05.04.2016 Views

Modern Engineering Thermodynamics

You also want an ePaper? Increase the reach of your titles

YUMPU automatically turns print PDFs into web optimized ePapers that Google loves.

148 CHAPTER 5: First Law Closed System Applications<br />

5.2 SEALED, RIGID CONTAINERS<br />

One of the most innocuous technical incantations in basic thermodynamics is the phrase sealed, rigid container<br />

(or tank or vessel). This phrase is composed of the following three technical terms:<br />

1. Sealed means the system is closed.<br />

2. Rigid means the system has a constant volume, V = constant and dV = 0. Therefore, there is no moving<br />

boundary mechanical work (i.e., R pdV = 0).<br />

3. Container (sometimes tank or vessel) means the system boundary lies inside the enclosure, because the material<br />

we want to analyze is inside the enclosure.<br />

The following example illustrates a typical problem of this type.<br />

EXAMPLE 5.1<br />

Read the problem statement. A sealed, rigid container whose volume is 1.00 m 3 contains 2.00 kg of liquid water plus water<br />

vapor at 20.0°C. The container is heated until the temperature inside is 95.0°C. 1 Determine<br />

a. The quality in the container when the contents are at 20.0°C.<br />

b. The quality in the container when the contents are at 95.0°C.<br />

c. The heat transport of energy required to raise the temperature of the contents from 20.0 to 95.0°C.<br />

Solution<br />

Step 1. Identify and sketch the system. Take the system to be the material inside the closed, rigid container as shown in<br />

Figure 5.1. Since we do not know the type or amount of material making up the container itself, the container cannot<br />

be part of the system. Also, since all the unknowns pertain to the container’s contents, detailed knowledge of the<br />

container’s construction is not relevant to the solution.<br />

Sealed<br />

rigid<br />

container<br />

∀ = 1.00 m 3<br />

m = 2.00 kg<br />

System<br />

boundary<br />

Vapor<br />

Liquid<br />

FIGURE 5.1<br />

Example 5.1.<br />

Step 2. Identify the unknowns. Here, there are three unknowns: x 1 , x 2 , and 1 Q 2 .<br />

Step 3. It is a closed system. To fix the system’s states, we note that we are given the initial and final temperatures of the<br />

water, but to find other properties (such as quality), we need the value of one more independent property in each<br />

state. Notice, however, that we are given both the total volume and the total mass and these do not change during<br />

the change of state. Therefore, we can calculate the system’s specific volume in each state as<br />

v 1 = v 2 = V/m = 1:00 m 3 /2:00 kg ¼ 0:500 m 3 /kg<br />

Now we can write the states and process path as<br />

State 1<br />

T 1 = 20:0°C<br />

v 1 = 0:500 m 3 /kg<br />

Process:“Rigid and Sealed” means v1 = v2 constant<br />

ƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒ! State 2<br />

T 2 = 95:0°C<br />

v 2 = v 1 = 0:500 m 3 /kg<br />

Step 4. Identify the process connecting the system states. The process here is one of constant volume (the container was<br />

specified as rigid), and the process path has already been indicated in the state property value listing.<br />

Step 5. Write down the basic equations. The only basic equation we have thus far for closed systems is Eq. (4.20), the<br />

energy balance (EB) equation:<br />

1Q 2 − 1 W 2 = m½ðu 2 − u 1 Þ + ke 2 − ke 1 + pe 2 − pe 1 Š system

Hooray! Your file is uploaded and ready to be published.

Saved successfully!

Ooh no, something went wrong!