In this section, we consider energy balances for closed systems. In the next section, open systems will be treated. Figure 2.8 shows a schematic of a closed system that undergoes a process from initial state 1 to fi nal state 2. In this fi gure, the system, surroundings, and boundary are delineated. In a closed system, mass cannot transfer across the system boundary. There are two ways in which to catalog the amount of material in the system—
by mass or by moles. Each way can be convenient. For a pure species or a mixture of constant composition, the two forms are equivalent and can be interconverted using the molecular weight. When we address systems undergoing chemical reaction, care must be taken. While the total mass must be conserved, the number of moles or the mass of a particular component may change. In the absence of chemical reaction, the number of moles remains constant:
n15n2
Since mass cannot enter or leave a closed system, the changes in the energy within the system 1D 5final2initial2 are equal to the energy transferred from the 2.4 The First Law of Thermodynamics for Closed Systems ◄ 55
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56 ► Chapter 2. The First Law of Thermodynamics
surroundings by either heat or work. Figure 2.8 also illustrates the sign convention that we have defi ned for heat and work, namely, positive for energy transfer from the sur- roundings to the system. Writing down the fi rst law in quantitative terms, we get:7
c
change in energy in system
s 5 c
energy transferred from surroundings to system
s DU1 DEK1 DEP5 Q1W ('''')''''* (')'*
property: depends on depends only path on state 1 and 2
(2.12a)
The properties on the left-hand side of Equation (2.12a) depend only on the initial and fi nal states. They can be calculated using the real path or any hypothetical path we cre- ate. The terms on the right-hand side are process dependent and the real path of the system must be used.
Since the composition of a closed system remains constant (barring chemical reac- tion), we can rewrite Equation (2.12a) using intensive properties by dividing through by the total number of moles:8
Du1 DeK1 DeP5q1w (2.12b)
We often neglect macroscopic kinetic and potential energy. For this case, the extensive and intensive forms of the closed system energy balances become:
DU5Q1W (2.13a)
and, Du5q1w (2.13b)
respectively.
Figure 2.8 Illustration of closed system and sign conventions for heat and work. All three forms of energy are considered.
Surroundings +Q +W
Boundary System property ΔEK = EK,2 − EK,1
ΔU = U2 − U1
ΔEP = EP,2 − EP,1
7 Heat and work already refer to the amount of energy transferred; hence, it would be would be wrong to write them DQ or DW. We reserve the D for state function that depends just on the initial and fi nal state of the system.
8 We will sometimes write balance equations on a molar basis, utilizing the appropriate intensive thermodynamic properties. For example, internal energy will be u [J/mol]. You should be able to convert any equation to a mass basis that uses the corresponding specifi c property, for example, u^ 3J/kg4.
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2.4 The First Law of Thermodynamics for Closed Systems ◄ 57 Differential Balances
The fi rst law can also be written in differential form for each differential step in time during the process. Numerical solutions are obtained by integration of the resulting dif- ferential energy balance. Common forms of energy balance over a differential element can be written by analogy to the equations just presented.
dU1dEK1dEP5 dQ1 dW extensive 3J4 du1deK1deP5 dq1 dw intensive 3J/mol4 or, neglecting kinetic and potential energy,
dU5 dQ1 dW extensive 3J4 (2.14) du5 dq1 dw intensive 3J/mol4
We use the exact differential d with the energy terms to indicate that they depend only on the fi nal and initial states; in contrast, we use the inexact differential d with heat and work to remind us that we must keep track of the path when we integrate to get these quantities.
The energy balances above are often differentiated with respect to time, yielding:
dU dt 1 dEK
dt 1dEP
dt 5Q# 1W#
extensive 3W4 du
dt 1 deK
dt 1 deP
dt 5q# 1w# intensive 3W/mol4
where the rate of heat transfer and the rate of work [J/s or W] are denoted with a dot over the corresponding variable. Again, we often neglect kinetic and potential energy to give:
dU dt 5Q#
1W#
extensive 3W4 (2.15) du
dt 5q# 1w# intensive 3W/mol4
►EXERCISE Consider the six processes depicted in Figures 2.4 through 2.7. What is the heat transferred in each case. Can you explain the difference in relation to the effi ciency factor?
