RH in (7.16) must be CH 4 (methane)
Pg. 7.3 7.10 RH = propene = CH 2 =CH-CH 3 = C 3 H 6 so, R = C 3 H 5 so the sequence of reactions (7.16) to (7.19) is: C 3 H 6 + OH C 3 H 5 • + H 2 O C 3 H 5 • + O 2 C 3 H 5 O 2 C 3 H 5 O 2 • + NO C 3 H 5 O • + 2 C 3 H 5 O • + O 2 HO 2 • + C 2 H 3 CHO The end product is acrolein, CH 2 CHCHO. 7.11 A 20- μ m particle blown to 8000 m. From (7.24) its settling velocity is v = d 2 ρ g 18 η = (20x10 6 m) 2 x 1.5x10 6 g/ m 3 x 9.80m/s 2 18 x 0.0172g/m -s = 0.019m/s Time to reach the ground = 8000m 0.019m /s x 3600s/hr x 24hr/d = 4.87days Horizontal distance = 4.87 days x 10 m/s x 3600s/hr x 24hr/d x 10 -3 km/m = 4200 km 7.12 Residence time for 10- μ m particle, unit density, at 1000m: Settling velocity = v = d 2 g 18 = (10x10 6 2 x 10 6 3 x 9.80m/s 2 18 x 0.0172g/m -s = 0.00317m/s Residence time = τ = h v = 1000m 0.00317m/s x 3600s/hr = 87.6hrs 7.13 Settling velocity and Reynolds numbers: a. 1 μ m: v = d 2 g 18 = (1x10 6 2 x 10 6 3 x 9.80m/s 2 18 x 0.0172g/m -s = 3.2x10 5 m /s Re = air dv = 1.29x10 3 3 x 1x10 -6 m x 3.17x10 -5 m/s 0.0172 m/s = 2.4x10 6 b. 10 μ m: v = d 2 g 18 = (10x10 6 2 x 10 6 3 x 9.80m/s 2 18 x 0.0172g/m -s = 3.2x10 3 m/s = air = 1.29x10 3 3 x 10x10 -6 m x 3.17x10 -3 m/s 0.0172 m/s = 2.4x10 3 c. 20 μ m: v = d 2 g 18 = (20x10 6 2 x 10 6 3 x 9.80m/s 2 18 x 0.0172g/m -s = 0.0127m/s

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Pg. 7.4 Re = ρ air dv η = 1.29x10 3 g/ m 3 x 20x10 -6 m x 0.0127m/s 0.0172 m/s = 0.02 So for all of these particles, the Reynolds number is much less than 1, which means (7.24) is a reasonable approximation to the settling velocity.
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## This note was uploaded on 04/03/2008 for the course C E 370 taught by Professor Burgos during the Spring '08 term at Penn State.

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Chptr7PrblmSolns - SOLUTIONS FOR CHAPTER 7 7.1 From(1.9 mg...

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