In Excel jow do I solve for X in X^1.5?

[Shawn]

I have a formula for finding the flow over a horizontal weir. ( gpm = 3 *
weir L ins. * overflow ins. 1.5th) I want to set up an Excel formula to give
the result as the overflow inches. ("X" in "X^1.5") How do I do that?

 

[Rick Rothstein \(MVP - VB\)]

Show us the equation you have for calculating flow over a horizontal weir...
there is no way we can help you isolate the "X" in "X^1.5" unless we know
the rest of the equation.

Rick

 

[Bernard Liengme]

If I understand the question
In A1 type the word Length and in B1 enter the value of L
In A2 type the word Overflow and in B2 enter the value for O (1.5)
In A3 type the word Flow; in B3 type =3*B1*B2
(Press the ENTER key after you have completed typing in each cell)
If the formula is Flow=3*L*(O^1.5) use =3*B1*B2^1.5

Please return with clarification
best wishes

 

[Shawn]

Here is the formula set up to solve for MGD (Million Gallons per Day).
=3*936*A21^1.5*1440/1000000
936 is the length in inches.
A21 is a cell with the inches going over the weir.
1440/1000000 is to convert the gallons per minute to MGD.
What I want to do is set up another formula where the desired MGD could be
entered in a cell and the result would be the inches to let flow over the
weir.
(The practical application of all this is to set movable gates on a dam.)

 

[Shawn]

Sorry if I wasn't clear. What I'm trying to do is set up a formula based on
one given solving for the Overflow alone, not the Overflow to the 1.5th power.

 

[Rick Rothstein \(MVP - VB\)]

So then, your **equation** is this...

G = 3 * L * (I ^ 1.5) * 1440 / 1000000

where G is Million Gallons Per Day, L is Length and I is Inches (going over
the weir). Multiplying both sides by 1000000 and dividing both sides by L, 3
and by 1440 gives this rearrangement...

I ^ 1.5 = (1000000 * G) / (3 * 1440 * L)

Now, the 1.5 power is the same as the 3/2 power, so if we raise both sides
to the 2/3 power, we will have isolated I..

I = ((1000000 * G) / (3 * 1440 * L)) ^ 2/3

Assuming the Million Gallons Per Day value was stored in G1, and the Length
value was stored in L1, then your worksheet formula should be...

= ((1000000*G1)/(3*1440*L1))^(2/3)

Rick

 

[Shawn]

Thanks. I'm about to get off work now. I'll try it out when I come back
Monday.

 

[joeu2004]

Here is the formula set up to solve for MGD (Million Gallons per Day).
=3*936*A21^1.5*1440/1000000
936 is the length in inches.
A21 is a cell with the inches going over the weir.
1440/1000000 is to convert the gallons per minute to MGD.
What I want to do is set up another formula where the desired MGD
could be entered in a cell and the result would be the inches to let
flow over the weir.

I know nothing of the application, but is this simple algebra? If,
say, B21 is the MGD and you want A21 to be the inches going over the
weir, then:

A21^1.5 = B21*1000000/3/936/1440

log(A21^1.5) = log(B21*1000000/3/936/1440)

1.5*log(A21) = log(B21) + log(1000000) - log(3) - log(936) - log(1440)

log(A21) = (log(B21) + log(1000000) - log(3) - log(936) - log(1440)) /
1.5

A21: =10^((log(B21) + log(1000000) - log(3) - log(936) - log(1440)) /
1.5)

Check: In C21, write:

=3*936*A21^1.5*1440/1000000

Should approximately equal B21. Don't be surprised by some small
difference, the result of numerical approximations of the log's as
well as binary computer arithmetic anomalies. (I got an error of
about -7E-14.)

 

[joeu2004]

PS....

Here is the formula set up to solve for MGD (Million Gallons per Day).
=3*936*A21^1.5*1440/1000000
[....]
What I want to do is set up another formula where the desired MGD
could be entered in a cell and the result would be the inches to let
flow over the weir.

[...] is this simple algebra?  If, say, B21 is the MGD and you want A21
to be the inches going over the weir, then:

A21^1.5 = B21*1000000/3/936/1440

I certainly prefer Rick's approach of taking 2/3 power of both sides.
Not sure why he has two variables, where I have one. Perhaps Rick has
given this more thought than I have. But I would say:

A21: =(B21*1000000/3/936/1440) ^ (2/3)

Then I was going to point out a simplification of my own formula,
which I will now apply to the one above. That is equivalent to:

=B21^(2/3) * (1000000/3/936/1440)^(2/3)

The advantage of that is: (1000000/3/936/1440)^(2/3) can be computed
one time in some cell, say A1. Then we have:

A21: =A1 * B21^(2/3)

Again, check by writing in C21:

=3*936*A21^1.5*1440/1000000

 

[Rick Rothstein \(MVP - VB\)]

Here is the formula set up to solve for MGD (Million Gallons per Day).

=3*936*A21^1.5*1440/1000000
[....]
What I want to do is set up another formula where the desired MGD
could be entered in a cell and the result would be the inches to let
flow over the weir.

[...] is this simple algebra? If, say, B21 is the MGD and you want A21
to be the inches going over the weir, then:

A21^1.5 = B21*1000000/3/936/1440

I certainly prefer Rick's approach of taking 2/3 power of both sides.
Not sure why he has two variables, where I have one. Perhaps Rick has
given this more thought than I have. But I would say:

I thought you might like to use this formula in the future for other
weirs... the odds that the length of every weir you will deal with in the
future will be 936 is pretty slim<g>, so I used a variable for that term.
The remainder of your simplifications is fine... I didn't simplify the
constants because you didn't in your original posting.

Rick

 

[joeu2004]

[joeu2004 wrote:]

I certainly prefer Rick's approach of taking 2/3 power of both sides.
Not sure why he has two variables, where I have one.  Perhaps Rick
has given this more thought than I have.

I thought you might like to use this formula in the future for other
weirs... the odds that the length of every weir you will deal with in the
future will be 936 is pretty slim<g>, so I used a variable for that term.
The remainder of your simplifications is fine... I didn't simplify the
constants because you didn't in your original posting.

Y'mean the OP, not "you" (in response to my posting). Yup, you gave
this more thought than I did. I did not pay attention to the initial
posting by the OP. I simply jumped in the middle, where the OP had
replaced "L" with 936.

 

[Rick Rothstein \(MVP - VB\)]

I certainly prefer Rick's approach of taking 2/3 power of both sides.

Y'mean the OP, not "you" (in response to my posting).

Yes, the OP... sorry, I lost track of who I was responding to.

Yup, you gave this more thought than I did. I did not pay attention to
the initial posting by the OP. I simply jumped in the middle, where the
OP had replaced "L" with 936.

Actually, the OP did not replace L with 936... he hard-coded the 936 in, but
in a later posting explained that this was a length. I then generalized this
to a variable L so that he could calculate other weirs in the future using
the same formula.

Rick

 

[Shawn]

Thank you all. I tested the formula against an old table that dated bsck to
when the gates were installed in the '70's. That gave the flow over all 8
gates in 2.4 inch increments. The formula gives values only 2% greater. I
wanted a formula that we could use with our automation system.
Here is my actual formula as it will be used:
=(((1000000*B5)/((3*1440*936)*B6))^(2/3))/12
As I said, I'm using a formula for weirs applied to movable gates on a dam.
There are 8 gates, each of them is 78 feet across, and we actually look at
feet of water over the gates.
B5 is the desired MGD
B6 is the number of gates with water going over
"/12" is to convert from inches to feet going over
Thanks again for your help.