How Is the Efficiency of the Diesel Cycle Derived Using Compression Ratios?

AI Thread Summary
The discussion focuses on deriving the efficiency of the Diesel cycle in relation to the compression ratio V1/V2. Participants clarify that the efficiency can be expressed as η = W/Qh, where W represents work done and Qh is the heat added during the cycle. Key points of the cycle include isothermal expansion, fuel injection/combustion, power stroke, and exhaust, with specific attention to the work done during each phase. The adiabatic exponent, denoted as δ (gamma), is crucial for calculations involving work and heat transfer. The conversation emphasizes the need to compute heat inputs and outputs based on temperature changes and heat capacities for accurate efficiency calculations.
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Homework Statement



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

Homework Equations



e=\frac{W}{Q_{h}}
w= ∫pdV



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})
 
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What are points 1,2,3, and 4? What is ##\delta##?
 
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
 
xspook said:
1,2,3 and 4 are the strokes of the cycle. Well the different points on the picture.
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##.
 
From 1→2 is compression
From 2→3 is fuel injection/combustion
From 3→4 is the power stroke
From 4→1 is exhaust
 
xspook said:
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
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
 
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