# Air Pollution Questionnaire

Air Pollution Questionnaire

Air Pollution Questionnaire.

ME 401/5501 Spring 2020 FINAL EXAM Name **_______** 1 In Duct A, The gas flow rate is 5000 SCFM (standard cubic feet per minute), the gas temperature is 350 F and the pressure is -32 inches of water.

In-Duct B, the gas flow rate is 4000 ACFM (actual cubic feet per minute), the temperature is 400 F and the pressure is -35 inches of water. Calculate the total gas flow rate in Duct C where the two flows combine. Use a barometric pressure of 29.92 inches of Hg.

## Downwind distance = *__________m crosswind distance = ____________m*

*__________m crosswind distance = ____________m*

Duct C gas flow rate = *SCFM K= L = Version 20200422 M= N= ME 401/5501 Final Exam Spring 2020 2page 2For your student-specific values, find downwind and crosswind distance between source and receptor: Downwind distance = __________m Crosswind distance = ____________m Use the gaussian dispersion model to estimate ground-level concentration on an overcast day if i)the source emits 100 g/s ii)the effective stack height is 37 m iii)the wind speed is 3.7 m/s Concentration in ug/m3 USE THE FOLLOWING PARAMETERIZATIONS FOR 𝜎y AND 𝜎Z Version 20200422 ME 401/5501 Final Exam Spring 2020 3 Shown to the right are sampling results upstream and downstream of an air pollution control device. All measurements are at actual (i.e., rather than standard) conditions, with concentration values (in mg/m3) shown in bold and gas velocity (in feet per minute) shown in italics. Find the emission rate through the upstream and downstream cross-sections. *

*What is the control efficiency for the control device? The sampling subarea (for each of the 56 areas in both tables) is 0.L x 0.M square meters, where L and M depend on your student ID number.* Upstream emission rate lb/hr Downstream emission rate lb/hr Control Efficiency Version 20200422 Upstream 42 43 68 31 64 32 57 73 857 1169 1338 1508 1572 1314 1119 984 65 35 46 59 74 73 26 70 1130 1321 1517 1074 1219 1014 1183 1227 55 60 63 46 35 54 73 33 842 1184 1558 1234 914 894 1570 1337 53 71 74 36 73 74 56 67 928 1311 999 1441 1331 1449 817 992 40 71 35 31 64 29 27 35 894 1408 1546 1542 855 1546 1409 1451 61 59 43 34 43 50 47 59 1238 879 1551 1104 1039 1406 922 1540 41 68 29 43 40 72 64 61 1378 1579 1037 1242 1001 1305 1136 1491 Downstream % **For example, if KLMN is 4983, this would be 0.9 x 0.8 = 0.72 sq m page 3 16.5 10.6 9.3 12.2 8.5 17 13.5 12.8 1270 1316 1521 847 1346 1552 805 803 10.9 9.2 13.7 10.8 12.6 15.2 12.9 12.8 1036 1088 886 869 1040 917 1212 1231 15.2 12.4 12 16.1 11.8 16.5 8.4 13.5 1375 835 914 1377 1556 1108 1155 874 9.8 9.1 15.7 15.4 15.2 10.1 13.4 16.6 1543 1561 836 1512 1157 1202 1386 1160 7.6 13.7 15.8 11.6 8.4 13.6 12.2 11.0 1013 1486 1170 1459 1527 1179 962 1184 16.7 10.6 12.4 16.7 10.9 7.7 15 8.4 1102 1097 972 1474 1221 1533 895 1412 14.6 16.4 11.9 17.5 15 10.1 14.3 9.6 1285 1370 1406 932 1216 1466 1363 909

## C8h18 = 𝟏𝟐 × 𝟖 + 𝟏 × 𝟏𝟖 = 𝟗𝟔 + 𝟏𝟖 = 𝟏𝟏𝟒 𝒈/𝒎𝒐𝒍 molar mass

**ME 401/5501 Final Exam Spring 2020 page 4 4 Estimate the 50% cutpoint for a cyclone whose inlet is shown to the right. Assume that: ● ● ● ● the volumetric flow rate is 150 m3 per minute; the gas has the same properties as air at standard conditions the particle density is 2000 kg/m3; the number of turns made equals M * from your student ID. D50 = ____**microns Use this value in problem 5 Assume that the the pressure drop across the cyclone is K* velocity heads. What does the pressure drop equal? *

*For example, if KLMN is 4983, this would be 8 turns and 4 velocity heads ΔP = Pin - Pout =*Pa Version 20200422 ME 401/5501 Final Exam Spring 2020 page 5 5 F or this problem, you’ll use a particle size distribution (PSD) based on your student ID number. (Assume a particle density of 2000 kg/m3.) First, complete Columns I thru III below. Then, using a cyclone 50% cutpoint from problem 4 , complete Columns IV to VI as done during class and find an overall collection efficiency. Cols I thru III Lower End Upper End Cumulative Mass Distribution (%) 0 M NM.L* M L+K MN.L* L+K 20 LK.M* 20 50 100 Size Bin Size Bin Lower End Upper End Mass Fraction in Bin Col IV Col V Col VI Representative Diameter** Estimated Efficiency (%) Product of Columns III and V

