Example 2.3.1: Sensible Heat Change

In this example, an electric resistance heater is installed in an air duct. The heater is needed to heat 650 pounds per hour of air from 75 degrees fahrenheit to 95 degrees fahrenheit. The problem is to find the capacity of the heater in kW.

The process line is shown here on the psychrometric chart. The electric resistance of the heater is converted into heat. This problem is an example of a sensible change in the air stream as it is heated. On the psychrometric chart, it looks like a straight line along some line of constant humidity ratio. To solve, we will use the sensible heat equation where Qsis equal to the mass flow rate of the air times the specific heat of the air times the change in dry bulb temperature. In this application Qsis the heat required in Btuhs. The specific heat of the air is 0.24 Btus per pound-degree fahrenheit. Please note that the mass flow rate of the air has to be in pounds per hour for the units to work. So, applying the math we find that Q sensible is equal to 0.24 times 650 times 95 minus 75 degrees fahrenheit and this works out to 3120 Btuhs. To find the required capacity of the heater, we simply need to convert Btus per hour to kW using the conversion factor 1kW is equal to 3410 Btuh and we find that the required capacity is 0.91 kW. Our recommendation would be to provide a 1 kW heater.

This is example 2.3.2

In the previous example we used the sensible heat equation to find the capacity needed to heat up the air by 20 degrees fahrenheit and at an airflow rate of 650 pounds per hour. Recall from lesson 2.2 that in HVAC applications, the sensible heat for the change in enthalpy of water vapor is typically considered negligible and is ignored. For this example, we will determine how much air results in neglecting the enthalpy change of the water vapor.

To do this, we need to know something about the water vapor content of our air sample. For this example, we will assume the humidity ratio W is 0.008 pounds of water per pound of dry air. To find the sensible change of the moist air sample, taking into consideration both the dry air and the water vapor, our equation would be Qsis equal to the mass flow rate of the air times the specific heat of the air times the dry bulb temperature change of the air sample plus the mass flow rate of the water vapor times the specific heat of the water vapor times the change in dry bulb temperature of the water vapor. The mass flow rate of the air sample is 650 pounds per hour.

So what is the mass flow rate of the water vapor? The mass flow rate of the water vapor is the humidity ratio, which is pounds per water per pounds of dry air, times the pounds per air hour, which is in this example, 650 pounds per hour. So the mass flow rate of the water would be 0.008 times 650, which is 5.2 pounds of water vapor per hour. And now we can apply this to our equation, where we would have Qsis equal to 0.24 times 650 times 20, which is the sensible heat change of the dry air, plus 0.45 times 0.52 times 20, which is going to be the sensible heat of the water vapor. This comes up to 3169 Btuhs. Qsin kW using the same conversion factor as in our previous example would be 0.93 kW. In the previous example, by neglecting the water vapor in our sample, we had 0.91 kW, so the amount of air is approximately 2%. In this application, we would probably specify 1kW heater regardless.

Example 2.3.3

In the previous examples, we worked with the airflow rate in terms of pounds per hour, which is the mass flow rate. But common practice is to specify airflow rate in terms of CFM. What is the airflow rate of the air entering the heater in our previous example in CFM?

Well, to solve this, we actually need to look at the psychrometric chart because we see if we take 650 pounds per hour and multiply that by 1 hour per 60 minutes, we need to multiply that also by the specific volume of the air in cubic feet per pound of air to convert to CFM. The problem is, what is the specific volume of the air sample at what conditions?

Looking at the process line on the psychrometric chart, we see that the specific volume of the air entering condition looks to be about 13.6 cubic feet per pound of air. If we put this into our equation, we would have 650 pounds per hour times 1 hour per 60 minutes times 13.6 cubic feet per pound and this would come up to approximately 148 CFM. But looking at the process line on the psychrometric chart, we see that the specific volume of the air leaving the heater is greater than the specific volume of the air entering the heater. Therefore, the CFM is going to be different and in fact the CFM does change and is dependant upon the conditions. We can determine the CFM at the leaving conditions by doing the same equation but using the specific volume of the air at the leaving conditions, and this would be 650 pounds per hour times 1 hour per 60 minutes, and in this case, we would use 14.16 cubic feet per pound of air, and we find that the CFM of the air leaving the heater is about 153 CFM. This can actually be confusing, so it is important to understand the conditions at which CFM is specified. Manufactures of mechanical equipment know this, so in rating equipment you might see reference to standard air conditions. Standard air may be expressed as having the following conditions: specific volume of 13.3 cubic feet per pound, the density of 0.075 pounds per cubic foot, dry bulb temperature of 60 degrees, and pressure at 29.92 inches of mercury. It is important to always check what the manufacture means when they specify standard air because it actually may be different than these conditions. The specific volume and density of the air depends on the condition it exists, so the CFM can actually change. If we were to apply these conditions to our previous example, we would see that the 650 pounds per hour times 1 hour per 60 minutes times the specific volume of 13.3 would yield an airflow of 144 CFM.

Example 2.3.4

This is an example of latent heat addition in humidification. As noted in lesson 2.2, humidification can be shown on the psychrometric chart as a process line of constant dry bulb temperature with increasing humidity ratio as shown here. The psychrometric chart shows that the conditions of winter air may have very low humidity, which is probably too low for comfort conditions. Therefore, humidification is a common air conditioning process useful for colder climates and heating applications. The following example demonstrates the latent heat process.

