Water hammer is an important problem that occurs in some liquidcontrol systems. It is defined as hydraulic shock that occurs when anon-viscous liquid flowing in a pipe experiences a sudden changein velocity, such as the fast closing of a valve.
202 FLUID FLOW Straight pipe 90" elbow (welding type) e@. - - 0 90" elbow Tee - EXAMPLE 4-6 Water Flow in Pipe System The system of Figure 4-3 1 consists of 125 ft of unknown size Sch. 40 steel pipe on the discharge side of a centrifugal pump. The flow rate is 500gpm at 75°F. Although the tank is located above the pump, note that this elevation difference does not enter into the pipe size-friction drop calculations. However, it will become a part of selection of the pump for the service (see Chapter 5). For quick estimate follow these steps: From Table 4-6, select 6 fps as a reasonable and usually econom- ical water rate. From Table 4-21, a 6-in. pipe has a velocity of 5.55 fps at 500gpm and a head loss of 0.72pd1OOft. The 5-in. pipe has a velocity of 8.02fps and might be considered; however 5-in. pipe is not commonly stocked in many plants and the velocity is above usual economical pumping velocities. Use the 6-in. pipe (rough estimate). Linear feet of straight pipe, L = 125 ft. From Figure 4-24, the equivalent length of fitting is 6-in.- 90" ell 2 14 ft straight pipe (using medium sweep elbow to repre- sent a welding elbow). Note that this is given as 6.5ft from Figure 4-25. This illustrates the area of difference in attempting to obtain close or exact values. Three 90" ell = 3 (14) = Le, = 42 ft (conservative) One tee = 1 (12) = Le, = 12 (run of standard tee) One 6 in. open gate valve = (1) (3.5) = Le, = 3.5 One sudden enlargement in tank at d/d' = 0; = loft, Figure 4-25 Total Le, = 67.5 ft small. 4. Neglect expansion loss at entrance to tank, since it will be so 5. No orifice or control valves in system. 6. From Table 4-21, at 500gpm, loss= 0.72psi/100eq. ft Total pressure drop from face to discharge flange on pump to nozzle connection on tank: AP= (125+67.5) [0.720/100]+0 AP = 1.386psi AP = 1.386psi (2.31 ft/psi) = 3.20ft water Note that a somewhat more accurate result may be obtained by following the detailed loss coefficients given in Figures 4- 1 4 4 - 18. However, most preliminary engineering design calculations for this type of water system do not warrant the extra detail. Figure 4-31 Pipe system for pipe sizing calculations (Example 4-6) Previous Page
Transcript
202 FLUID FLOW
Straight pipe 90" elbow (welding type) e@. - -
0
90" elbow Tee
-
EXAMPLE 4-6 Water Flow in Pipe System
The system of Figure 4-3 1 consists of 125 ft of unknown size Sch. 40 steel pipe on the discharge side of a centrifugal pump. The flow rate is 500gpm at 75°F. Although the tank is located above the pump, note that this elevation difference does not enter into the pipe size-friction drop calculations. However, it will become a part of selection of the pump for the service (see Chapter 5). For quick estimate follow these steps:
From Table 4-6, select 6 fps as a reasonable and usually econom- ical water rate.
From Table 4-21, a 6-in. pipe has a velocity of 5.55 fps at 500gpm and a head loss of 0.72pd1OOft. The 5-in. pipe has a velocity of 8.02fps and might be considered; however 5-in. pipe is not commonly stocked in many plants and the velocity is above usual economical pumping velocities. Use the 6-in. pipe (rough estimate). Linear feet of straight pipe, L = 125 ft. From Figure 4-24, the equivalent length of fitting is 6-in.- 90" ell 2 14 ft straight pipe (using medium sweep elbow to repre- sent a welding elbow). Note that this is given as 6.5ft from Figure 4-25. This illustrates the area of difference in attempting to obtain close or exact values.
Three 90" ell = 3 (14) = Le, = 42 ft (conservative) One tee = 1 (12) = Le, = 12 (run of standard tee) One 6 in. open gate valve = (1) (3.5) = Le, = 3.5 One sudden enlargement in tank at d /d ' = 0; = loft,
Figure 4-25 Total Le, = 67.5 ft
small. 4. Neglect expansion loss at entrance to tank, since it will be so
5. No orifice or control valves in system. 6. From Table 4-21, at 500gpm, loss= 0.72psi/100eq. ft
Total pressure drop from face to discharge flange on pump to nozzle connection on tank:
AP= (125+67.5) [0.720/100]+0
A P = 1.386psi
A P = 1.386psi (2.31 ft/psi) = 3.20ft water
Note that a somewhat more accurate result may be obtained by following the detailed loss coefficients given in Figures 4- 1 4 4 - 18. However, most preliminary engineering design calculations for this type of water system do not warrant the extra detail.
Figure 4-31 Pipe system for pipe sizing calculations (Example 4-6)
Previous Page
4.37 FRICTION PRESSURE DROP FOR COMPRESSIBLE FLUID FLOW 203
4.36 WATER HAMMER [23]
Water hammer is an important problem that occurs in some liquid control systems. It is defined as hydraulic shock that occurs when a non-viscous liquid flowing in a pipe experiences a sudden change in velocity, such as the fast closing of a valve. The kinetic energy of the moving mass of liquid upon sudden stoppage or abrupt change of direction is transformed into pressure energy, thereby causing an abrupt pressure rise in the system, often resulting in severe mechanical damage [7].
