Formula Rewriting for Different Units Help

[royblaze]

Homework Statement

Hello. I have a small question... How does one correctly "modify" an equation to work in different units?

The equation below gives the boiling temperature of isopropanol as a function of pressure:

T = B / (A - log₁₀P) - C

Where T is in kelvin, P is in bar, and

A= 4.57795
B= 1221.423
C= -87.474

Obtain an equation that gives the boiling temperature in degrees F, as a function of ln(P), with P in psi.

It must be in the form

T = B' / (A' - ln(P)) - C', where A', B', C' are different numerical values from those given above.

Homework Equations

log₁₀x = ln x / ln (10)

ln(ab) = ln(a) + ln(b)

1 bar = 14.5 psi
1 atm = 1.013 bar = 14.7 psi

Kelvin to degrees F: ((T - 273.15) * 1.8) + 32

The Attempt at a Solution

I worked it out, but when I try a "test" value of 1 atm (14.7 psi, 1.013 bar) the temperature values do not match after conversion!

If overall, the temperature T is in kelvin, then change the RHS into degrees F.

T = ([B / (A - ln(P*14.5 psi/bar)/ln(10)) - C] - 273.15)*1.8 + 32

T = ([B / ((A - ln(P) + ln(14.5))/ln(10)) - C] - 273.15)*1.8 + 32

I now multiply the B/A fraction by ln(10)/ln(10), and factor in the 1.8.

T = ([1.8*ln(10)B / (ln(10)A - ln(P) - ln(14.5)) - 1.8C] - 1.8*273.15) + 32

T = [1.8*ln(10)B / (ln(10)A - ln(P) - ln(14.5)) - 1.8(C + 273.15)] + 32

Now, I define

1.8*ln(10)B = B'

ln(10)A - ln(14.5) = A'

1.8(C + 273.15) + 32 = C'

So now I have

T = B' / (A' - ln(P)) - C'

A' = 7.866970777
B' = 5062.374706
C' = 302.2168 (I accounted for the negative so that it works in the equation written as is)

But when I test a pressure of 1 atm (1.013 bar = 14.7 psi), I get a temperature of 355K. This should correspond to a degrees F of around 180, but I'm getting around 600 when I use my new formula with pressure 14.7 psi.

Any help?

 

[Joffan]

To convert from psi to bar, divide (not multiply) by 14.5

So A' = ln(10)A + ln(14.5)

 

[royblaze]

Thanks! :D Got it.