1、 ANSI/FCI 13-1-2016 AMERICAN NATIONAL STANDARD DETERMINING CONDENSATE LOADS TO SIZE STEAM TRAPS Fluid Controls Institute, Inc. Sponsor: Fluid Controls Institute, Inc. 1300 Sumner Ave Cleveland, Ohio 44115-2851 iiANSI/FCI 13-1-2016 AMERICAN NATIONAL STANDARD Determining Condensate Loads to Size Steam
2、 Traps Sponsor Fluid Controls Institute, Inc. AmericanNationalStandardAmerican National Standard implies a consensus of those substantially concerned with its scope and provisions. An American National Standard is intended as a guide to aid the manufacturer, the consumer, and the general public. The
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7、rior written permission of the publisher. Suggestions for improvement of this standard will be welcome. They should be sent to the Fluid Controls Institute, Inc. Printed in the United States of America iiiivCONTENTS PAGE Foreword v 1. Purpose 1 2. Scope .1 3. Definitions.2 4. General Formulas for De
8、termining Condensate Loads 2 Tables Table 1 Pipe Size Conversion .3 Table 2 Properties of Saturated Steam at Various Pressures 7 Table 3 Density of Air Atmospheric Pressure 9 Table 4 Specific Gravity and Specific Heat at 60 F for Select Materials .9 Table 5 U Values for Schedule 40 Carbon Steel Pipe
9、 Steam to Watery Solution 9 Table 6 U Values for Schedule 40 Carbon Steel Pipe Steam to Air .9 Table 7 Weight Per Foot of Steel Sch 40 however, it may also shorten steam trap life in addition to possible unnecessary steam loss as referenced above. If the equipment manufacturer lists the heat output
10、of the steam equipment that needs to be drained, the estimated condensate rate can be easily calculated. Generally, equipment manufacturers provide the BTU/hr output. In that case, divide the BTU/hr output by the operating steam pressure latent heat (Table 1) to estimate the condensate generation ra
11、te from that equipment. (For more exact calculation, the steam quality / wetness has to be considered to adjust for actual lowered latent heat available at the process). Example: 2,500,000 BTU/hr 30 psig steam When selecting the trap type, size, and required discharge capacity, consideration still h
12、as to be given to whether the steam pressure is steady state or modulating. The calculation shown in the example above is maximum condensate load at full pressure. When the heat output rates from the equipment are unknown, estimated calculations for heat load using other sources of data can be utili
13、zed. 1 2 3. DEFINITIONS 3.1 Steam trap An integral, self actuated valve which automatically vents air in the steam system and drains condensate from a steam containing enclosure while remaining tight to live steam. Most steam traps will also pass non-condensible gases while remaining tight to live s
14、team. 3.2 Condensate The fluid created after steam has given up its latent heat energy and becomes liquid phase. 3.3 Capacity The manufacturers rated capability of a steam trap to discharge condensate. Capacity is typically stated in the manufacturers product specifications, illustrated through eith
15、er charts or tables. 3.4 Drip Points Vertical piping pockets located along the steam main pipe where condensate is collected for the purpose of draining from the system. These points are commonly referred to as collecting legs, drip points/legs/pockets, or drain pockets. In the formulas contained he
16、re-in this standard these locations are referred to as drip points. 4. GENERAL FORMULAS FOR DETERMINING CONDENSATE LOADS 4.1 Glossary of Abbreviations and Terms A = area of heating surface, sq ft BTU = British thermal unit Cp = specific heat, BTU/lb (Table 4) CFM = air flow, cubic feet/minute d = de
17、nsity of air, lb/ft3(Table 3) D = diameter of dryer, ft EDR = effective direct radiation, sq ft ft = length, Feet G = volume, gallons GPM = flow, gallons/minute L = latent heat BTU/lb (Table 2, saturated steam latent heat value, adjust for wetness %) Ll = latent heat of lower pressure, BTU/lb Lot =
18、length of tubes, ft N = number of tubes P = sq ft of surface area per lineal foot of pipe (See Table 1) Q = condensate generated, lb/hr Qi = Incoming condensate flow, lb/hr R = rate of condensation, (lb/sq ft-hr), (typical 7 lb/sq ft-hr) s.g. = specific gravity 60F S = sensible heat, BTU/lb Sh = sen
19、sible heat, higher pressure, BTU/lb Sl = sensible heat, lower pressure, BTU/lb t = time, hours Ta = ambient air temperature, F 3 Tf = process material final temperature, F Ti = process material initial temperature, F Ts = steam temperature at pressure, F U = heat transfer coefficient, BTU/hr ft F W