The pressure that can develop from the shock wave can be destructive to the containing system hardware, particularly in long pipe. Examples of conditions that can develop water hammer are
1. start, stop, or an abrupt change in a pump’s speed 2. power failure 3. rapid closing of a valve (usually a control valve, which can slam
shut in one or two seconds).
The magnitude of this shock wave can be expressed [23, 241:
hwh = “w b w ) /g
4660 (u,) - - g m K z
For water:
(4-110)
(4-111)
where
h,, =maximum pressure developed by hydraulic shock, ft of
u, =reduction in velocity, ft/s (actual flowing velocity, ft/s) water
g = acceleration due to gravity, 32.2ft/s2 K,, =ratio of elastic modulus of water to that of the pipe
material (see table below)
B, = ratio of pipe diameter (ID) to wall thickness a , = velocity of propagation of elastic vibration in the
Some typical K,, values of watedmetal are [23]
discharge pipe, ft/s.
Metal Khs ~ ~~
Copper 0.017 Steel 0.100 Brass 0.017 Wrought iron 0.012 Malleable cast iron 0.012 Aluminium 0.03
The time interval t, required for the pressure wave to travel back and forth in the pipe is
t, = 2L/a,, s (4-112)
L = length of pipe, ft (not equivalent ft) When the actual abrupt closing of a device to stop the flow
has a time shorter than t,, then the maximum pressure, h,,, will be exerted on the closed device and line.
Note that the value of h,, is added to the existing static pressure of the system.
4.37 FRICTION PRESSURE DROP FOR COMPRESSIBLE FLUID FLOW
VAPORS AND GASES
In contrast to non-compressible fluids, the densities of gases and vapors cannot be considered constant during flow through pipes as represented by the ideal gas law
V = R T / P or P V = R T (4-113)
P M , p=- RT
(4-1 14)
EXAMPLE 4-7 Water Hammer Pressure Development
Time interval for pressure wave travel:
An 8-in. process pipe for transfemng 2000gpm of methanol of sp gr = 0.75 from the manufacturing plant site to a user plant loca- tion is 2000 ft long, and the liquid is flowing at 10.8 ftls.
Maximum pressure developed (preliminary solution) when an emergency control valve suddenly closes:
(4-1 10)
t , = 2 L / a , = 2(2000)/4175 = 0.95 s (4-112)
If the shutoff time for the valve (or a pump) is less than 0.95 s, the water hammer pressure will be:
h n h = (uw) g
h,, = 4175(10.8)/32.2 = 1400ft of methanol
Since methanol has many properties similar to water: = (1400)/[(2.31)/0.75)] = 454psi hydraulic shock
4660
(1 $. KhSB,) ”* a, = Then total pressure on the pipe system
= 4175ft/s 4660
[1+0.01 (24.7*)]’” - -
*For 8-in. standard pipe, B, = 7.981/0.322 = 24.78
= 454 + (existing pressure from process/or pump)
This pressure level would most likely rupture an 8-in. Sch. 40 pipe. For a more exact solution, refer to specialty articles on the subject.
204 FLUID FLOW
EXAMPLE 4-8 Pipe Flow System With Liquid of Specific Gravity Other Than Water
This is illustrated by a line sizing sheet (Figure 4-32). Figure 4-33 represents a liquid reactor system discharging
crude product similar to glycol through a flow control valve and orifice into a storage tank. The reactor is at 350psig and 280" F with the liquid of 0.93 specific gravity and 0.91 centipoise viscosity. There is essentially no flashing of liquid across the control valve.
Flow rate: 11,000 l b h GPM actual = 11,000/(60) (8.33) (0.93) = 23.7 Design rate = 23.7 (1.05) = 25 gpm
From Table 4-10, selected velocity = 6fps.
Estimated pipe diameter, d = (0.408 Q/U)'/'
= [(0.408)25/6]'/' = 1.3in.
Try ll/zin. (ID = 1.61), since 11/4in. (ID = 1.38) is not stocked in every plant. If it is an acceptable plant pipe size, then it should be considered and checked, as it would probably be as good pressure drop-wise as the llh-in. The support of 11/4-in. pipe may require shorter support spans than the ll/zin. Most plants prefer a minimum of 1 '/'-in. valves on pressure vessels, tanks, and so on. The valves at the vessels should be 11/2in. even though the pipe might be 11/4in. The control valve system of gate and globe valves could very well be 1'/4in. For this example, use llh-in. pipe, Sch. 40: Linear length of straight pipe, L = 254ft. Equivalent lengths of fittings, valves, and so on.
Estimated Fittings Type (From Figure 4-24)
10 8 4
11/f -90" elbows 4'(10) =40
IT/# - Gate valves 1'/2)) - Tees 3'(8) = 24
l'(4) = 4 68' (Rounded total to 75')
No expansion or contraction losses (except control valve). 5. Pressure drop allowance assumed for orifice plate = 5 psig.
Control valve loss will be by difference, trying to maintain minimum 60% of pipe friction loss as minimum drop through valve, but usually not less than 10 psi.
6. Reynolds number, Re = 50.6Qp/dp
50.6(25)[0.93(62.3)] (1.61)(0.91)
- -
(4-27)
= 50,025 (turbulent)
7. From Figure 4.13, E / d = 0.0012 for I1/2-in. steel pipe. From Figure 4-5, at Re = 50,025, read f = 0.021 or use Eq. (4-35).
