Thermodynamic processes: Difference between revisions

Revision as of 09:44, 30 March 2026

d s = C_{v} \frac{d T}{T} + R \frac{d ν}{ν}

(Eq. 1.8)

d ν = \frac{ν}{R} d s - C_{v} \frac{ν}{R T} d T = \frac{ν}{R} d s - \frac{C_{v}}{p} d T

(Eq. 1.9)

for an isentropic process ( $d s = 0$ ), $d ν < 0$ for positive values of $d T$ .

d s = C_{p} \frac{d T}{T} - R \frac{d p}{p}

(Eq. 1.10)

d p = - \frac{p}{R} d s + C_{p} \frac{p}{R T} d T = - \frac{p}{R} d s + C_{p} ρ d T

(Eq. 1.11)

for an isentropic process ( $d s = 0$ ), $d p > 0$ for positive values of $d T$ .

Since $ν$ decreases with temperature and pressure increases with temperature for an isentropic process, we can see from (Eq. 1.9) that $d ν$ 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 $d v = 0$ which implies

0 = \frac{ν}{R} (d s - C_{v} \frac{d T}{T}) \Rightarrow \frac{d T}{d s} = \frac{T}{C_{v}}

(Eq. 1.12)

and thus we can see that the slope of an isochore in a $T - s$ -diagram is positive and that the slope increases with temperature.

In analogy, we can see that an isobar ( $d p = 0$ ) leads to the following relation

0 = \frac{p}{R} (C_{p} \frac{d T}{T} - d s) \Rightarrow \frac{d T}{d s} = \frac{T}{C_{p}}

(Eq. 1.13)

and consequently isobars will also have a positive slope that increases with temperature in a $T - s$ -diagram. Moreover, isobars are less steep than ischores as $C_{p} > C_{v}$ .

@@ Line 14: / Line 14: @@
 {{NumEqn|<math>
 ds=C_v\dfrac{dT}{T}+R\dfrac{d\nu}{\nu}
-</math>}}
+</math>|label=eq_process_ds_a}}
 {{NumEqn|<math>
 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>.
-<math display="block">
+{{NumEqn|<math>
 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
-</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>.
@@ Line 42: / Line 42: @@
 -->
-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>
 =\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.
@@ Line 52: / Line 52: @@
 In analogy, we can see that an isobar (<math>dp=0</math>) leads to the following relation
-<math display="block">
+{{NumEqn|<math>
 =\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>.

Thermodynamic processes: Difference between revisions

Revision as of 09:44, 30 March 2026

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