20、= weight of solid material, lbs Wg = liquid weight, lbs/gallon Wh = width of dryer, ft. Note: For any formula requiring latent heat (L) use Table 2 Table 1 Pipe Size Conversion Table (Divide lineal feet of pipe by factor given for size and type of pipe to get square feet of surface) Pipe Size (in) I
21、ron Pipe Copper or Brass Pipe 4.55 7.63 3.64 5.09 1 2.90 3.82 1-1/4 2.30 3.05 1-1/2 2.01 2.55 2 1.61 1.91 2-1/2 1.33 1.52 3 1.09 1.27 4 .848 .954 4.2 Steam Main: Steam mains in various applications may operate in saturated or superheat conditions. When the steam main is superheated, the start-up loa
22、d may be high to bring the pipe to temperature, but then very little or no condensate is generated when operating at full superheat. In low steam velocity conditions, such as very low demand or in a (stagnant flow) collecting leg, flow reduces to the threshold where the heat loss of the main exceeds
23、 the BTUs of superheat available. Then, condensate will again be created and must be removed from the system. In instances where the superheated steam flow is stopped, the main or collecting leg can revert first to saturated steam, and then to wet steam, depending upon the amount of time for the flo
24、w cessation. For that reason, a steam trap is required at the drip points along the pipe with either saturated or superheated steam. Steam mains generally operate more than 90% of the time in the full operational mode. The condensate load produced during normal operation is known as “Running Load”,
25、whereas the condensate load generated during the time when the main is brought up to full operation is known as “Start-up Load”. Start-up times of the main should not be so short as to generate large amounts of condensate that could cause water hammer. Another issue with short start-up times is that
26、 large steam traps need to be used to handle the over-large start-up load, and then those traps could be grossly oversized in the normal operating mode. It is important that both Start-up Load and Running 4 Load are calculated to help properly size the steam trap for both conditions. In some instanc
27、es, steam traps can even be selected to support automatic start-up of the main, provided the steam is applied in a slow, controlled manner. Slow starting is needed so as to not create water hammer conditions or to overwhelm the steam traps capacity at the available pressure differential while the st
28、eam pressure builds in the line (Both Start-up Load and Running Load calculations are needed to properly select and size the trap). 4.2.1 Condensate Running Load Calculated from Heat Loss with Totally Insulated Pipe: QRLI = (BTU bare pipe heat loss per hr ft of pipe length) x (ambient temperature F
29、correction factor) x (distance between drip points) x (1-insulation efficiency fractional) / L (Tables 8 Tables 8 both the lowest and highest trap inlet pressure must be evaluated for sufficient differential pressure. This range of consideration is required due to the resultant pressure fluctuation
30、from modulation and to enable discharge without condensate back-up into the equipment. To evaluate further, under certain reduced-demand conditions, it is possible that supply steam modulates to a lower pressure than the return line backpressure and, in some instances, vacuum conditions inside the e
31、quipment can occur. In such cases, the simplest solution is to install a vacuum breaker to restore the steam space to atmosphere conditions and enable free drainage through a steam trap if sufficient head pressure is available. When backpressure is present and does not allow gravity drainage, a seco
32、ndary pressure drainer, SPD Type I or II, is required to discharge condensate. The vacuum breaker solution can be effective on gravity drainage applications because steam traps on modulating service with atmosphere restored can achieve a differential pressure from the inlet liquid head pressure ahea
33、d of the steam trap (Table 10). Without differential pressure, steam traps on equipment that modulates to 0 psig or vacuum cannot discharge against return line backpressure and must either drain by gravity into an atmospheric pressure condition or alternatively incorporate either the use of a pump (
34、SPD Type I or electric condensate pump) or substitute a closed-loop, pump-trap combination (SPD Type II). 12 13 Table 10 Minimum Head Pressure Available to a Steam Trap by Vertical Drop Elevation difference between steam trap and equipment outlet Available pressure at steam trap (psi) Elevation diff