8. Pressure drop per 100 ft of pipe:
APlOOft =0.0216fpQ2d5 (4-65)
= 0.0216 (0.021) (62.3) (0.93) (25)'/( 1.61)5
= 1.52psi/lOOeq.ft
9. Total pressure drop The control valve must be sized to take the residual pressure
drop, as long as it is an acceptable minimum. Pressure drop accounted for:
Total psi drop= (245+75)(1.52/100)+5 = 1Opsi
Drop required across control valve:
Reactor = 350 psig Storage = Opsig
Differential = 350 psi A P = 1Opsi (sys. friction)
Control Valve A P = 34Opsi
Note that this control valve loss exceeds 60% of this system loss, since the valve must take the difference. For other systems where this is not the situation, the system loss must be so adjusted as to assign a value (see earlier section on control valves) of approxi- mately 10-20 psi or 25-60% of the system other than friction losses through the valve. For very low pressure systems, this minimum value of control valve drop may be lowered with the sacrifice of sensitive control.
where
V = molar volume, ft3/(lbmol) p = density, lb/ft3
M, = molecular weight, lb/(lb mol) P = pressure, lb,/ft2 T = temperature, O R = (" F + 460) R = universal gas constant, 1545 (ft lb,) / [" R (lbmol)] .
The effect of a change in pressure or temperature on the density of a gas causes the determination of pressure loss experienced by flowing compressible fluids in conduits, orifices, and nozzles to differ from that for non-compressible fluids.
The first step in analyzing the flow of a compressible fluid is to determine which type of polytropic process is being considered,
especially whether it is an isothermal or an adiabatic process. In an isothermal process, the gas flows at a constant temperature through a pipe and an ideal gas follows the relationship
PV = constant (isothermal) (4-1 15)
To attain a constant temperature, the gas must exchange heat with its surroundings which act as a uniform temperature heat sink. The isothermal flow assumption is typically applied to most uninsu- lated lines carrying gases flowing at a low velocity. The lack of insulation and the low velocity allow the gas to approach thermal equilibrium with its surroundings. For example, most plant air and gas supply lines may be treated as isothermal fluids. However, this does not apply when the gas travels at high velocities and in insulated pipelines.
4.37 FRICTION PRESSURE DROP FOR COMPRESSIBLE FLUID FLOW 205
Total Pipe System Friction AP= [(329) (1.52/1 OO)] + 5' = 10 psi for friction; *orifice
Total Loss, Feet Liquid =350 (2.31 ft/psi) (1/0.93) =869ft of Liquid
Checked by: Date:
Figure 4-32 Line sizing sheet for example problem (Example 4-8).
Gases and vapors flowing through measuring devices such as orifices, through pipes at high velocities or in insulated pipelines cannot maintain a constant temperature. Thus the condition should not be treated as an isothermal process. This is because the effect of velocity acceleration due to changes in flow cross-sectional area and the effect of friction create changes in temperature. High gas velocities and insulation prevent the compressible fluid from attaining thermal equilibrium with its surroundings. When heat cannot be exchanged with the surroundings, compressible flow conditions may be assumed to be essentially an adiabatic process, and ideal fluids then follow the relationship
PVk = constant (adiabatic) (4-1 16)
where
Cp = specific heat at constant pressure C, = specific heat at constant volume
Comparing Eqs (4-115) and (4-116), it can be seen that the isothermal process is equivalent to having an exponent k of unity for the volume term. The isothermal and adiabatic processes are special cases of polytropic fluid flow process. The isothermal process repre- sents thermal equilibrium with the environment, while the adiabatic process illustrates no heat exchange with the environment. Actual flow processes can often be approximated by these extremes, but
k = specific heat ratio = C,/C, = Cp/(C, - R )
R = universal gas constant, 1544/M, (ft.lbf)/[" R.lbmol].
206 FLUID FLOW
Flow Orifice Gate Control Gate plate valve valve valve
I / I
Gate valve Globe valve
Gate valve
Reactor at 350 psig
Crude product storage tank at atmospheric
pressure
Figure 4-33
there are other situations which can be represented by the general relationship
Liquid flow system (Example 4-8)
PV" = constant (polytropic) (4-1 17)
where I I = polytropic coefficient. The exponent n can be substi- tuted for k in Eq. (4-1 16) and can have a value between zero and infinity, depending on the amount of heat added or removed from the flowing system. In the special polytropic process known as the adiabatic case. heat is neither added nor removed, so that n = k . In the isothermal process, sufficient heat is added to a flowing system to maintain it at constant temperature and n is unity. For other polytropic processes, the value of n is in the range
(Isothermal) 1 < n < k (adiabatic)
4.38 COMPRESSIBLE FLUID FLOW IN PIPES
The flow of compressible fluids (e.g., gases and vapors) through pipelines and other restrictions is often affected by changing conditions of pressure, temperature, and physical properties. The densities of gases and vapors vary with temperature and pressure (PV = constant). Conversely, in adiabatic flow, that is no heat loss (PVk = constant), a decrease in temperature occurs when pres- sure decreases, resulting in a density increase. At high pressures and temperatures, the compressibility factor can be less than unity, which results in an increase in the fluid density.
