Heat Transfer Project

1. Measuring Pipe Outflow
Temperatures for Hemp
Smoke Using Forced
Internal Convection
Alec Popichak

2. Background
• Disclaimer
• Pipe smoked non-traditionally for mathematical purposes
 Bowl never removed for consistent, laminar flow
 No water added so smoke is internal fluid

3. Introduction
• The objective of this study is to determine the temperature at which smoke
reaches the human body for 2 scenarios:
 Forced internal convection with constant surface temp and constant convection at
surface
 Internal fluid = smoke, external = air
• Constant surface temp: Freeze pipe and bring to equilibrium
Ts = Tfreezer
• Constant convection: Run fan over pipe during smoking process

4. Materials
• 49.1 cm tall cylindrical pipe (or any that can achieve fully developed thermal
flow)
• Viable thermometer
 Mine rated for -50 C to 300 C
• Ruler
• Protractor
• Hemp
• Strobe tachometer
 Mine was an iOS app
• Bladed fan
• Freezer/fridge

5. Methods
• Approximate pipe as 2 cylinders: pipe 1 = stem, pipe 2 = main chamber
• Scenario 1: constant heat flux at surface
 Attach thermometer to entrance of stem
 Run fan over pipe
 Measure temperature of smoke entering stem
• Scenario 2: constant surface temperature
 Freeze pipe
 Measure temperature of smoke entering stem

6. Methods Cont.
• Mathematically determine temperature of smoke leaving pipe 2
• Temperature of mass entering pipe 2 = temperature of mass leaving pipe 1

7. Methods Cont.
• Determine fluid properties
• ρ = .00033 g/cm3
• ϒ = 1.54*10-6 m2/s
• k = .026 W/mK
• Cp = 1 kJ/kgK
• α = k/ρCp

8. Thermal Fully Developed Region
• Measure pipe dimensions
 D
 L
• Measure fan dimensions
 rmax = 10.5 cm
 rmin = 2.25 cm
 θblade = 50°
• Using Re < 2300 for laminar flow
 Thermal fully developed region occurs when x/D ≥ .05ReDPr
• Q = V/t
• u = Q/A

9. Proving Fully Developed Flow
• Calculate Re = uD/ϒ
 Both were less than 2300
 Showed that flow was indeed laminar as hypothesized
• Calculate Pr = ϒ/α
 Showed that flow was thermal fully developed flow

10. Condition 1: Constant Convection at
Surface
• Find Nu = 4.36 = hD/kf
 Use to solve for h
• Know I need U, so find hext
• For Forced External Convection:
 NuD = .3 + (.62ReD
1/2Pr1/3)/(1+(.4/Pr)2/3)1/4 * (1+(ReD/282000)5/8)4/5
 Re = uD/ ϒ
 u = ufan = (rpm(rmax)+rpm(rmin)/2)cos θ
 Pr = ϒ/α
 α = kair/ ρairCpair
 NuD = hextD/kf

11. Condition 1 Cont.
• U = 1/(
1
ℎ
+
1
ℎ 𝑒𝑥𝑡
+
Δ 𝑥
𝑘 𝑔𝑙𝑎𝑠𝑠
)
• ΔTlm = (Tinf – Tm,o)- (Tinf – Tm,i)/(ln((Tinf – Tm,o)/ (Tinf – Tm,i)))
• mCp(Tm,o-Tm,i) = UAΔTlm
 Tm,o = Tinf-(Tinf-Tm,i)e-UDLπ/mCp
 m = Q*ρ
 Tinf = Troom
• Repeat for both pipes
 Tm,i2 = Tm,o1

12. Condition 2: Constant Surface Temp
• Nu = 3.66 = hD/kf
• hAΔTlm = mCp(Tm,o-Tm,i)
 Solve with Newton’s Method
 xn+1 = xn-f(xn)/f ’(xn)
 Tm,o = xn+1
• Repeat for both pipes

13. Results
Pipe 1 Pipe 2
V1 = 30.47 cm3 V2 = 780.90 cm3
L1 = 18.45 cm L2 = 49.1 cm
D1 = 1.45 cm D2 = 4.5 cm
A1 = 1.65 cm2 A2 = 15.90 cm2
t1 = .16s t2 = 3,43s
Q1 = 190.2656 cm3/s Q2 = 227.6676 cm3/s
u1 = 115.3125 cm/s u2 = 14.319 cm/s
Re1 = 1084.326 < 2300 Re2 = 417.8615 < 2300

14. Results Cont.
• Pr = .195
• Condition 1
 Nu = 4.36
 Tinf = 75°F
 Tm,i = 22.33°C
 Tm,i2 = 23.13°C Pipe 1 Pipe 2
h1 = 7.82 W/m2K h2 = 2.25 W/m2K
NuD1 = 9.912 NuD2 = 17.454
hext1 = 17.773 W/m2K hext2 = 10.085 W/m2K
Reext,1 = 383.66 Reext,2 = 1190.66
Prext = .705 Prext = .705
U1 = 5.376 U2 = 2.001

15. Results Cont.
• Tm,o1 = 23.13 °C
• Tm,o2 = 23.77 °C
• Final temperature reaching the human body is 23.77°C

16. More Results
• Condition 2
 Nu = 3.36
 h1 = 6.563 W/m2K h2 = 2.115 W/m2K
 Ts = -15.72°C
 Tm,i = 17.33°C
• Pipe 1:
• Tm,o = 2.73°C
Pipe 1 f(x) f '(x)
x
.118x-1.088*ln((-15.72-x)/-33.05)-
.9566 = 0 .118-(1.008/(15.72+x))
0 -0.148120496 0.048788804
1 -0.097219437 0.05292823
2 -0.042419548 0.056600451
5 0.141411298 0.065490347
3 0.015850856 0.059880342
2.5 -0.013694178 0.058285401
x2 x3 = Tm,o
2.7349504 2.733535306

17. Even More Results
• Pipe 2:
 Tm,i = 2.73°C
• Tm,o = 1.76°C
Pipe 2 f(x) f '(x)
x
.2218x-.20475*ln((-15.72-x)/-18.45)-
.40076 = 0 .2218-(.2047/(15.72+x))
0 -0.367973263 0.208778372
5 0.684481824 0.211920656
3 0.261665371 0.210865171
2 0.051105845 0.210248081
1 -0.158800568 0.209557177
1.5 -0.05393371 0.20991266
x2 x3 = Tm,o
1.756934049
1.75683018
2

18. Final Results
• Constant Surface Temp: Tm,o = 1.76°C
• Constant Convection at Surface: Tm,o2 = 23.77 °C

19. Conclusion
• Fan did not do much to cool smoke
 Expected
• Freezing made a noticeable difference
 Difference you could feel
• Complexities