**____***For example, if KLMN is 4983, these entries would be 38.9 %, 83.9%, and 94.8%*

**Use the Mass-Mean-Diameter as you did in Worksheet 20200413OVERALL Version 20200422 EFFICIENCY =***% ME 401/5501 Final Exam Spring 2020 page 6 6 Refer to Example 10-1 in the wet scrubber material. This is a similar problem, with all the same variables except for the following: ● ● ● Uncontrolled particulate emission rate = 1500 kg/hr Mass-median particle size = (L+K+N)/2 μm L/G ratio = 1-N/(L+M) L/m3 = 1-N/(L+M) x10-3 ● ● ● Gas flow rate = (10 + M) m3/s Particle density = M1/2 g/cm3 Gas velocity in throat = (100 – M – N) m/s Complete the following table, based on the variables and notation defined in the scrubber handout. Don’t forget to include units where appropriate. Cunningham factor, CCAero geo mean diameter, dpg Droplet diameter, ddInertial parameter, KpgCunningham factor, CCAero geo mean diameter, dpg Use the figure at the left and the “parameter for L/G ratio” to determine the overall collection efficiency. Version 20200422 Name: _1__True or False Answers: (i). T: Criteria pollutants are those for which New Source Performance have been established. (ii). T: Stoke’s law predates the concept the Reynold’s number. (iii). T: The higher the relative humidity, lower the air density. (iv). T: The basic Gaussin Dispersion model assumes a flat ground surface. (v). T: A box model is useful for indoor air quality. (vi). F: Impaction may be considered a special case of interception. Version 20190501 page 2 (vii). T: Potential temperature provides a more straight forward way to characterize atmospheric stability. (viii). F: The plume rise equation given in the textbook does not depend on atmospheric stability class. (ix). T: The horizontal dispersion coefficient 𝜎𝑦 is always greater than the corresponding vertical coefficient. (x). F: “Albedo” refers to a code of silence in organized crime. (xi). T: Below-ground sources are used to account for complete reflection of a plume in the Gaussian dispersion. (xii). T: The mixing height represents the height at which the potential temperature equals the actual temperature. (xiii). T: It’s common practice to model stack emission at a unit emission rate. (xiv). F: Dosage refers to the total mass of pollutant that a person would have been exposed to had they been at one spot during their entire life. (xv). F:Deposition velocity is defined as the deposition rate times the concentration. +2_The processing equipment shown is associated with the emission factor given. The capture efficiency of the two-hood system is (100 – K – L)%.*The first APCD has a control efficiency of (10M+L)% and the second APCD has a control efficiency of (100 – N) % . * For example, if KLMN is 9483, the capture efficiency would be (100 – 9 – 4) = 87%, and the efficiency of APCD1 and APCD2 would be (10×8+3) = 83% and (100 – 3) = 97%, respectively. The emission factor would be 0.384 lb/ton and the throughput is 9483 lb/hr. Thus, the processing emission rate is 0.384 lb/ton x 9483 lb/hr K= 8 Version 20190501 L= 7 M= 5 N= 2 page 3 We have K=8, L=7 M=5 and N=2 Emission factor = 0.257 lb/ton and Throughput = 8752 lb/hr Capture efficiency of two-hood system= (𝟏𝟎𝟎 − 𝑲 − 𝑳)% = (𝟏𝟎𝟎 − 𝟖 − 𝟕)% = 𝟖𝟓% = 𝟎. 𝟖𝟓 Control efficiency of APCD1, 𝜼𝟏 = (𝟏𝟎𝑴 + 𝑳)% = (𝟏𝟎 × 𝟓 + 𝟕)% = 𝟓𝟕% = 𝟎. 