Solving Rate of Combustion Problem

In summary, if you want to keep the temperature difference between the hot and cold sides of the fiberglass at 40 degrees Celsius, you need a minimum thickness of fiberglass of 92mm.
  • #1
daimlerpogi
5
0
Rate of combustion??

Homework Statement



1. The broader side of a rectangular fiberglass of dimensions 3m x 5m is brought near the flame. The flame is produced upon combustion of gasoline. If the rate of combustion is 0.002 g/s, determine the minimum thickness of the fiberglass so that the temperature difference between hot and cold sides of the fiberglass is 40 degree Celsius. The latent heat of combustion of gasoline is 46,000 J/g.

k = 0.04 W/m.C

Homework Equations



Fourier's Law of Conduction

H = kA(Th-Tc)/L
where:
H - Heat if conduction (Watts of J/s)
k - Thermal Conductivity
A - Area (meter squared)
Th - Hot side temperature in Celsius
Tc - Cold side temperature in Celsius
L - Thickness

Heat of Combustion
Qc = mLc
m - mass in kg
Lc - Latent heat of combustion

The Attempt at a Solution


I really don't know where to start. The rate of conduction that is given is supposed to be H right? But the unit is g/s instead of J/s. And also, there aren't any Th and Tc given. Using the heat of combustion formula, i supposed you can get the mass of gasoline in the given 0.002 g/s. This is the last problem I'm solving. (I solved the other 4 but can't solve this one)
 
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  • #2
Welcome to PF!

daimlerpogi said:
The broader side of a rectangular fiberglass of dimensions 3m x 5m is brought near the flame. The flame is produced upon combustion of gasoline.
If the rate of combustion is 0.002 g/s, determine the minimum thickness of the fiberglass so that the temperature difference between hot and cold sides of the fiberglass is 40 degree Celsius. The latent heat of combustion of gasoline is 46,000 J/g.

k = 0.04 W/m.C

I really don't know where to start. The rate of conduction that is given is supposed to be H right? But the unit is g/s instead of J/s. And also, there aren't any Th and Tc given. Using the heat of combustion formula, i supposed you can get the mass of gasoline in the given 0.002 g/s.

Hi daimlerpogi ! Welcome to PF! :smile:

You don't need either Th or Tc separately … the formula only needs the combination (Th - Tc), which is 40C.

If you're bothered about the units, just put them side-by-side until you see how to fit them like a jigsaw :biggrin:

In this case, they give you a g/s and a J/g, but you want a J/s. :frown:

ok … just multiply … (J/g)(g/s) = J/s ! :smile:
 
  • #3


determine the minimum thickness of the fiberglass so that the temperature difference between hot and cold sides of the fiberglass is 40 degree Celsius.

My bad, didn't notice that! Thanks tiny-tim!

I got it now!

(46000 J/g)(0.002 g/s) = 92 W

Thanks again! :D
 

Related to Solving Rate of Combustion Problem

1. How do you calculate the rate of combustion?

The rate of combustion is calculated by dividing the change in the amount of fuel or reactants by the change in time. This provides the rate at which the fuel is being consumed.

2. What factors affect the rate of combustion?

The rate of combustion can be affected by several factors including the type and amount of fuel, ambient temperature, oxygen availability, and the presence of catalysts.

3. Can the rate of combustion be controlled?

Yes, the rate of combustion can be controlled through various methods such as adjusting the amount of fuel, regulating oxygen supply, and using catalysts to speed up or slow down the reaction.

4. How does the rate of combustion impact engine performance?

The rate of combustion directly affects the power output of an engine. A faster rate of combustion results in more power being produced, while a slower rate can decrease performance.

5. What are some common challenges when solving rate of combustion problems?

Some common challenges when solving rate of combustion problems include accurately measuring and controlling the various factors that can affect the rate, as well as accounting for any variations or inconsistencies in the combustion process.

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