In this screencast we are going to look at an example of adiabatic cooling, which
is a process by which cold liquid can be used to cool a hot gas. Specifically we
are going to look at an example of adiabatic humidification of air by the evaporation of cold
water. In this example we have humid air that is at 40 degC and 20% relative humidity. That
enters an adiabatic spray chamber at a specified rate and the air is adiabatically humidified
to 50% relative humidity. We want to use the humidity charts to find the outlet temperature
and the rate the water is evaporating in the spray chamber. A flow chart of the spray chamber
is shown below. So here we have our hot air coming in at the specified conditions. We
also have cold water coming in that is sprayed into the chamber. Some of these liquid water
droplets are going to evaporate. We will call that rate of evaporation M dot H2O.
We have water leaving as well. So the difference between the water going in, and the water
going out must be the rate that water is evaporating, and the evaporation of water has the effect
of cooling air on the outlet side. As water evaporates it is going to humidify the air
to 50% relative humidity. Now the term adiabatic humidification is somewhat contradicting.
We are clearly cooling the air as the water evaporates, but if we draw a system boundary
between the spray chamber and the surroundings we can see where the term adiabatic applies.
There is no heat exchange between the spray chamber and the surroundings. So Q is equal
to 0, which is implied with the term adiabatic. All the heat transfer is between the water
and the cool air inside the chamber. So to find the outlet temperature of the humidified
air we will go to the humidity charts. To see how this process works. I have shown a
reprint of the humidity charts in SI units. We know the air going into the spray chamber
is 40 degrees, and that can be found on the x axis, were the dry bulb temperature is shown.
So this is the temperature that we normally measure with with a thermometer. We are at
40 degC going into the chamber. We also know the relative humidity is 20%. Which is along
this curve. So the air entering the spray chamber is represented by the intersection
of these 2 points at approximately right here. A coincidence of the air water system is that adiabatic humidification
occurs along the lines of constant wet bulb temperature. The lines of constant wet bulb
temperature are the diagonal lines that go across the chart. So here we follow our initial
condition at 20% relative humidity along the lines of constant wet bulb temperature until
we reach the relative humidity of 50%. So the air leaving the spray chamber has a relative
humidity of 50% at this point. If we drop a line down to the temperature scale we see that
the temperature leaving the spray chambers is about 30 degC. So the temperature of the
air entering the spray chamber has been decreased from 40 degC to 30 degC at the outlets due
to the evaporation of water. So next we will find the rate the water is evaporating. Here the humidity
chart is shown again with our humidification process highlighted, and if we want to know
the rate at which water is evaporating we know that is related to the change in moister
content, which is shown on the y-axis. The change in moister content is also referred
to as the change in absolute humidity. The rate of water evaporating is going to be proportional
to the change in absolute humidity. If we look at the units here. We see that it is
kg/kg of dry air. Another words it is kg of water per kg of dry air. So we need to multiply
the change in absolute humidity on a per kilogram of dry air basis, by the actual flow rate of dry air that
we have. Now we do not know what the flow rate of dry air is. We were given a volumetric
flow rate. There is another quantity on the chart that is going to help us, and that is
the humid volume. This is in units of m^3 of humid air per kg of dry air. So by looking
at the units if we want the mass flow rate of dry air we need to divided the volumetric
flow rate by the humid volume. So we need to find the absolute humidity at the inlet
and the outlet, as well as the humid volume. We can get the absolute humidity at the inlet
by drawing a horizontal line from our initial condition to the y-axis. It is a little bit
hard to read. We can see that this is close to 0.009 so I am going to use a value of 0.0092,
and again that is in units of kg of water per kg of dry air. We can do the same thing to
find the absolute humidity at the outlet. So drawing a horizontal line from our outlet
condition to the y-axis. It is a little bit over 0.013. So I will use a value of 0.0132.
The last thing we need is the humid volume, and specifically we need the humid volume
at the inlet condition, because that is were we know the volumetric flow rate of humid
air that is entering the spray chamber. We know the volumetric flow rate is 100 m^3/hr.
So if we start at the initial condition and follow the humid volume line, which are the
steep diagonal curves. The humid volume is found below the x-axis, and in this case
we are right on the humid volume line of 0.9 m^3/kg. With the 3 numbers that we need from
the humidity charts. I will scroll down to calculate the rate of water evaporation. So
if we cancel out some of the units here we see that the volume of humid air cancels outs,
as well as the mass of DA. So plugging in these values we get a calculated rate of evaporation
of 0.44