𝟓𝟕 Control efficiency of APCD2, 𝜼𝟐 = (𝟏𝟎𝟎 − 𝑵)% = (𝟏𝟎𝟎 − 𝟐)% = 𝟗𝟖% = 𝟎. 𝟗𝟖 a) What’s the processing emission rate? 𝒍𝒃 𝟎. 𝟐𝟓𝟕 × 𝟖𝟕𝟓𝟐 𝒉𝒓 = 𝟐𝟐𝟒𝟗. 𝟐𝟔𝟒 𝒍𝒃/𝒉𝒓 b) What is the stack emission rate (in g/s) of the pollutant? 2.072 g/s Solution: Pollutant capture rate by two-hood system = 𝟎. 𝟖𝟓 × processing emission rate = 𝟎. 𝟖𝟓 × 𝟐𝟐𝟒𝟗. 𝟐𝟔𝟒 𝒍𝒃/𝒉𝒓 = 𝟏𝟗𝟏𝟏. 𝟖𝟕𝟒𝟒 𝒍𝒃/𝒉𝒓 It’s clear that efficiency of APCD1 is 57%, it means 43% of the pollutant captured by two-hood system is not captured by APCD1. And efficiency of APCD2 is 98%, it means 2% of the pollutant entering in APCD2 is not captured by APCD2. It means 𝟎. 𝟒𝟑 × 𝟎. 𝟎𝟐 = 𝟎. 𝟎𝟎𝟖𝟔 or 0.86% of the pollutant captured by two-hood system is not collected in APCD1 and APCD2 and thus released into the atmosphere through stack. K= 8 Version 20190501 L= 7 M= 5 N= 2 page 4 𝒍𝒃 Hence, stack emission rate = 𝟎. 𝟎𝟎𝟖𝟔 × 𝟏𝟗𝟏𝟏. 𝟖𝟕𝟒𝟒 𝒉𝒓 = 𝟏𝟔. 𝟒𝟒𝟐 𝒍𝒃/𝒉𝒓 = 𝟏𝟔. 𝟒𝟒𝟐 c) What is the fugitive emission rate? 𝒍𝒃𝒎𝒂𝒔𝒔 𝒉𝒓 𝟒𝟓𝟑.𝟓𝟗 𝒈 × 𝟏 𝒍𝒃 𝒎𝒂𝒔𝒔 𝟏 𝒉𝒓 × 𝟑𝟔𝟎𝟎 𝒔 = 𝟐. 𝟎𝟕𝟐 𝒈/𝒔 337.39 lb/hr Solution: Fugitive emission rate = Processing emission rate – Capture rate by hood = 𝟐𝟐𝟒𝟗. 𝟐𝟔𝟒 𝒍𝒃/𝒉𝒓 – 𝟏𝟗𝟏𝟏. 𝟖𝟕𝟒 𝒍𝒃/𝒉𝒓 = 𝟑𝟑𝟕. 𝟑𝟗 𝒍𝒃/𝒉𝒓 d) How many lb per hour of the pollutant are collected by APCD #1? 1282.08 lb/hr Solution: Rate of pollutant collected by APCD#1=Rate of pollutant collected by two-hood system × efficiency of APCD#1= 𝟐𝟐𝟒𝟗. 𝟐𝟔𝟒 𝒍𝒃 × 𝒉𝒓 𝟎. 𝟓𝟕 = 𝟏𝟐𝟖𝟐. 𝟎𝟖 𝒍𝒃/𝒉𝒓 e) What is the combined control efficiency of APCDs #1 and #2? 99.14 % Solution. Combined efficiency = (𝟏 − 𝟎. 𝟎𝟎𝟖𝟔) = 𝟎. 𝟗𝟗𝟏𝟒 = 𝟗𝟗. 𝟏𝟒% Or combined efficiency = 𝟏 − (𝟏 − 𝟎. 𝟓𝟕)(𝟏 − 𝟎. 𝟗𝟖) = 𝟎. 𝟗𝟗𝟏𝟒 = 𝟗𝟗. 𝟏𝟒% _3__Recall the question in the first homework assignment about cooling an air basin. The approach is far more feasible when one attempts to cool process gas streams. Suppose we wish to cool an air stream from 475 to 225o F by spraying liquid water (70o F) into it. Refer to the problem in HW#1 for the heat of vaporization for water and consider the specific heat of air to be constant at 0.245 BTU/lbmass-F over the temperature range of interest. For every pound of air, how many pounds of water would be necessary? 0.0545**lb water/lb air Solution. Given specific heat of air over the temperature range of interest = 0.245 BTU/lbmass-F K= 8 Version 20190501 L= 7 M= 5 N= 2 page 5 The heat of vaporization of water at 70oF (From HW#1) = 1050 BTU/lbmass Heat released by 1 lbmass air in cooling from 475o F to 225o F = 𝒎𝒄∆𝒕 = 𝟏 𝐥𝐛𝐦𝐚𝐬𝐬 × 𝟎.𝟐𝟒𝟓𝐁𝐓𝐔 𝐥𝐛𝐦𝐚𝐬𝐬 −𝐅 × (𝟒𝟕𝟓 𝐅 − 𝟐𝟐𝟓 𝐅) = 𝟔𝟏. 𝟐𝟓 𝐁𝐓𝐔 Amount of heat absorbed by 1 lbmass water at 70o F in vaporization = 1050 BTU Since the final temperature of air would be 225o F, therefore, final temperature of water vapor present in the air would be 225o F. It means that water vapor at 70o F would get heated to 225o F and the heat absorbed in this process = 𝐜𝐩𝐰𝐚𝐭𝐞𝐫−𝐯𝐚𝐩𝐨𝐮𝐫 × ∆𝐭 = 𝟎. 𝟒𝟕𝟔𝟕 𝐥𝐛 𝐁𝐓𝐔 𝐦𝐚𝐬𝐬−𝐅 𝑩𝑻𝑼 × (𝟐𝟐𝟓𝑭 − 𝟕𝟎𝑭) = 𝟕𝟑. 𝟖𝟖𝟖𝟓 𝒍𝒃 𝒎𝒂𝒔𝒔 Therefore, the net amount of heat absorbed by water when it is sprayed in the air at 70o F and final temperature of the air is 225o F= 1050 BTU/lbmass + 73.8885 BTU/lbmass= 1123.8885 BTU/lbmass Let the mass of water required per pound of air in cooling from 475o F to 225o F = m According to the conservation of energy, Heat released from the air = Heat gain by the water 𝑩𝑻𝑼 𝟔𝟏. 𝟐𝟓 𝑩𝑻𝑼 = 𝟏𝟏𝟐𝟑. 𝟖𝟖𝟖𝟓 𝒍𝒃 𝒎𝒂𝒔𝒔 ×𝒎 𝟔𝟏.𝟐𝟓 𝒎 = 𝟏𝟏𝟐𝟑.𝟖𝟖𝟖𝟓 𝒍𝒃𝒎𝒂𝒔𝒔 = 𝟎. 𝟎𝟓𝟒𝟓 𝒍𝒃𝒎𝒂𝒔𝒔 Solve for m,**

