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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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Steam at 4 MPa and 350°C is expanded in an adiabatic turbine to 125kPa. What is the isentropic efficiency (percent) of this turb

guajiro [1.7K]

Answer:

$\eta_{turbine} = 0.603 = 60.3\%$

Explanation:

First, we will find actual properties at given inlet and outlet states by the use of steam tables:

AT INLET:

At 4MPa and 350°C, from the superheated table:

h₁ = 3093.3 KJ/kg

s₁ = 6.5843 KJ/kg.K

AT OUTLET:

At P₂ = 125 KPa and steam is saturated in vapor state:

h₂ = $h_{g\ at\ 125KPa}$ = 2684.9 KJ/kg

Now, for the isentropic enthalpy, we have:

P₂ = 125 KPa and s₂ = s₁ = 6.5843 KJ/kg.K

Since s₂ is less than $s_g$ and greater than $s_f$ at 125 KPa. Therefore, the steam is in a saturated mixture state. So:

$x = \frac{s_2-s_f}{s_{fg}} \\\\x = \frac{6.5843\ KJ/kg.K - 1.3741\ KJ/kg.K}{5.91\ KJ/kg.K}\\\\x = 0.88$

Now, we will find $h_{2s}$ (enthalpy at the outlet for the isentropic process):

$h_{2s} = h_{f\ at\ 125KPa}+xh_{fg\ at\ 125KPa}\\\\h_{2s} = 444.36\ KJ/kg + (0.88)(2240.6\ KJ/kg)\\h_{2s} = 2416.088\ KJ/kg$

Now, the isentropic efficiency of the turbine can be given as follows:

$\eta_{turbine} = \frac{h_1-h_2}{h_1-h_{2s}}\\\\\eta_{turbine} = \frac{3093.3\ KJ/kg-2684.9\ KJ/kg}{3093.3\ KJ/kg-2416.088\ KJ/kg}\\\\\eta_{turbine} = \frac{408.4\ KJ/kg}{677.212\ KJ/kg}\\\\\eta_{turbine} = 0.603 = 60.3\%$

3 0

3 years ago

Air at atmospheric pressure and at 300K flows with a velocity of 1.5m/s over a flat plate. The transition from laminar to turbul

Savatey [412]

Answer:3.47 m

Explanation:

Given

Temperature(T)=300 K

velocity(v)=1.5 m/s

At 300 K

$\mu =1.846 \times 10^{-5} Pa-s$

$\rho =1.77 kg/m^3$

And reynold's number is given by

$Re.=\frac{\rho v\time x}{\mu }$

$5\times 10^5=\frac{1.77\times 1.5\times x}{1.846\times 10^{-5}}$

$x=\frac{5\times 10^5\times 1.846\times 10^{-5}}{1.77\times 1.5}$

x=3.47 m

5 0

3 years ago

Which of the following requirement statements is an example of a breakdown of the accuracy standard?

const2013 [10]

Answer:

The automobile rental prices shall show all taxes (including a 6% state tax).

Explanation:

Im pretty sure

4 0

2 years ago

g Let the charges start infinitely far away and infinitely far apart. They are placed at (6 cm, 0) and (0, 3 cm), respectively,

irina1246 [14]

Answer:

a) V =10¹¹*(1.5q₁ + 3q₂)

b) U = 1.34*10¹¹q₁q₂

Explanation:

Given

x₁ = 6 cm

y₁ = 0 cm

x₂ = 0 cm

y₂ = 3 cm

q₁ = unknown value in Coulomb

q₂ = unknown value in Coulomb

A) V₁ = Kq₁/r₁

where r₁ = √((6-0)²+(0-0)²)cm = 6 cm = 0.06 m

V₁ = 9*10⁹q₁/(0.06) = 1.5*10¹¹q₁

V₂ = Kq₂/r₂

where r₂ = √((0-0)²+(3-0)²)cm = 3 cm = 0.03 m

V₂ = 9*10⁹q₂/(0.03) = 3*10¹¹q₂

The electric potential due to the two charges at the origin is

V = ∑Vi = V₁ + V₂ = 1.5*10¹¹q₁ + 3*10¹¹q₂ = 10¹¹*(1.5q₁ + 3q₂)

B) The electric potential energy associated with the system, relative to their infinite initial positions, can be obtained as follows

U = Kq₁q₂/r₁₂

where

r₁₂ = √((0-6)²+(3-0)²)cm = √45 cm = 3√5 cm = (3√5/100) m

then

U = 9*10⁹q₁q₂/(3√5/100)

⇒ U = 1.34*10¹¹q₁q₂

5 0

3 years ago

Chlorine is one of the important commodity chemicals for the global economy. Before the advent of large scale

artcher [175]

The composition of gas in the feed, the percentage conversion and the

theoretical yield are combined to give the product stream composition.

Response:

The composition of gas in the product stream are;

HCl: 0.4 kmol/h, Cl₂: 1.6 kmol/h, H₂O: 1.6 kmol/h, O₂: 0.5 kmol/h

<h3>How can percentage conversion give the contents of the product stream?</h3>

The amount of oxygen used = 30% exceeding the theoretical amount

Number of moles of hydrochloric acid = 4 kmol/h

Percentage conversion = 80%

Required:

The composition of the gas in the product feed.

Solution;

The given reaction is; 4HCl + O₂ $\longrightarrow$ 2Cl₂ + 2H₂O

$Percentage \ conversion = \mathbf{ \dfrac{Moles \ of \ limiting \ reactant \ reacted}{Moles \ of \ limiting \ reactant \ supplied \ in \ the \, feed}}$