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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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The drag coefficient of a car at the design conditions of 1 atm, 25°C, and 90 km/h is to be determined experimentally in a large

SIZIF [17.4K]

Answer: 0.288

Explanation:

Given

Pressure of the car, P = 1 atm

Temperature of the car, T = 25° C

Speed of the car, v = 90 km/h = 90*1000/3600 = 25 m/s

Height of the car, h = 1.25 m

Width of the car, b = 1.65 m

Force acting on the far, F = 220 N

Drag coefficient, C(d) = ?

Using our table A-9, we can trace that the density of air ρ, at the given temperature and pressure of 25 °C and 1 atm, is 1.184 kg/m³

Area = h *b

Area = 1.25 * 1.65

Area = 2.0625 m²

Now we solve for the drag coefficient using the formula

C(d) = F / (1/2 * ρ * A * v²)

C(d) = 220 / (0.5 * 1.184 * 2.0625 * 25²)

C(d) = 220 / (1.221 * 625)

C(d) = 220 / 763.125

C(d) = 0.288

Therefore, the drag coefficient is 0.288

3 0

3 years ago

Sum of Numbers Design a function that accepts an integer argument and returns the sum of all the integers from 1 up to the numbe

cestrela7 [59]

Answer:

# define main to get an argument from the user, call the function, and display the final values

def main():

# ask an argument from the user.

number = int(input('Enter a positive integer: '))

# solve the integral thing of the number by putting the user's argument in the addition() function

finalSum = addition(number)

# display the final value integral thing

print("The integral of this number is ", finalSum, ".")

# define an addition function to find the integral

def addition(num):

# make an if-statement to test filter things out of your calculations. In this case, 1.

if num == 1:

#if 1, return 1. because 1 + (1 - 1) = 1

return 1

# this is where the recursion actually is. This is the part that calculates

else:

return num + addition(num-1)

# call your main function

main()

Explanation:

This is written in Python. A link to the output can be found below.

https://Sum-of-Numbers.hufflepuffler07.repl.run

4 0

3 years ago

A resistor, an inductor, and a capacitor are connected in series to an ac source. What is the condition for resonance to occur?.

vaieri [72.5K]

Answer:if power factor =1 is possible for that.

Explanation:when pf is unity. means 1.

6 0

1 year ago

Vocabulary review for shop

kaheart [24]

Hazard is a possible source of danger

5 0

3 years ago

A compressible clay layer has a thickness of 3.8 m. After 1.5 yr, when the clay is 50% consolidated, 7.3 cm of settlement has oc

grigory [225]

The amount of settlement that would occur at the end of 1.5 year and 5 year are 7.3 cm and 13.14 cm respectively.

<h3>How to determine the amount of settlement?</h3>

For a layer of 3.8 m thickness, we were given the following parameters:

U = 50% = 0.5.

Sc = 7.3 cm.

For Sf, we have:

Sf = Sc/U

Sf = 7.3/0.5

Sf = 14.6

Therefore, Sf for a layer of 38 m thickness is given by:

Sf = 14.6 × 38/3.8

Sf = 146 cm.

At 50%, the time for a layer of 3.8 m thickness is: $t_{50}$ = 1.5 year.

At 50%, the time for a layer of 38 m thickness is:

$t_{50}$ = 1.5 × (38/3.8)²

$t_{50}$ = 150 years.

For the thickness of 38 m, U₂ is given by:

$\frac{U_1^2}{U_2^2} =\frac{(T_v)_1}{(T_v)_2} = \frac{t_1}{t_2} \\\\U_2^2 = U_1^2 \times [\frac{t_2}{t_1} ]\\\\U_2^2 = 0.5^2 \times [\frac{1.5}{150} ]\\\\U_2^2 = 0.25 \times 0.01\\\\U_2=\sqrt{0.0025} \\\\U_2=0.05$

The new settlement after 1.5 year is:

Sc = U₂Sf

Sc = 0.05 × 146

Sc = 7.3 cm.

For time, t₂ = 5 year:

$U_2^2 = U_1^2 \times [\frac{t_2}{t_1} ]\\\\U_2^2 = 0.5^2 \times [\frac{5}{150} ]\\\\U_2^2 = 0.25 \times 0.03\\\\U_2=\sqrt{0.0075} \\\\U_2=0.09$

The new settlement after 5 year is:

Sc = U₂Sf

Sc = 0.09 × 146

Sc = 13.14 cm.

Read more on clay layer here: brainly.com/question/22238205

8 0

2 years ago