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According to the North American Combustion Handbook, Vol 1, p. 12.

v Adiabatic flame temperature of propane/air is 3572 °F / 1967 °C

But that doesn't seem right, perhaps the answer to this very simple question is a bit more complicated.  So let’s take a look at some general facts and variables that determine how hot a propane fueled, Bunsen burner flame is.

 

Let's begin with the combustion reaction itself.


 

C3H8 +  5O2  ⇒ 3CO2(g) + 4H2O(g)   -2020 kJ/mol

 

Theoretical combustion is the ideal combustion process during which a fuel is burned completely. To achieve the maximum heat of combustion, for every 1 mole of propane (C3H8)  that enters the reaction 5 moles of oxygen (5O2) must be available.  A complete combustion occurs when all the carbon (C) has formed (CO2) molecules and all the hydrogen (H) has formed (H2O). 


 

1) Fuel rich mixtures, excess propane or insufficient quantities of oxygen will result in a diminished temperature.  If there are unburned molecules in the exhaust gas such as C, Hand CO the combustion process is incomplete.  Unreacted propane and a yellow flame will result.


 

2) Fuel lean mixtures, insufficient quantities of propane or excess oxygen will also result in a diminished flame temperature.  A short, sharp inner cone, deeper purple, hissing, popping unstable flame will result.


 

3) As the quantity of propane and oxygen reacting increases so does the amount of heat.  However, a larger flame does not burn at a higher temperature but it does provide more heat given the same fuel-air mixture.


 

When consulting textbooks for the temperature of a propane flame one will discover that there is a rather wide range of reported temperatures.

This variance can be due to a number of factors.

1)  Instrumentation Error:  Measuring the temperature can effect the flame temperature and not all thermocouples are as accurate one would like.

2)  Where in the flame was the temperature measured?

3) Fuel flow is it laminar (smooth) or turbulent? (bunsen burners use a laminar flow)

4)  Is the fuel premixed?  The propane and oxygen molecules must collide if they are to react.  (fuel is premixed in the bunsen burner tube)

The Flame


 

v Adiabatic Temperature

 

Flame temperatures are reported as adiabatic meaning that there is no loss of heat energy to the environment or to external objects. This is a purely fictional scenario but convenient when computing enthalpies of reactions.  How much of the heat energy of the flame is absorbed by the burner head or radiated out to the environment?

 

v Heat Loss


a)  Minerals and metals act as a heat sink.  When placed in a flame they reduce the temperature of the flame.  The amount of temperature change (ΔT) is a function of the quantity of heat transferred (q), the specific heat of the sample and it’s mass.   (ΔT) = (specific heat)(mass) / q


 

b)  Flame temperature can also be affected by the volume of the sample. As the volume of the sample increases, its surface area increases which promotes the radiation of energy to the environment.


 

So perhaps the question of how hot the Bunsen burner flame is, really isn’t clear cut.  However you would be on safe ground to assume that if the melting point of the sample metal is at or near 1500 °C that an acorn size sample of the metal will not melt under a Bunsen burner flame.

 

Avogadro's Reflections
On
Hydrometers

  A hydrometer is an instrument whose function is based on Archimedes principle. This principle states that a body (the hydrometer) immersed in a fluid is buoyed up by a force equal to the weight of the displaced fluid. The hydrometer measures the weight of the liquid displaced by the volume of the hydrometer.

 

 Specific Gravity is a dimensionless unit defined as the ratio of density of the material to the density of water. If the density of the substance of interest and the reference substance (water) are known in the same units (e.g., both in g/cm3 or lb/ft3), then the specific gravity of the substance is equal to its density divided by that of the reference substance (water =1 g/cm3), hence

 

Specific Gravity = Density g/cm3
                             1 g/cm3

Herein lies the equality between specific gravity and density,
the dimensions drop out!

 

The greater the density, the tighter or closer the molecules are packed inside the substance.

Therefore, the greater the density / specific gravity of a liquid the higher a hydrometer will be buoyed by it.

 

Hydrometer

 

Fill your hydrometer jar about ¾ with the liquid you wish to test. Insert the hydrometer slowly. Do not drop it in! Now give it a spin with your thumb and index finger, this will dislodge any bubbles that may have formed. Once the hydrometer comes to a rest, observe the plane of the liquid surface. Your eye must be horizontal to this plane. The point at which this line cuts the hydrometer scale is your reading.

