Tuesday, 30 July 2013

In my previous post, we became familiar with the information that can be gleaned from a single point on the psychrometric chart and saw that air with a temperature of 21.0 C db and a humidity of 50% RH, has the following properties.

Temperature db = 21.0 C

Humidity = 50% RH

Moisture Content = 0.0078 kg/kg

Specific Volume = 0.844 m3/kg

Temperature wb = 14.7 C

Specific enthalpy = 41.0 kJ/kg

So what happens when the air is heated? We will look at the example of heating the air from the point 21.0C db, 50% RH to a new temperature of 31.0C db (an increase of 10.0C db).

The first thing to consider is where our point will move to. By definition, it will be somewhere on the 31.0C db line on the chart. The one thing that will not change is the moisture content of the air because the process of heating does not add or remove moisture from the air.

The new point will have a temperature of 31.0C db and a moisture content of 0.0078 kg/kg.

As in the previous post, we can find the specific enthalpy of the new point. It is 51.25 kJ/kg.

This is an increase in enthalpy of 51.25 - 41.0 = 10.25 kJ/kg from the original point and the new point.

A typical use of this information would be to calculate how much power is required to heat in an air system – perhaps to calculate the required capacity of a heating coil in an air handling unit (AHU). The change in enthalpy is given in kJ/kg so we must first find the mass of the air flowing through our heating system. Normally the air flowing through a system is given as a volume and we shall use a volume of 4.0 m3/s for this example. So we are looking at a system where air is being blown from a fan with a temperature of 21.0C db, 50% RH, over a heating coil, which heats the air to 31.0C db.

We know that the specific volume of the air before the heating coil is 0.844 m3/kg. The mass of air passing through the heating coil is 4.0/0.844 = 4.74 kg/s.

The energy input is 10.25 kJ/kg so the energy input for 4.74 kg is

10.25 x 4.74 = 48.6 kJ

Our system has an air flow rate of 4.74 kg/s so the power input is 48.6 kJ/s.
1.0 kJ/s = 1.0 kW, so the power required is 48.6 kW.

It should be noted that the specific volume of the air once it has been heated is 0.873 m3/kg. The volume of air at this point is 4.74 x 0.873 = 4.14 m3/s (rather than 4.0 m3/s). This is not surprising because warmer air has a bigger volume than cooler air for the same mass.

When making these calculations, it is important to calculate the mass of air using the correct air condition. For instance, if we had calculated the mass of air after it had been heated, our calculations would not be correct.

The last thing to point out is that heating the air does not alter the moisture content, but it does reduce the relative humidity. From our chart, the humidity of the air after the heater is 27% RH.


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