Automotive Power Comparison: 5 Parts Air v 800 PSI w/ Diesel

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The discussion compares two engine configurations: one with 5 parts air at 400 psi compression and another with 1 part air at 800 psi compression. Engine 1 has a compression ratio of 10.59 and thermal efficiency of 54.4%, while Engine 2 has a higher compression ratio of 17.37 and thermal efficiency of 62.6%. Despite Engine 2's greater efficiency, calculations indicate that Engine 1 can produce approximately 13 times more power due to its larger displacement and air volume per cycle. The confusion arises from the term "parts of air," which participants clarify refers to the volume of air drawn in relative to engine displacement. Ultimately, Engine 1 is favored for power output, while Engine 2 is noted for its efficiency.
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What is better for power? 5 parts air at 400 psi compression 10 inch bore by 30 inch stroke compared to 800 psi compression with 1 part air with 10 inch bore with 10 inch stroke? Same air fuel ratio in both With diesel? Which will be more power and which will be more efficient? And why?
 
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Compression ratio:
r = \left(\frac{P}{P_0}\right)^\frac{1}{1.4}
where P_0 = 14.7 psi.

Engine 1 (P = 400 psi): r = 10.59.
Engine 2 (P = 800 psi): r = 17.37.

Diesel cycle thermal efficiency:

bfe6abacc1fdef67d1fc3dfcc72448eb.png


If we assume \alpha = 2 (cut-off ratio) and \gamma= 1.4, then:
\eta_{th} = 1 - \frac{1.17}{r^{0.4}}
Engine 1: \eta_{th} = 0.544.
Engine 2: \eta_{th} = 0.626.

Engine 2 is more efficient.

Air volume per cycle (at atmospheric pressure):
[Here, I'm not sure what you mean by «5 parts air» and «1 part air»; I'm assuming you mean one engine's rpm is 5X faster than the other one or one has 5 cylinders and the other one has 1 cylinder]
V \propto ND^2S
and the energy per cycle is:
E \propto \eta_{th}V
\frac{E_1}{E_2} = \frac{\left(\eta_{th}ND^2S\right)_1}{\left(\eta_{th}ND^2S\right)_2} = \frac{0.544 \times 5 \times 10^2 \times 30}{0.626 \times 1 \times 10^2 \times 10} = 13
Engine 1 should produce 13 times more power than engine 2.
 
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jack action said:
Compression ratio:
r = \left(\frac{P}{P_0}\right)^\frac{1}{1.4}
where P_0 = 14.7 psi.

Engine 1 (P = 400 psi): r = 10.59.
Engine 2 (P = 800 psi): r = 17.37.

Diesel cycle thermal efficiency:

bfe6abacc1fdef67d1fc3dfcc72448eb.png


If we assume \alpha = 2 (cut-off ratio) and \gamma= 1.4, then:
\eta_{th} = 1 - \frac{1.17}{r^{0.4}}
Engine 1: \eta_{th} = 0.544.
Engine 2: \eta_{th} = 0.626.

Engine 2 is more efficient.

Air volume per cycle (at atmospheric pressure):
[Here, I'm not sure what you mean by «5 parts air» and «1 part air»; I'm assuming you mean one engine's rpm is 5X faster than the other one or one has 5 cylinders and the other one has 1 cylinder]
V \propto ND^2S
and the energy per cycle is:
E \propto \eta_{th}V
\frac{E_1}{E_2} = \frac{\left(\eta_{th}ND^2S\right)_1}{\left(\eta_{th}ND^2S\right)_2} = \frac{0.544 \times 5 \times 10^2 \times 30}{0.626 \times 1 \times 10^2 \times 10} = 13
Engine 1 should produce 13 times more power than engine 2.
By parts of air i mean volume of air.
 
I still don't understand what you mean with that.. Volume has specific units, so giving us "parts of air" when you also mention bore, stroke, and compression pressure is all a bit confusing.
 
+1

The volume of air drawn in is usually roughly equal to the displacement of the engine. Engine 2 has three times the displacement of Engine 1.
 
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