page 0 Prepared by Associate Prof. Dr. Mohamad Wijayanuddin Ali Chemical Engineering Department Universiti Teknologi Malaysia
page 1 Isothermal flow of gas in a pipe with friction is shown in Figure 13. For this case the gas velocity is assumed to be well below the sonic velocity of the gas. A pressure gradient across the pipe provides the driving force for the gas transport. As the gas expands through the pressure gradient the velocity must increase to maintain the same mass flowrate. The pressure at the end of the pipe is equal to the pressure of the surroundings. The temperature is constant across the entire pipe length. Isothermal flow is represented by the mechanical energy balance in the form shown in Equation 44. The following assumptions are valid for this case.
page 2 is valid for gases, and form Equation 23, assuming constant f, and since no mechanical linkages are present. A total energy balance is not required since the temperature is constant.
page 3 Figure 13 Isothermal, non-choked flow of gas through a pipe.
page 4 Applying the above assumptions to Equation 44, and, after considerable manipulation (58) (59) (60) (61)
page 5 where G is the mass flux with units of mass/(area time), and, (62) The various energy terms in Equation 62 have been identified. A more convenient form of Equation 62 is in terms of pressure instead of Mach numbers. This form is achieved by using Equations 58 through 60. The result is (63) kinetic energy compressibility pipe friction
page 6 A typical problem is to determine the mass flux, G, given the pipe length ( L ), inside diameter ( d ), and upstream and downstream pressures ( P 1 and P 2 ). The procedure is as follows. 1.Determine the Fanning friction factor, f, using Equation 27. This assumes fully developed turbulent flow at high Reynolds number. This assumption can be checked later, but is usually valid. 2.Compute the mass flux, G, from Equation 63. Levenspiel has shown that the maximum velocity possible during the isothermal flow of gas in a pipe is not the sonic velocity as in the adiabatic case. In terms of the Mach number, the maximum velocity is (64)
page 7 This result is shown by starting with the mechanical energy balance and rearranging it into the following form. (65) the quantity -( dP / dL )--> when Ma --> 1/ . Thus, for choked flow in an isothermal pipe, as shown in Figure 14, the following equations apply.
page 8 (66) (67) (68) (69) (70) where G choked is the mass flux with unit of mass/(area/time), and (71)
page 9 Figure 14 Isothermal, choked flow of gas through a pipe. The maximum velocity reached is a/ .
page 10 For most typical problems the pipe length ( L ), inside diameter ( d ), upstream pressure ( P 1 ), and temperature ( T ) are known. The mass flux, G, is determined using the following procedure. 1.Determine the Fanning friction factor, f, using Equation 27. This assumes fully developed turbulent flow at high Reynolds number. This assumption can be checked later, but is usually valid. 2.Determine Ma 1 from Equation 71. 3.Determine the mass flux, G, from Equation 70.
page 11 The vapor space above liquid ethylene oxide (EO) in storage tanks must be purged of oxygen and then padded with 81 psig nitrogen to prevent explosion. The nitrogen in a particular facility is supplied from a 200 psig source. It is regulated to 81 psig and supplied to the storage vessel through 33 feet of commercial steel pipe with and ID of 1.049 inches. In the event of a failure of the nitrogen regulator, the vessel will be exposed to the full 200 psig pressure from the nitrogen source. This will exceed the pressure rating of the storage vessel. To prevent rupture of the storage vessel it must be equipped with a relief device to vent this nitrogen. Determine the required minimum
page 12 mass flow rate of nitrogen through the relief device to prevent the pressure from rising within the tank in the event of a regulator failure. Determine the mass flowrate assuming : a.an orifice with a throat diameter equal to the pipe diameter, b.an adiabatic pipe, and c.an isothermal pipe. Decide which result most closely corresponds to the real situation. Which mass flowrate should be used?
page 13 a.The maximum flowrate through the orifice occurs under choked conditions. The area of the pipe is The absolute pressure of the nitrogen source is The choked pressure from Equation 39 is, for diatomic gas
page 14 Choked flow can be expected since the system will be venting to atmospheric conditions. Equation 40 provides the maximum mass flowrate. For nitrogen, = 1.4 and The molecular weight of nitrogen is 28 lb m /lb-mole. Without any additional information, assume a unit discharge coefficient, C o = 1.0. Thus,
page 15 b.Assume adiabatic, choked flow conditions. For commercial steel pipe, from Table 1, = 0.046 mm. The diameter of the pipe in mm is (1.049 in)(25.4 mm/in) = 26.6 mm. Thus, From Equation 27
page 16 For nitrogen, = 1.4. The upstream Mach number is determined from Equation 57. with Y 1 given by Equation 46. Substituting the number provided,
page 17 This equation is solved by trial and error for the value of Ma. The results are tabulated below. This last value looks very close. Then from Equation 46
page 19 The pipe outlet pressure must be less than 49.4 psia to insure choked flow. The mass flux is computed using Equation 56
page 20 c.For the isothermal case, the upstream Mach number is given by Equation 71. Substituting the numbers provided The solution is found by trial and error
page 21 The choked pressure is, from Equation 67 The mass flowrate is computed using Equation 70.
