Engineering a Better Air Bag

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Engineeringa Better Air Bag

DataTable 1: Model Air Bag

Activity

Data and Calculations

Volume of 6 × 9 inch bag

1.2 liters

Room temperature in Kelvin

The temperature in Celsius was 22°C. In Kelvin, this was (22+273)= 295K

Room pressure in atm

The barometric pressure was 30.12 inches. In atmospheres (atm) this was = 1.01 atm

Moles of CO2 required to inflate bag at room temperature and pressure

The Ideal Gas Law is given by PV=nRT where P = pressure (1.01 atm), V = volume (1.2 L), R = Gas constant (0.0821 L atm mol-1 K-1), T = temperature (295K), n = number of moles of CO&shy2. Rearranging the equation gives n = PV/RT

n = 0.05 moles of CO2

Balanced equation for the reaction of NaHCO3 and CH3COOH to produce CO2

The reaction follows the equation:

NaHCO3 (s) + CH3COOH (aq) &gt CO2 (g) + H2O (l) + CH3COONa (aq)

Mass of NaHCO3 needed for the reaction (84.0 g/mol)

The molar ratio in the equation above is 1:1 for each of the components. Therefore there will be the same number of moles for each component. The number of moles of NaHCO3 is thus 0.05 moles. The mass will therefore be:

Mass = molar mass x number of moles

Mass = (84.0 g/mol) x 0.05 mol

Mass = 4.2 g

Volume of vinegar required (0.833 M acetic acid)

Since the molar ratio is 1:1, the number of moles of acetic acid that will react is 0.05 moles. The volume will be given by:

Volume = number of moles / Molarity

Volume = 0.05 mol / 0.833 mol/L

Volume = 0.06 L

DataTable 2: Model Air Bag

Trial #

NaHCO3

(grams)

Vinegar

(mL)

Observations

1

4.2 g

60 mL

The bag inflates the slowest.

2

4.5 g

65 mL

The bag fills up with CO2 at a quicker rate compared to the initial experiment that used 4.2 g of NaHCO3 and 60 mL of vinegar.

3

4.7 g

70 mL

On the three experiments carried out, this bag inflates the quickest.

DataTable 3: 80-L Driver-Side AirBag

Activity

Calculations

Moles of CO2 required to inflate 80-L driver-side air bag at room temperature and pressure

The Ideal Gas Law is given by PV=nRT where P = pressure (1.01 atm), V = volume (80 L), R = Gas constant (0.0821 L atm mol-1 K-1), T = temperature (295K), n = number of moles of CO&shy2. Rearranging the equation gives n = PV/RT

n = 3.303 moles of CO2

Balanced equation for the reaction of NaHCO3 and CH3COOH to CO2

The balanced equation is:

NaHCO3 (s) + CH3COOH (aq) &gt CO2 (g) + H2O (l) + CH3COONa (aq)

Mass of NaHCO3 needed for the reaction

(84.0 g/mol)

The molar ratio in the equation above is 1:1 for each of the components. Therefore there will be the same number of moles for each component. The number of moles of NaHCO3 is thus 3.303 moles. The mass will therefore be:

Mass = molar mass x number of moles

Mass = (84.0 g/mol) x 3.303 mol

Mass = 277.45 g

Volume of vinegar required

(0.833 M acetic acid)

Since the molar ratio is 1:1, then the number of moles of acetic acid that will react is 3.303 moles. The volume will be given by:

Volume = number of moles / Molarity

Volume = 3.303 mol / 0.833 mol/L

Volume = 3.965 L

Data Table 4: 160-LFront Passenger-Side Air Bag

Activity

Calculations

Moles of CO2 required to inflate 160-L front passenger-side air bag at room temperature and pressure

The Ideal Gas Law is given by PV=nRT where P = pressure (1.01 atm), V = volume (160 L), R = Gas constant (0.0821 L atm mol-1 K-1), T = temperature (295K), n = number of moles of CO&shy2. Rearranging the equation gives n = PV/RT

n = 6.672 moles of CO2

Balanced equation for the reaction of NaHCO3 and CH3COOH to CO2

The equation is:

NaHCO3 (s) + CH3COOH (aq) &gt CO2 (g) + H2O (l) + CH3COONa (aq)

Mass of NaHCO3 needed for the reaction

(84.0 g/mol)

The molar ratio in the equation above is 1:1 for each of the components. Therefore there will be the same number of moles for each component. The number of moles of NaHCO3 is thus 6.672 moles. The mass will therefore be:

Mass = molar mass x number of moles

Mass = (84.0 g/mol) x 6.672 mol

Mass = 560.45 g

Volume of vinegar required (0.833 M acetic acid)

Because the molar ratio of the reaction is 1:1, the number of moles of acetic acid that will react is 6.672 moles. Therefore, the volume of vinegar will be given by:

Volume = number of moles / Molarity

Volume = 6.672 mol / 0.833 mol/L

Volume = 8.00 L

Engineeringa Better Air Bag Questions

  1. Let’s consider an actual automotive airbag:

  1. Using the air pressure you recorded in step 3 on page 8, calculate the moles of nitrogen required to fill a 160 liter passenger-side air bag. Show all calculations.

