The results are summarized in the following table:
Pchoked Qm
Case (psi4 (Ibm/s)
Orifice 113.4 4.16
Adiabatic pipe 49.4 1.81
Isothermal pipe 62.0 1.76
A standard procedure for these types of problems is to represent the discharge through the pipe as an orifice. The results show that this approach results in a large result for this case.
The orifice method always produces a larger value than the adiabatic pipe method, ensuring a conservative safety design. The orifice calculation, however, is easier to apply, requiring only the pipe diameter and the upstream supply pressure and temperature. The configurational de- tails of the piping are not required, as in the adiabatic and isothermal pipe methods.
Also note that the computed choked pressures differ for each case, with a substantial dif- ference between the orifice and the adiabaticlisothermal cases. A choking design based on an orifice calculation might not be choked in reality because of high downstream pressures.
Finally, note that the adiabatic and isothermal pipe methods produce results that are rea- sonably close. For most real situations the heat transfer characteristics cannot be easily deter- mined. Thus the adiabatic pipe method is the method of choice; it will always produce the larger number for a conservative safety design.
4-7 Flashing Liquids
Liquids stored under pressure above their normal boiling point temperature present substantial problems because of flashing. If the tank, pipe, or other containment device develops a leak, the liquid will partially flash into vapor, sometimes explosively.
Flashing occurs so rapidly that the process is assumed to be adiabatic. The excess energy contained in the superheated liquid vaporizes the liquid and lowers the temperature to the new boiling point. If m is the mass of original liquid, C,, the heat capacity of the liquid (energylmass deg), T , the temperature of the liquid before depressurization, and T, the depressurized boil- ing point of the liquid, then the excess energy contained in the superheated liquid is given by
This energy vaporizes the liquid. If AH, is the heat of vaporization of the liquid, the mass of liq- uid vaporized m , is given by
152 Chapter 4 Source Models
The fraction of the liquid vaporized is
Equation 4-86 assumes constant physical properties over the temperature range To to Tb. A more general expression without this assumption is derived as follows.
The change in liquid mass m resulting from a change in temperature T is given by
Equation 4-87 is integrated between the initial temperature To (with liquid mass m) and the final boiling point temperature Tb (with liquid mass m - m,):
where and are the mean heat capacity and the mean latent heat of vaporization, re- spectively, over the temperature range To to Tb. Solving for the fraction of the liquid vaporized, fv = m,lm, we obtain
Example 4-6
One lb, of saturated liquid water is contained in a vessel at 350°F. The vessel ruptures and the pres- sure is reduced to 1 atm. Compute the fraction of material vaporized using (a) the steam tables, (b) Equation 4-86, and (c) Equation 4-90.
Solution
a. The initial state is saturated liquid water at To = 350°F. From the steam tables P = 134.6 psia,
H = 321.6 Btu/lb,.
The final temperature is the boiling point at 1 atm, or 212°F. At this temperature and under saturated conditions
4-7 Flashing Liquids 153
H,,, = 1150.4 Btullb,, HIiquid = 180.07 Btu llb,.
Because the process occurs adiabatically, Hfina, = Hinitial and the fraction of vapor (or quality) is computed from
that is, 14.59% of the mass of the original liquid is vaporized.
b. For liquid water at 212'F C, = 1.01 Btullb, OF, AH, = 970.3 Btullb,.
From Equation 4-86
C,(T, - T,) (1.01 Btullb, "F)(350 - 212)"F
f"
= - -A H , 970.3 Btullb,
i c. The mean properties for liquid water between To and T, are
-
C, = 1.04 Btullb, OF, A H , = 920.7 Btullb,.
Substituting into Equation 4-90 gives f,, = 1 - e x p [ - G ( ~ , - T,)I=]
= 1 - exp[-(1.04 Btullb, "F)(350
-
212)"F/(920.7 Btullb,)]= 1 - 0.8557
Both expressions work about as well compared to the actual value from the steam table.
