01 POWER ISLAND / 01 CCPP / V. Ganapathy-Industrial Boilers and HRSG-Design (2003)
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7
Fluid Flow, Valve Sizing,
and Pressure Drop Calculations
7.01Sizing flow meters; discharge coefficients for orifices, venturis, and nozzles; permanent pressure drop across flow meters; correcting steam flow readings for different operating conditions
7.02Sizing orifices for water flow measurement
7.03Sizing orifices for steam flow measurement
7.04Significance of permanent pressure drop in flow meters; cost of permanent pressure drop across flow meters
7.05Converting pitot tube readings to air velocity, flow in ducts
7.06Sizing safety valves for boilers; ASME Code procedure
7.07Relieving capacities for steam service; orifice designations for safety valves; relating set and accumulated inlet pressures
7.08Selecting safety valves for boiler superheater; actual and required relieving capacities
7.09Relieving capacities of a given safety valve on different gases
7.10Relieving capacity of safety relief valve for liquid service
7.11Determining relieving capacity of a given safety valve on air and steam service
7.12Sizing control valves; valve coefficient Cv
7.13Calculating Cv for steam service; saturated and superheated steam; critical and noncritical flow
7.14Calculating Cv for liquid service
Copyright © 2003 Marcel Dekker, Inc.
7.15On cavitation: recovery factors
7.16Selecting valves for laminar flow
7.17Calculating pressure loss in water line; determining friction factor for turbulent flow; equivalent length of piping; viscosity of water
7.18Pressure loss in boiler superheater; estimating friction factor in smooth tubes; pressure drop in smooth tubing; Reynolds number for gases
7.19Determining pressure drop under laminar conditions; pressure drop in fuel oil lines; effect of temperature on specific volume, viscosity of oils
7.20Pressure drop for viscous liquids; friction factor under turbulent conditions
7.21Calculating flow in gpm and in lb=h for fuel oils; expansion factors for fuel oils at different temperatures
7.22Pressure loss in natural gas lines using Spitzglass formula
7.23Calculating pressure drop of flue gas and air in ducts; friction factors; equivalent diameter for rectangular ducts; Reynolds number estimation
7.24Determining Reynolds number for superheated steam in tubes; viscosity of steam; Reynolds number for air flowing over tube bundles.
7.25Determining flow in parallel passes of a superheater
7.26Equivalent length of piping system; equivalent length of valves and fittings
7.27Pressure drop of air and flue gases over plain tube bundles; friction factor for in-line and staggered arrangements
7.28Pressure drop of air and flue gases over finned tube bundles
7.29Factors influencing boiler circulation
7.30Purpose of determining circulation ratio
7.31Determining circulation ratio in water tube boilers
7.32Determining circulation ratio in fire tube boilers
7.33Determining steam flow in blowoff lines
7.34Sizing boiler blowdown lines
7.35Stack height and friction losses
7.36Flow instability in evaporators
7.01a
Q:
How are flow meters sized?
A:
The basic equation for pressure differential in head meters (venturis, nozzles, orifices) is [1]
p r
W ¼ 359YC d2 h ð1aÞ q
d o
1 b4
Copyright © 2003 Marcel Dekker, Inc.
where
W ¼ flow of the fluid, lb=h
Y ¼ expansion factor, which allows for changes in density of compressible fluids (for liquids Y ¼ 1, and for most gases it varies from 0.92 to 1.0)
Cd ¼ a coefficient of charge do ¼ orifice diameter, in.
r ¼ density of fluid, lb=cu ft
b ¼ ratio of orifice to pipe inner diameter ¼ do=di h ¼ differential pressure, in. WC
Cd may be taken as 0.61 for orifices and 0.95–0.98 for venturis and nozzles. It is a complicated function of Reynolds number and orifice size. The permanent pressure drop, Dp, across a flow meter is important, because it means loss in power or additional consumption of energy. It is the highest for orifices [2]:
DP ¼ h ð1 b2Þ |
ð1bÞ |
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For nozzles, |
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D |
P |
¼ |
h |
1 |
b2 |
ð |
1c |
Þ |
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þ b2 |
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1 |
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and for venturis it depends on the angle of divergence but varies from 10% to 15% of h. Q7.04 discusses the significance of permanent pressure drop and the cost associated with it.
