Isentropic relations: Difference between revisions

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=== First law of thermodynamics ===
=== First law of thermodynamics ===


First law of thermodynamics:
First law of thermodynamics:


<math display="block">
{{NumEqn|<math>de=\delta q - \delta w</math>}}
de=\delta q - \delta w
</math>


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 display="block">
{{NumEqn|<math>
de=Tds-pd\left(\frac{1}{\rho}\right)
de=Tds-pd\left(\frac{1}{\rho}\right)
</math>
</math>|label=eq-first-law-b}}


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 display="block">
{{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>|label=eq-dh}}


Eliminate $de$ in Eqn. \ref{eq:firstlaw:b} using Eqn. \ref{eq:dh}
Eliminate <math>de</math> in {{EquationNote|label=eq-first-law-b}} using {{EquationNote|label=eq-dh}}


<math display="block">
{{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>}}


<math display="block">
{{NumEqn|<math>
ds=\frac{dh}{T}-\frac{dp}{\rho T}
ds=\frac{dh}{T}-\frac{dp}{\rho T}
</math>
</math>}}


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 display="block">
{{NumEqn|<math>
ds=C_p\frac{dT}{T}-R\frac{dp}{p}
ds=C_p\frac{dT}{T}-R\frac{dp}{p}
</math>
</math>|label=eq-ds}}


Integrating Eqn. \ref{eq:ds} gives
Integrating {{EquationNote|eq-ds}} gives


<math display="block">
{{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>|label=eq-ds-b}}


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 display="block">
{{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>|label=eq-ds-c}}


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 {{EquationNote|label=eq-ds-c}} is obtained by using <math>de=C_v dT</math> in {{EquationNote|label=eq-first-law-b}}, which gives


<math display="block">
{{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>|label=eq-ds-d}}


Again, for a calorically perfect gas, we get
Again, for a calorically perfect gas, we get


<math display="block">
{{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>|label=eq-ds-e}}


=== Isentropic Relations ===
=== Isentropic Relations ===


Adiabatic and reversible processes, i.e., isentropic processes implies <math>ds=0</math> and thus Eqn. \ref{eq:ds:c} reduces to
Adiabatic and reversible processes, i.e., isentropic processes implies <math>ds=0</math> and thus {{EquationNote|label=eq-ds-c}} reduces to


<math display="block">
{{NumEqn|<math>
\frac{C_p}{R}\ln\left(\frac{T_2}{T_1}\right)=\ln\left(\frac{p_2}{p_1}\right)
\frac{C_p}{R}\ln\left(\frac{T_2}{T_1}\right)=\ln\left(\frac{p_2}{p_1}\right)
</math>
</math>}}


<math display="block">
<math display="block">
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</math>
</math>


<math display="block">
{{NumEqn|<math>
\frac{p_2}{p_1}=\left(\frac{T_2}{T_1}\right)^{\gamma/(\gamma-1)}
\frac{p_2}{p_1}=\left(\frac{T_2}{T_1}\right)^{\gamma/(\gamma-1)}
</math>
</math>|label=eq-isentropic-a}}


In the same way, Eqn. \ref{eq:ds:e} gives
In the same way, {{EquationNote|label=eq-ds-e}} gives


<math display="block">
{{NumEqn|<math>
\frac{\rho_2}{\rho_1}=\left(\frac{T_2}{T_1}\right)^{1/(\gamma-1)}
\frac{\rho_2}{\rho_1}=\left(\frac{T_2}{T_1}\right)^{1/(\gamma-1)}
</math>
</math>|label=eq-isentropic-b}}
 


Eqn. \ref{eq:isentropic:a} and Eqn. \ref{eq:isentropic:b} constitutes the isentropic relations
Eqn. {{EquationNote|eq-isentropic-a}} and Eqn. {{EquationNote|eq-isentropic-b}} constitutes the isentropic relations


<math display="block">
{{NumEqn|<math>
\frac{p_2}{p_1}=\left(\frac{\rho_2}{\rho_1}\right)^{\gamma}=\left(\frac{T_2}{T_1}\right)^{\gamma/(\gamma-1)}
\frac{p_2}{p_1}=\left(\frac{\rho_2}{\rho_1}\right)^{\gamma}=\left(\frac{T_2}{T_1}\right)^{\gamma/(\gamma-1)}
</math>
</math>}}