4 replies to this topic

[mellotango]

Hi, and thanks for your expert opinions in advance. I'm working on a theorical scenario of a flue gas funnel. Assuming hot air is rising from a funnel at a constant rate. Would the hot air rise faster if water droplets were sprayed at the top of the funnel, hence lowering the air temperature at the top part of the funnel? Would this lead to an increased air pressure differential at the bottom of the funnel vs the top of the funnel, leading to hot air rising faster, and hence, creating a more powerful updraft?

[TS1979]

If the flue gas is on natural draft, the flue gas flowing rate is driven by the density difference between the inside the funnel and the outside atmosphere. When you spary the water at the top of the funnel there will be two effects on the flue gas inside the funnel:

• The flue gas temperautre will be reduced. Reduced flue gas temperature will increase the flue gas density.
• Adding water droplets which may evaporate to steam will also increase the stream density.

Therefore, the density difference between the flue gas inside the funnel and outside atmosphere will be reduced and the updraft will be reduced instead of increased!

[mellotango]

If the flue gas is on natural draft, the flue gas flowing rate is driven by the density difference between the inside the funnel and the outside atmosphere. When you spary the water at the top of the funnel there will be two effects on the flue gas inside the funnel:

• The flue gas temperautre will be reduced. Reduced flue gas temperature will increase the flue gas density.
  
• Adding water droplets which may evaporate to steam will also increase the stream density.

Therefore, the density difference between the flue gas inside the funnel and outside atmosphere will be reduced and the updraft will be reduced instead of increased!

Thx for the very informative reply. Now, comes the hard part... Assuming we have small flue gas funnel: 40ft long x 1.8m diameter. Waste heat at the base of the funnel is supplied at 5000 kwh. At this rate, what would the air velocity be at the top of the funnel? (Assuming air temperature at 25C) We know that air expands with heat, so this extra air volume may be used to generate mechanical energy. And if the air velocity was used to turn a fan turbine at the top of the funnel (assuming turbine efficiency of 25%), what would the useful electrical output in kwh that can be generated by the turbine?

Edited by mellotango, 14 November 2012 - 11:03 AM.

[kkala]

Evaporation of water droplets will produce steam of lower density than stack gases at same temperature (steam MW=18, versus roughly MW= 29 for stack gases, close to air). Nevertheless resulting cooling will increase stack gas density compared to situation before spraying water, as pointed out by TS1979 in post No 2. Let us also indicate it by an example (rough figures).
Assume 100 kg/h of stack gases, temp=150 oC. Their density is 29/22.4*273/(273+150)=0.836 kg/m3. Sprayed with 1 kg/h of water (20 oC), temperature will lower to t oC through evaporation. By rough enthalpy balance (0 for 20 oC, liquid H2O), 100*0.24*(150-20) = 100*0.24*(t-20) + 1*585 + 1*1*(t-20) kcal/h (x), so t = 121.4. Volume of resulting gas (100/29+1/18)*22.4*(273+121.4)/273=113.4 m3, so density=101/113.4=0.891 m3 (higher than 0.836 kg/m3).

(x) For instance, see B Bhatt and S Vora, Stoiciometry, Tata McGraw-Hill (1976), para 6.6.1.2 and Appendix II.

thanks for all the information guys

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