In the Classroom
Discharge of Vessels Thermodynamic Analysis Jaime Wisniak Department of Chemical Engineering, Ben-Gurion University of the Negev, Beer-Sheva, Israel
The processes of filling and discharging of gases from a vessel occur very frequently in engineering applications. Their analysis from the thermodynamic viewpoint provides an interesting didactic application of the first and second laws of thermodynamics as well as of the linkage between open and closed systems. Both processes are essentially unsteady and from a physical viewpoint, the discharge of a pressurized vessel is similar to that of the charging of a vessel. However, from the thermodynamics viewpoint there is little similarity in the relations that result from applying the conservation of energy principle to the two processes. The problem of discharging is unlike the charging of a vessel from a pressurized line in that the properties of the fluid crossing the control surface are continuously changing owing to the decrease in mass within the control volume (1). For this reason we start our analysis by selecting the vessel as our control volume (CV) and write the first law in its general differential form:
δQ + δW = e 2+ P2v2 m 2dϑ – e 1+ P1v1 m 1dϑ + dE cv (1) where the energy balance is performed during the time period dϑ. Indexes 1 and 2 correspond to the entering and leaving streams and the specific energy e represents all the possible forms of energy: e = u + KE + PE
(2)
where KE and PE are the kinetic and potential energies, respectively. Dividing by dϑ we can write eq 1 as rate of change
Q + W = e 2 + P2v2 m 2 – e 1 + P1v1 m 1 + E cv
(3)
If we assume that no material is entering and that the ? vessel is rigid (constant volume V) then m? 1 and W are zero; if in addition, we neglect the kinetic and potential energy changes associated with either the control volume or the mass leaving the vessel through a suitable valve or opening, then all the energy is internal energy. The presence or absence of appreciable heat transfer effects will depend on the magnitude of the temperature difference between the system and the environment and the properties of the vessel wall. Under these assumptions eq 1 becomes
δQ = h 2dm 2 + dUcv = h 2dm 2 + d(mu)cv
(4)
where m is the mass present in the control volume at the given time. In the real situation the properties of the fluid change continuously from the bottom of the vessel to the exit port. If we now treat the internal fluid as a uniform system whose properties have been averaged over those of the real, continuous system (assumption of uniform state), then the material that is leaving represents the material inside the vessel at the same time and h 2 = h and dm2 = {dm. We can then transform eq 4
as follows:
δQ = {hdm + d(mh – PV )cv
(5)
δQ = mdh – VdP
(6)
δQ = V v dh – VdP
(7)
since V = vm. Integrating eq 7 between the initial and final states we get f
Q=V i
dh – V ∆P v
(8)
Equation 8 is deceptive in its simplicity because it indicates that the heat effect can be easily calculated if the relation between v and h is known. Usually this will not be the case because during discharge both pressure and temperature will simultaneously change in a manner that cannot be described theoretically. The solution of eq 8 will be simpler for the cases where one of the variables (such as temperature or pressure) is held constant. A reasonable first approximation made sometimes (2) for solving eq 4 is to assume that the enthalpy of the fluid leaving stays constant as the average ha between the initial enthalpy and the final enthalpy inside the vessel (assumption of uniform flow). This assumption is based on the fact that when the pressure range is not too large or the gas does not deviate too much from ideal behavior the derivatives (∂h / ∂v)T and (∂h / ∂P)T are generally a weak function of the pressure (actually, for an ideal gas their value is zero). Hence
ha =
hi + h f 2
(9)
Equation 4 can then be integrated to yield
Q = h am 2 + m f u f – m iu i
(10)
Again, it should be understood that eq 10 will yield only an approximate solution. Let us now analyze the solution of eq 8 for some particular cases of practical interest. 1. Adiabatic Discharge If the heat effect is slow or can be neglected, dQ = 0 and eq 4 becomes
h 2dm 2 + dUcv = 0
(11)
Assuming uniform state eq 11 is transformed as follows:
hdm = d(mu)
Vol. 74 No. 3 March 1997 • Journal of Chemical Education
(12)
301
In the Classroom du = dm m h–u
(13)
Again, since the control volume remains constant we write V = mv and get
dm = { dv v m
(14)
Substituting in eq 13 yields
du + Pdv = 0
(15)
Equation 15 is the general expression of the first law for the adiabatic discharge of a rigid vessel. If we have an ideal gas, eq 15 becomes h* – u* = Pv = RT
(16)
which integrates to
c*v dT = { dv v RT T2 v = v1 T1 2
R c* v
(17)
(18)
Using the relation γ = cp* / c*v we can write eq 18 as
T2 v = v1 T1 2
(19)
Equation 19 indicates that the temperature of the gas decreases during the adiabatic discharge of an ideal gas. The observant reader will notice immediately that eqs 18 and 19 have the same structure as those describing the adiabatic expansion of an ideal gas in a closed system (constant mass). We will address to this point later. 2. Isothermal Discharge In this situation eq 8 can be easily integrated numerically when the state properties h and v are available in a tabular form (e.g., steam tables). 3. Isobaric Discharge Addition of heat may be sufficient for maintaining isobaric conditions during the discharge process. Solution of eq 8 is similar to that for the isothermal case. Second Law Analysis of the Discharge Process Consider now the expression of the second law for a reversible change in a closed system:
A pressurized tank contains 1 kg of steam at 673.15 K and 10 MPa. Mass is allowed to flow from the tank until the pressure drops to 0.50 MPa. During the discharge process heat is added to keep the steam at constant temperature. Calculate the heat requirement of the process. We will first get the required answer using eq 8. First we calculate the volume of the tank from the initial conditions. The specific volume is 0.02641 m3 ?kg {1 (1) so that V = (1) (0.02641) m3. Using the steam tables (1) we plot 1/v against h (at 673.15 K) and calculate the area under the curve between the initial and the final conditions, as shown in Figure 1. From eq 8 we have 5 Q = (0.02641) 3178.6 – 5 – 100 103 = 334.8 kJ 10
From Figure 1 we see that the variation of the enthalpy with 1/v is essentially linear so that use of eqs 9 and 10 will yield the same numerical answer.