The Ideal Gas Law is given byPV=nRT whereP = pressure (1.01 atm), V = volume (160 L), R = Gas constant (0.0821L atm mol-1K-1),n = number of moles of N&shy2.Assuming the nitrogen gas is released at a temperature of 22°C(295K), the Ideal Gas Law can be rewritten as n= PV/RT

n= 6.672 moles of N2

  1. Using the reaction from page 5 of the procedure, calculate the grams of sodium azide required to produce the required amount of nitrogen. Show all calculations.

The reaction in an airbag thatinvolving sodium azide that leads to the releasing of nitrogen gasfollows the equation:

2NaN3(s) &gt2Na (s) +3N2 (g)

The molar ratio of NaN3to N2 is2:3. Therefore, for every 3 moles of N2there are 2 moles ofNaN3.However, since the reaction led to the production of 6.672 moles ofN2,then the number of moles of NaN3will be given by:

Moles of NaN3= (2×6.672)/3

Moles of NaN3= 4.448 moles.

Since NaN3has a molar mass of65 g/mol, then the mass of NaN3required to fill the160 L airbag will be given by:

Mass = Molar mass x Number ofmoles

Mass = (65 g/mol)/ 4.448 mol

Mass = 14.61 g of NaN3

  1. Do you think air bag manufacturers design the bags to inflate to a) slightly under atmospheric pressure, b) approximately equal to atmospheric pressure or c) over atmospheric pressure? Explain the reasoning for your choice. If you use any sources of information to answer this question, cite them using APA format.

Airbag manufacturers designairbags to inflate to a maximum pressure that is under theatmospheric pressure so that it can easily deflate when impacted by aperson inside a car, and by doing so absorb as much force as possiblefrom the impact and reduce chances of serious injury to the person.If it inflated to an atmospheric pressure higher than the atmosphericpressure then it would not deflate after being impacted by a personinside the car, and it would therefore not absorb any impact forcesfrom the person’s body, thereby translating to serious injuriessustained by the person (Barbat, Li, &amp Prasad, 2013).

  1. Describe at least 2 factors that you think air bag manufacturers take into account, if any, when designing systems for different parts of the world. Explain why these factors might be important. If you use any sources of information to answer this question, cite them using APA format.

Airbag manufacturers consider thespeed with which the airbag will deploy and fill up with N2gas. When a caraccident occurs, a driver can hit the steering wheel in around 23milliseconds (Deng, et al., 2013). Therefore, airbag manufacturerschoose those chemical compounds that when set off by the car’scrash sensor can combust or react to rapidly generate gases withwhich to fill the airbag in a time that does not exceed thisduration. In the case of a person sitting on the passenger side ofthe car the duration before he hits the dashboard is longer since thedistance to the dashboard is higher than that to the steering wheel.In such a case airbag designers often place bigger airbags on thepassenger side than on the driver’s side to achieve the same result(Deng, et al., 2013).

Another key aspect considered byairbag manufacturers in their design is the manner with which theenergy absorbed from a person’s impact with an airbag can bedissipated. To quickly absorb and dissipate as much energy aspossible following an impact, airbag manufacturers design theirairbags to have small vents through which the gases can escape afterimpact. As the gases escape the airbag deflates, thereby creating asoft cushion that slows down the person’s momentum (Deng, et al.,2013).

References

Barbat, S., Li, X., &amp Prasad,P. (2013). Bumper and Grille Airbags Concept for Enhanced VehicleCompatibility in Side Impact: Phase II. TrafficInjury Prevention,14(1), 30-39.

Deng, X., Potula, S., Grewal, H.,Solanki, K., Tschopp, M., &amp Horstemeyer, M. (2013). Finiteelement analysis of occupant head injuries: Parametric effects of theside curtain airbag deployment interaction with a dummy head in aside impact crash. AccidentAnalysis and Prevention,232-241.

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