*4__Recall the problem involving CO 2 tailpipe emissions. In this problem, you’ll see one way to develop an emission factor estimate. a) Consider the complete combustion of octane and complete the stoichiometric equation: C8H18 + 25/2 O2 → 8 CO2 + 9 H2O b) Given that the density of octane is 0.69 kg/liter = 5.8 lbmass/gallon, determine how many grams of CO2 are developed by combusting 1 gallon of octane K= 8 Version 20190501 L= 7 M= 5 N= 2 page 6 8123.104 grams CO2 per gallon Solution: Molar mass of C8H18 = 𝟏𝟐 × 𝟖 + 𝟏 × 𝟏𝟖 = 𝟗𝟔 + 𝟏𝟖 = 𝟏𝟏𝟒 𝒈/𝒎𝒐𝒍 Molar mass of O2 = 𝟏𝟔 × 𝟐 = 𝟑𝟐 𝒈/𝒎𝒐𝒍 Molar mass of CO2= 𝟏𝟐 + 𝟏𝟔 × 𝟐 = 𝟒𝟒 𝒈/𝒎𝒐𝒍 It’s given that the density of octane is 5.8 lbmass/gallon. Therefore, mass of 1 gallon octane 𝟒𝟓𝟑.𝟓𝟗 𝒈 is 5.8 lbmass. = 𝟓. 𝟖 𝒍𝒃𝒎𝒂𝒔𝒔 × 𝟏 𝒍𝒃 𝒎𝒂𝒔𝒔 = 𝟐𝟔𝟑𝟎. 𝟖𝟐𝟐 𝒈 Number of moles of octane in 1 gallon of octane = 𝟐𝟔𝟑𝟎.𝟖𝟐𝟐 𝒈 𝟏𝟏𝟒 𝒈 𝒎𝒐𝒍𝒆𝒔 = 𝟐𝟑. 𝟎𝟕𝟕 𝒎𝒐𝒍𝒆𝒔 It’s clear from the Stoichiometric Equation that the 1 mole of C8H18 produces 8 moles of 𝒈 CO2 i.e. 𝟖 𝒎𝒐𝒍 × 𝟒𝟒 𝒎𝒐𝒍 = 𝟑𝟓𝟐 𝒈 Therefore, mass of CO2 produced by 23.077 moles of octane 𝒈 = 𝟐𝟑. 𝟎𝟕𝟕 𝒎𝒐𝒍 × 𝟑𝟓𝟐 = 𝟖𝟏𝟐𝟑. 𝟏𝟎𝟒 𝒈 𝒎𝒐𝒍 Hence, 8123.104 g CO2 is produced by the combustion of 1 gallon of octane. c) The Department of Transportation estimated that the average light-duty vehicle 20.0 mpg in 2000 (source). Estimate a CO2 emission factor using the information from part (b). How does it compare to the value given in the textbook’s problem? 406.1552 grams/mile Solution: The calculated value of carbon dioxide emission is 8123.104 g/gallon. It’s given that mileage of a car is 20.0 mpg. K= 8 Version 20190501 L= 7 M= 5 N= 2 page 7 Therefore, CO2 emission factor = 𝒈 𝟖𝟏𝟐𝟑.𝟏𝟎𝟒 𝒈𝒂𝒍𝒍𝒐𝒏 𝒎𝒊𝒍𝒆𝒔 𝟐𝟎.𝟎 𝒈𝒂𝒍𝒍𝒐𝒏 = 𝟒𝟎𝟔. 𝟏𝟓𝟓𝟐 𝒈/𝒎𝒊𝒍𝒆 _5__As plant engineer, you asked three of your new hires to check the air flows in a manifold. They decided to divide up the work and each measured the flow in a duct. Unfortunately, they didn’t coordinate their activities and came back with the information given below. What’s the flow rate in Duct X? Assume that the standard pressure is 1013 mbar.*Duct Temp. Gage Pressure A 625 oF -28.5 in H2O B 198 oC -58.7 mm Hg C 852 oR -87.5 Flow Rate 7980 acfm 8 m3/s (actual) kPa 2578 acfm where K, L, M, and N depend on your student ID number. For example, if KLMN = 4983 , then the pressures in Ducts A, B, and C would be -34.8 in H2O, -84.9 mm Hg, -49.8 kPa, respectively; the temperature in Duct C would be 483 oR; and the flow rate in Ducts B and C would be 4 Nm3/s and 3894 acfm, respectively. 