# Given zeo fuel mass 4920 trip fuel

Classified in Chemistry

Written at on English with a size of 32.08 KB.

**Combustion is a major source of air pollution emissions. The complete combustion of hydrocarbon fuel produces only CO _{2} and H_{2}O as products, according to the following stoichiometry:**

**CH _{4} + 2 O_{2}**

**CO**

_{2}+ 2 H_{2}Oa) Mass of **air** **required** for complete combustion of 500 kg of CH_{4}, assuming O_{2} = 21% of air.

b) Mass of CO_{2} produced from the above combustion

c) Calculate the **mass** of air required for combustion of 500 kg of benzene (C_{6}H_{6}).

a) Mass of O_{2 }required =

Mass of air required= 2000/0.23=8696 kg

b) Mass of CO_{2 }produced=

c)

Mass of oxygen required=

Mass of air required= 1538/0.23= 6686 kg

Mass of CO_{2 }produced = (6 x 44)/78 x 500 = 1692 kg

**Fuel oil contains the following components (by weight); C- 88.3%, H- 9.5%, S- 1.6%, ash- 0.1%**

- Calculate the stoichiometric air required if 1 kg fuel oil is combusted completely.
- Determine the SO
_{2}**emission**from this process.

For C, O_{2} required =

Air required = 2.36/0.23 = 11.24 kg

For H,O_{2} required =

Air required = 0.76/0.23 = 3.30 kg

For S,O_{2} required =

Air required = 0.016/0.23= 0.069 kg

Total amount of air required = 10.26+3.30+0.069 = 13.6 kg

SO_{2} emitted = (64/32)*0.016*1 = 0.032 kg

Industrial representatives claim that the NAAQS for SO_{2} is so low that one exceeds it if one strikes a simple wooden match in a modest sized room. Is this true?

- Calculate the concentration expected for striking such a match in a room that is 15 ft by 15ft by 8ft. A typical 2-inch wooden match contains 2.5 mg of sulfur.
- Compare the resulting concentration to the annual average SO
_{2}NAAQS standard. The NAAQS for SO_{2}(annual average) is 80 mg/m^{3}.

S + O_{2}® SO_{2}

Mass of SO_{2} = **64** * 2.5 = 5 mg

32

Conc. Of SO_{2} = ** 5 mg ** * ** ft ^{3} **

(15 * 15 * 8) ft^{3} 0.028 m^{3}

= 0.099 mg/m^{3}

= 99 mg/m^{3} > 80 mg/m^{3}

So the conc. Of SO_{2} will exceed the NAAQS standard.

Calculate the total emission **rate** for CO in kg/yr, assuming CEMS data for CO as 1600 hr/yr for the period and has an exhaust gas temperature of 180°C. The CEM output for concentration of CO is 54.9 (ppm_{vd}) from a furnace firing waste fuel oil and the oxygen content is 10.5 % by volume, gas flow rate is 8.66 m^{3}/s and the production rate of product is 276 ton/hr.

**For CO _{®}**M = 12 + 16 = 28

P = 1 **atm**, T = 180°C = 453K, Q = 8.66 m^{3}/s, R = 0.0821 L.Atm/mol. °K

** E _{CO}** =

**Y**

_{i }* P * M * Q**R * T**

=**54.9 x 10 ^{-6} mol/mol * 1 atm x 28 g/mol * 8.66 m^{3}/s**

0.0821 L. Atm/mol.°K * 453°K * m^{3}/1000L

= 0.358 g/s * kg/1000 g * 3600 sec/hr = **1.29** **kg/hr**

**E**= 1.29 x 1600 =_{CO }total**2064 kg/yr**

**Estimate the emission rate of particulate matter (PM _{10}) if 2.5 kg of PM_{10} is emitted from a steam tube dryer in a canning and by-product manufacturing facility, for each ton of raw fish processed. The overall control efficiency of the pollutant is 40%. It is assumed that the facility operates for 10 hrs/d and 260 days/year, and the activity rate is 5 ton/hr.**

Emission Factor = 2.5 kg/ton

Efficiency= 40%=0.4

Operating hour=10*260=2600hr/year

Activity rate= 5 ton/hr

Emission rate=

**Calculate the mole fraction of H _{2}S in water that is in contact with air at 20°C and contains H_{2}S with partial pressure of 0.05 atm.**

**k _{H} of H_{2}S at 20°C is 4.83 x 10^{2} atm/mol fraction.**

0.05 = 4.38*10^{2}*C

C=1.1*10^{-4} mol fraction

**Typical person at rest produces 0.007 ml/min of CO and the typical human diet in wealthy countries like Canada is about 600 g/day of food (dry basis). Calculate the emission factor for CO by human (g CO/g food).**

**What is the emission factor for automobiles, in lb CO/ton of fuel burned? The permitted CO emission for new cars is 3.4 g/mile. Assume that the fuel economy is 25 mile/gal and the gasoline density is 0.72 kg/L.**Emission factor for CO

