Training to control valve

1. DTL flowrate and tank volume
calculation
Shandong Pengbo Safety & Environment Protection Services Limited
Nov. 2021

2. 3
DTL flow rate
Control valve Cv equation
Inherent characteristic
Tank volume calculation

3. DTL flow rate
PART
1

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Introduction
The percent of valve opening (DCS, local actual opening)
The percent of maximum flow coefficient (Cv)
The inherent characteristic
Differential pressure across the valve
Static head

5. Control Valve CV equation
PART
2

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Control Valve CV equation
Generalized Form of the Control Valve CV Equation
Q – flow rate across the valve
SG – const.
∆P=P1-P2, upstream-downstream
Q=Cv∗ ∆P/𝑺𝑮


7. Inherent characteristic
PART
3

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Inherent characteristic
Typical Control Valve Inherent Characteristic Curves
The most common characteristic curves
are the quick opening, linear, and
equal percentage curves. Valve
manufactures can also design valves
with modified linear, modified equal
percentage, parabolic, and square root
characteristics, as shown in the graph.

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Inherent characteristic
The CV of the valve at each position will not change
when installed in an actual system. However, the
installed characteristic curve of valve position vs.
percent of maximum flow rate will be different than
the inherent characteristic curve because the
differential pressure across the valve will change with a
change in valve position. How the curve shifts is
determined by the shape of the pump curve and the
amount of static head in the system.

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Inherent characteristic
Typical piping system
It can be used to show how the inherent
curve is changed by installing the valve
in an actual system and to create an
installed characteristic curve for a
control valve.
Consider the typical piping system in
Figure consisting of a Supply Tank,
Pump, Flow Control Valve (FCV), Heat
Exchanger (HX), and a Product Tank. The
pump curve has been specified for the
pump and CV data has been entered
for the FCV with an equal percentage
characteristic and a maximum CV of 225.

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Inherent characteristic
The installed characteristic curve for the FCV is obtained by giving the FCV a manual position from
5% to 100% and entering the resulting flow rates into an Excel spreadsheet, calculating the
percentage of maximum flow rate and graphing the result.
Named CV KV Actural CV Valve opening DP (bar) Q(M³/H) Q(BPD)
300 258 0.0% 0.0% 1 0
258 2.0% 10.0% 1 5.379672 806.95078
258 8.0% 20.0% 1 21.51869 3227.8031
258 10.0% 30.0% 1 26.89836 4034.7539
258 12.5% 32.5% 1 33.62295 5043.4424
258 20.0% 40.0% 1 53.79672 8069.5078
258 28.0% 50.0% 1 75.31541 11297.311
258 30.0% 52.5% 1 80.69508 12104.262
258 34.0% 55.0% 1 91.45442 13718.163
258 40.0% 60.0% 1 107.5934 16139.016
258 50.0% 65.0% 1 134.4918 20173.77
258 60.0% 70.0% 1 161.3902 24208.523
258 80.0% 80.0% 1 215.1869 32278.031
258 95.0% 90.0% 1 255.5344 38330.162
258 100.0% 100.0% 1 268.9836 40347.539

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Inherent characteristic
Named CV KV Actural CV Valve opening DP (bar) Q(M³/H) Q(BPD)
300 258 0.0% 0.0% 1 0
258 2.0% 10.0% 1 5.379672 806.95078
258 8.0% 20.0% 1 21.51869 3227.8031
258 10.0% 30.0% 1 26.89836 4034.7539
258 12.5% 32.5% 1 33.62295 5043.4424
258 20.0% 40.0% 1 53.79672 8069.5078
258 28.0% 50.0% 1 75.31541 11297.311
258 30.0% 52.5% 1 80.69508 12104.262
258 34.0% 55.0% 1 91.45442 13718.163
258 40.0% 60.0% 1 107.5934 16139.016
258 50.0% 65.0% 1 134.4918 20173.77
258 60.0% 70.0% 1 161.3902 24208.523
258 80.0% 80.0% 1 215.1869 32278.031
258 95.0% 90.0% 1 255.5344 38330.162
258 100.0% 100.0% 1 268.9836 40347.539
0.0%
10.0%
20.0%
30.0%
40.0%
50.0%
60.0%
70.0%
80.0%
90.0%
100.0%
0.0% 10.0% 20.0% 30.0% 40.0% 50.0% 60.0% 70.0% 80.0% 90.0% 100.0%
valve opening vs. Cv
0
50
100
150
200
250
300
0.0% 20.0% 40.0% 60.0% 80.0% 100.0%
valve opening vs. flow rate

13. Tank volume calculation
PART
4

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Tank volume calculation
TS,S1,S2,S3, Diesel tank
Chemical tank, Sum pit, water tank
GDP lube oil tank

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Tank volume calculation
A = 𝜋𝑅2
V = H ∗ 𝜋𝑅2
R _ semi diameter
H _ height
A _ area
V _ volume
𝜋 _ 3.14
H
R
H
L
S
V = L*S*H

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Tank volume calculation
H H
R
TS
V = H ∗ 𝜋𝑅2

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Tank volume calculation
V = [R²arccos(1-h/R)-(R-h) 2Rℎ − ℎ²]𝐻
D /m H /m h/% V'
/m 3
2.
4 11.
76 0 0.
0
2.
4 11.
76 10 2.
8
2.
4 11.
76 20 7.
6
2.
4 11.
76 30 13.
4
2.
4 11.
76 40 19.
9
2.
4 11.
76 50 26.
6
2.
4 11.
76 60 33.
3
2.
4 11.
76 70 39.
8
2.
4 11.
76 80 45.
6
2.
4 11.
76 90 50.
4
2.
4 11.
76 100 53.
2
FQS S1
D _ Diameter: 2.4m
H: 11.76m
h
H H
R

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Questions & comments?

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1- How much is the flow rate when DTL is 50% and TL pressure is 5bar?
Q = Cv∗ ∆P/𝑺𝑮, Cv=258,SG=0.92
Q = ?
2- How much of demulsifier left in the new chemical tank when the level is 50%?
R = 0.8m
H = 2.2m
H
L
R
R

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Thanks！