7.01b
Q:
The differential pressure across an orifice of a steam flow meter shows 180 in. WC when the upstream conditions are 1600 psia and 900 F. The steam flow was calibrated at 80,000 lb=h under these conditions. Because of different plant load requirements, the steam parameters are now 900 psia and 800 F. If the differential pressure is 200 in. WC, what is the steam flow?
A:
From Eq. (1a),
r
p h W / rh / v
where
W¼ steam flow, lb=h
v ¼ specific volume, cu ft=lb
h ¼ differential pressure, in. WC r ¼ density, lb=cu ft
Copyright © 2003 Marcel Dekker, Inc.
From the steam tables (see the Appendix),
n1 ¼ 0:4553 cu ft=lb at 1600 psia; 900 F n2 ¼ 0:7716 cu ft=lb at 900 psia; 800 F
h1 ¼ 180, h2 ¼ 200, and W1 ¼ 80,000. We need to find W2.
W2 |
¼ r |
¼ |
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80;000 |
180 |
0:7716 |
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1:235 |
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200 |
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0:4553 |
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Hence W2 ¼ 64,770 lb=h.
7.02
Q:
Determine the orifice size to limit the differential pressure to 100 in. WC when 700 lb=s of water at 60 F flows in a pipe of inner diameter 18 in. The density of water is 62.4 lb=cu ft.
A:
Equation (1a) is not handy to use when it is required to solve for the orifice diameter do. Hence, by substituting for b ¼ do=di and simplifying, we have
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p |
2 |
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W ¼ 359Cd Ydi2 |
p |
1rhbb4 |
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ð2Þ |
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This equation is easy to use either when orifice size is needed or when flow |
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2 |
= |
p |
4 |
is a function of b and |
through a given orifice is required. The term b |
1 b |
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can be looked up |
from |
Table 7.1. Substituting |
for W ¼ 700 3600; |
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Cd ¼ 0:61; Y ¼ 1; r ¼ 62:4, and h ¼ 100, we have |
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¼ |
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FðbÞ |
700 |
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3600 |
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359 |
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0:61 |
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1 |
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182p62:4 100 |
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or
FðbÞ ¼ 0:45
From Table 7.1, by interpolation, we note that b ¼ 0.64. Thus the orifice diameter do ¼ 0.64 18 ¼ 11.5 in.
7.03
Q:
What size of orifice is needed to pass a saturated steam flow of 26,480 lb=h when the upstream pressure is 1000 psia and line size is 2.9 in. and the differential is not to exceed 300 in. WC?
Copyright © 2003 Marcel Dekker, Inc.
TABLE 7.1 |
FðbÞ Values for Solving |
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Eq. (2) |
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FðbÞ ¼ b2=p |
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b |
1 b4 |
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0.3 |
0.09 |
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0.4 |
0.162 |
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0.5 |
0.258 |
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0.6 |
0.39 |
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0.7 |
0.562 |
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0.8 |
0.83 |
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A:
Using Eq. (2) and substituting Y ¼ 0:95; r ¼ 1=v ¼ 1=0:4456 ¼ 2:24 lb/cu ft, and di ¼ 2.9, we have
W ¼ |
26;480 |
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2:92 |
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ð |
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¼ |
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359 |
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0:61 |
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0:95 |
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F b |
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p2:24 300 |
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Hence
FðbÞ ¼ 0:58
From Table 7.1, b ¼ 0.71. Hence
do ¼ 0:71 2:9 ¼ 2:03 in:
7.04
Q:
What is the significance of a permanent pressure drop across the flow measurement device? 1.3 million scfh of natural gas with a specific gravity of 0.62 at 125 psia is metered using an orifice plate with a differential head of 100 in. WC. The line size is 12 in. What are the operating costs involved? Assume that electricity costs 20 mills=kWh.
