Hello,

I am very new to control valves. I'm still learning the basics. I cannot really understand equal percentage characteristic. I have found this explanation in control valve handbook:

"In the equal-percentage flow characteristic, equal increments of valve travel produce equal percentage changes in the existing flow".

I must admit it is hard for me to understand this. For this case, flow characteristic is quadratic (exponential) shape. How is that equal percentage. Does that sentence means that if valve is opened by 10% that flow will also increase by 10%? That shouldn't be the case.

I know that flow is proportional to a square root of differential pressure on the valve. This may have to do with quadratic shape function, but still cannot fully understand this.

Thanks

I am very new to control valves. I'm still learning the basics. I cannot really understand equal percentage characteristic. I have found this explanation in control valve handbook:

"In the equal-percentage flow characteristic, equal increments of valve travel produce equal percentage changes in the existing flow".

I must admit it is hard for me to understand this. For this case, flow characteristic is quadratic (exponential) shape. How is that equal percentage. Does that sentence means that if valve is opened by 10% that flow will also increase by 10%? That shouldn't be the case.

I know that flow is proportional to a square root of differential pressure on the valve. This may have to do with quadratic shape function, but still cannot fully understand this.

Thanks

Hi,

I think what is confusing is that where this equal percentage is coming from. As you said, the relation btw valve travel and

flow is logarithmic, not linear. As to equal percentage:

When valve is 20% open, an increase of X percent in the flow happens,with respect to the valve when it is 10% open .

When valve is 30% open, an increase of X percent in the flow happens,with respect to the valve when it is 20% open happens.

So, same INCREMENT in valve travel gives EQUAL PERCENTAGE increase in flow.

(Flow,next - Flow,prev)/Flow,prev = Equal(X)

As to what you described as 10%travel increase/10%flow increase, it is linear characteristic.

Hope this helps.

Regards,

Anri

I think what is confusing is that where this equal percentage is coming from. As you said, the relation btw valve travel and

flow is logarithmic, not linear. As to equal percentage:

When valve is 20% open, an increase of X percent in the flow happens,with respect to the valve when it is 10% open .

When valve is 30% open, an increase of X percent in the flow happens,with respect to the valve when it is 20% open happens.

So, same INCREMENT in valve travel gives EQUAL PERCENTAGE increase in flow.

(Flow,next - Flow,prev)/Flow,prev = Equal(X)

As to what you described as 10%travel increase/10%flow increase, it is linear characteristic.

Hope this helps.

Regards,

Anri

Hi Mike,

First of all please note that when referring to a valve's flow characteristics, its always the convention to talk about the valve's INHERENT characteristic i.e., the flow characteristics of the valve at constant pressure drop. In other words, a valve's characteristics refers to a graphical plot of the Valve's Cv vs lift(%). When a valve is installed in a process, the actual flow characteristics of the valve from 0 to 100% valve lift will be also determined by how the pressure drop varies with the flow in addition to the variation of Cv with %lift.

A valve having Equal percentage characteristic (inherent) will produce equal % increments in Cv for equal % increments in the valve lift. The % increment in Cv depends on the rangeability of the valve. Valves with high rangeability produce higher % increments in flow rate. The relation between the valve lift (%) and the Cv at that lift is given by

Cv = (R)^x * Cvmax / R

where,

x=valve lift expressed in fraction (full close = 0, full open = 1)

Cvmax = Cv at full open (100% lift)

R = Rangeability of the valve

For example a valve having a rangeability of 50, will produce 48% changes in the flow for every 10% change in the lift.

So in short, equal increments in valve lift (10%) anywhere in the control range produces equal increments in Cv (48%).

Hope this helps.

Ronald Deepak

jronalddeepak@gmail.com

First of all please note that when referring to a valve's flow characteristics, its always the convention to talk about the valve's INHERENT characteristic i.e., the flow characteristics of the valve at constant pressure drop. In other words, a valve's characteristics refers to a graphical plot of the Valve's Cv vs lift(%). When a valve is installed in a process, the actual flow characteristics of the valve from 0 to 100% valve lift will be also determined by how the pressure drop varies with the flow in addition to the variation of Cv with %lift.

A valve having Equal percentage characteristic (inherent) will produce equal % increments in Cv for equal % increments in the valve lift. The % increment in Cv depends on the rangeability of the valve. Valves with high rangeability produce higher % increments in flow rate. The relation between the valve lift (%) and the Cv at that lift is given by

Cv = (R)^x * Cvmax / R

where,

x=valve lift expressed in fraction (full close = 0, full open = 1)

Cvmax = Cv at full open (100% lift)

R = Rangeability of the valve

For example a valve having a rangeability of 50, will produce 48% changes in the flow for every 10% change in the lift.

So in short, equal increments in valve lift (10%) anywhere in the control range produces equal increments in Cv (48%).

Hope this helps.

Ronald Deepak

jronalddeepak@gmail.com

Thank you very much. That was the explanation I needed.

The main application for using an equal percentage valve characteristic is where it is controlling the flow generated from a pump. As the valve opens the inlet pressure falls. The =% characteristic increases the port area as the valve opens up. The result is linear flow.

Common mistake is to use this type of valve in a process where the inlet pressure remains constant. That creates some interesting tuning problems. If you have that problem add a function generator within the logic on the output to the valve to compensate for the incorrect characteristic

Common mistake is to use this type of valve in a process where the inlet pressure remains constant. That creates some interesting tuning problems. If you have that problem add a function generator within the logic on the output to the valve to compensate for the incorrect characteristic

Allan,

can you explain in more details what kind of tuning problems can be experienced? In other words, how can I recognize wrong valve's type behavior? Do you have any example?

Thanks.

can you explain in more details what kind of tuning problems can be experienced? In other words, how can I recognize wrong valve's type behavior? Do you have any example?

Thanks.

The following tutorials address solutions for non-linear valves:

"How Valve Performance Affects the Control Loop", from Chemical Engineering magazine:

http://www.expertune.com/r2p.asp?f=AList&l=articles/ValvePerf.pdf

(registration required)

"Linearize Your Process for Optimal Control Performance at Any Production Rate", from Control Magazine:

http://www.expertune.com/r2.asp?f=AList&l=MarketCONTROL97.html

3-minute WMV, "Linearize Your Process":

http://www.expertune.com/present.asp?name=Linear

(registration required)

...and many more articles on the effects of control valves on PID Loop Optimization:

http://www.expertune.com/r2.asp?f=AList&l=ArticlesFull.html#PID OptValves

-George

"How Valve Performance Affects the Control Loop", from Chemical Engineering magazine:

http://www.expertune.com/r2p.asp?f=AList&l=articles/ValvePerf.pdf

(registration required)

"Linearize Your Process for Optimal Control Performance at Any Production Rate", from Control Magazine:

http://www.expertune.com/r2.asp?f=AList&l=MarketCONTROL97.html

3-minute WMV, "Linearize Your Process":

http://www.expertune.com/present.asp?name=Linear

(registration required)

...and many more articles on the effects of control valves on PID Loop Optimization:

http://www.expertune.com/r2.asp?f=AList&l=ArticlesFull.html#PID OptValves

-George

The exponent in the inherent flow characteristic curve is constant (equal) ergo equal percent.

The nirvana is an installed curve with constant gain.

The nirvana is an installed curve with constant gain.

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