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xxMikexx [17]
2 years ago
5

4. If a hot wire is shorted to ground, what will usually happen?

Engineering
1 answer:
barxatty [35]2 years ago
5 0

Answer:

nothing will happen

Explanation:

except for whatever that wire feeds will lose power

help support our cause by spreading this flag all over brainly

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Sharon is designing a house in an area that receives a lot of rainfall all year. Which material should she use to stick the wood
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4. The instant the ignition switch is turned to the start position,
geniusboy [140]

Answer:

D. Both pull-in and hold-in windings are energized.

Explanation:

The instant the ignition switch is turned to the start position, "Both pull-in and hold-in windings are energized." This is because the moment the ignition switch is turned to the start position, voltage passes through to the S terminal of the solenoid.

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2 years ago
Diffusion of Ammonia in an Aqueous Solution Ammonia (A)-water (B) solution ta 278 K and 4 mm thick is in contact with an organic
Tom [10]

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3 years ago
What type of engineer makes sure equipment is safe and operational
zvonat [6]

Answer:

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3 years ago
Air enters the compressor of an air-standard Brayton cycle with a volumetric flow rate of 60 m3/s at 0.8 bar, 280 K. The compres
natima [27]

Answer:

a) The Net power developed in this air-standard Brayton cycle is 43.8MW

b) The rate of heat addition in the combustor is 84.2MW

c) The thermal efficiency of the cycle is 52%

Explanation:

To solve this cycle we need to determinate the enthalpy of each work point of it. If we consider the cycle starts in 1, the air is compressed until 2, is heated until 3 and go throw the turbine until 4.

Considering this:

h_{i} =T_{i}C_{pair}=T_{i}1.005\frac{KJ}{Kg K}

\mu_{comp}=\frac{h_{2S}-h_{1}}{h_{2}-h_{1}}

\mu_{comp}=\frac{h_{3}-h_{4}}{h_{3}-h_{4S}}

G_{m} =\frac{PMG_{v}}{TR} =59.73\frac{Kg}{s}

Now we can calculate the enthalpy of each work point:

h₁=281.4KJ/Kg

h₂=695.41KJ/Kg

h₃=2105KJ/Kg

h₄=957.14KJ/Kg

The net power developed:

P_{net}=P_{Tur}-P_{Comp}=G_{m}((h_{3}-h_{4})-(h_{2}-h_{1}))

The rate of heat:

Q=G_{m}(h_{3}-h_{2})

The thermal efficiency:

\mu_{ther}=\frac{P_{net}}{Q}

3 0
2 years ago
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