Example 2
(20)
Consider the same pressurized tank as in example 1, except that it is well insulated. We inquire about the final temperature and the amount of steam that is left in the tank when the pressure drops to 0.50 MPa.
If the process is reversible T ds = du + P dv
(21)
Equation 21 is a general expression that allows calculation of the entropy change for a simple compressible substance. For a reversible adiabatic process T ds = 0 and eq 21 becomes du + P dv = 0
(22)
which is identical with eq 15 and hence we face an apparent contradiction: the same expression of the first law describes both a process involving an open unsteady system (eq 15) and a process involving a closed system (eq 22). We are forced to conclude that for the reversible discharging process
302
(23)
Example 1
(γ – 1)
δQ = du + P dv
ds = 0
Therefore, for any fluid that flows from a pressurized vessel under the stated conditions (adiabatic), the specific entropy within the control volume (vessel) remains constant. The total entropy of the control volume decreases because the amount of mass in it decreases. The overall process is highly irreversible, so that the net entropy change of the control volume plus its environment must be positive. The fact that the discharging process is isentropic is not surprising, because it is adiabatic and quasistatic (assumptions) and all internal dissipative effects and finite gradients have been ruled out. That is, the process within the vessel is adiabatic and internally reversible. The apparent anomaly of open vs. closed system can be easily resolved if we consider the initial material present in the adiabatic discharging vessel divided into two parts by an imaginary partition. The lower portion contains exactly the amount of material that will be left at the end of the discharge process. Hence this material behaves like a closed system having a frictionless border that moves against a constant external pressure. Application of the first law to this situation yields exactly the expression described by eq 15. The different procedures will now be illustrated with the following examples.
A simple procedure is to use the second law analysis results: that during discharge the specific entropy remains constant. For the initial conditions we have si = 6.2120 kJ/ kg?K (1), so that in the final conditions Pf = 0.50 MPa and sf = 6.2120 kJ/kg?K. Inspection of the steam tables indicates that at 0.50 MPa the entropy of the saturated vapor is 6.8212 kJ/kg?K, so that the final condition is wet saturated steam at T = 425.05 K. The quality is
Journal of Chemical Education • Vol. 74 No. 3 March 1997
x = 6.2120 – 1.8607 = 0.877 6.8212 – 1.8607
In the Classroom
Figure 1. Graphical solution of example 1.
and the specific volume
v = (0.877) (0.3749) + (0.123) (0.0010926) = 0.329 m 3/kg Hence, the amount of steam that is left in the tank is
m f = 0.02641 = 0.0803 kg 0.32874 What happens if we use the first law approach? Since the process is adiabatic eq 8 yields
∆P =
dh v
(24)
Equation 24 cannot be solved by trial and error unless we have experimental data on the variation of h with v along the path of the process. If we assume that we have uniform flow conditions, then we can solve the problem using eq 10 (with Q = 0):
hi + h f m 2 + m f u f – m iu i = 0 2
(8a)
where m2 = m i – m f and
m f = vV f
(25)
Equation 8a is solved by trial and error by assuming the final temperature and determining the corresponding value of mf. This value is confronted against that obtained from eq 25. The final state may be superheated or saturated steam. In the latter case solution of eq 8a will yield the quality of the steam. Execution of the procedure for the system in question indicates that the final state is wet steam with quality x = 0.9435. This result is different from that obtained with the second law approach because in an adiabatic system the specific enthalpy is changing with both pressure and temperature and hence the approximation of an average value for the exit enthalpy is not very good. Literature Cited 1. Wark, W. Thermodynamics; McGraw-Hill: New York, 1989. 2. Van Wylen, G.; Sonntag, R.; Borgnakke, C. Fundamentals of Classical Thermodynamics, 4th ed.; John Wiley and Sons: New York, 1994.
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