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View Full Version : Show that (curl h/curl p)at contt temp=-cp(curl T/curlP)at constt h?


anzyy
Mar 1, 2011, 08:55 AM

Unknown008
Mar 1, 2011, 09:22 AM
I'm having a hard time guessing what curl h and curl p as well as cp mean... :(

jcaron2
Mar 1, 2011, 11:09 AM
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.

jcaron2
Mar 1, 2011, 12:57 PM
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 (http://en.wikipedia.org/wiki/Bridgman's_thermodynamic_equations):

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

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

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