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3 years ago

Chapter 19: Diesel Engine Operation and Diagnosis -Chapter Quiz

Engineering

1 answer:

Llana [10]3 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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A well-insulated, rigid tank has a volume of 1 m3and is initially evacuated. A valve is opened,and the surrounding air enters at

DiKsa [7]

Answer:

0.5 kW

Explanation:

The given parameters are;

Volume of tank = 1 m³

Pressure of air entering tank = 1 bar

Temperature of air = 27°C = 300.15 K

Temperature after heating = 477 °C = 750.15 K

V₂ = 1 m³

P₁V₁/T₁ = P₂V₂/T₂

P₁ = P₂

V₁ = T₁×V₂/T₂ = 300.15 * 1 /750.15 = 0.4 m³

$dQ = m \times c_p \times (T_2 -T_1)$

For ideal gas, $c_p$ = 5/2×R = 5/2*0.287 = 0.7175 kJ

PV = NKT

N = PV/(KT) = 100000×1/(750.15×1.38×10⁻²³)

N = 9.66×10²⁴

Number of moles of air = 9.66×10²⁴/(6.02×10²³) = 16.05 moles

The average mass of one mole of air = 28.8 g

Therefore, the total mass = 28.8*16.05 = 462.135 g = 0.46 kg

∴ dQ = 0.46*0.7175*(750.15 - 300.15) = 149.211 kJ

The power input required = The rate of heat transfer = 149.211/(60*5)

The power input required = 0.49737 kW ≈ 0.5 kW.

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3 years ago

Which rigid motion maps the solid-line figure onto the dotted-line figure?

Agata [3.3K]

I would love to answer but unfortunately there is no picture.

5 0

3 years ago

The roof of a refrigerated truck compartment consists of a layer of foamed urethane insulation (t2 = 21 mm, ki = 0.026 W/m K) be

lakkis [162]

Answer:

Tso = 28.15°C

Explanation:

given data

t2 = 21 mm

ki = 0.026 W/m K

t1 = 9 mm

kp = 180 W/m K

length of the roof is L = 13 m

net solar radiation into the roof = 107 W/m²

temperature of the inner surface Ts,i = -4°C

air temperature is T[infinity] = 29°C

convective heat transfer coefficient h = 47 W/m² K

solution

As when energy on the outer surface at roof of a refrigerated truck that is balance as

Q = $\frac{T \infty - T si }{\frac{1}{hA}+\frac{t1}{AKp}+\frac{t2}{AKi}+\frac{t1}{aKp}}$ .....................1

Q = $\frac{T \infty - Tso}{\frac{1}{hA}}$ .....................2

now we compare both equation 1 and 2 and put here value

$\frac{29-(-4)}{\frac{1}{47}+\frac{2\times0.009}{180}+\frac{0.021}{0.026}} = \frac{29-Tso}{\frac{1}{47}}$

solve it and we get

Tso = 28.153113

so Tso = 28.15°C

3 0

3 years ago

Why will screws never replace nails

cupoosta [38]

Answer:

because people have different opinions on nails and screws

Explanation:

3 0

3 years ago

A 0.40-m3 insulated piston-cylinder device initially contains 1.3 kg of air at 30°C. At this state, the piston is free to move.

Setler79 [48]

Answer:

(a) The Final Temperature is 315.25 K.

(b) The amount of mass that has entered 0.5742 Kg.

(d) The entrophy generation is 0.0398 kJ/kgK.

Explanation:

Explanation is in the following attachments.

6 0

3 years ago