Process of cooling ventilation with the help of example

Last updated on May 15th, 2024 at 08:04 am

The moving air can be utilized as a heat conveying medium. Warm air heating is usually combined with a mechanical ventilation system. Cooling ventilation can also provide a cooling effect simply by replacing the warm inside air with cooler outside air.

In cold climates, the need for cooling rarely arises. In warm climates the intention is to keep the indoor air cooler than the outdoor air, thus there can be no cooling by ventilation.

Process of cooling ventilation with the help of example
Process of cooling ventilation with the help of example

It can, however, be used quite successfully in a situation where the out-door air is at a comfortable temperature or just below that, but there is a significant internal heat gain (e.g. in a meeting room or a dance hall) which would cause indoor overheating.

As an example, let us assume that the out-door temperature is 18°C; the indoor temperature has risen to 28°C and there is an internal heat gain of 5 kW, which would cause a further increase of indoor temperature.

The temperature difference (AT) is 28°C-18°C=10degC. the specific heat of air is 1300 J/m³ deg. Using the ventilation heat loss equation
Q = 1300 xVXAT
5000 =1300xVx10

We must provide a ventilation rate of:
v=5000/1300 X 10=0.385m3/s

With an air velocity of 2 m/s the necessary duct cross-sectional area should be:

0.385m’/s/2m/s=0.192m², e.g. 0.30×0.64m

A fan giving the above ventilation rate can be selected from catalogues.

Note. The above is an approximate calculation only: a mechanical engineer

would make allowances for frictional losses in ducts (velocity and pressure gradients)

Another form of cooling by air movement is physiological cooling by directing an air stream of substantial velocity at the body’s surface.

This is achieved by table-top or ceiling-mounted fans (punkahs), which do not provide air exchange, but generate air movement.

What do you mean evaporative cooling? What is the purpose of this?

The evaporation cooling of water absorbs a significant amount of heat. The latent heat of evaporation, at normal temperatures, is around 2400kj/kg of water.

This phenomenon can be successfully utilized for the cooling of air when the air itself is dry so that the moisture does not cause inconvenience- and it may even improve the conditions. This is likely to be the case in hot-dry climates.

In a mechanical installation, a very fine spray of water may be put across an air intake duct to achieve maximum surface contact between air and water.

(it must be followed by a set of ‘eliminator plates’ which would trap and drain away any small droplets of water carried by the fast-moving air stream.) it may serve three purposes:

1. Washing the air, that is water droplets will stick to dust particles, which can thus no longer remain in suspension; they fall down and are washed away by the surplus water.

2. evaporative cooling,
3. humidification, i.e. the increase of relative humidity

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In a warm- humid climate the first of these functions could be performed; the second one only to a limited extent ( as the air is already humid, it will not take on much extra moisture, especially if its temperature is lowered at the same time)- but the third one would definitely be undesirable: it would increase the humidity, which is already too high.

Such water sprays can be utilized in warm-humid climates only as a preliminary treatment of air if it is to be subsequently dehumidified.

Write a short note on mechanical cooling?

The simplest example of mechanical cooling is the domestic refrigerator, shown diagrammatically in figure 2.1. A suitable gas, the refrigerant, is circulated in a closed circuit by a compressor.

This is most often a gas called Freon (CF,CI): at least in small installations in large plants, such as cold storage buildings, ammonia (NH) or carbon dioxide (CO₂) is often used: the former is toxic, therefore any leakage may be troublesome; the latter requires very high pressure.

The circuit consists of two coils:

a. the warm coil or condenser, and
b. the cold coil or evaporator.

The two coils are connected on one side through a compressor and on the other side through a pressure release valve.

The warm coil is thus kept under high pressure and the cold coil is under negative pressure.

The refrigerant is in a liquid state under compression and in a gaseous state under low pressure. Without changing the heat content, compression increases the temperature; expansion decreases it.

When liquefying, the refrigerant releases its latent heat of evaporation, and when evaporating, it absorbs a similar amount of heat.

The cycle can be described as follows:
a. compressor

(1) increases pressure
(ii) no change in heat content
(iii) temperature from, say, 0°C to 30°C

b. condenser

(i) no change in pressure
(ii) in condensation latent heat is released and dissipated into the environment
(iii) temperature from 30 to, say, 26°C

c. pressure valve

(1) admits liquid only above a set pressure, thus guarantees a low pressure in
evaporator
(ii) no change in heat content
(iii) temperature from 26 to, say, -4°C

d. evaporator

(1) no change in pressure
(ii) in evaporation latent heat absorbed
(iii) heat is taken from the environment
(iv) temperature from -4 to, say, 0°C

If the evaporator coil is placed into an air supply duct (instead of into a refrigerator cabinet) the air blown across it will be cooled

What are the various problems associated with cooling?

If the air in a space is to be cooled, the space must be fully enclosed, otherwise, the cooled inside air and the warm outside air would mix.

If doors and windows are closed, the fresh air needed by the occupants must be supplied mechanically. Thus cooling must be combined with some form of the mechanical ventilation system.

If the outside air is at a high temperature (30°C DBT) and of medium humidity (60%), this will be conditioned at the point of air intake.

If it is to be cooled to 18°C DBT, its RH will increase. It will actually reach a saturation point at 21.5°C, so with further cooling,

some moisture will condense, and at the end, we will have an 18°C DBT air of 100RH. As the AH at 21.5°C is 16 g/kg, and at 18°C and 100RH it is 13 g/kg, 3 g of moisture will condense out of every kg of air passing through the cooler.

This condensate may be drained away, but the 100 %RH of the supplied air would be the cause of acute discomfort.

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