A:
The first step is to size the orifice. Use a molecular weight of 0.62 29 ¼ 18 to compute the density. (The molecular weight of any gas ¼ specific gravity 29.) From Q5.03
125
r ¼ 18 492 359 520 15 ¼ 0:39 lb=cu ft
Copyright © 2003 Marcel Dekker, Inc.
(A temperature of 60 F was assumed.) The density at standard conditions of 60 F, 15 psia, is
492
r ¼ 18 359 520 ¼ 0:047 lb=cu ft
Hence mass flow is
W ¼ 1:3 106 0:047
p
¼ 359 0:61 122 0:39 100 FðbÞ
FðbÞ ¼ 0:31
From Table 7.1, b ¼ 0:55, so b2 ¼ 0:3. The permanent pressure drop, from Q7.01, is
DP ¼ ð1 b2Þh ¼ ð1 0:3Þ 100 ¼ 70 in. WC |
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The horsepower consumed in developing this head is |
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DP |
ð3Þ |
HP ¼ scfh ð460 þ tÞ P 107 |
It was assumed in the derivation of Eq. (3) that compressor efficiency was 75%. Substitution yields
HP ¼ 1:3 106 |
520 |
70 |
¼ 38 |
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107 125 |
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The annual cost of operation is
38 0:746 8000 0:02 ¼ $4535
(8000 hours of operation was assumed per year; 0.746 is the factor converting horsepower to kilowatts.)
7.05
Q:
Often, pitot tubes are used to measure air velocities in ducts in order to compute the air flow. A pitot tube in a duct handling air at 200 F shows a differential of 0.4 in. WC. If the duct cross section is 4 ft2, estimate the air velocity and the flow rate.
A:
It can be shown [3] by substituting r ¼ 40=ð460 þ tÞ that for a pitot,
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2:85 |
p |
ð |
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V |
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h ð460 þ tÞ |
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4 |
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Copyright © 2003 Marcel Dekker, Inc.
where |
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V ¼ velocity, fps |
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h ¼ differential pressure, in. WC |
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t |
¼ |
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p |
F |
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air or flue gas temperature, |
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The air flow rate in |
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V |
¼ |
2:85 |
0:4 660 ¼ 46 fps |
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¼ |
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acfm will be 46 |
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60 |
11,040 acfm. The flow W in |
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lb=h ¼ 11,040 60 40=660 ¼ 40,145 lb=h. |
[W ¼ acfm 60 density, and |
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density ¼ 40=ð460 þ tÞ.] |
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7.06
Q:
How are safety valves for boilers sized?
A:
The ASME Code for boilers and pressure vessels (Secs. 1 and 8) describes the procedure for sizing safety or relief valves. For boilers with 500 ft2 or more of heating surface, two or more safety valves must be provided. Boilers with superheaters must have at least one valve on the superheater. The valves on the drum must relieve at least 75% of the total boiler capacity. Superheater valves must relieve at least 20%. Boilers that have reheaters must have at least one safety valve on the reheater outlet capable of handling a minimum of 15% of the flow. The remainder of the flow must be handled by valves at the reheater inlet.
If there are only two valves for a boiler, the capacity of the smaller one must be at least 50% of that of the larger one. The difference between drum pressure and the lowest valve setting may be at least 5% above drum pressure but never more than the design pressure and not less than 10 psi. The range between the lowest boiler valve setting and the highest set value is not to be greater than 10% of the set pressure of the highest set valve. After blowing, each valve is to close at 97% of its set pressure. The highest set boiler valve cannot be set higher than 3% over the design pressure.
The guidelines above are some of those used in selecting safety valves. For details the reader should refer to the ASME Code [4].
7.07
Q:
How are the capacities of safety valves for steam service determined?
Copyright © 2003 Marcel Dekker, Inc.
A:
The relieving capacities of safety valves are given by the following expressions. ASME Code, Sec. 1 uses a 90% rating, whereas Sec. 8 uses a 100% rating [5].