The condition of high pressure drop (AP) in compressible flow frequently occurs in venting systems, vacuum distillation equip- ment, and long pipelines. Some design situations involve vapor flows at very high velocities resulting in A P > 10% of the upstream pressure. Such cases are vapor expanding through a valve, high speed vapor flows in narrow pipes, and vapors flowing in process lines under vacuum conditions. A P , in many cases, is critical and requires accurate analysis and design. For instance, the inlet pipe I P , of a safety relief valve should not exceed 3% of the relief valve set pressure (gauge) at its relieving capacity for stable oper- ation. This limit is to prevent the rapid opening and closing of the valve, resulting in lowered fluid capacity and subsequent damage of the valve seating surfaces. Conversely, the tail pipe or vent line of a relief valve should be designed such that AP < 10% of the relieving pressure, that is set pressure + over pressure in gauge. Figure 4-34 shows a typical tail pipe and relief valve connected to a heat exchanger.
4.39 MAXIMUM FLOW AND PRESSURE DROP
Determining the maximum fluid flow rate or pressure drop for process design often has the dominant influence on density. As
6" -i/-
Shell'fluid in Tube fldd out
Figure 4-34 pipe.
pressure decreases due to piping and component resistance, the gas expands and its velocity increases. A limit is reached when the gas or velocity cannot exceed the sonic or critical velocity. Even if the downstream pressure is lower than the pressure required to reach sonic velocity, the flow rate will still not increase above that evaluated at the critical velocity. Therefore, for a given pressure drop (AP), the mass discharge rate through a pipeline is greater for an adiabatic condition (Le., insulated pipes, where heat transfer is nil) than the rate for an isothermal condition by as much as 20%. There is, however, no difference if the pipeline is more than 1000 pipe diameters long [25]. In practice, the actual flows are between the two conditions, and the inflow rates are often below 20% even for lines less than 1000 pipe diameters.
Fluid flow through a heat exchanger, relief valve, and tail
4.40 SONIC CONDITIONS LIMITING FLOW OF GASES AND VAPORS
The flow rate of a compressible fluid in a pipe with a given upstream pressure will approach certain maximum rate that it cannot exceed even with reduced downstream pressure. The maximum possible velocity of a compressible fluid in a pipe is sonic (speed of sound) velocity, as:
(4-1 I S )
us = [(Cp/C.) (32.2)(Z)(1544/Mw) (460+t)]"* (4-119)
= 68.1 [(CJC.) P' /P] ' '~ , ft /s (4-120)
where
k = ratio of specific heat at constant pressure to specific heat at constant volume (CJC,)
Cp = specific heat at constant pressure C, = specific heat at constant volume R = individual gas constant = M R / M , = 1544/M,,
M , = molecular weight of the gas M R = universal gas constant = 1544
T = temperature of gas, R = (460 + t ) t = temperature, " F
4.40 SONIC CONDITIONS LIMITING FLOW OF GASES AND VAPORS 207
P’ = pressure, psia us = sonic or critical velocity of flow of a gas, ft/s VI = specific volume of fluid, ft3/lb (m3/kg) at T and P’ g = acceleration of gravity = 32.2 ft/s2 Z = compressibility factor.
In SI units,
(4-121)
where
R = Individual gas constant = 8314/Mw, J/kgK M , = molecular weight of the gas y = ratio of specific heat at constant pressure to specific
at constant volume = C,/C, T = absolute temperature, K = 273 + t t = temperature, O C
P‘ = pressure, N/m2 p‘ = pressure, bar 3 = specific volume of fluid, m3/kg W = fluid flow rate, kg/h p = density of fluid, kg/m3 at T and P’ (or p’).
heat
Thus the maximum flow in a pipe occurs when the velocity at the exit becomes sonic. The sonic location may be other than the exit, can be at restrictive points in the system, at expansion points in the system (e.g., an expander from a 2-in. to 3-in. pipe or a 6 in.-pipe into a larger header), or at control/safety relief valves.
A recommended compressible fluid velocity for trouble-free operation is u 5 0 . 6 ~ ~ . The design criteria for compressible fluid process lines (e.g., carbon steel) are shown in Table 4-22.
Shock waves are stationary, however if they were to travel, it would be at sonic velocity and exhibit a near discontinuity in pressure, density, and temperature, and a great potential exists for damage from such waves [26]. A discussion of shock waves is beyond the scope of this chapter.
Velocity considerations are important in rotating or recip- rocating machinery systems, because if the compressible fluid velocity exceeds the speed of sound in the fluid, shock waves can be set up and the results of such conditions are much different than the velocities below the speed of sound. The ratio of the actual fluid velocity to its speed of sound is called the Mach number [27].
The velocity of sound at 68 F in air is 1126 ft/s. For any gas, the speed of sound is
(4-122)
where
k = ratio of specific heat of gas, at constant pressure to that at constant volume,
This sonic velocity occurs in a pipe system in a restricted area (e&, valve, orifice, venturi) or at the outlet end of the pipe (open-ended), as long as the upstream pressure is high enough. The physical properties in the above equations are at the point of maximum velocity.