13552.117 Solution. Assumptions: Standard Atmospheric temperature = 𝟔𝟖𝒐 𝑭 = 𝟐𝟗𝟑 𝑲 = 𝟐𝟎𝒐 𝑪 = 𝟓𝟐𝟖𝒐 𝑹 K= 8 Version 20190501 L= 7 M= 5 N= 2 page 8 Standard Atmospheric pressure =1 atm = 1013 mbar =1.013 bar= 1 atm = 14.696 psi = 407 inches water = 29.92 inches Hg = 760 mm Hg 1 cubic meter = 35.29 cubic feet Strategy: We will convert the volumetric flow in each of A, B, and C duct into scfm. 𝑻 𝑷 Formula: 𝑸𝒔𝒕𝒅 = 𝑸𝒂𝒄𝒕 × 𝑻𝒔𝒕𝒅 × 𝑷𝒂𝒄𝒕 𝒂𝒄𝒕 𝒔𝒕𝒅 Calculations Duct A: Given: 𝑸𝑨,𝒂𝒄𝒕 = 𝟕𝟗𝟖𝟎 𝒂𝒄𝒇𝒎 , 𝑻𝑨,𝒂𝒄𝒕 = 𝟔𝟐𝟓𝒐 𝑭 = (𝟒𝟔𝟎 + 𝟔𝟐𝟓)𝒐 𝑹 = 𝟏𝟎𝟖𝟓𝒐 𝑹 𝒑𝑨𝒈 = −𝟐𝟖. 𝟓 𝒊𝒏. 𝑯𝟐 𝑶 𝒑𝑨 = 𝒑𝒃 + 𝒑𝑨𝒈 = 𝟒𝟎𝟕 𝒊𝒏. 𝑯𝟐 𝑶 − 𝟐𝟖. 𝟓 𝒊𝒏. 𝑯𝟐 𝑶 = 𝟑𝟕𝟖. 𝟓 𝒊𝒏. 𝑯𝟐 𝑶 Using the formula: 𝑻 𝑷 𝟓𝟐𝟖 𝑸𝑨,𝒔𝒕𝒅 = 𝑸𝑨,𝒂𝒄𝒕 × 𝑻𝒔𝒕𝒅 × 𝑷 𝑨 = 𝟕𝟗𝟖𝟎 × 𝟏𝟎𝟖𝟓 × 𝒂𝒄𝒕 𝒔𝒕𝒅 𝟑𝟕𝟖.𝟓 𝟒𝟎𝟕 𝒔𝒄𝒇𝒎 = 𝟑𝟔𝟏𝟏. 𝟒𝟐𝟓 𝒔𝒄𝒇𝒎 Duct B: Given: 𝑸𝑩,𝒂𝒄𝒕 = 𝟖 𝒎𝟑 /𝒔 , 𝑻𝑩,𝒂𝒄𝒕 = 𝟏𝟗𝟖𝒐 𝑪 = (𝟐𝟕𝟑 + 𝟏𝟗𝟖)𝑲 = 𝟒𝟕𝟏𝑲 𝒑𝑩𝒈 = −𝟓𝟖. 𝟕 𝒎𝒎 𝑯𝒈 𝒑𝑩 = 𝒑𝒃 + 𝒑𝑩𝒈 = 𝟕𝟔𝟎 𝒎𝒎 𝑯𝒈 − 𝟓𝟖. 𝟕 𝒎𝒎 𝑯𝒈 = 𝟕𝟎𝟏. 𝟑 𝒎𝒎 𝑯𝒈 Using the formula: 𝑻 𝑷 𝟐𝟗𝟑 𝑸𝑩,𝒔𝒕𝒅 = 𝑸𝑩,𝒂𝒄𝒕 × 𝑻 𝒔𝒕𝒅 × 𝑷 𝑩 = 𝟖 × 𝟒𝟕𝟏 × 𝑩,𝒂𝒄𝒕 𝑸𝑩,𝒔𝒕𝒅 = 𝟒. 𝟓𝟗𝟐 Or 𝒎𝟑 𝒔 × 𝒔𝒕𝒅 𝟑𝟓.𝟐𝟗 𝒇𝒕𝟑 𝟏 𝒎𝟑 𝟕𝟎𝟏.𝟑 𝟕𝟔𝟎 𝒎𝟑 /𝒔 = 𝟒. 𝟓𝟗𝟐 𝒎𝟑 /𝒔 𝟔𝟎 𝒔 × 𝟏 𝒎𝒊𝒏 = 𝟗𝟕𝟐𝟑. 𝟏𝟎𝟏 𝒔𝒄𝒇𝒎 Duct C: Given: 𝑸𝑪,𝒂𝒄𝒕 = 𝟐𝟓𝟕𝟖 𝒂𝒄𝒇𝒎 , 𝑻𝑨,𝒂𝒄𝒕 = 𝟖𝟓𝟐𝒐 𝑹 𝒑𝑪𝒈 = −𝟖𝟕. 𝟓 𝒌𝑷𝒂 𝒑𝑨 = 𝒑𝒃 + 𝒑𝑨𝒈 = 𝟏𝟎𝟏. 𝟑 𝒌𝑷𝒂 − 𝟖𝟕. 𝟓 𝒌𝑷𝒂 = 𝟏𝟑. 𝟖 𝒌𝑷𝒂 Using the formula: K= 8 Version 20190501 L= 7 M= 5 N= 2 page 9 𝑻𝒔𝒕𝒅 𝑸𝑪,𝒔𝒕𝒅 = 𝑸𝑪,𝒂𝒄𝒕 × 𝑻 𝑪,𝒂𝒄𝒕 𝑷𝑪 ×𝑷 𝒔𝒕𝒅 𝟓𝟐𝟖 𝟏𝟑.𝟖 = 𝟐𝟓𝟕𝟖 × 𝟖𝟓𝟐 × 𝟏𝟎𝟏.𝟑 𝒔𝒄𝒇𝒎 = 𝟐𝟏𝟕. 𝟔𝟒𝟒 𝒔𝒄𝒇𝒎 Let the density of the air at standard condition be ρ Applying the conservation of mass, we have 𝒎̇𝑿 = 𝒎̇𝑨 + 𝒎̇𝑩 + 𝒎̇𝑪 𝝆𝑸𝑿,𝒔𝒕𝒅 = 𝝆𝑸𝑨,𝒔𝒕𝒅 + 𝝆𝑸𝑩,𝒔𝒕𝒅 + 𝝆𝑸𝑪,𝒔𝒕𝒅 Divide each side by ρ 𝑸𝑿,𝒔𝒕𝒅 = 𝑸𝑨,𝒔𝒕𝒅 + 𝑸𝑩,𝒔𝒕𝒅 + 𝑸𝑪,𝒔𝒕𝒅 𝑸𝑿,𝒔𝒕𝒅 = 𝟑𝟔𝟏𝟏. 𝟒𝟐𝟓 𝒔𝒄𝒇𝒎 + 𝟗𝟕𝟐𝟑. 𝟏𝟎𝟏 𝒔𝒄𝒇𝒎 + 𝟐𝟏𝟕. 𝟔𝟒𝟒 𝒔𝒄𝒇𝒎 Plug in values, 𝑸𝑿,𝒔𝒕𝒅 = 𝟏𝟑𝟓𝟓𝟐. 𝟏𝟏𝟕𝒔𝒄𝒇𝒎*6__For your student-specific values, find downwind and crosswind distance between source and receptor: Downwind distance = 226.44 m Crosswind distance = 70.55 m Use the gaussian dispersion model to estimate ground-level concentration if the source emits 100 g/s. Stability Class Wind Speed (m/s) Effective Stack Hgt. (m) C 5 87 D 4 E D σY(m) σZ (m) Ground Level Concentration (ug/m3) 23 14 𝟕. 𝟑𝟔𝟔 × 𝟏𝟎−𝟕 0 15 8 1.043 3 25 11 6 𝟑. 𝟏𝟗𝟏 × 𝟏𝟎−𝟖 5 35 15 8 𝟓. 𝟖𝟏𝟖 × 𝟏𝟎−𝟓 Note: Values of Plume Dispersion coefficients for the downwind distance 0.2 km are taken from the table which is available online at the link http://courses.washington.edu/cee490/DISPCOEF4WP.htm K= 8 Version 20190501 L= 7 M= 5 N= 2 page 10 Concentration of pollutant at ground level is given by the relation. 𝟐 𝑸 𝒚 𝑯 𝟐 𝒄= 𝒆𝒙𝒑 [– (𝟎. 𝟓) ( ) ] 𝒆𝒙𝒑 [– (𝟎. 