1. Smoke Stream

2. Question:
 A glass jar is covered with cellophane.
A tightly folded paper tube of length 4-5
cm is inserted hermetically into the jar
through the cellophane cover. The tube
is oriented horizontally.
 If one burns out outside end of the tube
the dense smoke flows into the jar.
 Explore this phenomenon.

3. Introduction
 We conventionally see smoke rise.
 The tricky bit:
– Since smoke rises, how do can it flow
horizontally into
the jar?
– Smoke must sink

4. Introduction
 Why smoke rises:
– Convection
 Rising air column
carries the smoke particles up.
– Mass of particle ≈ 1.7 × 10-12 – 2.8 × 10-7 g

5. Theory
 Assuming smoke particles consist of
12C:
– ρ smoke particles ≈ 2200 kg/m3
– ρ air = 1.1839 kg/m3 @ 25°C
 ρ smoke particles > ρ air
– in the absence of convection, smoke
should sink naturally.

6. Evidence

7. Experimentation
Paper tube
Glass beaker
Retort stand
Cling wrap

8. Experimentation
 to record footage of the
phenomenon for analysis:
– Effect of the number of layers of paper.
– Effect of diameter / length of tube
– Mechanism driving smoke into the jar
– Theoretical vs. Actual flow speed in tube

9. Experimentation
 Variable:
– Number of layers of paper in each tube.
• 1, 2, 3 & 4
 Constants:
– Diameter of the paper tube (5.0mm).
– Type of paper used
– Length of paper tube (10.0cm)
• 6 cm protrusion outside the beaker

11.  At least 2 layers are required
– When layers > 2, inner layer burns slower
than outer layer
– Smoke in jar is from inner tube
Results: Variation in layers
Fast-burning
outer layer
Slow-burning
inner layer

12. Results: Smoke Flow
 Smoke flow in jar is
laminar
– Reynolds no. < 2000
– No vortices or eddies
Turbulent flow Laminar flow

13. Effect of length & ⌀
 15cm tube used in separate test
– Length of tube has no significant effect
– Smoke exits from opposite end almost
immediately
 Diameter of tube = 12.0 mm:
– smoke remains in tube, flow extremely
slow.

14. The really tricky bit:
What is the mechanism driving smoke
into the jar?
Patm air
Phot air
Pressure Gradient?
Smoke?

15. Proposed mechanism:
Patm air
Hot air expands
Inside tube
Expanded air
pushes both ways
Expanding air drives
Smoke into jar
smoke
Patm air

16. Proposed mechanism:
 Such a mechanism supports results.
– Flow speed not significantly affected by
length
– Larger diameter, less (slower) smoke flow
• Air has more room to expand
 Determine if theoretical flow speed
fits actual values.

17. Calculations
Volumetric flow
rate
 C6H10O5 + O2  5C + 5H2O + CO2
(incomplete combustion of cellulose)
 Find rate of combustion of paper
 Use eqn. to find rate of expulsion of CO2 &
H2O gas
 To find rate of expansion of gas
 Volumetric flow rate
of air into the tube.

18. Calculations (Cont’d)
 C6H10O5 + O2  5C + 5H2O + CO2
 Determine average flow speed
Average flow speed

19. Expanded air
pushes both ways smoke
Assumptions
 Expanded air splits equally on each side
 Paper is 100% cellulose; Incomplete
combustion
 Smoke is a suspension that travels
along with air
 Temperature of burning paper
is at 400°C
(slightly above flash point of paper)

20. Calculating rate of combustion
 Expt. to derive rate of combustion of paper:
 Avg. rate = 0.0186 gs-1
= 0.0186 gs-1 ∕ 162.0 gmol-1
= 0.000115 mols-1
Trial Initial mass/
g
Final mass/
g
Δ mass/
g
Time/ s Rate of
combustion/ gs-1
1 0.480 0.210 0.270 15.50 0.0174
2 0.461 0.259 0.202 11.16 0.0181
3 0.341 0.045 0.296 15.37 0.0192
4 0.336 0.201 0.165 8.41 0.0196

21. Calculating Volumetric Flow Rate
 C6H10O5 + O2  5C + 5H2O + CO2
 Rate of combustion of O2 =
0.000115mols-1
 Rate of formation of CO2 & H2O gas
= 0.000115 mols-1 × 6
= 0.000689 mols-1
initial temp. = 400°C , exit temp. = 50°C
 Ideal gas law: PV=nRT

22. Calculating Volumetric Flow Rate
 Volumetric flow rate into tube
= 0.0000380 m3s-1 ∕ 2
= 0.0000190 m3s-1
 Volumetric flow rate at exit
=0.00000913 m3s-1
 Average volumetric flow rate
= 0.0000141 m3s-1

23. Limitations
 Only expansion of combusted gas
considered
– (ignore heating of surrounding air)
 ‘Leakage’ of air not considered.
 Drops in pressure caused by drag not
factored.
 Only average flow speed is derived
– Instantaneous flow speed cannot be
calculated as we do not know how temp.
varies along tube
‘leaks’

24. Average Flow speed:
 Theoretical
– Vnarrow tube ≈ 0.7 ms-1
– Vwide tube ≈ 0.15ms-1
 Actual
– Vnarrow tube ≈ 0.5ms-1
– Vwide tube (N.A)

25. Computed Reynolds number
 Narrow tube (⌀ 5.0mm)
– Re ≈ 83
 Wide tube (⌀ 12.0mm)
– Re ≈ 34
 Reynolds no. are ultra-low, in the order
of 101
 Flow in tube fits laminar flow as
observed.

26. Conclusion:
 Mechanism has been proposed for the
phenomenon.
– Volumetric expansion of air
– Cooling of gas produced
 However, other factors are also
involved:
– ‘leakage of air’
– Change in pressure due to drag
 Hagen Poiseuille equation

27. Thank You