Reading Pressure accurately on a P-H semi-log diagram 3

[jkal321]

Hi,

I feel I should know how to read a logarithmic graph but I seem to have been wrong all along and I need someone to correct me.
I am taking the PE exam in HVAC soon. I have been going over the refrigeration cycle problems in a reference book I have.

Whenever I measure the pressure of the diagram, I am always off from the answer which is approximate and close enough to the answer.
But one answer, I was way off because I read the pressure wrong or not close enough to the instructor.

As an example, I have attached a marked up chart showing the location: [ The area in between 200-400 psi ] accordingly to the graph is divided into 4 divisions but logarithmic divisions, the pressure was read by the instructor [320 psi], while I read it [350 psi] which made a big difference in the answer.

Could you please help and if there is a rule of thumb to read these value correctly vs. the linear scales.


Thank you,

 

[Latexman]

Excellent question! As a matter of fact, this exact same question is what led to the start of my "Handy Dandy Reference Materials" 3-ring notebook back in 1979! Now, many notebooks later, it consumes most of a 3 ft. bookshelf. Here's my "Log Graph Linear Interpolation":

Good luck,
Latexman

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[LiteYear]

Nice guide Latexman. Very useful if you have a rule and a calculator. In the absence of those it's nice to have a bit of intuition.

Best way I've found is to find the closest two gridlines and figure out their value. Then find where the mid value of those two are - it's always 70% of the way from the lower to the upper (log(5)). Then, if necessary, note that the value of the point halfway between the grid lines is 32% of the base for that division - if you're on a nice base 10 division (say from 10 to 100) that's easy to do in your head (it'll be 32). Otherwise the midpoint between two grid values a and b is a^(1/2)*b^(1/2) which is a lot harder to do in your head.

Looking at lots of log graphs helps too of course.

But anyway, in answer to your question, your instructor is wrong. Not only does the value lie on a grid line, which can be easily determined to be the 350 grid line, even if the grid lines were wrong, measuring between the 200 and 400 lines gives 350:

Using pixels on my screen as the unit, the red line is 242/295 of the way from 200psia towards 400psia. The value is therefore 200^(53/295)*400^(242/295) = 353 which is close enough to 350.

By the way, a handy formula I derived in writing this post:

y = x^(1-r)*z^r

where y is the point on the log axis between x and z at a (linear) ratio of r.

 

[IRstuff]

A measurement in Visio results in 0.8079 for the fraction, so an alternate equation:

200 * 10^{log(400/200)*0.8079} = 350.1 or LB * (UB/LB)^0.8079

I find this a simpler form to understand, but some manipulation shows that it's equivalent to the LiteYear's equation

TTFN
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[jkal321]

Thank you all for your answers. I don't want to bore you further on this topic. I just have a few couple of questions to your responses:

Latexman, Attachment very nice. See the marked up attachment and please confirm, when I print your pdf file and measure your dimensions, they come up short and that in return will differ from your answers listed. Could you correct me and clarify. Unless otherwise the printed scale is not same as the original.

LiteYear, I like your approach because it is more practical if a ruler is not handy. But the 32% and 70% ratios, I couldn't follow. Especially that you used your screen pixels to come up with 242, 295 in your last equation to end up with 353. Am I correct?

IRstuff: Your equation is very straight forward. But you used your program to come with 0.8079. Is there another way to come up with this ratio?

 

[Latexman]

the printed scale is not same as the original.



Good luck,
Latexman

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[IRstuff]

You can use a ruler, or pixels, as LiteYear did.

TTFN
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[jkal321]

Clear. Thank you again for the replys and answers.