𝟓) ( ) ] 𝝅𝒖𝝈𝒚 𝝈𝒛 𝝈𝒚 𝝈𝒛 Where u is average speed of the plume, y is crosswind distance and H is the stack height. Q is the emission rate from the stack. Plug in the values, we have Ist row: u= 5 m/s, H = 87 m and 𝝈𝒚 = 𝟐𝟑 𝒎 and 𝝈𝒛 = 𝟏𝟒 𝒎 while x and y are calculated above. 𝒈 𝒄= 𝟏𝟎𝟎 𝒔 𝟑. 𝟏𝟒 × 𝟓 𝒎 × 𝒔 𝟐𝟑 𝒎 × 𝟏𝟒 𝒎 𝒆𝒙𝒑 [– (𝟎. 𝟓) ( 𝟕𝟎. 𝟓𝟓𝒎 𝟐 𝟖𝟕 𝒎 𝟐 ) ] 𝒆𝒙𝒑 [– (𝟎. 𝟓) ( ) ] 𝟐𝟑 𝒎 𝟏𝟒 𝒎 𝒈 𝒄 = 𝟕. 𝟑𝟔𝟔 × 𝟏𝟎−𝟏𝟑 𝒎𝟑 = 𝟕. 𝟑𝟔𝟔 × 𝟏𝟎−𝟕 𝝁𝒈/𝒎𝟑 𝑰𝑰𝒏𝒅 𝒓𝒐𝒘 : u = 4 m/s, H=0, 𝝈𝒚 = 𝟏𝟓 𝒎 and 𝝈𝒛 = 𝟖 𝒎 𝒈 𝟕𝟎. 𝟓𝟓𝒎 𝟐 𝟎𝒎 𝟐 𝒄= 𝒆𝒙𝒑 [– (𝟎. 𝟓) ( ) ] 𝒆𝒙𝒑 [– (𝟎. 𝟓) ( ) ] 𝒎 𝟏𝟓 𝒎 𝟖𝒎 𝟑. 𝟏𝟒 × 𝟒 𝒔 × 𝟏𝟓 𝒎 × 𝟖 𝒎 𝟏𝟎𝟎 𝒔 𝒈 𝒄 = 𝟏. 𝟎𝟒𝟑 × 𝟏𝟎−𝟔 𝒎𝟑 = 𝟏. 𝟎𝟒𝟑 𝝁𝒈/𝒎𝟑 IIIrd row: u= 3 m/s, H = 25 m and 𝝈𝒚 = 𝟏𝟏 𝒎 and 𝝈𝒛 = 𝟔 𝒎 while x and y are calculated above. 𝒈 𝒄= 𝟏𝟎𝟎 𝒔 𝟑. 𝟏𝟒 × 𝟑 𝒎 𝒔 × 𝟏𝟏 𝒎 × 𝟔 𝒎 𝒆𝒙𝒑 [– (𝟎. 𝟓) ( 𝟕𝟎. 𝟓𝟓𝒎 𝟐 𝟐𝟓 𝒎 𝟐 (𝟎. ) ] 𝒆𝒙𝒑 [– 𝟓) ( ) ] 𝟏𝟏 𝒎 𝟔𝒎 𝒈 𝒄 = 𝟑. 𝟏𝟗𝟏 × 𝟏𝟎−𝟏𝟒 𝒎𝟑 = 𝟑. 𝟏𝟗𝟏 × 𝟏𝟎−𝟖 𝝁𝒈/𝒎𝟑 IVth row: : u = 5 m/s, H=35 m, 𝝈𝒚 = 𝟏𝟓 𝒎 and 𝝈𝒛 = 𝟖 𝒎 𝒈 𝟕𝟎. 𝟓𝟓𝒎 𝟐 𝟑𝟓 𝒎 𝟐 𝒄= 𝒆𝒙𝒑 [– (𝟎. 𝟓) ( ) ] 𝒆𝒙𝒑 [– (𝟎. 𝟓) ( ) ] 𝒎 𝟏𝟓 𝒎 𝟖𝒎 𝟑. 𝟏𝟒 × 𝟓 × 𝟏𝟓 𝒎 × 𝟖 𝒎 𝟏𝟎𝟎 𝒔 𝒔 𝒄 = 𝟓. 𝟖𝟏𝟖 × 𝟏𝟎−𝟏𝟏 K= 8 Version 20190501 L= 7 M= 5 𝒈 𝒎𝟑 N= 2 = 𝟓. 𝟖𝟏𝟖 × 𝟏𝟎−𝟓 𝝁𝒈/𝒎𝟑 page 11 Using K=8, L=7, m=5 and N=2, We have coordinates of source as (-8,-7) and coordinates of receptor is (5,2). Where 1 unit = 15 m as shown in above graph. Now distance between the source and the receptor is 𝒅 = √(𝟓 + 𝟖)𝟐 + (𝟐 + 𝟕)𝟐 = √𝟏𝟑𝟐 + 𝟗𝟐 = 𝟓√𝟏𝟎 And the direction of the line segment joining source to receptor from the x-axis is 𝜷 = 𝒕𝒂𝒏−𝟏 ( 𝟐+𝟕 𝟗 ) = 𝒕𝒂𝒏−𝟏 ( ) = 𝟑𝟒. 𝟔𝟗𝟓𝒐 𝟓+𝟖 𝟏𝟑 And wind direction is 𝜶 = 𝟓𝟐𝒐 from the X-axis The angle between the direction of wind and the line joining the source and the receptor is 𝜽 = 𝜶 − 𝜷 = 𝟓𝟐𝒐 − 𝟑𝟒. 𝟔𝟗𝟓𝒐 = 𝟏𝟕. 𝟑𝟎𝟓𝒐 K= 8 Version 20190501 L= 7 M= 5 N= 2 page 12 𝒐) Now the downwind distance 𝒙 = 𝒅 𝒄𝒐𝒔 𝜽 = 𝟓√𝟏𝟎 𝒄𝒐𝒔 (𝟏𝟕. 𝟑𝟎𝟓 = 𝟏𝟓. 𝟎𝟗𝟔 𝒖𝒏𝒊𝒕𝒔 But 1 unit = 15 m Therefore, 𝒙 = 𝟏𝟓. 𝟎𝟗𝟔 × 𝟏𝟓 𝒎 = 𝟐𝟐𝟔. 𝟒𝟒 𝒎 Cross wind distance is 𝒚 = 𝒅 𝒔𝒊𝒏 𝜽 = 𝟓√𝟏𝟎 𝒔𝒊𝒏 (𝟏𝟕. 𝟑𝟎𝟓𝒐 ) = 𝟒. 𝟕𝟎𝟑 𝒖𝒏𝒊𝒕𝒔 Or 𝒚 = 𝟒. 𝟕𝟎𝟑 × 𝟏𝟓 𝒎 = 𝟕𝟎. 𝟓𝟒𝟓 𝒎 = 𝟕𝟎. 𝟓𝟓 𝒎 Note: Handwritten solution is also attached with this document. K= 8 Version 20190501 L= 7 M= 5 N= 2 page 13 K= 8 Version 20190501 L= 7 M= 5 N= 2 ME 401/5501 HW#5 Name*due 3/16/2020 1 Turner (of “Turner’s Workbook”) put together an early computer screening model that considered 49 combinations of wind and stability for a single point source. Which of the 49 combinations cannot occur, based on Table 5.1 of the textbook? 