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2 months ago

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A four-cylinder, four-stroke internal combustion engine has a bore of 3.7 in. and a stroke of 3.4 in. The clearance volume is 16

% of the cylinder volume at bottom dead center and the crankshaft rotates at 2400 RPM. The processes within each cylinder are modeled as an air-standard Otto cycle with a pressure of 14.5 lbf/in. 2 and a temperature of 60 8 F at the beginning of compression. The maximum temperature in the cycle is 5200 8 R.
Based on this model,
1- Write possible Assumptions no less than three assumptions
2- Draw clear schematic for this problem
3- Determine possible Assumptions no less than three assumptions
4- Draw clear schematic for this problem.
5- calculate the net work per cycle, in Btu, and the power developed by the engine, in horsepower.

1 answer:

iogann1982 [368]2 months ago

8 0

1) The three possible assumptions are

a) All processes are internally reversible

b) Air, as the working fluid, circulates in a closed-loop

cycle

c) Combustion is represented as a heat-adding process

2) Diagrams are included

5) The net work per cycle is 845.88 kJ/kg

The horsepower produced is approximately 45374 hP

Explanation:

1) The three valid assumptions are

a) All processes are internally reversible

b) Air, the working fluid, moves continuously in a closed-loop

cycle

c) The combustion process is set as a heat addition step

2) Diagrams for illustration are provided

5) The cylinder bore diameter measures 3.7 in., equating to 0.09398 m

The stroke length is 3.4 in., approximately 0.08636 m.

The cylinder volume is calculated as v₁= 0.08636 ×(0.09398²)/4 = 5.99×10⁻⁴ m³

The clearance volume amounts to 16% of the cylinder volume = 0.16×5.99×10⁻⁴ m³

The clearance volume, v₂ is 9.59 × 10⁻⁵ m³

p₁ is 14.5 lbf/in.² = 99973.981 Pa

T₁ equals 60 F = 288.706 K

$\dfrac{T_{2}}{T_{1}} = \left (\dfrac{v_{1}}{v_{2}} \right )^{K-1}$

For the Otto cycle's T-S diagram,

T₂ calculates to 288.706* $6.25^{0.393}$ = 592.984 K

The peak temperature is T₃ = 5200 R = 2888.89 K

$\dfrac{T_{3}}{T_{4}} = \left (\dfrac{v_{4}}{v_{3}} \right )^{K-1}$

T₄ resolves to 2888.89 / $6.25^{0.393}$ = 1406.5 K

Work performed, W = $c_v$ ×(T₃ - T₂) - $c_v$ ×(T₄ - T₁)

0.718×(2888.89 - 592.984) - 0.718×(1406.5 - 288.706) = 845.88 kJ/kg

The power generated in the Otto cycle = W×Cycle per second

= 845.88 × 2400 / 60 = 33,835.377 kW = 45373.99 ≈ 45374 hP.

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Answer:

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Explanation:

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