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# Derivatives

## Derivatives

(OP)
Hello All.

How do I solve the power of e , I circled in image below? what is Xdx?

Thank you.

### RE: Derivatives

(OP)
Sorry I meant Ddx (not Xdx)

### RE: Derivatives

The "…al Profile" in the top LH corner of your image is consistent with Greg's impression of what the formula is for.  However several things about that formula seem a bit odd, not just the dx.  Could you please give us an explanation of what it is supposed to be for, and where it comes from.  Plus full definitions of all the terms, and the units in which the terms are to be expressed.

### RE: Derivatives

(OP)
This formula is for steady state thermal profile of a subsea pipe. I know the units are mismatch.I was looking to understand dx to solve the exponential value but I only had pipe length and no other varying parameter for x. But I believe GregLocock is right and I will proceed on my calcs with that. Thank you so much.

### RE: Derivatives

The fact that the exponent is not non-dimensional is what I was hinting at with my "several things about that formula seem a bit odd".  "A bit odd" is a massive understatement:  be VERY careful if you intend to proceed without digging further into the formula's provenance.

### RE: Derivatives

Looks like a really good problem for SMath... units and derivative... maybe try coding it up...

Dik

### RE: Derivatives

Actually the units look ok to me assuming (dx) has units of length (maybe it is shorthand for a small delta-X as opposed to a calculus symbol which would not be properly applied without another differential or integral symbol somewhere). I don't know what the physical problem is - I'm just looking at the units.

The numerator and denominator of the exponential argument both have units of watts per degrees Kelvin (W/K), so the argument ends up dimensionless.
Numerator = U D dx = (W/m^2*k) * (m) *(m) = W / K
Denominator = m-dot * cp = (kg/sec) * (Joule/[kg*K]) = Joule / [sec*K]) = W / K

=====================================
(2B)+(2B)' ?

### RE: Derivatives

So Cp is not a dimensionless coefficient (despite being presented as such in the hand annotations)?

### RE: Derivatives

Oh, I see what you're saying. Yes, the handwriter apparently forgot to label those units. I missed that he missed that. That's a good thing to question / clarify.

Cp is apparently referring to specific heat capacity (at a constant pressure) which would have SI units J/[kg*K] (I don't know what specific units were intended to accompany the value listed 1800)

=====================================
(2B)+(2B)' ?

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