Sarahisme Posted September 14, 2005 Share Posted September 14, 2005 hey heres the question & heres my answer, but i got stuck so yep any help would be greatly apprectiated! ok here is what i have got so far... [math] \epsilon = 1 - \frac{Q_{out}}{Q_{in}} [/math] [math] Q_{out} \ = \ |Q_{d->a}| \ = \ C_v|T_a - T_d| \ = \ C_v(T_d - T_a) [/math] [math] Q_{in} \ = \ Q_{b->c} = C_p(T_c - T_b) [/math] so [math] \epsilon \ = \ 1 - \frac{C_v(T_d - T_a)}{C_p(T_c - T_b)} [/math] but [math] \frac{C_p}{C_y} \ = \ \gamma [/math] So [math] \epsilon = 1 - \frac{T_d - T_a}{ \gamma (T_c - T_b)} [/math] then using PV = nRT [math] \epsilon = 1 + \frac{1}{ \gamma} \frac{P_aV_a - P_dV_d}{P_cV_c - P_bV_b} [/math] now using the fact that [math] V_a = V_d [/math] and [math] P_c=P_b [/math] [math] \epsilon = 1+ \frac{1}{ \gamma} \frac{V_a(P_a - P_d)}{P_c(V_c-V_b)} [/math] now dividing top and bottom by [math] V_aP_c [/math] [math] \epsilon = 1 + \frac{1}{ \gamma} \frac{\frac{P_a}{P_c} - \frac{P_d}{P_c}}{ \frac{V_c}{V_a} - \frac{V_b}{V_a}} [/math] now [math] PV^{ \gamma } = constant[/math] so [math] P_cV_c^{ \gamma } = P_dV_d^{ \gamma } [/math] then [math] \frac{P_d}{P_c} = \frac{V_c^{ \gamma }}{V_d^{ \gamma }} = ( \frac{V_c}{V_d} )^{ \gamma } [/math] and similarly [math] \frac{P_a}{P_b} = \frac{V_b^{ \gamma }}{V_a^{ \gamma }} = ( \frac{V_b}{V_a} )^{ \gamma } [/math] so [math] \epsilon = 1 + \frac{1}{ \gamma} \frac{( \frac{V_b}{V_a} )^{ \gamma } - ( \frac{V_c}{V_d} )^{ \gamma }}{ \frac{V_c}{V_a} - \frac{V_b}{V_a}} [/math] this is where i get stuck, any suggestions guys n' gals? Sarah Link to comment Share on other sites More sharing options...
Sarahisme Posted September 15, 2005 Author Share Posted September 15, 2005 anyone? lol, or is thermodynamics not a favourite around here Link to comment Share on other sites More sharing options...
Sarahisme Posted September 19, 2005 Author Share Posted September 19, 2005 anyone going to help me at all here? Link to comment Share on other sites More sharing options...
Sarahisme Posted September 23, 2005 Author Share Posted September 23, 2005 nevermind, i think what i've got is correct! Link to comment Share on other sites More sharing options...
Ivan Posted September 24, 2005 Share Posted September 24, 2005 Long time no see thermodynamic equations... I'm little bit rusty. However it seem to me that everything is correct. Fine example of diesel cycle. Link to comment Share on other sites More sharing options...
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