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2 years ago

6

6. You are an electrician working in an industrial plant. You are building a motor control cabinet that contains six motor start

ers and six pilot lamps. All control components operate on 120 volts AC. Two of the motor starters have coil currents of 0.1 amperes each and four have coil currents of 0.18 amperes each. The six pilot lamps are rated at 5 watts each. The supply room has control transformers with the following rating (in volt-amperes):
75, 100, 150, 250, 300, and 500.

Which of the available control transformers should you choose to supply the power for all the control components in the cabinet?

1 answer:

Sveta_85 [38]2 years ago

8 0

Answer:

It's indeed safer to suggest a 150 VA transformer. Following table however is the clarification given.

Explanation:

For 2 motors with 0.1 A, the Power will be:

P = $2\times 120\times 0.1$

= $24 \ W$

For 4 motors with 0.18 A, the Power will be:

P = $4\times 120\times 0.18$

= $86.4 \ W$

As we know, for 6 pilot lamps, the power is "5 W".

So,

The total power will be:

⇒ P = $24+86.4+5$

= $115.4 \ W$

Now,

Consider the power factor to be "0.95"

VA of transformer is:

= $PF\times Power$

= $115.4\times 0.9$

= $109.63 \ VA$

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Consider insulation on a circular pipe For the same thickness and type of insulation, the thermal resistance of the insulation i

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Answer:

b). The same for all pipes independent of the diameter

Explanation:

We know,

$R_{conduction}=\frac{ln(\frac{r_{2}}{r_{1}})}{2\pi LK}$

$R_{convection}=\frac{1}{h(2\pi r_{2}L)}$

From the above formulas we can conclude that the thermal resistance of a substance mainly depends upon heat transfer coefficient,whereas radius has negligible effects on heat transfer coefficient.

We also know,

Factors on which thermal resistance of insulation depends are :

1. Thickness of the insulation

2. Thermal conductivity of the insulating material.

Therefore from above observation we can conclude that the thermal resistance of the insulation is same for all pipes independent of diameter.

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2 years ago

A platinum resistance temperature sensor has a resistance of 120 Ω at 0℃ and forms one arm of a Wheatstone bridge. At this tempe

oksian1 [2.3K]

Answer : 9.36ohms/ temperature

Explanation:

Expression for the variation of resistance of platinum with temperature

Rt= Ro(1+*t)

Rt= resistance @ t°C

Ro= resistance @ 0°C

*= temperature coefficient of resistance

Calculate the change in resistance by putting 120ohms for Ro,

0.0039/K for *

20°C for t

Using this formula:

Rt = Ro(1+*t)

Rt- Ro = Ro*t

= (120ohms)(0.0039/K)(20°C)

= 9.36ohms/K

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2 years ago

Universal Containers has a junction object called "Job Production Facility". With 2 master-detail relationships to the Job and P

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Answer:

C. The user will not be able to see the junction object records or the field values.

Explanation:

For the profile of the user to give access permission such as create and read to the job without granting access permission to the production facility object, the value of the field or records of the junction object will not be seen by the user. This is one of the necessary criteria or principle for the universal container with a junction object.

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2 years ago

A fatigue test was conducted in which the mean stress was 46.2 MPa and the stress amplitude was 219 MPa.

sleet_krkn [62]

Answer:

a)σ₁ = 265.2 MPa

b)σ₂ = -172.8 MPa

c) $Stress\ ratio =-0.65$

d)Range = 438 MPa

Explanation:

Given that

Mean stress ,σm= 46.2 MPa

Stress amplitude ,σa= 219 MPa

Lets take

Maximum stress level = σ₁

Minimum stress level =σ₂

The mean stress given as

$\sigma_m=\dfrac{\sigma_1+\sigma_2}{2}$

$2\sigma_m={\sigma_1+\sigma_2}$

2 x 46.2 = σ₁ + σ₂

σ₁ + σ₂ = 92.4 MPa --------1

The amplitude stress given as

$\sigma_a=\dfrac{\sigma_1-\sigma_2}{2}$

$2\sigma_a={\sigma_1-\sigma_2}$

2 x 219 = σ₁ - σ₂

σ₁ - σ₂ = 438 MPa --------2

By adding the above equation

2 σ₁ = 530.4

σ₁ = 265.2 MPa

-σ₂ = 438 -265.2 MPa

σ₂ = -172.8 MPa

Stress ratio

$Stress\ ratio =\dfrac{\sigma_{min}}{\sigma_{max}}$

$Stress\ ratio =\dfrac{-172.8}{265.2}$

$Stress\ ratio =-0.65$

Range = 265.2 MPa - ( -172.8 MPa)

Range = 438 MPa

8 0

2 years ago

The 8-mm-thick bottom of a 220-mm-diameter pan may be made from aluminum (k = 240 W/m ⋅ K) or copper (k = 390 W/m ⋅ K). When use

Artemon [7]

Answer:

For aluminum 110.53 C

For copper 110.32 C

Explanation:

Heat transmission through a plate (considering it as an infinite plate, as in omitting the effects at the borders) follows this equation:

$q = \frac{k * A * (th - tc)}{d}$

Where

q: heat transferred

k: conduction coeficient

A: surface area

th: hot temperature

tc: cold temperature

d: thickness of the plate

Rearranging the terms:

d * q = k * A * (th - tc)

$\frac{d * q}{k * A} = th - tc$

$th = \frac{d * q}{k * A} + tc$

The surface area is:

$A = \frac{\pi * d^2}{4}$

$A = \frac{\pi * 0.22^2}{4} = 0.038 m^2$

If the pan is aluminum:

$th = \frac{0.008 * 600}{240 * 0.038} + 110 = 110.53 C$

If the pan is copper:

$th = \frac{0.008 * 600}{390 * 0.038} + 110 = 110.32 C$

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2 years ago