In screencast we'll take a look at the differences between the inherent characteristics of valves
and the installed characteristics of valves, why they are important, and how to conduct
a calculation with them. So the inherent characteristics represent the characteristics of the valve
really as if the valve was by itself. In other words, minimal piping and no other unit operations.
And the important part about this in terms of the valve equation is that delta p is a
constant regardless of the amount of flow going through your control valve. This would
really represent buying it off the shelve and just running the valve, the problem is that
often times when we are going to want to use control valves, or valves of any kind, there
not going to be just by themselves in a little bubble, they are going to be in line with
potentially long amounts of piping, various unit operations such as heat exchangers and
reactors and separators and the like. Because of the presence of these other unit operations
the way the valve acts by itself is not going to be the way it works when it is put in a
process, and the represents the installed characteristics. And the important point issue
with the valve being the process is that delta p is now a function of your flow rate q. When
you have a system your fluid is going to start at some particular pressure, lets say five
psi, and then its going to end at some separate pressure, lets say one psi, its open to the
environment. So therefore as it goes through its going to go through its going to go through
a pressure drop, and this pressure drop can be caused by friction losses of some sort.
These friction losses can come from the various pieces of equipment. If I had just a control
valve in this start to end the valve would take the entire pressure drop, so therefore
inherently the delta p for the valve would be four psi. However from an installed standpoint
the pressure drop would most likely be less than four psi because it goes through other
pieces of equipment, heat exchangers, etc. the pressure is going to be dissipated because
of other aspects of a mechanical energy balance, with one example being friction, also velocity
changes, etc. Another important issue here is that inherent characteristics and installed
characteristics have a great affect on the selection of valve that you use. So if we
look at our two common types of valves, linear valves and equal percentage valves, the definitions
have been established for systems with a constant pressure drop, so therefore for a linear valve
the flow rate is directly proportional to the lift, and for equal percentage it is related
so that there is an equal percentage change in flow for a change in the lift, which represents
different things, but the question represents which one do we want? We desire to have a
valve that has a gain that is as close to constant as possible, so what we would want
to have here is we would want either df dl, where f is the valve characteristic, or that
dq dl, where q is the flow rate, is approximately constant. f and q are proportional to each
other, just remembering the design equation for a control valve. So at first glance this
looks like our choice is really simple because by definition linear valves have their flow
rate directly proportional to the lift, so therefore dq dl there will always be a constant.
That is only true for the inherent valve, therefore if your system is operating such
that your pressure drop across the valve is either constant or approximately constant,
than you are going to want a linear valve. For processes where the pressure drop across
the valve is not constant because of the installed equipment aside from the valve, whats shown
to happen is that the equal percentage valve, which by nature is non linear for an inherent
system, becomes much more linear for an installed valve. So to see this difference between inherent
and installed characteristics we will look at a system where we are trying to flow two
hundred gpm of water through a control valve with a particular Cv, and we want to see how
open our valve needs to be for both a linear valve and a equal percentage valve for two
situations. One where we have a pressure drop of 10 psi constant, independent of the flow
rate of water, and then we have one where the pressure across the valve can be represented
by the function provided, which shows delta p as a function of q. So to do this we can
start with the valve equation here and recognize that our goal here is to find the openness
of the valve, and the openness of the valve is the lift, represented by a cursive l, and
this is related to the valve characteristic f. We want to solve for f, and this will leave
us with f equals q over Cv multiplied by the square root of gs, the specific gravity, divided
by the pressure drop across the valve. So for a linear valve, f is just equal to l.
