Thermodynamic processes: Difference between revisions

From Flowpedia
← Older edit
Nian (talk | contribs)
Bureaucrats, Interface administrators, Administrators
580 edits
VisualWikitext
No edit summary
No edit summary
 
(11 intermediate revisions by the same user not shown)
Line 1: Line 1:
[[Category:Compressible flow]]
[[Category:Compressible flow]]
[[Category:Thermodynamics]]
[[Category:Thermodynamics]]<!--
--><noinclude><!--
-->[[Category:Compressible flow:Topic]]<!--
--></noinclude><!--


__TOC__
--><nomobile><!--
 
-->__TOC__<!--
=== Specific Heat Relations ===
--></nomobile><!--
 
--><noinclude><!--
For thermally perfect and calorically perfect gases
-->{{#vardefine:secno|1}}<!--
 
-->{{#vardefine:eqno|23}}<!--
<math display="block">
--></noinclude><!--
\begin{aligned}
-->
&C_p=\frac{dh}{dT}\\
{{NumEqn|<math>
&C_v=\frac{de}{dT}
\end{aligned}
</math>
 
From the definition of enthalpy and the equation of state <math>p=\rho RT</math>
 
<math display="block">
h=e+\frac{p}{\rho}=e+RT
</math>
 
Differentiate Eqn. \ref{eq:enthalpy} with respect to temperature gives
 
<math display="block">
\frac{dh}{dT}=\frac{de}{dT}+\frac{d(RT)}{dT}
</math>
 
Inserting the specific heats gives
 
<math display="block">
C_p=C_v+R
</math>
 
Dividing Eqn. \ref{eq:specificheat:b} by <math>C_v</math> gives
 
<math display="block">
\frac{C_p}{C_v}=1+\frac{R}{C_v}
</math>
 
Introducing the ratio of specific heats defined as
 
<math display="block">
\gamma=\frac{C_p}{C_v}
</math>
 
Now, inserting Eqn. \ref{eq:gamma} in Eqn. \ref{eq:specificheat:c} gives
 
<math display="block">
C_v=\frac{R}{\gamma-1}
</math>
 
In the same way, dividing Eqn. \ref{eq:specificheat:b} with <math>C_p</math> gives
 
<math display="block">
1=\frac{C_v}{C_p}+\frac{R}{C_p}=\frac{1}{\gamma}+\frac{R}{C_p}
</math>
 
and thus
 
<math display="block">
C_p=\frac{\gamma R}{\gamma-1}
</math>
 
=== Isentropic Relations ===
 
First law of thermodynamics:
 
<math display="block">
de=\delta q - \delta w
</math>
 
For a reversible process: <math>\delta w=pd(1/\rho)</math> and <math>\delta q=Tds</math>
 
<math display="block">
de=Tds-pd\left(\frac{1}{\rho}\right)
</math>
 
Enthalpy is defined as: <math>h=e+p/\rho</math> and thus
 
<math display="block">
dh=de+pd\left(\frac{1}{\rho}\right)+\left(\frac{1}{\rho}\right)dp
</math>
 
Eliminate $de$ in Eqn. \ref{eq:firstlaw:b} using Eqn. \ref{eq:dh}
 
<math display="block">
Tds=dh-\cancel{pd\left(\frac{1}{\rho}\right)}-\left(\frac{1}{\rho}\right)dp+\cancel{pd\left(\frac{1}{\rho}\right)}
</math>
 
<math display="block">
ds=\frac{dh}{T}-\frac{dp}{\rho T}
</math>
 
Using <math>dh=C_p T</math> and the equation of state <math>p=\rho RT</math>, we get
 
<math display="block">
ds=C_p\frac{dT}{T}-R\frac{dp}{p}
</math>
 
Integrating Eqn. \ref{eq:ds} gives
 
<math display="block">
s_2-s_1=\int_1^2 C_p\frac{dT}{T}-R\ln\left(\frac{p_2}{p_1}\right)
</math>
 
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">
s_2-s_1=C_p\ln\left(\frac{T_2}{T_1}\right)-R\ln\left(\frac{p_2}{p_1}\right)
</math>
 
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 display="block">
s_2-s_1=\int_1^2 C_v\frac{dT}{T}-R\ln\left(\frac{\rho_2}{\rho_1}\right)
</math>
 
Again, for a calorically perfect gas, we get
 
<math display="block">
s_2-s_1=C_v\ln\left(\frac{T_2}{T_1}\right)-R\ln\left(\frac{\rho_2}{\rho_1}\right)
</math>
 
=== Isentropic Relations ===
 
Adiabatic and reversible processes, i.e., isentropic processes implies <math>ds=0</math> and thus Eqn. \ref{eq:ds:c} reduces to
 
<math display="block">
\frac{C_p}{R}\ln\left(\frac{T_2}{T_1}\right)=\ln\left(\frac{p_2}{p_1}\right)
</math>
 
<math display="block">
\frac{C_p}{R}=\frac{\gamma}{\gamma-1}
</math>
 
<math display="block">
\frac{\gamma}{\gamma-1}\ln\left(\frac{T_2}{T_1}\right)=\ln\left(\frac{p_2}{p_1}\right)\Rightarrow
</math>
 
