HG/T 20570.15-1995 Setting of flow limiting orifice plates in pipelines
Some standard content:
Industry Standard of the People's Republic of China
International General Design System and Method
HG/T 20570—95
Process System Engineering Design
Technical Regulations
1996—05—02 Issued
1996-09—01
Ministry of Chemical Industry of the People's Republic of China
Industry Standard of the People's Republic of China
Technical Regulations for Process System Engineering Design
HG/T 20570—95
Editing Unit: Process System Design Technology Center of the Ministry of Chemical Industry Approving Department: Ministry of Chemical Industry
Implementation Date: September 1, 1996
Engineering Construction Standard Editing Center of the Ministry of Chemical Industry
Setting of Pipeline Flow Limiting Orifice Plate
HG/T 20570.15—95
Compiled by: China Global Chemical Engineering Corporation Approved by: Chemical Industry Department
Effective Date: September 1, 1996 Prepared by:
China Global Chemical Engineering Corporation Lv Wenpu
Reviewed by:
China Global Chemical Engineering Corporation Wang Qingyu
Chemical Industry Department Process System Design Technology Center Station
Application Scope
1.0.1 The flow limiting orifice is set in the pipeline to limit the flow rate of the fluid or reduce the pressure of the fluid. The flow limiting orifice is used in the following areas:
1.1.1 The process material needs to be depressurized and the precision requirement is high. 1.1.2 When a large pressure drop is required upstream and downstream of the pipeline valve, in order to reduce the erosion of the valve by the fluid, when the throttling through the orifice does not produce a gas phase, an orifice plate can be connected upstream of the valve. 1.0.1.3 Where the fluid needs to flow at a small flow rate and continuously, such as the flushing pipeline of the pump, the bypass pipeline of the hot standby pump (waiting flow protection pipeline), analytical sampling tube and other places. 1.0.1.4 Where a pressure drop is required to reduce noise or friction, such as the wire release system. 1.0.2 Unless otherwise specified, the pressures in this regulation are absolute outputs. 445
Z.1 Classification
2 Key points for classification and selection
The flow-limiting orifice plate is divided into a single-hole plate and a multi-hole plate according to the number of stop holes in the plate: it can be divided into a single-hole plate and a multi-hole plate according to the number of grids. 2.0.2 Key points for selection
2. 2. 1, steam
In order to avoid blockage in the flow-limiting orifice plate, the pressure behind the flow-limiting orifice plate (\) cannot be less than the pressure before the plate (P, about 55%, that is, P0.55P. Therefore, when P<0.55P:, a single plate cannot be used, and multiple plates must be selected. The number of plates must ensure that the pressure of each plate is greater than 55% of the pressure before the plate. 2.0.2.2 Liquid
(1) When the liquid pressure drop is less than or equal to 2.5MPa, a single plate plate is selected. (2) When the liquid pressure is greater than 2.5MPa, a multi-plate orifice plate is selected to make the pressure barrier of the orifice plate less than 2.5MPa.
2.0.3 Determination of the number of holes
2.0.3.1 For pipes with a nominal diameter less than or equal to 15mm, a single-hole plate is used: when the nominal diameter is greater than 15mm, an orifice plate is used.
2 .0.3.2 The aperture of the multi-hole plate (d.) can generally be selected from 12.5mm, 20mm, 25mm, and 40mm. When calculating the multi-hole plate, first calculate the aperture () according to the single-hole plate. Then use formula (2..3) to calculate the number of holes in the selected multi-hole plate (N),
N is the number of holes in the hole-by-hole limiting bottle plate, m;
is the aperture of the single-hole limiting bottle, m; d is the aperture of the multi-hole limiting bottle, m. The diameter of the orifice plate, m. 446
3.0.1 Single-plate orifice plate
3.1.1 Body-mounted orifice
3 Calculation method
(1) Single-plate orifice plate for gas and steam is calculated according to formula (3.6.11: W=43.7BaP.
