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3 years ago

When your fixing a car, what is the first thing you want to do?

Engineering

1 answer:

Shtirlitz [24]3 years ago

6 0

Answer:

Changing oil.

Explanation:

You need to regularly check and change your car’s oil to ensure smooth running of the vehicle and to prolong the lifespan of its engine.

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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

Bad White [126]

Answer:

the net work per cycle $\mathbf{W_{net} = 0.777593696}$ Btu per cycle

the power developed by the engine, W = 88.0144746 hp

Explanation:

the information given includes;

diameter of the four-cylinder bore = 3.7 in

length of the stroke = 3.4 in

The clearance volume = 16% = 0.16

The cylindrical volume $V_2 = 0.16 V_1$

the crankshaft N rotates at a speed of 2400 RPM.

At the beginning of the compression , temperature $T_1$ = 60 F = 519.67 R

and;

Otto cycle with a pressure = 14.5 lbf/in² = (14.5 × 144 ) lb/ft²

= 2088 lb/ft²

The maximum temperature in the cycle is 5200 R

From the given information; the change in volume is:

$V_1-V_2 = \dfrac{\pi}{4}D^2L$

$V_1-0.16V_1= \dfrac{\pi}{4}(3.7)^2(3.4)$

$V_1-0.16V_1= 36.55714291$

$0.84 V_1 =36.55714291$

$V_1 =\dfrac{36.55714291}{0.84 }$

$V_1 =43.52040823 \ in^3 \\ \\ V_1 = 43.52 \ in^3$

$V_1 = 0.02518 \ ft^3$

the mass in air ( lb) can be determined by using the formula:

$m = \dfrac{P_1V_1}{RT}$

where;

R = 53.3533 ft.lbf/lb.R°

$m = \dfrac{2088 \ lb/ft^2 \times 0.02518 \ ft^3}{53.3533 \ ft .lbf/lb.^0R \times 519 .67 ^0 R}$

m = 0.0018962 lb

From the tables of ideal gas properties at Temperature 519.67 R

$v_{r1} =158.58$

$u_1 = 88.62 Btu/lb$

At state of volume 2; the relative volume can be determined as:

$v_{r2} = v_{r1} \times \dfrac{V_2}{V_1}$

$v_{r2} = 158.58 \times 0.16$

$v_{r2} = 25.3728$

The specific energy $u_2$ at $v_{r2} = 25.3728$ is 184.7 Btu/lb

From the tables of ideal gas properties at maximum Temperature T = 5200 R

$v_{r3} = 0.1828$

$u_3 = 1098 \ Btu/lb$

To determine the relative volume at state 4; we have:

$v_{r4} = v_{r3} \times \dfrac{V_1}{V_2}$

$v_{r4} =0.1828 \times \dfrac{1}{0.16}$

$v_{r4} =1.1425$

The specific energy $u_4$ at $v_{r4} =1.1425$ is 591.84 Btu/lb

Now; the net work per cycle can now be calculated as by using the following formula:

$W_{net} = Heat \ supplied - Heat \ rejected$

$W_{net} = m(u_3-u_2)-m(u_4 - u_1)$

$W_{net} = m(u_3-u_2- u_4 + u_1)$

$W_{net} = m(1098-184.7- 591.84 + 88.62)$

$W_{net} = 0.0018962 \times (1098-184.7- 591.84 + 88.62)$

$W_{net} = 0.0018962 \times (410.08)$

$\mathbf{W_{net} = 0.777593696}$ Btu per cycle

the power developed by the engine, in horsepower. can be calculated as follows;

In the four-cylinder, four-stroke internal combustion engine; the power developed by the engine can be calculated by using the expression:

$W = 4 \times N' \times W_{net$

where ;

$N' = \dfrac{2400}{2}$

N' = 1200 cycles/min

N' = 1200 cycles/60 seconds

N' = 20 cycles/sec

W = 4 × 20 cycles/sec × 0.777593696

W = 62.20749568 Btu/s

W = 88.0144746 hp

8 0

3 years ago

In javaWrite a program that simulates flipping a coin to make decisions. The input is how many decisions are needed, and the out

Pavel [41]

Answer:

// Program is written in Java Programming Language

// Comments are used for explanatory purpose

import java.util.*;

public class FlipCoin

{

public static void main(String[] args)

{

// Declare Scanner

Scanner input = new Scanner (System.in);

int flips;

// Prompt to enter number of toss or flips

System.out.print("Number of Flips: ");

flips = input.nextInt();

if (flips > 0)

{

HeadsOrTails();

}

public static String HeadsOrTails(Random rand)

{

// Simulate the coin tosses.

for (int count = 0; count < flips; count++)

{

rand = new Random();

if (rand.nextInt(2) == 0) {

System.out.println("Tails"); }

else {

System.out.println("Heads"); }

rand = 0;

}

7 0

3 years ago

) A shaft encoder is to be used with a 50 mm radius tracking wheel to monitor linear displacement. If the encoder produces 256 p

andrey2020 [161]

Answer:

number of pulses produced = 162 pulses

Explanation:

give data

radius = 50 mm

encoder produces = 256 pulses per revolution

linear displacement = 200 mm

solution

first we consider here roll shaft encoder on the flat surface without any slipping

we get here now circumference that is

circumference = 2 π r .........1

circumference = 2 × π × 50

circumference = 314.16 mm

so now we get number of pulses produced

number of pulses produced = $\frac{linear\ displacement}{circumference}$ × No of pulses per revolution .................2

number of pulses produced = $\frac{200}{314.16}$ × 256

number of pulses produced = 162 pulses

5 0

3 years ago

Two dogbone specimens of identical geometry but made of two different materials: steel and aluminum are tested under tension at

makkiz [27]

Answer:

$\dot L_{steel} = 3.448\times 10^{-4}\,\frac{in}{min}$

Explanation:

The Young's module is:

$E = \frac{\sigma}{\frac{\Delta L}{L_{o}} }$

$E = \frac{\sigma\cdot L_{o}}{\dot L \cdot \Delta t}$

Let assume that both specimens have the same geometry and load rate. Then:

$E_{aluminium} \cdot \dot L_{aluminium} = E_{steel} \cdot \dot L_{steel}$

The displacement rate for steel is:

$\dot L_{steel} = \frac{E_{aluminium}}{E_{steel}}\cdot \dot L_{aluminium}$

$\dot L_{steel} = \left(\frac{10000\,ksi}{29000\,ksi}\right)\cdot (0.001\,\frac{in}{min} )$

$\dot L_{steel} = 3.448\times 10^{-4}\,\frac{in}{min}$

7 0

3 years ago

Read 2 more answers

Who has good grades in school like A's B's dm me

siniylev [52]

Answer:

I do

Explanation:

4 0

3 years ago

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