HEATER FOR WINDTUNNEL

[wsuengineer]

Problem: How to heat air from 30ºF to 275ºF in a wind tunnel. Initially the air must be heated then maintained. Heater coil/medium cannot be in windtunnel (test particles would collect on them). Thus air must be first heated then introduced into the wind tunnel.

Specifications:
Volume Air in Wind tunnel = 110 cu. ft.
Volume Air rate = 4000 cfm
Velocity = 2164 fpm

Ideas: Electric duct heater --> 310 kw heater! 380 load amps! Too high of a load.

What are the options?

 

[MintJulep]

Fire.

 

[wsuengineer]

Here is my new idea. Heat the air incrementally until desired setpoint then close off the duct heater section and run windtunnel. Using an electric duct heater sized at 20 kw instead of 310kw. Maintain heat loss with heat trace cable system. See attached photo.

 

[IRstuff]

Your math looks a little wonky. Energy required to raise 110 cf of air is about 510 kJ, divided by 310 kW, gives 1.6 seconds, which is essentially impossible.

20 kW heater gives a time value of 25 s, which still seems a trifle fast. The big question is how long can you take to heat the air up, and how much heat loss do you have, and how much heat exchanger area do you have? You need to look at the actual heat exchange mechanism. Your air velocity should result in something like 25 W/m^2-K, so with 10 m^2 of heat exchanger, you'd only get about 1168 W transferred at the start, evne with a 590ºF temperature delta, and it gets worse as the air temperature goes up.

Do I have the numbers right? It looks like your active length of tunnel is about 60 ft long with a cross-sectional area of 1.8 sq ft? You could be losing as much as 16 kW of heat, so lots of insulation around the tunnel would need to be installed.

TTFN

FAQ731-376

 

[wsuengineer]

The total length of the tunnel is 120 ft of 18 inch duct. Looking at 210 cu. ft of air now.

I calculated the heat loss to be about 10,000 btu/hr using 1 ft thick fiberglass insulation with outdoor temperature of 30 degrees and inside temp of 275.

The KW requirement for a heater can be found using the following equation: 1.08 x CFM x delta T (btu/hr)

CFM: 4000

convert btu/hr to kw by dividing by 3412.

I used an equation from several heater unit companies.

 

[ChrisConley]

Is the 4000 cfm outdoor air, or recirculated air?

If it is recirculated air, if you have a heat loss of 10,000 Btu/hr. And you have a heater capable of doing say 15,000 Btu/hr you will eventually be able to heat the wind tunnel up to temp.

If you have 4000 cfm of 'cold' air coming in, then you will need much more heat.

Fire.