W ¼ 45APaKsh |
ð5aÞ |
W ¼ 50APaKsh |
ð5bÞ |
where |
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W ¼ lb=h of steam relieved
A ¼ nozzle or throat area of valve, in.2
Pa ¼ accumulated inlet pressure ¼ Ps ð1 þ accÞ þ 15, psia (The factor acc is the fraction of pressure accumulation.)
Ps ¼ set pressure, psig
Ksh ¼ correction factor for superheat (see Fig. 7.1)
The nozzle areas of standard orifices are specified by letters D to T and are given in Table 7.2. For saturated steam, the degree of superheat is zero, so Ksh ¼ 1. The boiler safety valves are sized for 3% accumulation.
7.08
Q:
Determine the sizes of valves to be used on a boiler that has a superheater. The parameters are the following.
Total steam generation ¼ 650,000 lb=h
Design pressure ¼ 1500 psig
Drum operating pressure ¼ 1400 psig
Steam outlet temperature ¼ 950 F
Pressure accumulation ¼ 3%
Superheater outlet operating pressure ¼ 1340 psig
A:
The set pressure must be such that the superheater valve opens before the drum valves. Hence the set pressure can be 1500 7 60 7 40 ¼ 1400 psig (60 is the pressure drop and 40 is a margin). The inlet pressure Pa ¼ 1:03 1400 þ 15 ¼ 1457 psia. From Fig. 7.1, Ksh ¼ 0:79.
A ¼ |
W |
130;000 |
¼ 2:51 in:2 |
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¼ |
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45KshPa |
45 0:79 1457 |
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We used a value of 130,000 lb=h, which is 20% of the total boiler capacity. A |
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K2 orifice is suitable. This relieves |
(2.545=2.51) 130,000 ¼ 131,550 lb=h. |
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The drum valves |
must relieve 650,000 7 131,550 ¼ 518,450 lb=h. About |
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260,000 lb=h may be handled by each drum valve if two are used. Let the first
Copyright © 2003 Marcel Dekker, Inc.
FIGURE 7.1 Correction factors for superheat.
valve be set at 1475 psig, or
Pa ¼ 1:03 1475 þ 15 ¼ 1535 psia
and the next at Pa ¼ 1575 psia. |
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Area of first valve: A ¼ |
260;000 |
¼ 3:76 in:2 |
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45 |
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1535 |
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Area of second valve: A ¼ 260;000 ¼ 3:67 in:2 45 1575
Use two M2 orifices, which each have an area of 3.976 in.2 Relieving capacities are
33::76 þ |
3:67 |
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260;000 ¼ 556;000 lb=h |
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976 |
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3:976 |
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which exceeds our requirement of 520,000 lb=h.
Copyright © 2003 Marcel Dekker, Inc.
TABLE 7.2 |
Orifice |
Designation |
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Type |
Area (in.2) |
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D |
0.110 |
E |
0.196 |
F |
0.307 |
G |
0.503 |
H |
0.785 |
J |
1.287 |
K |
1.838 |
K2 |
2.545 |
L |
2.853 |
M |
3.600 |
M2 |
3.976 |
N |
4.340 |
P |
6.380 |
Q |
11.05 |
R |
16.00 |
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7.09a
Q:
How is the relieving capacity of safety valves for gaseous service found?
A:
The expression used for estimating the relieving capacity for gases and vapors [6] is
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r |
ð Þ |
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W |
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CKAPa |
MW |
6 |
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T |
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where |
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C ¼ a function of the ratio k of specific heats of gases (Table 7.3) K ¼ valve discharge coefficient, varies from 0.96 to 0.98
Pa ¼ accumulated inlet pressure ¼ Psð1 þ accÞ þ 15; psia Ps ¼ set pressure, psig
MW ¼ molecular weight of gas T ¼ absolute temperature, R
7.09b
Q:
A safety valve is set for 100 psig for air service at 100 F and uses a G orifice. What is the relieving capacity if it is used on ammonia service at 50 F, pressure being the same?
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