simple air flow problem 1

[ddace]

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?

 

[prex]

As the specific heat of air is close to 1000 J/kg°C you simply get
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

http://www.xcalcs.com

Online tools for structural design

 

[25362]

The heat rejection is quite impressive. At the present cost of electricity, this means the generator would be throwing away more than USD 15 an hour. Although the general geometric dimensions of the generator and its hot parts as well as the overall energy output were not given, one can appreciate from the lost heat it seems quite a big unit.

It appears it would be impossible to keep the temperature of the room at 40^oC 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 10^oC, to estimate its flow rate we can use the balancing equation:

205,000 W(=J/s) = (m)*(Cp)*(10) = (m)*1010 J/(kg*^oC)*(10^oC) = (30)*(A)*(T_gen-T_surroundings)

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)*(T_gen-T_surroundings) = 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.

 

[ddace]

I am now worried about this continuous basis information.

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 ?

 

[lilling]

I guess that the mobile generator may consume 205 kW fuel. often they are watercooled and the exhaust gases are of course blown out, only a fraction will be left in the room as losses. Ask the supplier of the generator abou this figure. My guess is 40kW. If the outdoor air is 30 C. Then the air flow will be 40 kW/(40-30)= 4 kg/s

 

[25362]

The continuity of the operation refers to around-the-clock operation, since the night outside air may be much cooler than 30 deg C, meaning a smaller more amenable cooling air flow rate.

A heat release of 205 kW, means 8.2% of the generated energy. Kindly refer to lilling's comments for clarity.