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

In summary, the 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).
  • #1
xspook
19
0

Homework Statement



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

Homework Equations



e=[itex]\frac{W}{Q_{h}}[/itex]
w= ∫pdV



The Attempt at a Solution



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

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} = [itex]\frac{e}{1-δ}[/itex]*cv[itex]^{-δ+1}[/itex] = [itex]\frac{1}{1-δ}[/itex]*(P_{2}V_{2}-P_{1}V_{1})

and

W_{3-1} = [itex]\frac{e}{1-δ}[/itex]*cv[itex]^{-δ+1}[/itex] = [itex]\frac{1}{1-δ}[/itex]*(P_{4}V_{4}-P_{3}V_{3})
 
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  • #2
What are points 1,2,3, and 4? What is ##\delta##?
 
  • #3
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[itex]_{v}[/itex](T[itex]_{1}[/itex]-T[itex]_{2}[/itex]
for the compression stroke...but I guess that is incorrect
 
  • #4
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##.
 
  • #5
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
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[itex]_{v}[/itex](T[itex]_{1}[/itex]-T[itex]_{2}[/itex]
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
 

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

1. What is the Diesel cycle?

The Diesel cycle is a thermodynamic cycle used in diesel engines to convert heat energy into mechanical work. It involves four processes: intake, compression, power and exhaust.

2. How does the efficiency of the Diesel cycle compare to other cycles?

The efficiency of the Diesel cycle is typically higher than that of other cycles, such as the Otto cycle used in gasoline engines. This is because the Diesel cycle operates at a higher compression ratio, leading to better utilization of heat energy.

3. What factors affect the efficiency of the Diesel cycle?

The efficiency of the Diesel cycle is affected by several factors, including compression ratio, air-fuel ratio, engine speed, and combustion temperature. Higher compression ratios and leaner air-fuel mixtures generally result in higher efficiency.

4. Can the efficiency of the Diesel cycle be improved?

Yes, the efficiency of the Diesel cycle can be improved through various methods such as turbocharging, intercooling, and exhaust gas recirculation. These techniques help to increase the amount of air and reduce the amount of fuel needed for combustion, resulting in improved efficiency.

5. How does the efficiency of the Diesel cycle impact the performance of a diesel engine?

The efficiency of the Diesel cycle plays a crucial role in the overall performance of a diesel engine. A higher efficiency means that the engine is able to convert more heat energy into mechanical work, resulting in better fuel economy and power output. This is why diesel engines are commonly used in heavy-duty vehicles that require high levels of efficiency and power.

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