TABLE 4-22 Recommended Fluid Velocity and Maximum AP for Carbon Steel Vapor Lines
Type of Service Recommended Maximum, Velocity (ft/s) AP
(psi/lOOft)
General Recommendation Fluid pressure in psig Su b-atmospheric O < P 5 5 0 5 0 < P 1 1 5 0 150 < P 5 200 200 < P 5 500 P > 500 Tower overhead Pressure (P z 50 psia) Atmospheric Vacuum (P < 10 psia) Compressor piping suction Compressor piping discharge Gas lines with battery limits Refrigerant suction lines Refrigerant discharge lines
3. Exhaust steam lines ( P > atmosphere) Leads to exhaust header
Relief valve, entry point at silencer
2. High pressure steam lines
4. Relief valve discharge
40-50 60-1 00
75-200 100-250
15-35 35-60
200 250
167 117
0.5 vs vs
0.18 0.15 0.30 0.35 1 .o 2.0
0.2-0.5
0.05-0.1 0.5 1 .o
0.5
0.25 0.40 1 .o 1.5
1 .o 0.5 0.5
1.5
With a high velocity vapor flow, the possibility of attaining critical or sonic flow conditions in a process pipe should be investi- gated. These occur whenever the resulting pressure drop approaches the following values of AP as a percentage of the upstream pressure [6] where the properties are evaluated at the condition of sonic flow.
This applies regardless of the downstream pressure for a fixed upstream pressure. This limitation must be evaluated separately from pressure drop relations, as it is not included as a built-in limitation.
Sonic velocity will be established at a restricted point in the pipe or at the outlet, if the pressure drop is great enough to establish the required velocity. Once the sonic velocity has been reached, the flow rate in the system will not increase, as the velocity will remain at this value even though the fluid may be discharging into a vessel at a lower pressure than that existing at the point where sonic velocity is established. AP can be increased by continuing to lower the discharge pressure. But no additional flow rate will result. The usual pressure drop equations do not hold at the sonic velocity, as in an orifice. Conditions or systems exhausting to atmosphere (or vacuum) from medium-to-high pressures should be examined
208 FLUID FLOW
TABLE 4-23 Approximate k Values for Some Common Gases (68" F, 14.7 psia)
Gas
Chemical Approximate
Symbol Weight k(Cp/&) Formula or Molecular
Acetylene (ethyne) CZHZ 26.0 1.30 Air - 29.0 1.40 Ammonia N H3 17.0 1.32 Argon A 39.9 1.67
for critical flow, otherwise the calculated pressure drop may be in error.
All flowing gases and vapors (compressible fluids) including steam (which is a vapor) are limited or approach a maximum in mass flow velocity or rate, that is lb/s or lb/h (kgis or kg/h), through a pipe depending upon the specific upstream or starting pressure. This maximum rate of flow cannot be exceeded regardless of how much the downstream pressure is further reduced. The mean velocity of fluid flow in a pipe by continuity equation is [4]
0.0509Wv 0.0509W u = or ~ , f t ls (4- 124)
d2 Pd2
In SI units,
354wv 354w or u=- pd2 ' m/s ?J= ~
d2 (4- 125)
where
d = internal pipe diameter, in. (mm) W = rate of flow, l b h (kgh) v = specific volume of fluid, ft3/lb (m3/kg) p = fluid density, lb/ft3(kg/m3).
This maximum velocity of a compressible fluid in a pipe is limited by the velocity of propagation of a pressure wave that travels at the speed of sound in the fluid 141. This speed of sound is specific for each individual gas or vapor and is a function of the ratio of specific heats of the fluid. The pressure reduces and the velocity increases as the fluid flows downstream through the pipe, with the maximum velocity occurring at the downstream end of the pipe. When or if the pressure drop is great enough, the discharge or exit or outlet velocity will reach the velocity of sound for that fluid.
If the outlet or discharge pressure is lowered further, the pres- sure upstream at the origin will not detect it because the pressure
wave can only travel at sonic velocity. Therefore, the change in pressure downstream will not be detected upstream. The excess pressure drop obtained by lowering the outlet pressure after the maximum discharge has been reached takes place beyond the end of the pipe [4]. This pressure is lost in shock waves and turbulence of the jetting fluid. See [26, 28-30] for further expan- sion of shock waves and detonation waves through compressible fluids.
In the case of a high pressure header, the flow may be sonic at the exit. Therefore, it is often necessary to check that the outlet pressure of each pipe segment is not critical. If P, is less than terminal P2, the flow is subcritical. If, however, P, is greater than P2, then the flow is critical. Although, it may be impractical to keep the flow in high pressure subheaders below sonic, Mak [31] suggests that the main flare header should not be sized for critical flow at the outlet of the flare stack. This would obviate the undesir- able noise and vibration resulting from sonic flow. Crocker's [32] equation for critical pressure can be expressed as:
P,=-(-) G RT ' I2 ,psia 11,400d k [k + 13
or
P, = 2.45 10-3 (3 ( &)0'5
where
R = (z) 29sp gr , molar gas constant
Molecular weight of gas Molecular weight of air
sp gr =
2 = compressibility factor d = internal pipe diameter, in. M , = fluid molecular weight G = fluid flow rate, lb/h T = fluid temperature, ' R p = fluid density, lb/ft3.