w Fluid flow rate, kg/h:
(Y([-
()*1 (3.0.1-1)
C—orifice flow coefficient + from Re and &/D values, refer to Figure 6.0.1: d, - orifice diameter, m +
pipe inner diameter, m;
P, - orifice pressure, T;
Pa - orifice start force or limit flow repulsion, whichever is greater, Pa: M - year:
Z--compression coefficient, according to the fluid relative pressure (P.) and relative temperature, refer to the gas compression coefficient diagram;
T-. -Fluid temperature before orifice, K;
…Adiabatic index, -C
Cr——Fluid constant positive heat capacity, kJ/kg·K>: C,—Fluid constant volume heat capacity, kJ/(kg-K). (2) Recommended value of critical limiting pressure (P,) for cannon and steam: F,-0.58P
Overpotential steam and polyatomic gas rate: 1.0-0.55Pair and diatomic gas = 0.
In the above three formulas, 1 is the pressure before the orifice. 3.0.1.2 Liquid
The single plate emulsion of liquid is calculated according to formula (3.0.1-2>: AP
Q 128. 45 -c- de.7
(3. 0.1—2)
Q--the accumulated flow under T working state·n\/h orifice plate flow coefficient, from Re value and, /D refer to Figure 6.0.1 to obtain: d. Orifice plate aperture, m:
--the pressure drop through the orifice, Pa
7--the relative density under T working state+ (compared with the case of 4℃ water).3.0.2 Orifice plate
3.0.2.1 Gas, steam
(t) First calculate the total number of orifice plates and the pressure before and after each orifice (see the figure below) P
For superheated steam in this example:
P,-0. 55F.
P+=(0.55)\P
n=ig(g/P,)/tgD.55
-3.85g(a/,)
(3. 0. 2 1)
is rounded to an integer and then redistributed to each plate's front and rear pressures, and the formula (3.0.2-2) is used to obtain the pressure behind the base plate:
P-(P/P).P
-total number of plates:
P,·pressure before the first plate of a multi-plate orifice plate.Pa: P,-pressure behind the last plate of a multi-plate orifice plate; PaFm pressure behind the mth plate in a multi-plate orifice, Pa#
(H.0.2-2)
《(2) According to the pressure before and after each plate, calculate the aperture of each orifice plate·Let the calculation method be applied to a single-plate orifice plate. After the sample is counted, the pressure before and after each plate can be re-distributed. 3.0.2.2 Filter body
(1) First calculate the total number of orifice plates (yuan) and the pressure before and after each orifice plate 45
Calculate n according to formula (3.0.2-3) and then calculate n as an integer, and then add the orifice plates in sequence to evenly distribute the pressure before and after each plate with an integer (n):
In the formula, n and P have the same meanings as before.
(2) Calculate the diameter of each orifice plate. The calculation method is the same as the single orifice plate calculation method. 3.0.3 Gas-liquid two-phase flow
(3.0. 23)
First calculate the specific flow rate according to the gas and liquid flow using their respective formulas, and then calculate the two-phase flow orifice diameter using the following formula:
=two-phase orifice plate diameter, m;
dc\-liquid orifice plate diameter, m:
,-gas phase orifice plate diameter, m.
3.4 Identification of orifice plate with flow limiting effect
(3.0.3)bZxz.net
Calculate the orifice diameter (d.) according to formula (3.3.1-1) or formula 13.0.12) or formula (3.0.3). According to the value of 4/D, the critical flow rate K is changed according to Table 6.0.2 (, when the pressure ratio P/P before and after the orifice plate is changed, the fluid flow rate can be limited to a certain value, indicating that the designer has made a decision. If the pressure is not reduced, the pipe diameter of the pipeline needs to be adjusted, and then the pipe diameter needs to be adjusted again until the requirements are met. 4 Calculation example
4,0.1 A stream of exhaust gas is discharged from the orifice plate to the fuel gas pipe. The gas composition is as follows: Composition
Gas flow rate is 3466kg/h, gas absolute pressure is 10.3MP=, temperature is 57℃, and gas density before decompression is 1.305× 10\5mPa·s+After decompression, the absolute pressure of the gas is 2.UMPa, and the inner diameter of the pipe before decompression is D 38.1mm. Calculate the size of the flow-limiting orifice.
