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Aliun [14]
3 years ago
8

When checking for a no-star concern, you notice that an engine has no spark Technician A says to turn on the ignition engine (en

gine off) and, using a DMM, find out how much battery voltage is available at the negative terminal of the ignition coil. Technician B says the DMM reading should factuate indicating that the primary coal current is being turned on and off. who is correct?
Technician B
Technician A
Both Technicians A and B
Neither Technician A nor B​
Engineering
1 answer:
lbvjy [14]3 years ago
5 0

Answer:

Technician B

Explanation:

Technician B is correct in his argument. This is because according to what he said, as the computer pulses stimuli the coil will turn on and off, promoting an increase in the voltage that will cause the fluctuation. Technician A is incorrect because the procedure he indicated imposes that the voltage is checked at the negative terminal and not at the positive.

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A heat engine receives 6 kW from a 250oC source and rejects heat at 30oC. Examine each of three cases with respect to the inequa
taurus [48]

Answer:

Explanation:

Given

T_h=250^{\circ}C\approx 523\ K

T_L=30^{\circ}C\approx 303\ K

Q_1=6 kW

From Clausius inequality

\oint \frac{dQ}{T}=0  =Reversible cycle

\oint \frac{dQ}{T}  =Irreversible cycle

\oint \frac{dQ}{T}>0  =Impossible

(a)For P_{out}=3 kW

Rejected heat Q_2=6-3=3\ kW

\oint \frac{dQ}{T}= \frac{Q_1}{T_1}-\frac{Q_2}{T_2}

=\frac{6}{523}-\frac{3}{303}=1.57\times 10^{-3} kW/K

thus it is Impossible cycle

(b)P_{out}=2 kW

Q_2=6-2=4 kW

\oint \frac{dQ}{T}= \frac{Q_1}{T_1}-\frac{Q_2}{T_2}

=\frac{6}{523}-\frac{4}{303}=-1.73\times 10^{-3} kW/K

Possible

(c)Carnot cycle

\frac{Q_2}{Q_1}=\frac{T_1}{T_2}

Q_2=3.47\ kW

\oint \frac{dQ}{T}= \frac{Q_1}{T_1}-\frac{Q_2}{T_2}

=\frac{6}{523}-\frac{3.47}{303}=0

and maximum Work is obtained for reversible cycle when operate between same temperature limits

P_{out}=Q_1-Q_2=6-3.47=2.53\ kW

Thus it is possible

6 0
4 years ago
2. What is the most obvious elements of design?<br> O color<br> O shape<br> O line<br> O texture
muminat

Answer:

the moat obvious element of design is shape

3 0
3 years ago
"How would you direct traffic in the result of a catastrophic earthquake?"
11Alexandr11 [23.1K]

Answer:

In the result of a earthquake, you should direct people into areas that are safe, such as places where the ground has not been broken up, and then tell people to get out of their cars and be careful. You should also have them carry any valuable items out of the car(if they have time), and then wait until the initial shocks and aftershocks are over, and then tell them to get back into their cars and then drive carefully and safely.

8 0
3 years ago
Read 2 more answers
Steam at 40 bar and 500o C enters the first-stage turbine with a volumetric flow rate of 90 m3 /min. Steam exits the turbine at
a_sh-v [17]

Answer:

(a) 62460 kg/hr

(b) 17,572.95 kW

(c) 3,814.57 kW

Explanation:

Volumetric flow rate, G = 30 m³ / 1 min => 90 / 60 => 1.5

Calculate for h₁ , h₂ , h₃

h₁ is h at P = 40 bar, 500°C => 3445.84 KJ/Kg

Specific volume steam, ц = 0.086441 m³kg⁻¹

h₂ is h at P = 20 bar, 400°C => 3248.23 KJ/Kg

h₃ is h at P = 20 bar, 500°C => 3468.09 KJ/Kg

h₄ is hg at P = 0.6 bar from saturated water table => 2652.85 KJ/Kg

a)

Mass flow rate of the steam, m = G / ц

m = 1.5 / 0.086441

m = 17.35 kg/s

mass per hour is m = 62460 kg/hr

b)

Total Power produced by two stages

= m (h₁ - h₂) + m (h₃ - h₁)

= m [(3445.84 - 3248.23) + (3468.09 - 2652.85)]

= m [ 197.61 + 815.24 ]

= 17.35 [1012.85]

= 17,572.95 kW

c)

Rate of heat transfer to the steam through reheater

= m (h₃ - h₂)

= 17.35 x (3468.09 - 3248.23)

= 17.35 x 219.86

= 3,814.57 kW

8 0
3 years ago
"Water is flowing in a metal pipe. The pipe OD (outside diameter) is 61 cm. The pipe length is 120 m. The pipe wall thickness is
Katarina [22]

Answer:

Total wight =640.7927 KN

Explanation:

Given that

do= 61 cm

L =120

t= 0.9 cm

That is why inner diameter of the pipe

di= 61 - 2 x 0.9 cm

di=59.2 cm

Water density ,ρ = 1 kg/L = 1000 kg/m³

Weight of the pipe ,wt = 2500 N/m

wt = 2500 x 120 N = 300,000 N

The wight of the water

wt ' = ρ V g

wt'=1000\times \dfrac{\pi}{4}\times (0.61^2-0.0592^2)\times 9.81\times 120 N

wt'=340792.47 N

That is why total wight

Total wight = wt + wt'

Total wight =300,000+ 340792.47 N

Total wight =640,792.47 N

Total wight =640.7927 KN

7 0
4 years ago
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