2 The two commonly used wind speed profiles are The logarithmic profile u(z) = uk* ln (z/z o ) ( k = 0.4) and the power fit (u2 /u1 ) = (z 2 /z 1 )p For the following pairs of measurements, fit each to both the log-profile and power fit and estimate speed at a 30-m height Wind speed (mph) at height Set Log profile 2.2 m 4.8 m 4.8 m 2.2 m Friction velocity (mph) Roughness height z0(m) Estimated 30-m Speed (mph) I 11.0 3.9 1.43 3.64 27.7 II 16.0 3.8 1.73 6.26 44.7 III 17.0 4.6 1.65 6.36 46.1 IV 13.0 7.0 0.89 3.08 27.1 V 14.0 4.8 1.46 4.72 35.6 Refer to March 14 email dropping this part 3 Using Turner’s workbook, estimate the “factor of safety” in treating the P-G dispersion coefficients as representing 1-hr averages rather than approximately 5 Your supervisor asks you, “How conservative are they?” How would you answer? Based on the table shown above, you could tell him that 1-hr averages are about 64% too high. Using the equation and “probably about 10 minutes” you might estimate more like 35% too high. A good answer might be “about 50% too high.” 4 An exhaust gas at 100o C containing CO at 1.6 % by volume leaves a 22-m high stack. The exit velocity is 14 m/s and the stack diameter is 0.75 m.Volumetric flow is 0.785*(0.75)2 *14 = 6.2 m3 /s At 1.6%, CO leaves at 0.099 m3 = 99 liters per sec a**__________*** What is the concentration of CO (in mg/m3) at 25 o C and 29 in Hg?15,000 mg/m3 Stack exit pressure is essentially atmospheric (which is about 29 in Hg in our area**), so need only correct for temperature: 99 liters * (298/373) = 79 liters (std). That volume corresponds to 3.2 g-moles, or 3.2 * 28 = 91 grams of CO. As a result, the emission rate is 9 1 g/s and the CO concentration is 1 4,600 mg/m3 (std cond**) or 18,300 mg/m3 (actual cond) at exit. b* What’s the emission rate (in g/s)91 g/s c_Use SCREEN3 from the SCRAM website to estimate the highest downwind hourly concentration in both ppm and ug/m3. How far downwind does the maximum occur? 2.0 ppm ( 2330 ug/m3) at 500 m downwind SEE NEXT PAGE FOR SCREEN3 OUTPUT ** Values would be within ~3% had you referenced 29.92 as standard pressure. 03/16/20 17:06:07 *** SCREEN3 MODEL RUN *** *** VERSION DATED 13043 *** HW5 PROBLEM 4 SIMPLE TERRAIN INPUTS: SOURCE TYPE = EMISSION RATE (G/S) = STACK HEIGHT (M) = STK INSIDE DIAM (M) = STK EXIT VELOCITY (M/S)= STK GAS EXIT TEMP (K) = AMBIENT AIR TEMP (K) = RECEPTOR HEIGHT (M) = URBAN/RURAL OPTION = BUILDING HEIGHT (M) = MIN HORIZ BLDG DIM (M) = MAX HORIZ BLDG DIM (M) = POINT 91.0000 22.0000 0.7500 14.0000 373.0000 293.0000 1.2000 RURAL 0.0000 0.0000 0.0000 THE REGULATORY (DEFAULT) MIXING HEIGHT OPTION WAS SELECTED. THE REGULATORY (DEFAULT) ANEMOMETER HEIGHT OF 10.0 METERS WAS ENTERED. BUOY. FLUX = 4.141 M