Solution: Connect formula (3.0.2-1) to calculate the number of orifices in the part Total number of plates - 3.85g/)
—3.85tg(2.0/10.3)
Connect n=3
Compare formula (3.0.2-21 to calculate:
F-(a/pkpm-l
P =(2.0/10.3)^10.3-5,96MPF-(2.0/10.3)*5,96-3,45MPa
F,-(2.0/10.3)3.45-2.00MPa
Calculate the first orifice plate according to formula (3.0.1--1): aperture
Known F-10.3X10Pa
W=3466kg/h
T-330K
Calculate Z and is the value
P, (MPa)
Mixed gas: 7=72.65K12.15MF
According to the mixed gas=1.4
Compare temperature: T: 330/71.66: 4.6
Compare jade force; F =10.3/2.16=4.77
According to the gas pressure coefficient diagram, we can get 1. Prove Ar
Mass flow rate, G3456/(3600×0.7R5x0.03812)844.9kg/msViscosity: ±=1.306×10-mPa9
D0.0381X844.
1.305X10-5
=3466/43.78×X10.3×1G=
ds-9. 256X10 -/C
Assume C-0.60, get =12.4mm
108X330
Take d=12.5mmd/D=12.5/38.10.32810.3
From Figure 6.0.1, we can find C0.60[0.60, which shows that the obtained value is =12.5mm. The second plate:
Compared with the pressed piece -5.96/3.16-276
Assuming T, unchanged, based on PT, set gas positive shrinkage coefficient diagram, check Z=1.04-1.4 for tube calculation, assuming that the annual viscosity of the gas remains unchanged, then Re-2.5X106 will have the data to substitute for the ideal formula to get =3466/43.78×C×5.96×10°
d8-1.557X10 1/0
Assume C-0,61. Get da-0.01398m, take d,=16mm.z./)=0,42 Figure 6.0.1.Re-2.5×10%
dp/D:=0.42
get C-0.61. This shows that taking::tm is effective. 451
The third plate:
The stop pressure P,-3.45/2.16=1.597
Assuming that T is constant, check the compression coefficient diagram of the period FT and get the gas pressure coefficient Z.1.0: Take a 1.4.
Assuming that the gas station is constant, then Re-2.5×106=3456/13.78×CX3.45×10°
Get 4,=2.61×104/C
Assume (=0.63.d,=0.020351m
Take 4=20mm, ds/D=0.525
Check Figure 6.0,1: K=2.5×10%d/D=0,525Get C-0 63, this means that the alliance 4.=20mm angle effect. 0. 4
4.0.z, the basic decarbonation solution is known, the flow rate is 1150m/h, the pressure will decrease with the orifice, the absolute pressure before the pressure decrease is P=2.06MPa, the absolute pressure after the pressure decrease is F-0.74MFa, the inner diameter of the pipeline is D=509mm, the melting temperature is 115C. The temperature is 0.56×10-mPa-5+phase density=1.24. Find the filling diameter of this flow-limiting orifice: Solution, 4P-2.0G—0. 74-1.2MPa2.5MPa Therefore, a single plate flow limiting orifice is selected.
The base flow rate of the liquid G (kg/m\) is:
360G×0.785×0.509z-1947.7kg/m*s1150×1240
G.509×1947,1.77×10
Use formula (3.0.1-2):
Q=128. 45XCXd,
1150 - 128. 45XtXd,
=8. 66X10-5/C
11.32×10%
d,/D=C.12/0.509=0.2358
2C-0.595—0.12m
From Figure 6.0.1, we can find that C-0.595, ℃ value is selected appropriately, which means that 4=0.12m is effective (single hole, single report 2
If a multi-hole plate is selected, the hole diameter is 0.02m. Then the total number is, W=（0.12*/(0.02>\36个45202m. Then the total number is, W = (0.12*/(0.02>\36 45202m. Then the total number is, W = (0.12*/(0.02>\36 452
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