**A gas kitchen stove consumes 3000 kcal/h of fuel. The emission factor for NO _{2} for that kind of stove is about 61 µg/kcal. The air infiltration rate is 3000 ft^{3}/h and the outdoor concentration 20 µg/m^{3}; there is no destruction of NO_{2}; estimate the steady state concentration when the stove has run long enough to reach steady sate.**

Emission rate = E =

Q = 3000 ft^{3}/h = 84.95 m^{3}/hr, k=0

C_{a} = 20 / m^{3 }

C =

C = 2174 /m^{3}

A bar with volume 500 m^{3} has 50 smokers in it, each smoking two cigarettes per hour. An individual cigarette emits, among other things, about 1.4 mg of formaldehyde (HCHO). Formaldehyde converts to carbon dioxide with reaction rate coefficient k = 0.40 h^{-1}. Fresh air enters the bar at the rate of 1000 m^{3}/hr, and stale air leaves at the same rate. Assuming complete mixing,

a) Estimate the steady state concentration of formaldehyde in the air at 25°C and 1 atm of pressure.

b) How does the result compare with the threshold for eye irritation of about 0.05 ppm?

k=0.4 h^{-1}, V = 500 m^{3}, Q=1000 m^{3}/hr, C_{a} = 0.0

MW_{HCHO} = 30 g/mol

Emission rate = E = 50*2*1.4 = 140 mg/hr

C=

C (mg/m^{3}) = C_{ppm}* PM/RT

0.117 = C_{ppm} * (1*30) / (0.08206*298)

C_{ppm} = 0.095 ppm

**The sidestream smoke from one cigarette releases about 0.1mg of benzo(a) pyrene (BaP). In an apartment, with fresh air entering through holes and cracks (infiltration) at an average rate of 120 m ^{3}/hr. The potency factor inhalation route for benzo(a) pyrene is 6.11 (mg/kg-d)^{-1} as per USEPA.**

**What would be the steady-state indoor concentration of BaP if one cigarette per hour is smoked? (Assume that BaP is a conservative pollutant).****What would be the incremental cancer risk to a non-smoking roommate who spends 8 hr/day for 1 year in this apartment? (Assume an inhalation rate of 20m**^{3}/day).- Input rate = output rate

1 cig/hr x 0.1 mg/cig = 120m^{3}/hr x C mg/m^{3}

C = 0.1/120 = 0.00083 mg/m^{3}

b) Living for a year with a smoker:

Chronic Daily Intake, (CDI) = **0.00083 mg/m ^{3 }x20m^{3}/day x 1 day/24hr x 8 hr/day x 365 day**

70 kg x 365 d/yr x 70 yr

= 1.1 x 10^{-6} mg/kg-d

R = 1.1 x 10^{-6} mg/kg-d x 6.11 (mg/kg-d)^{-1}

= 7 x 10^{-6}

**A Container of Carbon Tetrachloride (CCl _{4}) is left open in an unventilated cabinet. An individual opens the doors and is exposed to it. What is the carbon tetrachloride concentration in ppm that the employee is exposed to? (The vapour pressure of (CCl_{4}) at 20°C is 0.12atm).**

**Solution:**

Vapour pressure of CCl_{4 }= 0.12 atm

P_{1} V_{1} = n_{1} R T_{1 } for CCl_{4}

P_{2} V_{2} = n_{2} R T_{2} for air

P_{1 }= 0.12 atm P_{2 }= 1 atm

V_{1} = V_{2 } T_{1 }= T_{2 }

n_{1}/n_{2 }= P_{1 }/P_{2 }

Concentration of CCl_{4 }= n_{1}/n_{2 }= P_{1 }/P_{2 }

= (0.12 atm / 1atm) x 10^{6}

= 120000 ppm

Therefore the employee will be exposed to 120000 ppm of CCl_{4}

**Ethyl alcohol (C _{2}H_{5}OH) is accidentally spilled on the floor of a store room in which the temperature is 23°C, 1 atm. The average air velocity is 2 m/s and the spill is 2 m in diameter. Estimate the rate of evaporation in kg/hr.**

**Calculate the vapor composition of 50 mol% benzene and 50 mol% toluene in equilibrium with air in a closed container. Vapor pressures of benzene and toluene are 1.45 psi and 0.42 psi, respectively. **

**(b)Aqueous solution of toluene with mole fraction x _{tol}= 9.77 x 10^{-5}; Toluene vapor pressure @ 25 ^{0}C =29.8 mm Hg; Henry’s law constant = 3.72 x 10^{7} N/m^{2}. Calculate the partial pressure of toluene over the aqueous solution at 25 ^{0}C using Raoult’s law and Henry’s Law.**y

_{benzene}= x

_{benzene}= 0.5

y_{toulene} = x_{toluene }= 0.5

y_{air} = 1-0.049-0.014= 0.937

P_{i}= x_{i }H= (3.72 x 10^{7} N/m^{2}) (9.77 x 10^{-5}) = 3634 N/m^{2}

P_{j} = x_{j} P_{v,j}= (9.77 x 10^{-5}) ( 29.8/760 atm) (10^{5} Nm^{-2} atm ^{-1}) = 0.382 N/m^{2}