A water humidifier in a warm air duct system increases the moisture content from 0.004 to 0.016 pounds of water per pound of air. If the air flow rate is 5,000 CFM and the dry bulb temperature is at 95 degrees dry bulb, then use the psychrometric chart to determine the heat required in this latent process. From the psychrometric chart, we can determine the enthalpy of the air entering and leaving the humidifier. The enthalpy of the air entering the humidifier is approximately 26.8 Btus per pound. The enthalpy leaving the humidifier is at 38.8 Btus per pound. By applying the enthalpy equation, which is Q is equal to the mass flow rate of the air multiplied by the change in enthalpy we can determine the heat required for this process. Please note that this equation to get our Q in Btuhs, the flow rate of the air is in pounds per hour which is the mass flow rate. So just as in our previous examples, we have to determine what the mass flow rate is for 5,000 CFM.

Note that I have decided to use the conditions for standard air to determine the mass flow rate based on this 5,000 CFM as follows: 5,000 CFM multiplied by 60 minutes per hour times the specific volume of air as one pound per 13.3 cubic feet times the change in enthalpy. In doing the math, this comes out to 270,677 Btu/hr.

This is example 2.3.5: Combined Sensible and Latent Heat Process

Air conditioning for occupant comfort often requires that the air be both cooled sensibly and dehumidified. To determine the total heat removed for this process, both the sensible and latent components must be considered. Q total is going to be the sensible component plus the latent component. If our airflow rate is given in CFM, then our sensible component Qsis going to be some constant multiplied by the CFM times the change in temperature. The latent component is going to be a different constant times CFM times the change in humidity ratio. As you can see, this gets pretty complicated right off the bat, so there is actually an easier way to do this using the psychrometric chart and our enthalpy equation.

The enthalpy equation can be used for all processes, whether they are latent, sensible or a combination of the two. Let’s consider the following example: and air conditioning unit has a cooling coil that cools and dehumidifies 400,000 CFM of air from 78 degrees fahrenheit and at 50% relative humidity to 62 degrees fahrenheit dry bulb and 56 degrees wet bulb. In this process, the air is going to be both cooled and dehumidified. This can be shown on the psychrometric chart using the enthalpy equation. Find the sensible, latent and total capacity of the cooling coil in units of Btuhs and also tons of refrigeration.

Let’s consider the total change first, from the entering and leaving conditions. The Q total is going to be the mass flow rate of the air times the enthalpy of the air change as it goes through this process. The mass flow rate of the air is 40,000 CFM times 60 minutes per hour times 1 over 13.3 cubic feet, which is the specific volume of the air sample. Please note that for this example, I decided to use the standard air conditions to convert to the mass flow rate.

So the total change is going to be the enthalpy change as it goes through this process. The entering enthalpy is h1 as shown on the psychrometric chart to be about 30 Btus per pound. At the leaving conditions, the enthalpy is 23.8. This gives us a total change of approximately 6.2 Btus per pound and applying the enthalpy equation, we get 1,118,797 Btuhs. Using the conversion factor for tons of refrigeration as 1 ton equals to 12,000 Btuhs, we see that the total cooling capacity of this cooling coil is approximately 93 tons. Let’s look at what the latent change is for this example.

Using the enthalpy equation, we can determine the latent change by determining the change in enthalpy for the latent process. The entering condition is the change as 30 Btus per pound but the enthalpy change for the latent process is shown here as 26.8 Btus per pound in the leaving conditions. So this gives a change in enthalpy of about 3.2 Btus per pound. Applying this into our enthalpy equation, we get 577,444 Btuhs, or about 48 tons of refrigeration due to the latent change. Of course, the same application can be done to determine the sensible change of the air sample. In this example, the enthalpy change due to the sensible change can be found where h1 is 26.8 Btus per pound off of the psychrometric chart, and h2 is 23.8 Btus per pound. So applying the enthalpy equation, the sensible change is approximately 541,353 Btus per hour, or about 45 tons of refrigeration. You will find that the sensible change plus the latent change will equal the total change and the total cooling capacity.

Example 2.3.6

For the previous example, how much moisture will condense out of the cooling coil? Well, to determine this we need to determine the change in moisture content or humidity ratio as it goes through this process. The humidity ratio entering the coil at the entering conditions can be found on the psychrometric chart as about 0.0102 pounds of water per pound of air. The humidity ratio leaving the coil is about 0.0082. To find the rate of condensation in pounds per water, we multiply the mass flow rate of the air by the change in humidity ratio. Here we see that the mass flow rate of the air is equal to the 40,000 CFM times 60 minutes per hour times 1 pound per 13.3 cubic feet, and in doing the math, we will find that we will get approximately 360 pounds of water per hour that condenses out of this cooling coil. So how much is this in gpm? Well, I’m not going to solve that but I will let you figure that out. I will give you a hint, and that is to use 8.33 pounds per gallon and of course, 60 minutes per hour, and apply that to the 360 pounds per hour and see what you get in terms of gallons per minute.