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9966 [12]
3 years ago
7

PLEASE HURRY TIMED TEST

Engineering
2 answers:
ladessa [460]3 years ago
5 0

Answer:

1

Explanation:

Vladimir [108]3 years ago
3 0
EXHAUST VALVE OPENS....... 1...... AAAAAAA
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What does a peak flow meter allow you to assess?
Alex Ar [27]

Answer:

  peak flow and any engineering considerations related thereto

Explanation:

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3 0
3 years ago
Chapter 19: Diesel Engine Operation and Diagnosis -Chapter Quiz
Llana [10]

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?

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6 0
3 years ago
The water in a large lake is to be used to generate electricity by the installation of a hydraulic turbine-generator. The elevat
ankoles [38]

Answer:

a) 75%

b) 82%

Explanation:

Assumptions:

\text{The mechanical energy for water at turbine exit is negligible.} \\ \\ \text{The elevation of the lake remains constant.}

Properties: The density of water \delta = 1000 kg/m^3

Conversions:

165 \  ft \  to \  meters  = 50 m  \\ \\7000 \ lbm/s \  to  \ kilogram/sec = 3175 kg/s \\ \\1564 \ hp \  to \  kilowatt = 1166 kw \\ \\

Analysis:

Note that the bottom of the lake is the reference level. The potential energy of water at the surface becomes gh. Consider that kinetic energy of water at the lake surface & the turbine exit is negligible and the pressure at both locations is the atmospheric pressure and change in the mechanical energy of water between lake surface & turbine exit are:

e_{mech_{in}} - e_{mech_{out}} = gh - 0

Then;

gh = (9.8 m/s^2) (50 m) \times \dfrac{1 \ kJ/kg}{1000 m^2/s^2}

gh = 0.491 kJ/kg

\Delta E_{mech \ fluid} = m(e_{mech_{in}} - e_{mech_{out}} ) \\ \\ = 3175 kg/s \times 0.491 kJ/kg

= 1559 kW

Therefore; the overall efficiency is:

\eta _{overall} = \eta_{turbine- generator} = \dfrac{W_{elect\ out}}{\Delta E_{mech \fluid}}

= \dfrac{1166 \ kW}{1559 \ kW}

= 0.75

= 75%

b) mechanical efficiency of the turbine:

\eta_{turbine- generator} = \eta_{turbine}\times   \eta_{generator}

thus;

\eta_{turbine} = \dfrac{\eta_{[turbine- generator]} }{\eta_{generator}} \\ \\ \eta_{turbine} = \dfrac{0.75}{0.92} \\ \\ \eta_{turbine} = 0.82 \\ \\ \eta_{turbine} = 82\%

6 0
3 years ago
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