HVAC system design is always tricky and complicated. Many of us want to know how to calculate the supply air condition and amount or volume flow rate of air required for a specific application and will end up in confusion ( at least me if not everyone !!). We know that several parameters like the temperature of supply air, the volume of air, sensible heat load of space, the latent heat load of space...and so on are involved in the calculations but which parameter has to be determined and which has to be referred from some standards makes me/us land in the said confusion state.

In this article am explaining in a very simple manner on how toÂ calculate supply air condition and volume flow rate for cooling with an example case of a small auditorium conditioning.

**Details Available with us:**

- Auditorium Dimensions
**:Â**L15 m x W15 m x H4 mÂ - No of maximum occupants: 100
- Comfort temperature required: 21 ÂºCÂ - Dry bulb temperature
- Comfort relative humidity required: 50 %Â
- Cooling load estimated: 10 kWÂ Â
- Latent load estimated: 3 kWÂ

(Both cooling load and latent loads are calculated considering many parameters and calculations like envelope type, the orientation of auditorium, solar heat gains of different elements, thermal load due to occupants, equipment etc)

**1st Step:**Â Considering a ceiling-mounted supply throwing down, a supply temperature differential of 5K is a reasonable optimum first choice.Â

Q = m x cp x Î”t

10 = m x 1â‹…026 x 5

m = 1.95 kg/sec

QÂ Â =Â Cooling load estimated

Cp = Specific heat capacity of air in kJ/kg K

Î”tÂ = Temperature differential between supply air and room airÂ

mÂ =Â mass flow rate of air in kg/s

**2nd Step:**Since the density of air at the desiredÂ temperature of 21ÂºC and 50% RHÂ is 1.1856 kg/m3, the mass flow rate of air in cubic meter per sec isÂ

Volume flow rate: = mÂ = 1.95 / 1.1856

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 1.64 m3/s

As per ventilation requirement standards, 100 occupants require around 10 l/s per person outdoor air for comfort considering the area as non smoking (reference CIBSE Guide A, table 1.5)

100 Ã— 10 = 1000 l/s = 1.0 m3/s

Just as a cross-check, The volume flow rate will, therefore, be more than sufficient to supply the outdoor air requirement. Some recirculation can be used to achieve the required volume flow rate ofÂ around 1.64 m3/s

Now let's cross-check our airflow rate volume with air changes rate standard too. The air change rate is nothing but the volume of air in cubic meter per hour to the volume of conditioned space. For our application, the standard (CIBSE B2 Table 3.6/CIBSE B, Table 2.14: design requirements: assembly halls and auditoria.)Â normally saysÂ 6 - 10 air changes per hour.

Air change rate = (1.64 x 3600) / ( 15 x 15 x 4)

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â =Â 6.6 -

**within limit hence the value obtained is optimum****3rd Step:**Now its time to calculate humidity addition or reduction to make the air delivered at desired 50% RH.

Ql = m x hfg x Î”g

3 = 1.95 Ã— 2450 Ã— Î”g

Î”g = 0.000628 kg/kg

Ql = Latent heat load

mÂ =Â Â mass flow rate of air in kg/s

hfg = Latent heat of vaporisation in kJ/kg

Î”g =Â moisture content differential between supply air and room air, ie (gr â€“ gs)

Now we understand that the humidity difference in supply air to that desired air asÂ 0.000628 kg/kg

And from the psychrometric chart, it is evident thatÂ air at 21ÂºC and 50% saturation has the humidity ratio of 0.007542 kg/kg

So the required supply air humidity ratio should be 0.007542 -Â 0.000628 =Â 0.006914 kg/kg

And since our desired temperature is 21ÂºC and considered temperature difference in step 1 as 5K, now the supply air temperature can be derived as 21 - 5 = 16ÂºC

Hence our designed final values are like

**16ÂºC DBT,Â 0.006914 kg/kg humidity ratio andÂ 1.64 m3/s**
Pl. inform me supply air condition & valume rate for a room of L10. 5 ft. X W 7.5 ft.

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