(4-126)
(4- 127)
(4-128)
4.41 THE MACH NUMBER, MA
The Mach number, Mu, is the velocity of the gas divided by the velocity of the sound in the gas and can be expressed as:
Mu = u /u , (4-129)
The exit Mach number for com ressible isothermal fluid has been shown to be Mu # 1, but 1 , Jd where k is the ratio of the fluid- specific heat capacitites at constant pressure to that at constant volume. Table 4-23 shows the k values for some common gases. The following cases are such:
1. Mu < 1/&, the flow is subsonic 2. Mu = I/&, the flow is sonic 3. Mu > 1/&, the flow is supersonic
(4- 129a) (4- 129b) (4- 129c)
4.43 FLOW RATETHROUGH PIPELINE 209
For expanding gas flow, v2 # v l ; with horizontal pipe, Z , = Z,. Hence, the differential form of Bernoulli's equation can be expressed as
Case 3 is produced under certain operating conditions in the throt- tling processes (e.g., a reduction in the flow cross-sectional area). Kirkpatrick [33] indicates that there is a maximum length for which continuous flow is applied for an isothermal condition, and this corresponds to Ma = l/&. The limitation for isothermal flow, however, is the heat transfer required to maintain a constant temper- ature. Therefore, when Ma < 1/&, heat must be added to the stream to maintain constant temperature. For M u > l / & heat must be rejected from the stream. Depending on the ratio of specific heats, either condition could occur with subsonic flow. Therefore, to maintain isothermal flow during heat transfer, high temperatures require high Mach numbers and low temperatures require low Mach numbers.
d P 1
P s c -_ - - -vdv+e, (4- 132)
Where p is constant and the velocity head is
" ef = K-
2SC (4-133)
The mass flowrate through the pipe is
G = pvA 4.42 MATHEMATICAL MODEL OF COMPRESSIBLE
The derivations of the maximum flowrate and pressure drop of Compressible isothermal flow are based on the following assump- tions:
ISOTHERMAL FLOW or
G - = p v = constant A
(4- 134)
Because both density and velocity change along the pipeline, Eq. (4- 134) can be expressed in differential form as:
pdv + vdp = 0 (4-135)
In expanding gas flow, the pressure and density ratio are constant:
1. Isothermal compressible fluid. 2. No mechanical workdone on or by the system. 3. The gas obeys the perfect gas laws. 4. A constant friction factor along the pipe. 5. Steady state flow or discharge unchanged with time.
Figure 4-35 illustrates the distribution of fluid energy with work done by the pump and heat added to the system. Table 4-6 gives friction factors for clean commercial steel pipes with flow in zones of complete turbulence.
P d P _ - _ - P dP
and, with respect to initial condition, P, and p,
P d P PI - - _ - - - - P dP Pi
From Eq. (4- 135)
(4-136)
(4-137) 4.43 FLOW RATE THROUGH PIPELINE
Bernoulli's equation for the steady flow of a fluid can be expressed as:
dv = -Ed, P
(4-138)
(4- 130) and
d P dp P P _ - - - (4-139) where a (the dimensionless velocity distribution) = 1 for turbulent
or plug flow. Assuming no shaft work is done (e.g., SW, = 0), then Eq. (4-130) becomes therefore
(4-13 1) P2 - P, v 2 - v 2 g + - ( z 2 - z , ) + ef = 0
P 2gc gc (4-140)
Substituting Eq. (4-139) into Eq. (4-138) gives 1 Point 1 'Point 2
I I I I I I
+q (heat added to fluid)
(4-141)
Therefore
dv = -v ($) (4- 142) I ! P. I I I L Z
Substituting Eq. (4-142) into Eq. (4-132) gives
(4-143) nltlIm I - pump on fluid) . 1
(4-144) Figure 4-35 Energy aspects of a single-stream piping system.
210 FLUID FLOW
Rearranging Eq. (4-144) gives 4.44 PIPELINE PRESSURE DROP (AP)
- d P = - p
From the mass flowrate, G = p u A
G u = - PA
If A P due to velocity acceleration is relatively small compared
with the frictional drop, then In (2) may be neglected. Therefore
Eq. (4-153) becomes
(4- 145)
0.5 (4-146) G = 1 3 3 5 . 6 d 2 [ L (p,)] P; -Pi
KTotal
Substituting Eq. (4-146) into Eq. (4-145) gives
G2 K d P
g c A 2 p 2 p ( ? - P )
In terms of the initial conditions of pressure and density, that is substituting Eq. (4-140) into Eq. (4-147) yields
Putting C = 1335.6d2
- G2 - - -~ p1 P12-P22
(4-147) -dP = -
c2 KTotal
(4- 155)
(4- 1 56)
(4- 157)
That is,
P1G2KTotal (4-158) (4-148)
PI2 - P22 = PI c2
Integrating Eq. (4-148) between points 1 (inlet) and 2 (outlet) gives
which is Therefore, the pressure drop
(4-159)
(4- 160)
(4-161) Therefore, for maximum flowrate through the pipe,
G = [ ( A2p1gc ) (y)]'" where KTotal is the total resistance (loss) coefficient due to friction, fittings, and valves.
(4-151) In SI units, KTotal + In (2)
0.5
G = 2.484 x 10-4d2 [ ( 1 L
KTotal = fD - + K , (pipe fittings + valves) (4-152) (4-154) D K , (pipe fittings + valves) is the sum of the pressure loss coef-
ficient for all the fittings and valves in the line. Expressing the maximum fluid rate in pounds per hour, Eq. (4-151) becomes
Putting C , = 2.484 x 10-4d2
G 2 p, P,2-P22 _ - -~ C? - ~ T o r a l PI
0.5
G = 1335.6d2 [ ( p 1 ) (y)] , lb/h That is, KTotal f In (2)
P12 - P22 = PI G2KTotal PIC,*
(4- 153)
(4-162)
(4-163)
(4-164)
In SI units, (4-165)
0 5
G = 2.484 x
d 2 [ ( (p')2i (p')2 )] , kg/s Therefore, the pressure drop
A P % PI - P2 (4- 166) (4-154)
That is, where d = pipe internal diameter, mm
p1 = upstream gas density, kg/m3 p ; = upstream gas pressure, bara p ; = downstream gas pressure, bara.