So for the linear valve for part (a) we can just note that l will equal q over Cv multiplied
by the square root. And here our flow rate is 200 gpm, we have our value for the valve
110 gpm per psi to the one half, and then we have the square root of the specific gravity,
which for water will be 1, divided by the pressure drop across the valve, which is 10
psi. And when we do this we find that our valve openness is 0.575
We can do this exact same equation for part (b), but the only difference here is that
the pressure drop has changed. So instead of it being 10 psi, it is now 10 minus .03
times 200. we end up with .909. So therefore in our case here the valve has to be much
more open for this, mainly because of the fact that there is less of a driving force
present because of the reduced pressure drop, which if we were to do the actual calculation would
represent that our delta p is now 4 psi. So analysis for the equal percentage valve, the
major difference here is that f does not equal just the lift, in fact its a relationship
of r raised to the l minus 1, where r is just a parameter of the equal percentage valve,
which is usually somewhere in the range of 20 to 50. So if we wanted to use this equation
we would have to say the r raised to the l minus 1 equals q over Cv times the square
root of gs over the pressure drop of the valve. What we then do is take either the natural
log or the log of both sides. Here taking advantage of the rule of logs we can write
this as l minus one log of r equals the log of q over Cv times the square root of gs over
delta pv. After doing some relatively basic algebra we can solve for l as l equals one
plus the log term we just had on the right, divided by the log of r. For this analysis
we'll assume that r is 40. So for part (a) when the pressure drop is 10 we'll have an
answer of 0.850. For part (b), again remembering the fact that here where delta pv is now equal
to 4, without showing the math we end up with an answer of 0.974. The valve is much more open
for the equal percentage case which makes sense based on the definition of how equal
percentage valve works, but you'll also notice the fact that the difference between the lifts in
part (a) and part(b) is much less significant in the case of the equal percentage
valve versus the linear valve. so to determine which one of these valves is of interest what
we would need to do is not focus on just one flow rate but focus on trying to find pairs
of the lift and the flow rate for a variety of flow rates, and when trying to select a
valve here our two criteria would be 1.) is the valve gain relatively constant over the
flow range of interest, and 2.) is the valve nearly fully open or nearly fully closed over
this range. the reason why one is important is because if the gain is not that constant
and gets too large in places what will happen is the flow can tend to oscillate, which means
that small changes can lead to very large flow rates. However if the valve gain gets
to small in comparison what this means is the fact that the process will be sluggish
because it won't be able to respond as quickly to change because of the size of the gain.
Additionally you are going to want the valve to be away from being fully open and fully
closed.The reason why is because of the fact that the valves don't operate as well from
a reproducibility standpoint when you are on the edges, whether it be too closed or
too open. So what we are going to do now is show a graph of how these systems work, both
from an inherent standpoint and an installed standpoint. So if we look at the graph of
these flow rates versus lift what you'll notice here is that there are (a) is a very big difference
between the inherent and installed characteristics of both valves, the dashed lines representing
the installed characteristics for both, and we have the blue representing the linear and
the red representing the equal percentage, and what you see is a very big difference
in their trajectories. Another important difference to realize here is the fact that the linear
curve has become much less linear, and has a significant amount of curvature to it, and similarly what we see with
the equal percentage curve as we go from inherent to installed is we actually see the fact that
the curve has flattened out a bit, making it more linear. So therefore this shows that
in some situations we would actually chose a equal percentage valve because of being
in this more linear range. So if we look at where we are interested in, is we're interested
in the flow rate up to approximately 200 gpm, and what we see here this valve Cv value is
a very poor choice once this is installed. The reason why is the fact that in both cases
we are operating very close to a fully open valve, something that we found by carrying
out the earlier calculations. Therefore if I were to choose a linear equal percentage
valve here, I really wouldn't aim for either one because we would want a different valve.
We would want a valve with a higher Cv value, one that would allow for a higher flow rate
to go through, which recall the definition of Cv represents, in the units of gpm per
psi to the one half, it represents how much water flows through a valve when it is fully
open, and the pressure drop across the valve is 1 psi, with the flow being water. So here
we want to choose a higher Cv, but also if I look at this aside from that is I am in
a range where the equal percentage output where I am actually relativly quite linear
and not terribly flat, two positive things in terms of selecting a valve, where as for the linear
installed since the slope is so small it would be bit sluggish to respond because in order
to make a change in flow rate it would involve a significant change in the valve. You have
to be very careful in how you select a valve because for example in both of these processes
if I am running it with the aforementioned process where I have that delta p relationship,
these valves will be unable to handle flow rates much greater than 200 gpm. So therefore if
this process can fluctuate, say between 150 and 250 gpm, this can cause significant problems
for your process, mainly because caused by the pressure drop issue. So this screencast
hopefully showed the importance of inherent vs installed characteristics and how that
can be really important in terms of selecting the type of valve you want to use for your
process, both from a Cv standpoint, and from a linear versus equal percentage standpoint.