Question

Problema anexată, vă rog. Mulțumesc!

Answer 1

$Q_{23}=0=>p_2V_2^{\gamma}=p_3V_3^{\gamma}=>p_2=p_3(\frac{V_3}{V_2})^{\gamma}\\p_3=p_1;V_2=V_1=>p_2=p_1(\frac{V_3}{V_1})^{\gamma}|:p_1=>\frac{p_2}{p_1}= (\frac{V_3}{V_1})^{\gamma}=>(\frac{V_3}{V_1})^{\gamma}= \delta=>\frac{V_3}{V_1}=\delta^{\frac{1}{\gamma}}=>V_3=\delta^{\frac{1}{\gamma}}V_1\\p_1=p_3=const=>\frac{V_2}{T_2}=\frac{V_3}{T_3}=>\frac{V_3}{V_1}=\frac{T_3}{T_1}=> \frac{T_3}{T_1}=\delta^{\frac{1}{\gamma}}$

$V_1=V_2=const=>\frac{p_1}{T_1}=\frac{p_2}{T_2}=>\frac{p_2}{p_1}=\frac{T_2}{T_1}=>\frac{T_2}{T_1}=\delta=>T_2=\delta T_1$

$Q_{12}=\nu C_v(T_2-T_1)=\frac{\nu R(T_2-T_1)}{\gamma-1}=\frac{\nu R(\delta T_1-T_1)}{\gamma-1}=\frac{\nu RT_1(\delta -1)}{\gamma-1}>0=>Q_{12}=Q_p\\ Q_{23}=0$

$Q_{31}=\nu C_p(T_1-T_3)=\frac{\nu\gamma R(T_1-T_3)}{\gamma-1}=\frac{\nu\gamma R(T_1-\delta^{\frac{1}{\gamma}}T_1)}{\gamma-1}=\frac{\nu\gamma RT_1(1-\delta^{\frac{1}{\gamma}})}{\gamma-1}<0=>Q_{31}=Q_p$

$\eta=1-\frac{|Q_p|}{Q_c}=1-\frac{\frac{\nu\gamma RT_1(-1+\delta^{\frac{1}{\gamma}})}{\gamma-1}}{\frac{\nu RT_1(\delta -1)}{\gamma-1} }=1-\frac{\gamma (-1+\delta^{\frac{1}{\gamma}})}{\delta -1}$