Modern Engineering Thermodynamics

148 CHAPTER 5: First Law Closed System Applications
5.2 SEALED, RIGID CONTAINERS
One of the most innocuous technical incantations in basic thermodynamics is the phrase sealed, rigid container
(or tank or vessel). This phrase is composed of the following three technical terms:
1. Sealed means the system is closed.
2. Rigid means the system has a constant volume, V = constant and dV = 0. Therefore, there is no moving
boundary mechanical work (i.e., R pdV = 0).
3. Container (sometimes tank or vessel) means the system boundary lies inside the enclosure, because the material
we want to analyze is inside the enclosure.
The following example illustrates a typical problem of this type.
EXAMPLE 5.1
Read the problem statement. A sealed, rigid container whose volume is 1.00 m 3 contains 2.00 kg of liquid water plus water
vapor at 20.0°C. The container is heated until the temperature inside is 95.0°C. 1 Determine
a. The quality in the container when the contents are at 20.0°C.
b. The quality in the container when the contents are at 95.0°C.
c. The heat transport of energy required to raise the temperature of the contents from 20.0 to 95.0°C.
Solution
Step 1. Identify and sketch the system. Take the system to be the material inside the closed, rigid container as shown in
Figure 5.1. Since we do not know the type or amount of material making up the container itself, the container cannot
be part of the system. Also, since all the unknowns pertain to the container’s contents, detailed knowledge of the
container’s construction is not relevant to the solution.
Sealed
rigid
container
∀ = 1.00 m 3
m = 2.00 kg
System
boundary
Vapor
Liquid
FIGURE 5.1
Example 5.1.
Step 2. Identify the unknowns. Here, there are three unknowns: x 1 , x 2 , and 1 Q 2 .
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
water, but to find other properties (such as quality), we need the value of one more independent property in each
state. Notice, however, that we are given both the total volume and the total mass and these do not change during
the change of state. Therefore, we can calculate the system’s specific volume in each state as
v 1 = v 2 = V/m = 1:00 m 3 /2:00 kg ¼ 0:500 m 3 /kg
Now we can write the states and process path as
State 1
T 1 = 20:0°C
v 1 = 0:500 m 3 /kg
Process:“Rigid and Sealed” means v1 = v2 constant
ƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒ! State 2
T 2 = 95:0°C
v 2 = v 1 = 0:500 m 3 /kg
Step 4. Identify the process connecting the system states. The process here is one of constant volume (the container was
specified as rigid), and the process path has already been indicated in the state property value listing.
Step 5. Write down the basic equations. The only basic equation we have thus far for closed systems is Eq. (4.20), the
energy balance (EB) equation:
1Q 2 − 1 W 2 = m½ðu 2 − u 1 Þ + ke 2 − ke 1 + pe 2 − pe 1 Š system