Efficiency of a Diesel Engine

[akan]

Homework Statement

The figure shows the thermodynamic cycle of a diesel engine. This cycle differs from that of a gasoline engine in that combustion takes place isobarically. The compression ratio r is the ratio of maximum to minimum volume: r = V_1 /V_2. In addition, the so-called cutoff ratio is defined by r_c = V_3 /V_2.

Find an expression for the engine's efficiency, in terms of the ratios r and r_c and the specific-heat ratio y. Although your expression suggests that the diesel engine might be less efficient than the gasoline engine , the diesel's higher compression ratio more than compensates, giving it a higher efficiency (the ratio of the work delivered to the heat extracted during the combustion phase).

http://img89.imageshack.us/img89/5663/rw1867sf0.th.jpg http://g.imageshack.us/thpix.php

Homework Equations

W = (p1v1 - p2v2) / (y-1)
e = W / Q_h

The Attempt at a Solution

Q_h = p(V_3 - V_2)
W_12 = (p_1 V_1 - p_2 V_2) / (y - 1)
W_23 = p(V_3 - V_2) (though I think this one is not work delivered)
W_34 = (p_3 V_3 - p_4 V_4) / (y - 1)
W_41 = 0

W = (p_1 V_1 - p_2 V_2) / (y - 1) + (p_3 V_3 - p_4 V_4) / (y - 1)
= (p_1 V_1 - p_2 V_2 + p_3 V_3 - p_4 V_4) / (y - 1)

e = (p_1 V_1 - p_2 V_2 + p_3 V_3 - p_4 V_4) / [(y - 1) (p (V_3 - V_2))]

Since p_2 and p_3 are the same, this simplifies to...

e = (p_1 V_1 + p (V_3 - V_2) - p_4 V_4) / [(y - 1) (p (V_3 - V_2))]

Now, p_4 is not given and I cannot eliminate it so I don't know how to proceed from here. Please help.

 

[anathema]

Thank you for your post and for sharing your solution attempt. Your approach is on the right track, but there are a few corrections and clarifications that can be made.

Firstly, the work done by the engine is not just the work done during the combustion phase (W_23), but also includes the work done during the compression phase (W_12) and the expansion phase (W_34). So your expression for work done should be:

W = W_12 + W_23 + W_34

Secondly, the heat input (Q_h) should be equal to the work done during the combustion phase (W_23), since this is the only phase where heat is added to the system. So your expression for heat input should be:

Q_h = W_23 = p(V_3 - V_2)

Now, let's look at your expression for efficiency (e):

e = W / Q_h

= (W_12 + W_23 + W_34) / (p(V_3 - V_2))

= [(p_1 V_1 - p_2 V_2) / (y - 1) + p(V_3 - V_2) + (p_3 V_3 - p_4 V_4) / (y - 1)] / (p(V_3 - V_2))

= [(p_1 V_1 - p_2 V_2) + p(V_3 - V_2) + (p_3 V_3 - p_4 V_4)] / [(y - 1) p (V_3 - V_2)]

= [(p_1 V_1 + p V_3 - p V_2 - p_4 V_4)] / [(y - 1) p (V_3 - V_2)]

= [(p_1 V_1 + p (V_3 - V_2) - p_4 V_4)] / [(y - 1) p (V_3 - V_2)]

= [(p_1 V_1 + Q_h - p_4 V_4)] / [(y - 1) p Q_h]

= [(p_1 V_1 - p_4 V_4) + Q_h] / [(y - 1) p Q_h]

= [(p_1 V

 

[thomate1]

I would first like to commend the student for their attempt at finding an expression for the efficiency of a diesel engine. It is clear that they have a good understanding of the thermodynamic cycle and the relevant equations. However, there are a few points that could be improved upon.

Firstly, the equation for work done by the engine (W) should be the sum of the work done in each stage of the cycle, which includes not only the compression and expansion stages, but also the heat addition and rejection stages. So the correct expression for W should be:

W = (p1v1 - p2v2) + (p3v3 - p4v4)

Secondly, the efficiency (e) should be calculated as the ratio of the work done by the engine (W) to the heat added during the combustion phase (Q_h). The heat added during the combustion phase is given by:

Q_h = p(V3 - V2)

So the correct expression for efficiency should be:

e = W / Q_h = (p1v1 - p2v2 + p3v3 - p4v4) / (p(V3 - V2))

Now, to find an expression for the efficiency in terms of the ratios r and r_c and the specific-heat ratio y, we need to express the volumes in terms of these ratios. We know that:

r = V1 / V2

and

r_c = V3 / V2

Rearranging these equations, we get:

V1 = rV2

and

V3 = r_cV2

Substituting these values into the expression for efficiency, we get:

e = (p1rV2 - p2V2 + p3r_cV2 - p4V4) / (p(r_c - 1)V2)

Now, we can simplify this expression by noting that p2 = p3 and p4 = p1, since these pressures are the same in the compression and expansion stages. So we get:

e = (p1rV2 - p1V2 + p1r_cV2 - p1V4) / (p(r_c - 1)V2)

= p1(r + r_c - 1) / (p(r_c - 1)V2)

= p1(r + r_c - 1) / (pV3 - p