154 Chapter 4 Source Models
For flashing liquids composed of many miscible substances, the flash calculation is com- plicated considerably, because the more volatile components flash preferentially. Procedures are available to solve this problem.16
Flashing liquids escaping through holes and pipes require special consideration because two-phase flow conditions may be present. Several special cases need consideration.17 If the fluid path length of the release is short (through a hole in a thin-walled container), nonequi- librium conditions exist, and the liquid does not have time to flash within the hole; the fluid flashes external to the hole. The equations describing incompressible fluid flow through holes apply (see section 4-2).
If the fluid path length through the release is greater than 10 cm (through a pipe or thick- walled container), equilibrium flashing conditions are achieved and the flow is choked. A good approximation is to assume a choked pressure equal to the saturation vapor pressure of the flashing liquid. The result will be valid only for liquids stored at a pressure higher than the sat- uration vapor pressure. With this assumption the mass flow rate is given by
where
A is the area of the release,
C, is the discharge coefficient (unitless), p, is the density of the liquid (mass/volume), P is the pressure within the tank, and
PSat is the saturation vapor eressure of the flashing liquid at ambient temperature.
Example 4-7
Liquid ammonia is stored in a tank at 24'C and a pressure of 1.4 X lo6 Pa. A pipe of diameter 0.0945 m breaks off a short distance from the vessel (the tank), allowing the flashing ammonia to escape. The saturation vapor pressure of liquid ammonia at this temperature is 0.968 X 106 Pa, and its density is 603 kg/m3. Determine the mass flow rate through the leak. Equilibrium flashing con- ditions can be assumed.
Solution
Equation 4-91 applies for the case of equilibrium flashing conditions. Assume a discharge coeffi- cient of 0.61. Then
16J. M. Smith and H. C. Van Ness, Introduction to Chemical Engineering Thermodynamics, 4th ed. (New York: McGraw-Hill, 1987), p. 314.
"Hans K. Fauske, "Flashing Flows or: Some Practical Guidelines for Emergency Releases," Plantloper- ations Progress (July 1985), p. 133.
4-7 Flashing Liquids 155
For liquids stored at their saturation vapor pressure, P = P"; and Equation 4-91 is no longer valid. A much more detailed approach is required. Consider a fluid that is initially quies- cent and is accelerated through the leak. Assume that kinetic energy is dominant and that po- tential energy effects are negligible. Then, from a mechanical energy balance (Equation 4-l), and realizing that the specific volume (with units of volume/mass) v = llp, we can write
A mass velocity G with units of mass/(area time) is defined by
Combining Equation 4-93 with Equation 4-92 and assuming that the mass velocity is constant results in
Solving for the mass velocity G and assuming that point 2 can be defined at any point along the flow path, we obtain
Equation 4-95 contains a maximum, at which choked flow occurs. Under choked flow condi- tions, dGldP = 0. Differentiating Equation 4-95 and setting the result equal to zero gives
156
Solving Equation 4-97 for G, we obtain
The two-phase specific volume is given by
where
vfg is the difference in specific volume between vapor and liquid, v, is the liquid specific volume, and
f, is the mass fraction of vapor.
Differentiating Equation 4-99 with respect to pressure gives
But, from Equation 4-86,
d f , = -- C*
A H , dT' and from the Clausius-Clapyron equation, at saturation
d P - A H , d T T v k '
Substituting Equations 4-102 and z-101 into Equation 4-100 yields
- - dv - -- v:, TC,.
d P AH;
Chapter 4 Source Models
The mass flow rate is determined by combining Equation 4-103 with Equation 4-98:
Note that the temperature Tin Equation 4-104 is the absolute temperature from the Clausius- Clapyron equation and is not associated with the heat capacity.
Small droplets of liquid also form in a jet of flashing vapor. These aerosol droplets are readily entrained by the wind and transported away from the release site. The assumption that the quantity of droplets formed is equal to the amount of material flashed is frequently made.lx
IsTrevor A. Kletz, "Unconfined Vapor Cloud Explosions," in Eleventh Loss Prevention Symposium (New York: American Institute of Chemical Engineers, 1977).