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2 years ago

Chapter 19: Diesel Engine Operation and Diagnosis -Chapter Quiz

Engineering

1 answer:

Llana [10]2 years ago

6 0

Answer: See explanation

Explanation:

1. How is diesel fuel ignited in a warm diesel engine?

B. Heat compression

2. Which type of diesel injection produces less noise?

A. Indirect injection (IDI)

3. Which diesel injection system requires the use of a glow plug?

A. Indirect injection (IDI)

4. The three phases of diesel ignition include:

C. Ignition delay, repaid combustion, controlled combustion.

5. What fuel system component is used in a vehicle equipped with a diesel engine that is seldom used on the same vehicle when it is equipped with a gasoline engine?

D. Water-fuel separator

6. The diesel injection pump is usually driven by a _________________.

A. Gear off the camshaft

7. Which diesel system supplies high-pressure diesel fuel to all the injectors all of the time?

C. High-pressure common rail

8. Glow plugs should have high resistance when _____________and lower resistance when __________________.

B. Warm/cold

9. Technician A says that glow plugs are used to help start a diesel engine and are shut off as soon as the engine starts. Technician B says that the glow plugs are turned off as soon as a flame is detected in the combustion chamber. Which Technician is correct?

D. Neither Technicians A NOR B

10. What part should be removed to test cylinder compression on a diesel engine?

D. A glow plug

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Answer:

helping handdffsdthikhdawefvbhreryiohf

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2 years ago

The chart describes four people’s credit histories.

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Answer:

D). Eesha pays more than the minimum payment each month.

Explanation:

Eesha would be considered most creditworthy among the given persons as she not only pays on time but also repays more than the minimum amount assigned to pay each month. In order to test the creditworthiness of an individual, his ontime debt paying capability is tested at first followed by the past credit repayment history and the credit score. Except Eesha, all the given candidates have failed to make timely repayment of their debts and hence, they cannot be considered creditworthy.

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2 years ago

Read 2 more answers

On the first statistics exam, the coefficient of determination between the hours studied and the grade earned was 80%. The stand

True [87]

Answer:

$\left[\begin{array}{ccccc}&DF&SS&MS&F\\Regression&1&7200&7200&72\\Error&18&1800&100\\total&19&900\end{array}\right]$

Explanation:

Sample size, n=20

Degrees of freedom is 1

Number of degrees of freedom for error is n-2 hence 20-2=18

Total number of degrees of freedom is 18+1=19

Standard error estimate is $s_{y-x}=\sqrt {\frac {SSE}{n-2}}$

Here, $SSE=(n-2)s_{y-x}^{2}=(20-2)(10)^{2}=1800$

Coefficient of determination $r^{2}=\frac {SSE}{SS total}$

Here, $SSR=r^{2}(SSR+SSE)$

$SSR=\frac {r^{2}}{1-r^{2}} SSE=\frac {0.8}{1-0.8}(1800)=7200$

The total sum of squares is

SS total=SSR+SSE=7200+1800=9000

MSR=SSR=7200

$MSE=\frac {SSE}{n-2}=\frac {1800}{20-2}=100$

F value is given by

$F=\frac {MSR}{MSE}=\frac {7200}{100}=100$

The ANOVA table is then

$\left[\begin{array}{ccccc}&DF&SS&MS&F\\Regression&1&7200&7200&72\\Error&18&1800&100\\total&19&900\end{array}\right]$

4 0

3 years ago

Determine the maximum volume in gallons [gal] of olive oil that can be stored in a closed cylindrical silo with a diameter of 3

Olin [163]

Answer:

V=1601gal

Explanation:

Hello! This problem is solved as follows,

First we must raise the equation that defines the pressure at the bottom of the tank with the purpose of finding the height that olive oil reaches.

This is given as the sum due to the atmospheric pressure (1atm = 101.325kPa), and the pressure due to the weight of the olive oil, taking into account the above, the following equation is inferred.

P=Poil+Patm

P=total pressure or absolute pressure=26psi=179213.28Pa

Patm= the atmospheric pressure =101325Pa

Poil=pressure due to the weight of olive oil=0.86αgh

α=density of water=1000kg/m^3

g=gravity=9.81m/s^2

h= height that olive oil reaches

solving

P=Poil+Patm

P=Patm+0.86αgh

$h=\frac{P-Patm}{0.86\alpha g } =\frac{179214.28-101325}{(0.86)(1000)(9.81)} \\h=9.23m$ [/tex]

Now we can use the equation that defines the volume of a cylinder.

V= $V=\frac{\pi }{4} D^{2} h$

D=3ft=0.9144m

h=9.23m

solving

$V=\frac{\pi }{4} (0.9144)^{2} 9.23=6.06m^3$

finally we use conversion factors to find the volume in gallons

$V=6.06m^3\frac{1000L}{1m^3} \frac{1gal}{3.785L} =1601gal$

3 0

3 years ago

A crane is set up for steel erection at the site of a 5 story office building where each story is 15 feet tall. The new building

stepladder [879]

Answer:

127.58 ft

Explanation:

We need first to calculate the length from corner to corner of the story, L.

Since the length of each floor is 75 ft and its width 50 ft, and since each floor is a rectangle with diagonal, L, using Pythagoras' theorem, we have

L² = (75 ft)² + (50 ft)²

= 5625 ft² + 2500 ft²

L² = 8125 ft²

L = √(8125 ft²)

L = 90.14 ft

Since the crane is 5 ft off the southwesterly corner of the building, the working radius, R = L + 5 ft = 90.14 ft + 5 ft = 95.14 ft.(since the diagonal length of the floor plus the distance of the crane from the south westerly corner add to give the working radius)

The boom tip height, H = height of building h + clearance of boom, h'

h = height of each story, h" × number of stories, n

Since h" = 15 ft and n = 5

h = 15 ft × 5 = 75 ft

Also, h' = 10 ft

So, H = h + h'

H = 75 ft + 10 ft

H = 85 ft

So, the minimum boom length, L' = √(H² + R²)

substituting the values of the variables into the equation, we have

L' = √(H² + R²)

L' = √((85 ft)² + (95.14 ft)²)

L' = √(7225 ft² + 9051.6196 ft²)

L' = √(16276.6196 ft²)

L' = 127.58 ft

4 0

2 years ago