Updated spacing between @ latex expressions

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правка #10477 предыдущая #10182 GitHub c702ebe ← раньше
en/5.9.10.md +6 −6
@@ -75,23 +75,23 @@
and the state 4 is characterized by
$$P_4V_2=nRT_4$$
So, separating $P_4$ and putting it into above Poisson law
− $$T_4 = \frac{P_1V_1^\gamma}{nRV_2^{\gamma-1}} \quad(3)$$
+ $$T_4 = \frac{P_1V_1^\gamma}{nRV_2^{\gamma -1}} \quad(3)$$
Proceeding similarly, $T_3$ can be found. Applying Poisson Law for process $2-3$,
− $$P_2V_1^\gamma=P_3V_2^\gamma$$
+ $$P_2V_1^\gamma =P_3V_2^\gamma$$
State equation for state 3 is
$$P_3V_2=nRT_3$$
Separating $P_3$ and substituting into respective Poisson Law
− $$T_3 = \frac{P_2V_1^\gamma}{nRV_2^{\gamma-1}} \quad(4)$$
+ $$T_3 = \frac{P_2V_1^\gamma}{nRV_2^{\gamma -1}} \quad(4)$$
Substituting $(3)$ and $(4)$ into $(2)$, and as $C_v=iR/2$
− $$Q_{34} = \frac{i}{2}\frac{V_1^\gamma}{V_2^{\gamma-1}}\left(P_1-P_2\right) \quad(5)$$
+ $$Q_{34} = \frac{i}{2}\frac{V_1^\gamma}{V_2^{\gamma -1}}\left(P_1-P_2\right) \quad(5)$$
For the process $1-2$
$$Q_{12}=\Delta U_{12} = nC_v\left(T_2-T_1\right)$$
Taking account the state equations for states 1 and 2, temperatures can be found,
$$Q_{12}=nC_v\left(\frac{P_2V_1}{nR}-\frac{P_1V_1}{nR}\right)$$
$$Q_{12}=\frac{i}{2}V_1\left(P_2-P_1\right) \quad(6)$$
Putting $(5)$ and $(6)$ into $(1)$
− $$\eta = 1-\left(\frac{V_1}{V_2}\right)^{\gamma-1}$$
− Since $\gamma=C_p/C_v=1+2/i$ and $i=3$ (monoatomic gas),
+ $$\eta = 1-\left(\frac{V_1}{V_2}\right)^{\gamma -1}$$
+ Since $\gamma =C_p/C_v=1+2/i$ and $i=3$ (monoatomic gas),
$$\boxed{\eta = 1-\left(\frac{V_1}{V_2}\right)^{\frac{2}{3}}}$$
b) Let enumerate again the states following a clockwise way, this time starting in coordinate $(P_1,T_1)$. The efficience is
$$\eta = 1-\frac{|Q_c|}{Q_h} \quad(I)$$
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