35、erence between steam trap and equipment outlet Available pressure at steam trap (psi) 3” .12 20” .71 5” .18 24” .86 7” .25 28” 1.0 10” .36 32“ 32 12” .43 36” 1.3 14” .5 40” 1.4 18” .69 48” 1.8 To determine available inlet head pressure (psig) for all other elevations, divide height (inches) from the
36、 inlet of the steam trap vertically up to the drained equipment outlet by 28. The steam trap must have sufficient condensate flow capability to discharge condensate at the lowest available differential pressure when the steam valve modulates. 14 Note: In all of the examples that follow, no provision
37、 has been made for steam wetness. To approximate for typical wetness, multiply the calculated loads by 1.10 or 1.15. Example 1 Calculate Condensate Load for a Heat Exchanger Raising Flowing Water Temperature from 50 to 100F: Notes: Temperature Control can modulate to vacuum condition Vacuum breaker
38、installed Discharges to Vented Receiver Maximum Press: 20 psig Minimum Press: Vacuum (equalized to atmosphere by vacuum breaker installation) Flow Rate: 50 GPM Vertical Head: 28” Back Pressure: 0 psig Load Factor: 2 Q = (GPM x 500.4 x Tf-Ti) / L Calculate: Q = (50 x 500.4 x (100-50) / 939 = 1,332.27
39、 lbs/hr Load Factor Effect: 2 x 1,332.27 lbs/hr = 2,665 lbs/hr Trap must discharge a minimum of 2,665 lbs/hr with 1 psig differential (28” = 1 psig, Table 10) Example 2a Calculate Condensate Running Load of an Insulated Steam Main: Steam Pressure: 125 psig (constant) Insulation Efficiency: 90% Lengt
40、h Between Drip Points: 100 ft Pipe Diameter: 8” (Sch 40) Outside Air Temperature: 0F Load Factor: 2 QRLI = (BTU bare pipe heat loss per hr ft of insulated pipe) x (ambient temperature F correction factor) x (distance between drip points) x (1-insulation efficiency fractional) / L Bare Pipe Heat Loss
41、: 1,875 BTU/hr (Table 8) at 125 psig x 1.214 = 2,276 BTU/hr ft Loss Between Drips: 2,276 BTU/hr x 100 ft = 227,625 BTU/hr total bare pipe Calculate: Q = (227,625 x (1- .9) / 868 = 26.2 lbs/hr running load Load Factor Effect: 2 x 26.2 lbs/hr = 52.4 lbs/hr Note: If an automatic start-up condition, bot
42、h the Condensate Running load and also the Condensate Start-up load must be calculated, and the steam trap must be capable of handling the total (Example 4). Example 2b Calculate Condensate Running Load of an Insulated Steam Main: Steam Pressure: 200 psig (constant) Insulation Efficiency: 90% Length
43、 Between Drip Points: 200 ft Pipe Diameter: 6“ (Sch 40) Outside Air Temperature: 70F Load Factor: 2 15 QRLI = (BTU bare pipe heat loss per hr ft of insulated pipe) x (ambient temperature F correction factor) x (distance between drip points) x (1-insulation efficiency fractional) / L Bare Pipe Heat L
44、oss: 1,735 BTU/hr at 200 psig = 1,735 BTU/hr ft (Table 8) Loss Between Drips: 1,735 BTU/hr x 200 ft = 347,000 BTU/hr total bare pipe Calculate: Q = (347,000 x (1-.9) / 837 = 41.46 lbs/hr running load Load Factor Effect: 2 x 41.46 lbs/hr = 82.9 lbs/hr Note: If an automatic start-up condition, both th
45、e Condensate Running load and also the Condensate Start-up load must be calculated, and the steam trap must be capable of handling the total (Example 4). Example 3a Calculate Condensate Running Load of a Partially-Insulated Steam Main: Steam Pressure: 125 psig (constant) Insulation Efficiency: 90% L
46、ength of Pipe: 100 ft Insulated Pipe %: 80 Pipe Diameter: 8“ (Sch 40) Indoor Air Temperature: 0F Load Factor: 2 QRLPI= (BTU bare pipe heat loss per hr ft of insulated pipe) x (ambient temperature F correction factor) x (distance between drip points) x (1-insulation efficiency fractional) / L) + (BTU
47、 bare pipe heat loss per hr ft of uninsulated pipe) x (ambient temperature F correction factor) x (distance between drip points) / L Bare Pipe Heat Loss: 1,875 BTU/hr ft x 1.214 = 2,276 (Table 8) Insulated Pipe Running Load: Q = (2,276 x (100 x .8) x (1-.9) / 868 BTU/#) = 20.98 lbs/hr Uninsulated Pi
48、pe Running Load: Q = (2,276 x (100 x .2) / 868 BTU/# = 52.44 lbs/hr Total running load: Q (total) = 20.98 + 52.44 = 73.42 lbs/hr Load Factor Effect: 2 x 73.42 lbs/hr = 146.8 lbs/hr Example 3b - Calculate Condensate Running Load of a Partially-Insulated Steam Main: Steam Pressure: 200 psig (constant)
49、 Insulation Efficiency: 90% Length of Pipe: 200 ft Insulated Pipe %: 80 Pipe Diameter: 6” (Sch 40) Indoor Air Temperature: 70 F Load Factor: 2 QRLPI= (BTU bare pipe heat loss per hr ft of insulated pipe) x (ambient temperature F correction factor) x (distance between drip points) x (1-insulation efficiency fractional) / L) + (BTU bare pipe heat loss per hr ft of uninsulated pipe) x (ambient temperature F correction factor) x (distance between drip points) / L Bare Pipe Heat Lo
