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3 years ago

What is the function of the following: 1- Oil rings 2- Flywheel 3- Timing gears

Engineering

1 answer:

Tanya [424]3 years ago

4 0

Answer:

Explanation:

The functions of the following are as follows:

1). Oil Rings:

Regulates oil within cylinder walls.
Helps to keep cylinder walls lubricated
Prevent Heat transfer

Reduce friction between piston and cylinder

2). Fly Wheel:

It is used to store rotational energy
It resist any change in rotational speed due to its moment of inertia
It helps to control the orientation of the mechanical system

3). Timing gears:

It provides synchronization in the rotation of crankshaft and the camshaft so as to provide proper valve opening and closing time during each cylinder's stroke(intake and exhaust).

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Answer:

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3 years ago

Write an application that reads three integers, adds all three together, computes an average of the three entries, and computes

Natali5045456 [20]

Answer:

//The program prompts user to input three integers and it displays them, adds and gets their average

//begin

public class Test

{

public static void Main()

{

//input intergers

int[] score = new int[3];

int avg,rem,sum = 0;

for(int i=0;i<3;i++)

{

Console.WriteLine("Enter an integer score ");

score[i] = Convert.ToInt32(Console.ReadLine());

sum = sum + score[i];

}

avg = sum/3;

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4 years ago

Explain the working of McLeod gauge for measurement of very low pressure.

AfilCa [17]

Explanation:

McLeod gauge:

It is used to measure very low pressure of gas.It measure gas pressure by the help of mercury(Hg).it measure absolute pressure of gas.

McLeod gauge on the principle of Boyle's law.Boyle's law state that ,at constant temperature the pressure of gas is directly proportional to its volume.From Boyle's law

$P_1V_1=P_2V_2$

A sample of gas is taken from vacuum then its pressure is measured by help Hg.

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3 years ago

2. A thin vertical panel L = 3 m high and w = 1.5 m wide is thermally insulated on one side and exposed to a solar radiation flu

n200080 [17]

Answer: 383.22K

Explanation:

L = 3m, w = 1.5m

Area A = 3 x 1.5 = 4.5m2

Q' = 750W/m2 (heat from sun) ,

& = 0.87

Q = &Q' = 0. 87x750 = 652.5W/m2

E = QA = 652.5 x 4.5 = 2936.25W

T(sur) = 300K, T(panel) = ?

Using E = §€A(T^4(panel) - T^4(sur))

§ = Stefan constant = 5.7x10^-8

€ = emmisivity = 0.85

2936.25 = 5.7x10^-8 x 0.85 x 4.5 x (T^4(panel) - 300^4)

T(panel) = 383.22K

See image for further details.

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3 years ago

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2 years ago

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