Ask Me Help Desk - Show that (curl h/curl p)at contt temp=-cp(curl T/curlP)at constt h?

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show that (curl h/curl p)at contt temp=-cp(curl T/curlP)at constt h?

I'm having a hard time guessing what curl h and curl p as well as cp mean... :(

Does h mean enthalpy, p mean pressure, T, mean temperature, and cp mean specific heat at constant pressure? Also, is there a difference between p and P?

Is this what you're saying?

$\left (\frac{\bigtriangledown \times h}{\bigtriangledown \times p} \right )_T=C_p\left (\frac{\bigtriangledown \times T}{\bigtriangledown \times P} \right )_h$

If so, I don't think that makes any sense because the curl operation results in a vector. Vector division, as such, is not defined.

By the way, are you sure you don't mean the partial derivative, rather than curl?

$\left ( \frac{\partial h}{\partial P} \right )_T=c_p\left ( \frac{\partial T}{\partial P} \right )_h$

This can be shown with a few of Bridgman's equations:

$\left ({\partial P} \right )_T=-1$

$\left ({\partial P} \right )_H=-C_p$

$\left ({\partial H} \right )_T=-\left ({\partial T} \right )_H$

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