simple air flow problem
simple air flow problem
(OP)
Administration kills engineering. Design calculation ability is a muscle, if not trained it atrophies...
I am curently designing a mobile generator unit in a closed area.I need to determine the amount of air flow requiered to maintain an ambiant temp of no more than 40ºc.
Known:
desired ambiant temp: 40ºc
outside temp (cooling air) :could reach up to 30ºC
kw regected by components inside habitacle :205Kw
unknown:
air flow volume needed to maintain ambiant temp at no more than 40ºc in order to select exhaust fan and inlet damper/loovers?
air speed could reach up to 1000ft/min
I need reminding of practical equations/formulas and basic notions?
I am curently designing a mobile generator unit in a closed area.I need to determine the amount of air flow requiered to maintain an ambiant temp of no more than 40ºc.
Known:
desired ambiant temp: 40ºc
outside temp (cooling air) :could reach up to 30ºC
kw regected by components inside habitacle :205Kw
unknown:
air flow volume needed to maintain ambiant temp at no more than 40ºc in order to select exhaust fan and inlet damper/loovers?
air speed could reach up to 1000ft/min
I need reminding of practical equations/formulas and basic notions?





RE: simple air flow problem
Q=205000/1000/10=20.5 kg/s or some 16 m3/s.
This seems quite a huge amount (unless you are in a wind tunnel): an area of some 3 m2 is required to stay within your speed limit.
prex
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RE: simple air flow problem
It appears it would be impossible to keep the temperature of the room at 40oC only by convection with the imposed limitations on a continuous basis.
Even with radiation effects included the temperature of the hot surfaces become very large or, alternatively, the temperature difference between them and the surroundings are excessive, or both.
For example, htc for convection (20) at the given "wind" speeds, and for radiation (10), together would result in a combined htc of about 30 W/(m2*deg C).
Assuming an air heat up of 10oC, to estimate its flow rate we can use the balancing equation:
205,000 W(=J/s) = (m)*(Cp)*(10) = (m)*1010 J/(kg*oC)*(10oC) = (30)*(A)*(Tgen-Tsurroundings)
Then, m = 205,000/10100 = 20.3 kg/s, or some 36,000 cfm, which would need a generator cross-section of 36 ft2 to get a "front" air velocity of 1,000 fpm as a given maximum.
On the other hand, the product (A)*(Tgen-Tsurroundings) = 205000/30 = 6833, means either a very large area for the hot surface, or a very high temperature for this surface, or both.
A finned-tube chiller could be considered to remove this heat to the required level on a continuous basis.
I've stressed the continuity of the operation since nothing was said about it in ddace's presentation.
RE: simple air flow problem
But to answer the questions yes it is a large unit: 2.5 Mw
the 205kw regected includes every thing dissipated in the enclosure.
Please explain why convection is insufficient on a continous
basis ?
RE: simple air flow problem
RE: simple air flow problem
A heat release of 205 kW, means 8.2% of the generated energy. Kindly refer to lilling's comments for clarity.