(4-167)
Table 4-24 shows the conditions at the pipe exit as a function of the Mach number.
4.45 CRITICAL PRESSURE RATIO 211
TABLE 4-24 Conditions at the Pipe Exit as a Function of the Mach Number
y = ratio of specific heat at constant pressure to specific heat at constant volume
= (Cp/CJ. The maximum flux or maximum mass flow of the gas at sonic conditions is
Isothermal Flow Adiabatic Flow
1 G 2 k P + - - =constant P = p x constant - _ 2 P2 (k-1) P
Subsonic Flow
PZ P, x Ma, / M a , 3 J 2 + (k - 1) Mal Ma, 2 + (k - 1 ) Ma,’ (4-172)
T2 Tl where
W= mass flow rate, lb/s A = cross-sectional flow area, ft2 G = mass flux, lb/(ft2 s) Po= pressure at source condition, psia To= temperature at source condition, O R k = specific heat ratio constant.
Recently, Kumar [34] developed a method using thermodynamic principles to determine the status of flow (e.g., whether choking flow exists or not), ( A P / P o ) c r , and the flow rate. His method removes the use of plots as generated in Crane Manual [4], which have few limitations. For an adiabatic compressible fluid flow, he showed that
2 + ( k - 1 ) M a 1 2 2 + (k - 1) Ma2’ Tl
Ma, 2 + (k - 1) Ma,’ P2 P1 x Ma, / M a , d 2 + ( k - 1 ) M a 1 2
V2 vl x Ma,/Ma, Vl x P l I P 2
Critical or Sonic Flow
p2 Pl x Ma, x f i 2 + (k - 1) M a l 2 k+l
2+ (k- 1) Mal2 T2 Tl ” (k-I)
PZ p1 x Ma, x df (k+ 1) P1 M a 1 / 2 + (k - 1 ) M a l 2
L J “2 VS
(4- 1 73) Ma. Ma, Ma=,
and
[ O S ( y + 1) Mu1’]”’ r = ( ? ) = (4- 174)
cr [ 1 + 0.5 ( y - 1) Mu, ’ 1 The hypothetical Pipe length, L2is such that its inlet Ma, is the same as the exit from the actual pipe
The critical expansion factor is 4.45 CRITICAL PRESSURE RATIO
The maximum attainable mass flux for given supply conditions is when $ is a maximum, and is represented by
._ / K ( I + r ) (4-175)
(4-176)
rcr = { 2 [ K + 2 l n ( t ) ]
and the mass flow rate at critical condition is
Therefore, the pressure P, causing the maximum flux can be found by differentiating 4’ with respect to P and equating the result to zero, which gives
where
D = pipe internal diameter, mm L = pipe length, m
PA = ambient pressure, kPaa Po = stagnation upstream pressure, kPaa P I = pressure at inlet tip of the pipe, kPaa P2 = pressure at outlet tip of the pipe, kPaa
Mu, = Mach number at inlet tip of the pipe Mu, = Mach number at outlet tip of the pipe
y = isentropic ratio of specific heat at constant pressure to
K = loss coefficient Vo = specific volume at upstream stagnation point, m3/kg
Y,, = critical expansion factor, dimensionless
specific heat at constant volume
r = (P,/P0),, = overall critical pressure ratio, dimensionless
The critical pressure ratio P J P , is given by
(4-171)
For y = 1.4, P J P , = 0.528. For other values of y , the value of the critical pressure ratio lies in the approximate range 0.5 to 0.6 where P, = upstream pressure .~ P, = critical pressure W = mass flow-rate, kg/h.
212 FLUID FLOW
EXAMPLE 4-9 Case Study
The vapor (C3, C, and C,) from the debutanizer unit C1007 oper- ating at 17.6 bara in Figure 4-23 is cooled via an air cooler E-1031 to the accumulator V1008. The overhead gas line 8"-P10170- 3 IOIC-P is 84.7 m long, and the debutanizer boil-up rate is 17 kg/s (1738 m3/h). Calculate the pressure drop along the 8-in pipe to the air cooler condenser E-1031. The following are the other data obtained from the piping isometrics, piping data sheets, and fluid characteristics:
Operating temperature = 86 O C Fluid density = 35.2 kg/m3 Ratio of specific heats y = (C,/C\) = 1.1 1 Kinematic viscosity = 0.2cSt = 0.2 x 1O-'h/s2.
Fittings Number
90"ell ( r / R = 1.5) 5 Ball valve 2 Tee (straight thru) 3
SolLltion
Dynamic viscosity = density x kinematic viscosity
= 35.2 x 0.2 x
= 0.00704 x kg/ms
= 0.00704cP
The average molecular weight of the overhead vapor is ~ ~~~
Composition Molecular weight Percentage in the Average molecular (kglkmol) vapor phase (%) weight
Assume compressibility factor 2 = 0.958 An 8-in. (203.2 rnm) pipe size Sch. 40 (ID = 202.7 mm) Friction factor fT for %in. Sch. 40 CS material = 0.014 (Table 4-6) Assuming that the flow condition of the vapor through the 8-in.
Area of pipe pipe is isothermal.
