# Efficiency of Diesel cycle

1. Apr 15, 2013

### xspook

1. The problem statement, all variables and given/known data

Derive a formula for the efficiency of the Diesel cycle, in terms of the compression ratio V1/V2

2. Relevant equations

e=$\frac{W}{Q_{h}}$
w= ∫pdV

3. The attempt at a solution

Now I know I should have used e=1-$\frac{Q_{c}}{Q_{h}}$ to get it but he said it is possible using e=$\frac{W}{Q_{h}}$

I know that W_{4-1} is equal to zero and W_{2-3} is equal to P_{2}(V_{3}-V_{2})

what I don't know is how he got

W_{1-2} = $\frac{e}{1-δ}$*cv$^{-δ+1}$ = $\frac{1}{1-δ}$*(P_{2}V_{2}-P_{1}V_{1})

and

W_{3-1} = $\frac{e}{1-δ}$*cv$^{-δ+1}$ = $\frac{1}{1-δ}$*(P_{4}V_{4}-P_{3}V_{3})

2. Apr 15, 2013

### Staff: Mentor

What are points 1,2,3, and 4? What is $\delta$?

3. Apr 15, 2013

### xspook

1,2,3 and 4 are the strokes of the cycle. Well the different points on the picture. and δ is supposed to be gamma, which is the adiabatic exponent.

I thought I could use

W=∫pdV = C$_{v}$(T$_{1}$-T$_{2}$
for the compression stroke...but I guess that is incorrect

4. Apr 15, 2013

### Staff: Mentor

You have to be more specific. The numbering of points on a cycle is arbitrary. For each part of the cycle, state something like: 1→2 isothermal expansion at $T_2$.

5. Apr 15, 2013

### xspook

From 1→2 is compression
From 2→3 is fuel injection/combustion
From 3→4 is the power stroke
From 4→1 is exhaust

6. Apr 15, 2013

### Andrew Mason

The ideal diesel cycle consists of a constant pressure expansion (2-3) followed by an adiabatic expansion (3-4) followed by constant volume cooling (4-1) followed by adiabatic compression (1-2).

So heat goes in only from 2-3 and heat goes out only from 4-1. Since W = Qh-Qc and η = W/Qh = 1-Qc/Qh you just have to deal with the two parts in which heat flows (ie. 4-1 and 2-3).

Can you work out Qh and Qc? (hint: it involves temperature change and heat capacity).

AM