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Y_Kistochka [10]
3 years ago
14

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.

Engineering
1 answer:
abruzzese [7]3 years ago
8 0

Answer:

1) The three possible assumptions are

a) All processes are reversible internally

b) Air, which is the working fluid circulates continuously in a closed loop

cycle

c) The process of combustion is depicted as a heat addition process

2) The diagrams are attached

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

The power developed in horsepower ≈ 45374 hP

Explanation:

1) The three possible assumptions are

a) All processes are reversible internally

b) Air, which is the working fluid circulates continuously in a closed loop

cycle

c) The process of combustion is depicted as a heat addition process

2) The diagrams are attached

5) The dimension of the cylinder bore diameter = 3.7 in. = 0.09398 m

Stroke length = 3.4 in. = 0.08636 m.

The volume of the cylinder v₁= 0.08636 ×(0.09398²)/4 = 5.99×10⁻⁴ m³

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

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

p₁ = 14.5 lbf/in.² = 99973.981 Pa

T₁ = 60 F = 288.706 K

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

Otto cycle T-S diagram

T₂ = 288.706*6.25^{0.393} = 592.984 K

The maximum temperature = T₃ = 5200 R = 2888.89 K

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

T₄ = 2888.89 / 6.25^{0.393} = 1406.5 K

Work done, 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 developed in an Otto cycle = W×Cycle per second

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

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A vector AP is rotated about the Z-axis by 60 degrees and is subsequently rotated about X-axis by 30 degrees. Give the rotation
Musya8 [376]

Answer:

R = \left[\begin{array}{ccc}1&0&0\\0&cos30&-sin30\\0&sin30&cos30\end{array}\right]\left[\begin{array}{ccc}cos 60&-sin60&0\\sin60&cos60&60\0&0&1\end{array}\right]

Explanation:

The mappings always involve a translation and a rotation of the matrix. Therefore, the rotation matrix will be given by:

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The matrix is given by the following expression:

\left[\begin{array}{ccc}1&0&0\\0&cos30&-sin30\\0&sin30&cos30\end{array}\right]\left[\begin{array}{ccc}cos 60&-sin60&0\\sin60&cos60&60\0&0&1\end{array}\right]

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3 years ago
Write a C program that asks the user to enter two numbers, obtains the two numbers from the user and prints the sum, product, di
Bogdan [553]

Answer:

View Image

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Astronauts who landed on the moon during the Apollo 15, 16, and 17 missions brought back a large collection of rocks to the eart
stiks02 [169]

Answer: a) W(earth) = 935.62 lbs

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Explanation:

a) From Newton's law, W = mg. Whether on the moon or on earth. Although, the mass of the rocks everywhere is the same, that is, mass of rocks on the moon = mass of rocks on earth.

W(moon) = mg(moon)

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m = W(moon)/g(moon) = 154/5.3 = 29.06 lb.s2/ft

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