Transposition of Formula, hopefully a simple one

  • Thread starter cliopete
  • Start date
  • Tags
    Formula
In summary, the balance equation for an I/P system is Pout = (KcY/Ax)*Iin + FsZ/Ax, with Kc being the constant of proportionality and Fs representing the force exerted by the zero spring. To obtain a straight line response of 'A', suitable values for y and x must be determined. By setting the intercept (C) of the line to zero, the equation can be simplified to Pout = (KcY/Ax)*In. Using the given values of Pout and In from the graph, the equation can be rearranged to solve for Y, which is found to be equal to 50Ax/Kc. Since A and Kc are constants, there is
  • #1
cliopete
1
0

Homework Statement



The balance equation for a I/P system is:

Pout = (KcY/Ax)*Iin + FsZ/Ax

Where:

Kc = 2500
x+y = 0.1
Z = 0.075
A = 0.05

Kc is the constant of proportionality relating to the input coil and Fs is the force exerted by the zero spring.

a - Determine suitable values of y and x to give a straight line response of 'A' (shown on a graph I havnt put up here)

Homework Equations



Pout = (KcY/Ax)*Iin + FsZ/Ax

Y = Mx + C

The Attempt at a Solution



According to the graph, the intercept (C) of the line runs through zero, thus Its safe to discount the FsZ/Ax part of the equation, leaving me with:

Pout = (KcY/Ax)*In

(Y=mx)

According to the graph, reading values for Pout and In, these are:

@ Pout = 1.0bar then In = 20mA.

putting these values into the formula:

1.0 = (KcY/Ax)*0.02

Transposing gives

1.0/0.02 = KcY/Ax

thus

50 = KcY/Ax

and I know that A = 0.05 and Kc = 2500

50 = 2500*Y / 0.05*Ax

This is where I get a little confused as to the correct method of transposing to give x, I will run through the rest of my working now...

50*0.05 = 2500*Y / x (moving the 0.05 from the bottom of one side to the top - becoming a multiple)

thus

2.5 = 2500*Y/X

2.5/2500 = 0.001

0.001 = Y / X

as mentioned earlier on, x+y = 0.1
so logically, y = 100-x
inserting in the formula...

0.001 = 100 - x / x

shifting the 0.1 across..

0.001 -0.1 = - x / x

Heres the bit where I run into trouble, I am not sure how to deal with the -x / x in order to get it into one value,

Im a little rusty when I get a further into transposition but this is driving me up the wall as its the key moment in the question which unlocks the value for further working.

I tried multiplying each side by x to remove the fraction, ending up with:

-0.099 * (x) = -x * (x)

which I would then assume to give?

-0.099x = -x2 ?

Like I said short of going into see my tutor next week I am a little stuck, if it does square the negative x how do I remove the x on the other side? argh!

Looking back at the working I am not even entirely sure -0.099 is a correct value either but If someone could give guidance on how to deal with the x's that would be great...

thanks in advance...
 
Physics news on Phys.org
  • #2
cliopete said:

Homework Statement



The balance equation for a I/P system is:

Pout = (KcY/Ax)*Iin + FsZ/Ax

Where:

Kc = 2500
x+y = 0.1
Z = 0.075
A = 0.05

Kc is the constant of proportionality relating to the input coil and Fs is the force exerted by the zero spring.

a - Determine suitable values of y and x to give a straight line response of 'A' (shown on a graph I havnt put up here)

Homework Equations



Pout = (KcY/Ax)*Iin + FsZ/Ax

Y = Mx + C

The Attempt at a Solution



According to the graph, the intercept (C) of the line runs through zero, thus Its safe to discount the FsZ/Ax part of the equation, leaving me with:

Pout = (KcY/Ax)*In

(Y=mx)

According to the graph, reading values for Pout and In, these are:

@ Pout = 1.0bar then In = 20mA.

putting these values into the formula:

1.0 = (KcY/Ax)*0.02

Transposing gives

1.0/0.02 = KcY/Ax

thus

50 = KcY/Ax
I'm going to assume that your work up to here is correct. Your formatting makes it a little hard to read, since I have to parse KcY as meaning Kc * y, while Ax is A * x.

Since Y * Kc/(Ax) = 50, then Y * Kc = 50Ax, and Y = 50A x /Kc.

Assuming A and Kc are constants, there is a linear relationship between x and Y. (Why is Y capped by x not?)
cliopete said:
and I know that A = 0.05 and Kc = 2500

50 = 2500*Y / 0.05*Ax

This is where I get a little confused as to the correct method of transposing to give x, I will run through the rest of my working now...

50*0.05 = 2500*Y / x (moving the 0.05 from the bottom of one side to the top - becoming a multiple)

thus

2.5 = 2500*Y/X

2.5/2500 = 0.001

0.001 = Y / X

as mentioned earlier on, x+y = 0.1
so logically, y = 100-x
inserting in the formula...

0.001 = 100 - x / x

shifting the 0.1 across..

0.001 -0.1 = - x / x

Heres the bit where I run into trouble, I am not sure how to deal with the -x / x in order to get it into one value,

Im a little rusty when I get a further into transposition but this is driving me up the wall as its the key moment in the question which unlocks the value for further working.

I tried multiplying each side by x to remove the fraction, ending up with:

-0.099 * (x) = -x * (x)

which I would then assume to give?

-0.099x = -x2 ?

Like I said short of going into see my tutor next week I am a little stuck, if it does square the negative x how do I remove the x on the other side? argh!

Looking back at the working I am not even entirely sure -0.099 is a correct value either but If someone could give guidance on how to deal with the x's that would be great...

thanks in advance...
 

Related to Transposition of Formula, hopefully a simple one

1. What is transposition of formula?

Transposition of formula is the process of rearranging an algebraic equation to solve for a different variable. This is often done to isolate a specific variable and make it the subject of the equation.

2. Why is transposition of formula important?

Transposition of formula is important because it allows us to solve for different variables in an equation, making it a useful tool in problem-solving and understanding relationships between different quantities.

3. How do you transpose a formula?

To transpose a formula, you need to identify the variable you want to solve for and then use algebraic manipulations, such as applying inverse operations, to isolate that variable on one side of the equation. Remember to perform the same operation on both sides of the equation to maintain balance.

4. What are some common mistakes in transposing formulas?

Common mistakes in transposing formulas include forgetting to apply the same operation on both sides of the equation, making incorrect use of inverse operations, and forgetting to distribute a coefficient when isolating a variable.

5. Can you provide an example of transposing a formula?

Yes, for example, if we have the equation A = 2(B + C), and we want to solve for B, we can transpose the formula by first dividing both sides by 2 to isolate the parentheses: A/2 = B + C. Then, we can subtract C from both sides to isolate B: A/2 - C = B.

Back
Top