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- Automotive
- Thread starter Hybrid_engine
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- #2

Nidum

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https://nathbeke.files.wordpress.co...and-operating-characterstics-of-ic-engine.pdf

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[tex]r = \left(\frac{P}{P_0}\right)^\frac{1}{1.4}[/tex]

where [itex]P_0[/itex] = 14.7 psi.

Engine 1 ([itex]P[/itex] = 400 psi): [itex]r = 10.59[/itex].

Engine 2 ([itex]P[/itex] = 800 psi): [itex]r = 17.37[/itex].

Diesel cycle thermal efficiency:

If we assume [itex]\alpha = 2[/itex] (cut-off ratio) and [itex]\gamma= 1.4[/itex], then:

[tex]\eta_{th} = 1 - \frac{1.17}{r^{0.4}}[/tex]

Engine 1: [itex]\eta_{th} = 0.544[/itex].

Engine 2: [itex]\eta_{th} = 0.626[/itex].

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]

[tex]V \propto ND^2S[/tex]

and the energy per cycle is:

[tex]E \propto \eta_{th}V[/tex]

[tex]\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[/tex]

Engine 1 should produce 13 times more power than engine 2.

- #4

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By parts of air i mean volume of air.

[tex]r = \left(\frac{P}{P_0}\right)^\frac{1}{1.4}[/tex]

where [itex]P_0[/itex] = 14.7 psi.

Engine 1 ([itex]P[/itex] = 400 psi): [itex]r = 10.59[/itex].

Engine 2 ([itex]P[/itex] = 800 psi): [itex]r = 17.37[/itex].

Diesel cycle thermal efficiency:

If we assume [itex]\alpha = 2[/itex] (cut-off ratio) and [itex]\gamma= 1.4[/itex], then:

[tex]\eta_{th} = 1 - \frac{1.17}{r^{0.4}}[/tex]

Engine 1: [itex]\eta_{th} = 0.544[/itex].

Engine 2: [itex]\eta_{th} = 0.626[/itex].

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]

[tex]V \propto ND^2S[/tex]

and the energy per cycle is:

[tex]E \propto \eta_{th}V[/tex]

[tex]\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[/tex]

Engine 1 should produce 13 times more power than engine 2.

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- #6

CWatters

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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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