# Problem finding temperature gain of air through heated glass tube

I have a 20 cm long glass tube of outer diameter 6.4 mm and inner diameter 4.3 mm. Air at Ti = 20 °C is pushed through the glass tube at 2 litres per minute. The glass tube has a wire wound resistance heater on it which we can assume is perfectly thermally insulated from the ambient air. My aim is to find out the power needed to heat the air flowing through the tube to Te=650 °C at the tube exit.

After doing the calculation using

$\dot{Q}=\dot{m}c_p(T_e-T_i)&space;=\rho\dot{v}c_p(T_e-T_i)\\$

Where $\dot{Q}$ is the power supplied by the heater (W),

$\dot{m}$ is the mass flow rate (kg/s),

$\rho$ is the density of the air kg/m^3 at the bulk temperature $T_b=(T_i+T_e)/2$,

I get the power required as being 12.5 W. This seems to me to be too low. Unfortunately I do not know where I am going wrong. Does anybody have any idea where I am making a mistake in my calculation?

• I didn't check the math, but the approach is correct for the problem as defined. But some other questions come to mind. How hot does the inner wall of the tube have to be in order to actually transfer that heat? Given the low thermal conductivity of glass, how hot will the outer surface then be? Is there a reason that you really need glass? Ordinary glass will almost certainly shatter under such a thermal load - you'll need borosilicate at least, maybe vycor or even fused quartz.
– Mark
Aug 30 '17 at 0:54
• @Mark I am using quartz glass with a 1200°C maximum operating temperature. I calculated the temperature of the inner wall of the tube and found it to be ´766.6 °C´ at the exit. I made the assumption that the thermal resistance of quartz is negligible for the low wall thickness in this case. I need glass because I want to thermally dessociate certain nitrates in air samples. The nitrates would react with or stick to the walls of other materials which are not chemically inert.
– Buzz
Aug 30 '17 at 9:10

If I perform a back of the envelope calculation (using just the density of air at standard conditions, 1.225 kg/m3):

Converting to SI units of mass flow rate:

2 SLM of air -> 0.04 grams / sec = 0.00004 kg / sec
Cp = 1005 J / kg K
Te = 923.15 K
Ti = 293.15 K

P = 0.00004 * 1005 * (923.15-293.15) = 25.326 W (or J / s)


Your density is probably different by a factor of two or so: @ 350C, air density = 0.566 kg / m3. Everything looks in order to me.

Yes, the required power may seem low, but look at how little mass is flowing through the section that is being heated every second. 2 liters a minute sounds like a lot until you break it down into grams per second. Electrothermal heating is also a fairly lossless process, so if you perform this experiment, you will likely not need to increase this number by too much to account for inefficiencies.

• Thanks for your answer. Yes, I used 0.566 kg/m3 for the density. I guess you are right about the small quantities. My ultimate aim is to find out how much length of 1 mm outer diameter electrically insulated nichrome wire I would need to wind around the 20cm tube to get the 12 W using a 120V power suply. The nichrome has a resistance of 12 Ohms/m.
– Buzz
Aug 30 '17 at 9:34
• Buzz - I'd encourage you to open up a new question for that, but P = V^2 / R from Ohms Law. You'll also be limited by the maximum possible current of the power supply. Aug 30 '17 at 17:15
• Thanks. I realised that the issue was that I should have used the logarithmic mean temperature difference to relate the power required to the temperature since the inlet and exit temperatures differ by a large margin. In this case Q=hAdTln where Q is the power required, h is the convection heat transfer coefficient, A is the area of the tube inner surface, and dTln is the logarithmic temperature difference between the inlet and the outlet of the glass tube.
– Buzz
Aug 31 '17 at 9:15