Consider a piston–cylinder assembly containing 10.0 kg of water. Initially, the gas has a pres- sure of 20.0 bar and occupies a volume of 1.0 m3. The system undergoes a reversible process in which it is compressed to 100 bar. The pressure volume relationship during this process is given by:
Pv1.55const Example 2.4
Closed System Energy Balance
(Continued)
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58 ► Chapter 2. The First Law of Thermodynamics
(a) What is the initial temperature?
(b) Calculate the work done during this process.
(c) Calculate the heat transferred during this process.
(d) What is the fi nal temperature?
SOLUTION (a) We need two independent intensive properties to constrain the state of the system. Once values for these properties are determined, we can use the steam tables to fi nd other properties. The specifi c volume of the initial state can be determined as follows:
v^5 V m5 1.0
10.050.10 Bm3 kgR
The values in the steam tables are in units of Pa, so we convert 20 bar to 2 MPa. If we look at Table B.4 (Appendix B), we see that at P52 MPa to three signifi cant fi gures the volume at the saturation temperature is 0.100 Bm3
kgR. Therefore, looking at Table B.2, we see the temperature is 212.4°C (or slightly above).
(b) To calculate the work it is useful to draw a schematic of the process. We defi ne the initial state as state 1 and the fi nal state as state 2, as shown in Figure E2.4.
State 1 State 2
v1 = 0.1 [m3/ kg]
P1 = 20 [bar]
P2= 100 [bar]
T1= 212.4 [° C]
H2O
Process
Figure E2.4 Initial and final states of the expansion process.
Because this is a reversible process, we can write:
w^ 5 23PEdv^5 23Pdv^
To fi nd the upper limit on the integrand, we need to know the specifi c volume of the fi nal state, v^2. We can calculate v^2 from the equation in the problem statement:
Pv^1.55const5P1v^1.51 5P2v^1.52
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2.4 The First Law of Thermodynamics for Closed Systems ◄ 59
Solving for v^2 gives:
v^25 ¢P1v^1.51
P2 ≤
1/1.5
5 10.230.11.521/1.550.0342 Bm3 kgR Now we can solve for work:
w^ 5 23
0.0342 0.1
Pdv^5 23
0.0342 0.1
P1v^1.51
v^1.5 dv^52P1v^1.51 B 1 v^0.52
2 1 v^0.51
R
0.0342
0.1
5284 BkJ kgR The sign of work is positive because we are adding energy to the system during the compression.
(c) To fi nd heat, we can apply the fi rst law:
q5 Du2w
Because we solved for work in part (b), we need only to determine the internal energy in states 2 and 1 from the steam tables. In part (a), we found that state 1 is approximately saturated vapor at 20 bar. Looking at Table B.2, we fi nd:
u152600.3 BkJ kgR
For u2, we look at Table B.4 at a pressure of 10 MPa (100 bar). In this case, we need to interpolate:
u22u1T5500ºC2
u1T5550ºC22u1T5500ºC2 5 v22v1T5500ºC2 v1T5550ºC2 2v1T5500ºC2 Solving for u2,
u25u1T5500ºC2 13u1T5550ºC22u1T5500ºC2 4
B v22v1T5500ºC2 v1T5550ºC22v1T5500ºC2R
53045.8133144.523045.84B0.034220.03279 0.0356420.03279R 53094.6 BkJ
kgR Thus,
q5 Du2w5u22u12w5210 BkJ kgR
Since the value of q is positive, heat transfers from the system to the surroundings.
(d) To fi nd T2, we must also interpolate. From Table B.4,we get:
T25500ºC1 3550ºC2500ºC4B v22v1T5500ºC2 v1T5550ºC22v1T5500ºC2R 5500ºC1350ºC4B0.034220.03279
0.0356420.03279R 5525ºC
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60 ► Chapter 2. The First Law of Thermodynamics