 


Table 1970A  U.S. Internal Revenue Service

Temperature Correction Table for Alcohol Hydrometers
calibrated at 60 °F

Proof Reading

51 to 59 °F 61 to 70 °F 71 to 80 °F 81 to 90 °F
 

Add

Subtract

Subtract

Subtract

 10

0.08

0.13

0.16

0.19

20

0.12

0.16

0.20

0.23

 30

0.20

0.23

0.25

0.26

40

0.29

0.30

0.30

0.30

 50

0.37

0.35

0.35

0.35

60

0.42

0.39

0.38

0.38

 70

0.41

0.40

0.41

0.41

80

0.40

0.40

0.41

0.41

 90

0.38

0.39

0.39

0.40

100

0.37

0.37

0.37

0.38

 110

0.36

0.36

0.37

0.38

120

0.34

0.35

0.36

0.36

 130

0.33

0.34

0.35

0.35

140

0.32

0.33

0.34

0.34

 150

0.31

0.32

0.32

0.33

160

0.29

0.31

0.31

0.32

 170

0.28

0.29

0.29

0.30

180

0.25

0.26

0.27

0.28

 190

0.21

0.22

0.24

0.25

200

----

0.17

0.18

0.20

Size #
Top Dia.
inchs       mm
Bottom Dia.
inchs       mm
Length
inchs        mm
000
0.50
12.7
0.31
8.2
0.81
20.5
00
0.59
15.0
0.41
10.0
1.00
25.4
0
0.66
16.8
0.50
12.7
1.00
25.4
1
0.75
19.0
0.56
14.2
1.00
25.4
1 Blue
0.73
18.5
0.58
14.7
0.99
25.2
2
0.78
19.8
0.63
15.9
1.00
25.4
3
0.94
23.8
0.72
18.2
1.00
25.4
4
1.00
25.4
0.75
19.0
1.00
25.4
5

1.06

26.9

0.91

23.0
1.00
25.4
5.5
1.11
28.2
0.97
24.6
1.00
25.4
6
1.25
32.0
1.03
26.2
1.00
25.4
6 Blue
1.27
32.3
0.98
24.9
1.10
27.9
6.5
1.34
34.0
1.06
27.0
1.00
25.4
7
1.44
36.5
1.19
30.2
1.00
25.4
7 Blue
1.43
36.3
1.14
29.0
1.20
30.5
7.5
1.50
38.1
1.13
28.6
1.00
25.4
8
1.63
41.3
1.31
33.2
1.00
25.4
8.5
1.68
42.9
1.41
35.7
1.00
25.4
9
1.75
44.5
1.47
37.3
1.00
25.4
9.5
1.81
46.0
1.50
38.1
1.00
25.4
10
1.97
50.0
1.63
41.3
1.00
25.4
10.5
2.08
52.8
1.75
44.5
1.00
25.4
11
2.20
55.9
1.88
47.6
1.00
25.4
11.5
2.47
62.7
2.00
50.8
1.00
25.4
12
2.50
63.5
2.13
54.0
1.00
25.4
12 Blue
2.38
60.5
2.01
51.1
1.40
35.5
13
2.69
68.2
2.28
57.9
1.40
35.6
12.5
2.97
75.4
2.44
61.9
1.40
35.6
14
3.50
88.9
3.00
76.2
1.50
38.1
14 Blue
2.92
74.2
2.41
61.2
1.60
40.6
15
4.00
101.6
3.25
82.6
1.50
38.1
16
5.00
127.0
3.50
88.9
2.00
50.8

 

100% ethanol has a specific gravity of .785 which is lighter than water with a specific gravity of 1.0

 

A 50/50 mixture of water and ethanol (100 proof / 50%) will have the following specific gravity.

 

(.5L x 1.0) + (.5 L x 0.785) = 0.8925

 

A 75/25 mixture of water and ethanol (50 proof / 25%) will have the following specific gravity

 

(.75 L x 1.0) + (.25 L x 0.785) = 0.9463

 

As you can see the specific gravity of the mixture is inversely proportional to the alcohol concentration. As the alcohol concentration decreases the specific gravity increases and the hydrometer floats higher in the solution.

 

The alcohol hydrometer is calibrated in two scales % alcohol and proof.(1% alcohol = 2 proof ) The manufacturer used the specific gravity of alcohol at various concentrations to calibrate the instrument.

 

To determine the concentration of a 1 liter solution of alcohol and water using specific gravity.

 

1) Measure the specific gravity of the solution

 

Let X = unknown volume of water

Let (1-X) = unknown volume of alcohol

 

Then X + (1-X) = 1 Liter

 

Specific Gravity of water = 1.0

Specific Gravity of ethanol = 0.785

 

(X)(1.0) + (1-X)(0.785) = Sp. G of solution

 

Solve for X

 

Example: Assume the measured specific gravity is 0.9463

 

X (1.0) + (1-X)(0.785) = 0.9463

X + (0.785 - 0.785X) = 0.9463

0.215X + 0.785 = 0.9463

0.215X = 0.1613

X =0.75

 

(X)(100%) = the concentration of water

(0.75)(100%) = 75% water

 

1-X =0.25

(1-X)(100%) = the concentration of alcohol

(0.25)(100%) = 25% alcohol