Stuck on a formula

[SteveM]

Hello forum. I'm stuck trying to solve a problem. I hope I'm in the right forum. To keep it short, I'll post the problem. I am more than happy to provide more information.
The formula is a way to calculate flow rate based on an orifice size and pump pressure for heating oil nozzles.

x*(y/100)^.5

As an example, when
x=.75
y=145
The answer rounded to 2 decimal places is .90 Works fine.

So where I'm stuck...

What I am trying to figure out is a new formula that uses the previous result of .90, and if I change x to .65 get this new formula to provide an answer that calculates for y.

With the above numbers, when I change x to .65 in this new calculation, solving for y should equal 190. I can back into it with the first formula by consistently changing y until I get the .90 answer.

Trying to put is simpler terms, if I know the flow rate needed (.90 based on the first calculation) I want to know (figure out) if I select a smaller orifice (.65), what pump pressure will I need to maintain that flow rate (190).

Thanks for reading!
Steve

 

[blamocur]

If you need to keep the flow constant then [imath]x[/imath] and [imath]y[/imath] are interdependent -- can you figure out the formula for that dependency?

 

[SteveM]

Thanks for your reply. I had calculus in high school, in 1983! I appreciate your response but have no idea what you are talking about.
Once I use the first formula I get the flow rate .90 I'm trying to figure if I change the nozzle to .65, what pressure would I need to maintain that same .90 flow rate.
Obviously this doesn't work, but I can't figure out a better way to explain it but I'll keep trying.
I guess I'm basically asking if I take my know variables:
x*(y/100)^.5
.90=.75*(145/100)^.5
How would I solve for y (obviously not this way):
.90=.65*(y/100)^.5

Thanks again

 

[blamocur]

Let's try another hint: if you divide the orifice size by half (i.e. x = 0.45) how would you need to adjust the pressure y to maintain the same flow?

 

[blamocur]

Thanks for your reply. I had calculus in high school, in 1983! I appreciate your response but have no idea what you are talking about.
Once I use the first formula I get the flow rate .90 I'm trying to figure if I change the nozzle to .65, what pressure would I need to maintain that same .90 flow rate.
Obviously this doesn't work, but I can't figure out a better way to explain it but I'll keep trying.
I guess I'm basically asking if I take my know variables:
x*(y/100)^.5
.90=.75*(145/100)^.5
How would I solve for y (obviously not this way):
.90=.65*(y/100)^.5

Thanks again

Just curious: are you taking a class after a break of 38 years? If you are, what class is that?

 

[Dr.Peterson]

The formula is a way to calculate flow rate based on an orifice size and pump pressure for heating oil nozzles.

x*(y/100)^.5

As an example, when
x=.75
y=145
The answer rounded to 2 decimal places is .90 Works fine.

So where I'm stuck...

What I am trying to figure out is a new formula that uses the previous result of .90, and if I change x to .65 get this new formula to provide an answer that calculates for y.

I assume this is a real-life problem, for which it is appropriate just to give you the answer, with a nudge to help you recall the algebra needed for next time.

Evidently you have x = orifice size, y = pump pressure, and z = flow rate (in some units unstated), and you have learned a formula, [math]z = x\sqrt{\frac{y}{100}}[/math] You want to solve for y.

To do this, I would divide both sides by x, then square both sides, and then multiply by 100. The result is [math]y=100\left(\frac{z}{x}\right)^2[/math]
For x = 0.75 and z = 0.90, this gives [math]y=100\left(\frac{0.90}{0.75}\right)^2=144[/math] which is off because your z was rounded.

Change x to something else, and as long as the formula you started with still applies, you will get the new value of y.

 

[SteveM]

@blamocur Yes this is a real life problem, not taking any classes.
This is for nozzles used in heating oil-fired burners. Based on the attached chart, the goal was to accurately change to a smaller nozzle (orifice), and raise the pump pressure to maintain the same gallons per hour flow rate. Doing this allows the oil to atomize into smaller droplets, which allows the heating oil to burn cleaner. The chart only uses predetermined pump pressures.
The formula will be used in my little database app to do the calculations.

@Dr.Peterson Thank you very much! That was the formula I couldn’t figure out. Thank you for your solution and explanation.

 

[SteveM]

Unrelated, does this forum have a way to mark a post or answer as 'correct'? Some do, and that answer floats to the top. Thanks again.

 

[blamocur]

Unrelated, does this forum have a way to mark a post or answer as 'correct'? Some do, and that answer floats to the top. Thanks again.

Or, as in Stackoverflow, the answer with the most votes goes to the top. This seems to be important in forums where the same question is likely to be asked by more than one person -- the first person brings about a bunch of responses, after which the answers can be found by a search.

From what I've seen so far, which is not that much, most of the questions here are unique and are unlikely to be repeated by different participants. But I might be wrong.