<math display="block">
\frac{p_2}{p_1}=\left(\frac{T_2}{T_1}\right)^{\gamma/(\gamma-1)}
</math>
 
In the same way, Eqn. \ref{eq:ds:e} gives
 
<math display="block">
\frac{\rho_2}{\rho_1}=\left(\frac{T_2}{T_1}\right)^{1/(\gamma-1)}
</math>
 
Eqn. \ref{eq:isentropic:a} and Eqn. \ref{eq:isentropic:b} constitutes the isentropic relations
 
<math display="block">
\frac{p_2}{p_1}=\left(\frac{\rho_2}{\rho_1}\right)^{\gamma}=\left(\frac{T_2}{T_1}\right)^{\gamma/(\gamma-1)}
</math>
 
=== Flow Processes ===
 
<math display="block">
ds=C_v\dfrac{dT}{T}+R\dfrac{d\nu}{\nu}
ds=C_v\dfrac{dT}{T}+R\dfrac{d\nu}{\nu}
</math>
</math>|label=eq_process_ds_a}}


<math display="block">
{{NumEqn|<math>
d\nu=\dfrac{\nu}{R}ds-C_v\dfrac{\nu}{RT}dT=\dfrac{\nu}{R}ds-\dfrac{C_v}{p}dT
d\nu=\dfrac{\nu}{R}ds-C_v\dfrac{\nu}{RT}dT=\dfrac{\nu}{R}ds-\dfrac{C_v}{p}dT
</math>
</math>|label=eq_process_dnu}}


for an isentropic process (<math>ds=0</math>), <math>d\nu < 0</math> for positive values of <math>dT</math>.
for an isentropic process (<math>ds=0</math>), <math>d\nu < 0</math> for positive values of <math>dT</math>.


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


<math display="block">
{{NumEqn|<math>
dp=-\dfrac{p}{R}ds+C_p\dfrac{p}{RT}dT=-\dfrac{p}{R}ds+C_p\rho dT
dp=-\dfrac{p}{R}ds+C_p\dfrac{p}{RT}dT=-\dfrac{p}{R}ds+C_p\rho dT
</math>
</math>|label=eq_process_dp}}


for an isentropic process (<math>ds=0</math>), <math>dp > 0</math> for positive values of <math>dT</math>.
for an isentropic process (<math>ds=0</math>), <math>dp > 0</math> for positive values of <math>dT</math>.
Line 187: Line 43:
-->
-->


Since <math>\nu</math> decreases with temperature and pressure increases with temperature for an isentropic process, we can see from Eqn.~\ref{eqn:process:dnu} that <math>d\nu</math> will be greater at lower temperatures and thus isochores (lines of constant specific volume) will be closely spaced at low temperatures and more sparse at higher temperatures. For an isochore <math>dv=0</math> which implies
Since <math>\nu</math> decreases with temperature and pressure increases with temperature for an isentropic process, we can see from {{EquationNote|label=eq_process_dnu}} that <math>d\nu</math> will be greater at lower temperatures and thus isochores (lines of constant specific volume) will be closely spaced at low temperatures and more sparse at higher temperatures. For an isochore <math>dv=0</math> which implies


<math display="block">
{{NumEqn|<math>
0=\dfrac{\nu}{R}\left(ds-C_v\dfrac{dT}{T}\right) \Rightarrow \dfrac{dT}{ds}=\dfrac{T}{C_v}
0=\dfrac{\nu}{R}\left(ds-C_v\dfrac{dT}{T}\right) \Rightarrow \dfrac{dT}{ds}=\dfrac{T}{C_v}
</math>
</math>}}


and thus we can see that the slope of an isochore in a <math>T-s</math>-diagram is positive and that the slope increases with temperature.
and thus we can see that the slope of an isochore in a <math>T-s</math>-diagram is positive and that the slope increases with temperature.
Line 197: Line 53:
In analogy, we can see that an isobar (<math>dp=0</math>) leads to the following relation
In analogy, we can see that an isobar (<math>dp=0</math>) leads to the following relation


<math display="block">
{{NumEqn|<math>
0=\dfrac{p}{R}\left(C_p\dfrac{dT}{T}-ds\right) \Rightarrow \dfrac{dT}{ds}=\dfrac{T}{C_p}
0=\dfrac{p}{R}\left(C_p\dfrac{dT}{T}-ds\right) \Rightarrow \dfrac{dT}{ds}=\dfrac{T}{C_p}
</math>
</math>}}


and consequently isobars will also have a positive slope that increases with temperature in a <math>T-s</math>-diagram. Moreover, isobars are less steep than ischores as <math>C_p > C_v</math>.  
and consequently isobars will also have a positive slope that increases with temperature in a <math>T-s</math>-diagram. Moreover, isobars are less steep than ischores as <math>C_p > C_v</math>.  
Retrieved from "https://wiki.g3dflow.com/index.php?title=Thermodynamic_processes"