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# Solve for B when X = 0

Posted on 2014-02-26
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I need to try to solve, in reverse, an equation that boils down to the following:

(A+B) - (C x D) - E = X

What I need is a way to solve for "B", when X = 0

Help?
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Accepted Solution

phoffric earned 250 total points
>> (A+B) - (C x D) - E = 0
Add (C x D) + E to both sides of the equation to keep both sides balanced:
A+B =  (C x D) + E
Subtract A from both sides of the equation to keep both sides balanced:
B =  (C x D) + E - A
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Expert Comment

B = E + (CxD) -A
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Expert Comment

Then you should check your work by plugging in some numbers, which I'll pick at random:
A = 10
C = 2
D = -13
E = -44
===========================
Solve for B when X = 0
B =  (C x D) + E - A
B = (2* (-13)) + (-44) - 10
= -26 - 44 -10
B = -80
===========================
Plug numbers back into original to see if equations still balance:
(A+B) - (C x D) - E = X
(10 + (-80)) - (2 * (-13)) - (-44) = 0
(10 - 80) + 26 +44 = 0
-70 + 70 = 0
0 = 0 Checked
===========================
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Author Comment

That was quick, You folks are good! I will plug this in tomorrow and test it out.
Let me explain this more completely as a word problem and see if I stated this right.

"x" is a pressure measurement through a filter medium. When there is no flow, the pressure should be zero. but because "Term B" can change during the process (and is the only term that can) , when flow stops, sometimes a value other than zero is shown for X.

My intention is to detect when there is no flow and zero "x" by recalculating for B. I will then plug B into the pressure calculation and that should zero x.

Does that change anything?
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Expert Comment

If I understand you correctly, in your equation, (A+B) - (C x D) - E = X,
A, C, D, and E are constants that will not change. Only B and X change, but as they change, the equation holds true. If that is the case, then when X is 0, your recalculation of B should give you the correct new value for B.

You might note that if you let K = A - (C x D) - E, where K is a constant, then your equation becomes: B + K = X
Notice that as X increases or decreases by 1, then so must B increase or decrease by 1, respectively.

If you originally had a value for B and X (where X is not equal to 0), then as X goes to 0, then to keep the equation balanced, B = -K.
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Author Comment

Hmmm, to re-state... "A" and "C" can change, but I can measure and verify those in realtime. I can't measure B directly. Its essentially filter fouling and its value is not accessible nor predictable.
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Expert Comment

>> (A+B) - (C x D) - E = X
As long as this equation is known to be true under all of your experimental scenarios, then as X changes (and you can measure X), and as A, C, D, and E change, then you can calculate B to be (C x D) + E - A  to know what B must be for the equation to hold true. (It does not matter whether D and E are constants or whether you can measure them, as long as you know what they are when X becomes 0.)
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Author Comment

I can't measure X, X is the result of the math on all the other values. I just know that under certain known conditions, like when there is no flow through the filter, X MUST be zero and if it isn't, then B must have changed from the constant it used to be.
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Expert Comment

>> X MUST be zero and if it isn't, then B must have changed from the constant it used to be.
>> and if it isn't
OK, B is supposed to be a constant that you cannot measure directly, and you are trying to determine what the B constant is. You also note that you cannot measure X directly. There is the presumption that X must be zero when there is no flow.

Two things:
1) How can you determine the and if it isn't part of X not being zero since you cannot measure X directly?
2) Is there a chance that there is some small minimum pressure threshold that must be exceeded to have a flow? That is, if the pressure on one side of the filter is very small (but greater than zero), then that pressure is not enough to get through the resistance of the filter.
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Expert Comment

I'm getting confused from a fluid dynamics point of view. I'm thinking it might be some kind of a tangential flow filter, but the two terms you have mentioned don't normally have the same units. If x is a pressure measurement, why does B have the same units yet is described as "fouling"?

Could you please label each of the terms?
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Author Comment

Fair enough. I will try to do this from memory, the code is at work.
(A+B) - (C x D) - E = X
(WaterLevel + MediaThickness) - (PressureTransmitterRawUnscaledValue * ScalingMultipler) - PressureTransmitterElevationDifference = FilterHeadLoss

All are in feet. Media thickness seems to vary due to loss of media during wash cycle and is under water with no way to measure it directly.
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Assisted Solution

byundt earned 250 total points
OK. So this seems more like a sand filter. http://faculty.washington.edu/markbenj/CEE483/filter%20hl%20&%20breakthrough%20example.pdf

At zero flow, I'd expect the MediaThickness and FilterHeadLoss both to drop out of the equation. Downstream pressure ought to equal the water level.

And at constant flow under clean conditions, there ought to be a pressure drop equal to a factor (greater than 1) times the media thickness. The term you call MediaThickness would equal that factor less one under clean conditions. And under fouled conditions, it would have one flow resistance for the fouled portion of the filter, and the original flow resistance for the unfouled bottom portion of the filter media.
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Author Comment

You are close enough, its a dual media filter but the idea is the same. At present we "assume" a certain media thickness and that assumption generally never permits the FilterHeadLoss to reach zero. I hope to recalculate this value each time the filter is stopped.
An additional benefit of this would appear to be that I can indirectly "measure" the media thickness. At this moment that isn't supported very accurately by the calculation, (i.e. actual measurement made a few months back was 44 inches and calculated it resolves to 4.13 feet). I really can't visualize how this is truly possible anyway, instead I will have essentially a correction factor and thats all.  At least thats how it seems to me at this time. I am going to survey the installation and verify the elevation of all the "known" factors to narrow the error.

Do you agree with this idea?
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Author Comment

phoffric
1) How can you determine the and if it isn't part of X not being zero since you cannot measure X directly?

The intention it to change B to make X = 0. This assumes that there is zero headloss when there is no flow. Its an assumption, but one that is reasonable. I then need to change the media depth until X = 0. Its an Auto Zero function.

2) Is there a chance that there is some small minimum pressure threshold that must be exceeded to have a flow? That is, if the pressure on one side of the filter is very small (but greater than zero), then that pressure is not enough to get through the resistance of the filter.

It is a vertical filter so there is always pressure on both sides of the filter. There is a valve at the bottom that is either open or shut.  This valve determines when flow, therefore differential pressure (Headloss),  is even possible across this vertical media bed.
The point at which the valve is SHUT is when I want to zero the headloss (X) using media depth (B).

As I said in a previous post, Media depth may become nothing more than a compensation factor.  What I am asking for may not be possible.
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Expert Comment

If you close the valve underneath the filter, then the filter media is no longer a factor in the problem. It becomes strictly a water hydraulics problem. Pressure under the filter (measured in feet of water) will equal the water level.

The pdf file I linked to confirms this with its discussion of how to solve problem b (in the pdf).

The formula you are trying to plug into is only relevant when there is flow through the filter.

In other words, even if you could solve for B when x is zero, the answer would be a meaningless artifact of test conditions.
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Author Comment

Thats what I was starting to understand.  I did write some code today using the math to figure out what will create zero when the valve is shut. I store that meaningless number as an offset and use it in the headloss calcs when the valve opens. Only tried it for an hour but it seems to work so far. Time will reveal the truth.
But sadly, as you said, it doesn't represent the media depth, only the pressure transmitter vertical offset.
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Author Closing Comment

Thank you both for the Solution and the follow up.
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