# I/P convertor equation finding 2 unknowns

tommoturbo

## Homework Statement

Basically its all on the attached sheet as i dont know how to stick images in the post.

im having trouble finding two distances x and y (x+y= 100mm) just need a little help rearranging the equation on the attached word doc.

But ‘A’ passes through origin so c = 0.

m = 1/20 bar per mA

We know the values of Kc and A and hence can find x and y. Sorry but I cant fathom how to rearrange it as A=5cm^2 this multiplied by an unknown number in mm means it is tiny and hence once divided by the top figure will give a massive value for Po. When we know it is 0.05?

(N.B. be sure to have all measurements in S.I units.)

## Homework Equations

Po= (Kc*y)/A*x *Iin + (Fs*z)/A*x

A= 5cm^2
Z= 75mm
x+y = 100mm
Kc= 2500N A^-1
Po is in bar and is 0.2 at 4mA
Iin is mA and is 4 at 0.2Bar

## The Attempt at a Solution

i cant attach any files im afraid as it wont work?

any help greatly appreciated

Mentor

## Homework Statement

Basically its all on the attached sheet as i dont know how to stick images in the post.

im having trouble finding two distances x and y (x+y= 100mm) just need a little help rearranging the equation on the attached word doc.

But ‘A’ passes through origin so c = 0.

m = 1/20 bar per mA

We know the values of Kc and A and hence can find x and y. Sorry but I cant fathom how to rearrange it as A=5cm^2 this multiplied by an unknown number in mm means it is tiny and hence once divided by the top figure will give a massive value for Po. When we know it is 0.05?

(N.B. be sure to have all measurements in S.I units.)

## Homework Equations

Po= (Kc*y)/A*x *Iin + (Fs*z)/A*x

A= 5cm^2
Z= 75mm
x+y = 100mm
Kc= 2500N A^-1
Po is in bar and is 0.2 at 4mA
Iin is mA and is 4 at 0.2Bar

## The Attempt at a Solution

i cant attach any files im afraid as it wont work?

any help greatly appreciated

Attaching files should work. Is it real big or something? Also, it's better to post as a PDF -- less chance of transfering worms etc. You can get a free PDF converter from PrimoPDF if you don't have one already.

Mentor
Since x + y = 100, you can replace y with 100 - x so that your equation for Po involves only one variable.

You haven't provided enough information for me to figure out what you're doing. E.g., what units is Kc in? I don't understand N/A. A usually represents amperes, but that doesn't seem to be what you're doing.

tommoturbo
Hi guys thanks for the replies sorry its so vague i did it all on a word doc importing graphs and the equation etc thinking i could post it but the uploader wont work.

the units for the constant Kc are in Newtons but it has an A to the minus 1 after it.

ill try and post it from home it might not work on the pc here.

tommoturbo
HI guys here is the notes i have which should clear things up i hope

#### Attachments

• image003.bmp
63.3 KB · Views: 459
• ICP TMA2 Q9.doc
39 KB · Views: 148
Mentor
But ‘A’ passes through origin so c = 0.

m = 1/20 bar per mA

Po= (Kc*y)/A*x *Iin + (Fs*z)/A*x
This formula is different from the one in the attachments. This is the one in your attached files:
$$P_o = \left( \frac{K_c y}{A x}\right)I_{in} + \frac{F_s z}{Ax}$$

The way you wrote it above indicates that only A is in the denominators, not Ax.

This problem is a lot more confusing that it should be. For one thing, A is being used for three different purposes, making it difficult to figure out what A represents in a given situtation. One use for A is identifying one of the two straight lines (A and B). Another use is area. The third use is for amperes.

The equation above gives Po as a linear function of Iin. Since line "A" (better, line 1) goes through the origin, (Fs z)/(Ax) must be zero.

You are given that m = (1/20) mA, which means that the expression that multiplies Iin must be equal to 1/20 mA, so solve the equation (Kc y)/(Ax) = 1/20.

A= 5cm^2
Z= 75mm
x+y = 100mm
Kc= 2500N A^-1
Po is in bar and is 0.2 at 4mA
Iin is mA and is 4 at 0.2Bar

The notation N A^(-1) is the same as N/A, or newtons per ampere.