A = m- ( I D ) ' / 4
= 7~ (0.2027)'/4
= 0.0323 m'
The velocity of gas in the line is
G = ~ L ~ A
where A = pipe area, m2 G = mass flow rate, kg/s L' = fluid velocity. m / s p = fluid density, kg/m'.
or
u = G/(P A )
17 (35.2 x 0.0323)
- -
= 14.95 m/s
Sonic velocity of vapor in the line is
=/(0.958) (1.11) (SI (359.15),
= 238.63 m/s
The inlet Mach number M a , is
v 14.95 Ma - - = - us 238.63
= 0.0626
1 -
(4-1 21)
Therefore, the flow of gas through the pipe is subsonic, since the inlet Mach number is less than 1. Gas Reynolds number is
W Re = 354 -
dP - (354) (17) (3600)
(202.7) (0.00704) -
= 15,181,975 (fully turbulent)
= 15.2 x lo6
Relative pipe roughness is
E - D
1 ~
J5;c where
A =
A =
1 -- a- - -
0.000046 202.7
= 0.000226 - -
(4-28)
(4-35)
0 9
3.7
0'00027 + ( 6'7 = 6,32408 x
3.7 15,18 1,975
1 5.02 15,181.975
log (6.32408 x lo-,)
16.8098
f , = 0.00354
(continued)
4.45 CRITICAL PRESSURE RATIO 213
EXAMPLE 4-9--(continued) The Darcy friction factor fD = 4 f c
fD = 4 (0.00354)
= 0.0142
Therefore, the pressure drop
A P Z PI - P'
= 0.335 bar
(4-166)
Using Darby's 3-K method The process is assumed to be isothermal, therefore outlet tempera-
Re = 15,181 975 (turbulent) Nominal pipe diameter of 8 in., D, = 203.2 mm
ture T2 is
T2 = Tl, K (4-74)
T2 = 86" C
= (273.15 + 86) K
= 359.15 K Fittings n K1 nK, 6 n h Kd K,
Density of the vapor at the exit is 90" ell 5 800 4000 0.071 0.355 4.2 1.1543 ( r / D = 1.5) Ball valve 2 300 600 0.017 0.034 4.0 0.1069
Outlet pressure P, is The exit Mach number M a , = l/&
(4- 165) Ma, = 1 / m = 0.949
where Rearranging Eq. (4- 165) gives
Cl = 2.484 x d2 (4-162) [Pl? - PlG'KTotd]o'5 - P 2 = 0
PI C12
(4- 129b)
(4-168)
(4-163) Equation (4-168) involves a trial-and-error computation using a guess value for P2. This is substituted into Eq. (4-168) until the left side gives a value of zero. The Excel spreadsheet with a Solver add- in tool is the most convenient computational tool for Eq. (4-168). Equation (4-168) is set to zero using a guess value of P,. The procedure involves setting the quadratic Eq. (4-168) to zero; with a guess value of the outlet pressure and the Solver determine the outlet pressure after a set of iterations. The Excel spreadsheet (Example 4 - 9 . ~ 1 ~ ) has been developed to determine the pressure drop of a compressible isothermal flow fluid using Eq. (4-168). Figure 4-36 shows the snapshots of the Excel spreadsheet calcula- tion of Example 4-9.
= 2.174 G2 17'
C12 - (2.484 x _ -
x 202.7*)'
(4-165)
(17.6) (2.774) (8.5152) 35.2
= (17.6' -
= 17.26bara
214 FLUID FLOW
Figure 4-36 The Excel spreadsheet snapshot of Example 4-9.
4.45 CRITICAL PRESSURE RATIO 215
Figure 4-36(continued)
216 FLUID FLOW
Figure 4-36-(conrinued)
4.45 CRITICAL PRESSURE RATIO 217
EXAMPLE 4-10 Propane vapor at 90°F and an upstream pressure PI = 20psig flows at a rate of 24,000 lbk in an 800-ft long, 6-in. Sch. 40 horizontal carbon steel pipe. Under these conditions, the viscosity p, = 0.0094cP and the gas compressibility factor 2, = 0.958. Calculate the total pressure drop under isothermal flow conditions. Check for critical flow.
Solution Since the pipe is lcng, assume an isothermal condition for the compressible vapor flow.
The density of propane at 90" F and pressure of 20 psig is
p=- MWP 10.72 T
= ( li47; i::o) = 0.25891b/ft3
Velocity of the gas is The 6-in. Sch. 40 pipe size (ID = 6.065 in.) Area of pipe
A = 7~ (ID)*/4
= 7~ (0.5054)'/4
= 0.2006 ft2
The velocity of gas in the line is
G = p v A
where
A = pipe area, ft2 G = mass flow rate, lb/s
= fluid velocity, ft/s p = fluid density, lb/ft3.
or
U = G/(P A) 24,000
(0.2589 x 0.2006 x 3600) - -
= 128.36ft/~
Sonic velocity is
U s = /&q, ft/s
0.958 x 1.15 x 32.174 x 144 x 34.7 0.2589
- -
=827.11ft /~
(4-118)
Sonic flow would occur at the end of the pipe and not where the pressure is 20 psig. The inlet Mach number M a , is
M a , = - (4-129) U
us
128.36 827.11
- - - = 0.1552
Since the Mach number is less than 1, the flow of propane through the pipe is subsonic. Reynolds number Re:
24,000 6.065 x 0.0094
Re = 6.31
= 2,656,329
= 2.66 x lo6
Friction factor, f :
E 5.02 3.70 Re
--410g 1 -- JK
where
0.9 A = - + ( $ ) &ID
3.7 E = pipe roughness, ft D = pipe internal diameter, ft.