**4/S**3; MOM. FLUX = 21.651 M

**4/S**2. *** FULL METEOROLOGY *** *** SCREEN AUTOMATED DISTANCES *** *** TERRAIN HEIGHT OF DIST 0. M ABOVE STACK BASE USED FOR FOLLOWING DISTANCES *** CONC U10M USTK MIX HT PLUME (M) (UG/M**3) STAB (M/S) (M/S) (M) HT (M) ——- ———- —- —– —– —— —–500. 2329. 3 2.5 2.7 800.0 600. 2266. 3 2.0 2.2 640.0 700. 2163. 3 1.5 1.6 480.0 800. 2050. 3 1.5 1.6 480.0 900. 1895. 3 1.0 1.1 320.0 1000. 1854. 4 2.5 2.8 800.0 1100. 1803. 4 2.0 2.3 640.0 1200. 1749. 4 2.0 2.3 640.0 1300. 1686. 4 2.0 2.3 640.0 1400. 1617. 4 2.0 2.3 640.0 1500. 1579. 4 1.5 1.7 480.0 1600. 1537. 4 1.5 1.7 480.0 1700. 1492. 4 1.5 1.7 480.0 1800. 1445. 4 1.5 1.7 480.0 1900. 1397. 4 1.5 1.7 480.0 2000. 1349. 4 1.5 1.7 480.0 2100. 1302. 4 1.5 1.7 480.0 2200. 1256. 4 1.0 1.1 320.0 2300. 1236. 4 1.0 1.1 320.0 2400. 1221. 5 1.0 1.3 10000.0 2500. 1221. 5 1.0 1.3 10000.0 2600. 1219. 5 1.0 1.3 10000.0 2700. 1214. 5 1.0 1.3 10000.0 2800. 1207. 5 1.0 1.3 10000.0 2900. 1199. 5 1.0 1.3 10000.0 3000. 1189. 5 1.0 1.3 10000.0 3500. 1128. 5 1.0 1.3 10000.0 4000. 1071. 6 1.0 1.5 10000.0 4500. 1058. 6 1.0 1.5 10000.0 5000. 1036. 6 1.0 1.5 10000.0 5500. 1007. 6 1.0 1.5 10000.0 6000. 975.5 6 1.0 1.5 10000.0 6500. 942.4 6 1.0 1.5 10000.0 7000. 908.9 6 1.0 1.5 10000.0 7500. 873.7 6 1.0 1.5 10000.0 SIGMA Y (M) —–44.99 50.74 60.32 60.32 79.48 44.10 49.63 49.63 49.63 49.63 58.84 58.84 58.84 58.84 58.84 58.84 58.84 77.25 77.25 65.53 65.53 65.53 65.53 65.53 65.53 65.53 65.53 56.28 56.28 56.28 56.28 56.28 56.28 56.28 56.28 SIGMA Z (M) —–55.16 65.23 75.29 84.85 95.11 68.42 74.73 80.83 86.88 92.89 99.10 105.02 110.91 116.76 122.58 128.38 134.14 140.37 146.07 113.57 117.80 122.01 126.20 130.38 134.54 138.69 159.24 119.57 132.86 146.00 158.99 171.86 184.60 197.24 209.77 DWASH —-33.09 39.19 45.46 51.04 57.90 32.71 35.03 36.94 38.81 40.63 42.98 44.69 46.38 48.03 49.65 51.24 52.81 55.61 57.10 39.19 40.02 40.85 41.65 42.45 43.24 44.02 47.76 32.35 34.01 35.58 37.07 38.50 39.87 41.18 42.31 NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO 8000. 8500. 9000. 9500. 10000. 15000. 20000. 840.1 808.2 777.9 749.3 722.2 520.7 401.3 6 6 6 6 6 6 6 1.0 1.0 1.0 1.0 1.0 1.0 1.0 MAXIMUM 1-HR CONCENTRATION AT OR BEYOND 500. 2329. 3 2.5 DWASH= DWASH=NO DWASH=HS DWASH=SS DWASH=NA MEANS MEANS MEANS MEANS MEANS 1.5 1.5 1.5 1.5 1.5 1.5 1.5 10000.0 10000.0 10000.0 10000.0 10000.0 10000.0 10000.0 56.28 56.28 56.28 56.28 56.28 56.28 56.28 222.20 234.54 246.80 258.98 271.08 388.55 501.04 500. M: 2.7 800.0 44.99 55.16 NO CALC MADE (CONC = 0.0) NO BUILDING DOWNWASH USED HUBER-SNYDER DOWNWASH USED SCHULMAN-SCIRE DOWNWASH USED DOWNWASH NOT APPLICABLE, X

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