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OlgaM077 [116]
2 years ago
11

You are working as an electrical technician. One day, out in the field, you need an inductor but cannot find one. Looking in you

r wire supply cabinet, you find a cardboard tube with single-conductor wire wrapped uniformly around it to form a solenoid. You carefully count the turns of wire and find that there are 570 turns. The diameter of the tube is 8.10 cm, and the length of the wire-wrapped portion is 35.0 cm. You pull out your calculator to determine the following.
a. the inductance of the coil (in mH)
b. the emf generated in the coil if the current in the wire increases at the rate of 3.00 A/s (Enter the magnitude in mV.)
Engineering
1 answer:
telo118 [61]2 years ago
7 0

Answer:

a) the inductance of the coil is 6 mH

b) the emf generated in the coil is 18 mV  

Explanation:

Given the data in the question;

N = 570 turns

diameter of tube d = 8.10 cm = 0.081 m

length of the wire-wrapped portion l =  35.0 cm = 0.35 m

a) the inductance of the coil (in mH)

inductance of solenoid

L = N²μA / l

A = πd²/4  

so

L = N²μ(πd²/4) / l

L = N²μ(πd²) / 4l

we know that μ = 4π × 10⁻⁷ TmA⁻¹

we substitute

L = [(570)² × 4π × 10⁻⁷× ( π × (0.081)² )] / 4(0.35)

L =  0.00841549 / 1.4

L = 6 × 10⁻³ H    

L = 6 × 10⁻³ × 1000 mH

L = 6 mH

Therefore, the inductance of the coil is 6 mH

b)

Emf ( ∈ ) = L di/dt

given that; di/dt = 3.00 A/sec

{∴ di = 3 - 0 = 3 and dt = 1 sec}

Emf ( ∈ ) = L di/dt

we substitute

⇒ 6 × 10⁻³ ( 3/1 )

= 18 × 10⁻³ V

= 18 × 10⁻³ × 1000

= 18 mV  

Therefore, the emf generated in the coil is 18 mV  

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kow [346]

Answer:

i wish i knew what you meant by that

7 0
3 years ago
A single-degree-of-freedom mass-spring-damper system is observed during its free vibration and the displacement amplitude decays
AleksandrR [38]

Answer:

Logarithmic decrement is equal to 0.182

Explanation:

given,

amplitude decay = 9 dB          

number of cycles = 12 cycles        

mass of the system = 7 kg        

spring stiffness = 3000 N/m            

logarithmic decrement = ?                  

now,                                                      

logarithmic decreament = ln\ D^{\frac{1}{n}}

                                        = ln\ 9^{\frac{1}{12}}

                                        =ln (1.2)

                                        = 0.182

Hence, Logarithmic decrement is equal to 0.182

5 0
3 years ago
Should you recommend changing fluids based on the color or smell of the fluid?
iragen [17]

Answer:

hi

Explanation:

Should you recommend changing fluids based on the color or smell of the fluid?

Only if the color is brown and the fluid smells burned

Yes, always.

No. Stick with the manufacturer's recommendations.

Only if the fluid is red and smells sweet

3 0
2 years ago
Is the sample mean always the most frequently occurring value? If so, explain why. If not, give an example.
Ugo [173]

Answer:

No the sample mean is not the most frequently occurring data.

Explanation:

If we take a number of observation in an experiment the most frequently repeating value is known as mode of data.

Mean of data is the the value to which all the given observational values are closest.

Consider a given set of observations as

1,1,1,1,1,5,4,3,4

The Mean of the above data set is\bar{x}=\frac{1+1+1+1+1+1+5+4+3+4}{10}=2 While as the most often repeating value is 1 which is not equal to mean  

6 0
3 years ago
A shell-and-tube heat exchanger with two shell passes and eight tube passes exchanges heat between water and hot air. The air fl
lesya692 [45]

Answer:

overall heat transfer coefficient is U = 29.614 W/m².k

Explanation:

We are given;

Mass Flow rate of water;m' = 45,500 kg/h = 45500/3600 kg/s = 12.6389 kg/s

total surface area for the heat exchange; A = 925 m²

Water heated from 80°C to 150°C

Exhaust gases cool from 350°C to 175°C

specific heat of water;c = 4,236 J/kg °C

Let's first calculate net rate of heat transfer;

The formula for net rate of heat transfer is given as;

q = m'•c•Δt

Where;

m' is mass flow rate

c is specific heat

Δt is change in the temperature of water.

Thus;

q = 12.6389•4,236•(150 - 80)

q = 3747683.33 W

Now, the overall heat transfer coefficient is gotten from the formula;

q = U•A•ΔT_lm

Where;

U is the overall heat transfer coefficient

q is the net rate of heat transfer

ΔT_lm is logarithmic mean temperature difference

A is the total surface area for the heat exchange

f is the correction factor

ΔT_lm is calculated as;

ΔT_lm = [(350 - 150) - (175 - 80)]/[In((350 - 150)/(175 - 80))]

ΔT_lm = [(200 - 95)]/[In(200/95)]

ΔT_lm = 141.04°C

Since:q = U•A•f•ΔT_lm

Let's make U the subject;

U = q/(A•f•ΔT_lm)

From the chart i attached, f is estimated to be 0.97

Plugging in the relevant values to obtain;

U = 3747683.33/(925 x 0.97 x 141.04)

U = 29.614 W/m².k

5 0
3 years ago
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