diesel cycle

[Sarahisme]

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

[Sarahisme]

anyone? lol, or is thermodynamics not a favourite around here

[Sarahisme]

anyone going to help me at all here?

[Sarahisme]

nevermind, i think what i've got is correct!

[Ivan]

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.