Isentropic relations: Difference between revisions
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For a reversible process: <math>\delta w=pd(1/\rho)</math> and <math>\delta q=Tds</math> | For a reversible process: <math>\delta w=pd(1/\rho)</math> and <math>\delta q=Tds</math> | ||
<math | {{NumEqn|<math> | ||
de=Tds-pd\left(\frac{1}{\rho}\right) | de=Tds-pd\left(\frac{1}{\rho}\right) | ||
</math> | </math>|2}} | ||
Enthalpy is defined as: <math>h=e+p/\rho</math> and thus | Enthalpy is defined as: <math>h=e+p/\rho</math> and thus | ||
<math | {{NumEqn|<math> | ||
dh=de+pd\left(\frac{1}{\rho}\right)+\left(\frac{1}{\rho}\right)dp | dh=de+pd\left(\frac{1}{\rho}\right)+\left(\frac{1}{\rho}\right)dp | ||
</math> | </math>|3}} | ||
Eliminate $de$ in Eqn. \ref{eq:firstlaw:b} using Eqn. \ref{eq:dh} | Eliminate $de$ in Eqn. \ref{eq:firstlaw:b} using Eqn. \ref{eq:dh} | ||
<math | {{NumEqn|<math> | ||
Tds=dh-\cancel{pd\left(\frac{1}{\rho}\right)}-\left(\frac{1}{\rho}\right)dp+\cancel{pd\left(\frac{1}{\rho}\right)} | Tds=dh-\cancel{pd\left(\frac{1}{\rho}\right)}-\left(\frac{1}{\rho}\right)dp+\cancel{pd\left(\frac{1}{\rho}\right)} | ||
</math> | </math>|4}} | ||
<math | {{NumEqn|<math> | ||
ds=\frac{dh}{T}-\frac{dp}{\rho T} | ds=\frac{dh}{T}-\frac{dp}{\rho T} | ||
</math> | </math>|5}} | ||
Using <math>dh=C_p T</math> and the equation of state <math>p=\rho RT</math>, we get | Using <math>dh=C_p T</math> and the equation of state <math>p=\rho RT</math>, we get | ||
<math | {{NumEqn|<math> | ||
ds=C_p\frac{dT}{T}-R\frac{dp}{p} | ds=C_p\frac{dT}{T}-R\frac{dp}{p} | ||
</math> | </math>|6}} | ||
Integrating Eqn. \ref{eq:ds} gives | Integrating Eqn. \ref{eq:ds} gives | ||
<math | {{NumEqn|<math> | ||
s_2-s_1=\int_1^2 C_p\frac{dT}{T}-R\ln\left(\frac{p_2}{p_1}\right) | s_2-s_1=\int_1^2 C_p\frac{dT}{T}-R\ln\left(\frac{p_2}{p_1}\right) | ||
</math> | </math>|7}} | ||
For a calorically perfect gas, <math>C_p</math> is constant (not a function of temperature) and can be moved out from the integral and thus | For a calorically perfect gas, <math>C_p</math> is constant (not a function of temperature) and can be moved out from the integral and thus | ||
<math | {{NumEqn|<math> | ||
s_2-s_1=C_p\ln\left(\frac{T_2}{T_1}\right)-R\ln\left(\frac{p_2}{p_1}\right) | s_2-s_1=C_p\ln\left(\frac{T_2}{T_1}\right)-R\ln\left(\frac{p_2}{p_1}\right) | ||
</math> | </math>|8}} | ||
An alternative form of Eqn. \ref{eq:ds:c} is obtained by using <math>de=C_v dT</math> Eqn. \ref{eq:firstlaw:b}, which gives | An alternative form of Eqn. \ref{eq:ds:c} is obtained by using <math>de=C_v dT</math> Eqn. \ref{eq:firstlaw:b}, which gives | ||
<math | {{NumEqn|<math> | ||
s_2-s_1=\int_1^2 C_v\frac{dT}{T}-R\ln\left(\frac{\rho_2}{\rho_1}\right) | s_2-s_1=\int_1^2 C_v\frac{dT}{T}-R\ln\left(\frac{\rho_2}{\rho_1}\right) | ||
</math> | </math>|9}} | ||
Again, for a calorically perfect gas, we get | Again, for a calorically perfect gas, we get | ||
<math | {{NumEqn|<math> | ||
s_2-s_1=C_v\ln\left(\frac{T_2}{T_1}\right)-R\ln\left(\frac{\rho_2}{\rho_1}\right) | s_2-s_1=C_v\ln\left(\frac{T_2}{T_1}\right)-R\ln\left(\frac{\rho_2}{\rho_1}\right) | ||
</math> | </math>|10}} | ||
=== Isentropic Relations === | === Isentropic Relations === | ||