page 22 The results are summarized in the following table 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 will always produce a larger value than the adiabatic pipe method, insuring a conservative safety design. The orifice calculation, however, is easier to apply, requiring only the pipe
page 23 diameter and upstream supply pressure and temperature. The configurational details of the piping are not required, as in the adiabatic or isothermal pipe methods. Also note that the choked pressures computed differ for each case, with a substantial difference between the orifice and adiabatic/isothermal cases. A choking design based on an orifice calculation might not be choked in reality due to high downstream pressures. Finally,note that the adiabatic and isothermal pipe methods produce results that are reasonably close. For most real situations the heat transfer characteristics cannot be easily determined. Thus, the adiabatic pipe method is the method of choice; it will always produce the larger number for a conservative safety design.
page 24 Liquids stored under pressure above their normal boiling point temperature present substantial problems due to 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 lower the temperature to the new boiling point. If m is the mass of original liquid, C p the heat capacity of the liquid (energy/mass deg), T o the temperature of the liquid prior to depressurization, and T b the depressurized boiling point of the liquid, then the excess energy contained in the superheated liquid is given by (72)
page 25 This energy vaporizes the liquid. If Δ H v is the heat of vaporization of the liquid, the mass of liquid vaporized, m v is given by (73) The fraction of the liquid vaporized is (74) Equation 74 assumes constant physical properties over the temperature range T o to T b. A more general expression without this assumption is derived as follows.
page 26 The change in liquid mass, m, due to a change in temperature, T is given by (75) Equation 75 is integrated between the initial temperature T o (with liquid mass m ) and the final boiling point temperature T b (with liquid mass m – m v ), (76)
page 27 (77) where and are the mean heat capacity and mean latent heat of vaporization, respectively, over the temperature range T o to T b. Solving for the fraction of the liquid vaporized, f v = m v / m, (78)
page 28 One lb m of saturated liquid water is contained in a vessel at 350°F. the vessel ruptures and the pressure is reduced to 1 atm. Compute the fraction of material vaporized using : a.the steam tables, b.Equation 74, and c.Equation 78.
page 29 a.The initial state is saturated steam at T o = 350°F. from the steam tables, The final temperature is the boiling point at 1 atm, or 212°F. At this temperature, and saturated conditions,
page 30 Since the process occurs adiabatically, H final = H initial and the fraction of vapor (or quality) is computed from, 14.59% of the mass of the original liquid is vaporized.
page 31 b.For liquid water at 212°F, From Equation 74
page 32 c.The mean properties for liquid water between T o and T b are Substituting into Equation 78 Both expressions work about as well when compared to the actual value from the steam table.
page 33 For flashing liquids composed of any miscible substances, the flash calculation is complicated considerably, since the more volatile components will lash preferentially. Procedures are available to solve this problem. Flashing liquids escaping through holes and pipes require very special consideration since two-phase flow conditions may be present. Several special cases need consideration. If the fluid path length of the release is very short (through a hole in a thin-walled container), non-equilibrium 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.
page 34 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 liquids stored at a pressure higher than the saturation vapor pressure. With this assumption the mass flowrate is given by (79)
page 35 where A is the area of the release, C o is the discharge coefficient (unitless), t is the density of the liquid (mass/volume), P is the pressure within the tank, and P sat is the saturation vapor pressure of the flashing liquid at ambient temperature
page 36 Liquid ammonia is stored in a tank at 24°C and a pressure of 1.4 10 6 Pa. A leak of diameter 0.0945 m forms in the tank, allowing the flashing ammonia to escape. The saturation vapor pressure of liquid ammonia at this temperature is 0.968 10 6 Pa and its density is 603 kg/m ³. Determine the mass flowrate through the leak. Equilibrium flashing conditions can be assumed.
page 37 Equation 79 applies for the case of equilibrium flashing conditions. Assume a discharge coefficient of 0.61.
page 38 for liquids stored at their saturation vapor pressure, P = P sat, Equation 79 is no longer valid. For this case the choked, two- phase mass flowrate is given by (80) where v is the specific volume with units if (volume/mass). The two-phase specific volume is given by (81)
page 39 where v f g is the difference in specific volume between vapor and liquid, v f is the liquid specific volume, and f v is the mass fraction of vapor. Differentiating Equation 81 with respect to pressure, (82) But, from Equation 74, (83)
page 40 and, from the Clausius-Clapyron equation, at saturation, (84) Substituting Equations 84 and 83 into Equation 82 yields, (85) The mass flowrate is determined by combining Equation 85, with Equation 80. (86)
page 41 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.
page 42 Propylene is stored at 25°C in a tank at its saturation pressure. A 1 cm diameter hole develops in the tank. Estimate the mass flowrate through the hole. At these conditions, for propylene. Δ H v = 3.34 x 10 5 J/kg v fg = 0.042 m 3 /kg P sat = 1.15 10 6 Pa C p = 2.18 10 3 J/kg K
page 43 Equation 86 applies to this case. The area of the leak is Using Equation 86
page 44 The case for evaporation of volatile from a pool of liquid has already been considered in Chapter 3. The total mass flowrate from the evaporating pool is given by where Q m is the mass vaporization rate (mass/time), M is the molecular weight of the pure material, K is the mass transfer coefficient (length/time), A is the area of exposure, P sat is the saturation vapor pressure of the liquid, R g is the ideal gas constant, and T L is the temperature of the liquid.
page 45 For liquids boiling from a pool, the boiling rate is limited by the heat transfer from the surroundings to the liquid in the pool. Heat is transferred : 1. from the ground by conduction, 2. from the air by conduction and convection, and 3. by radiation from the sun and/or adjacent sources such as fire.