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3 years ago

Which type of load generates a magnetic field?

Engineering

1 answer:

kozerog [31]3 years ago

5 0

Answer:

inductive loads

Explanation:

Inductive: When a changing current passes through an inductor, it induces a magnetic field around itself. Turning the inductor into a coil increases the magnetic field. Examples of inductive loads include transformers, fans, electric motors and coils.

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A cylindrical resistor element on a circuit board dissipates 0.6 W of power. The resistor is 1.5 cm long, and has a diameter of

Burka [1]

Answer:

a. 51.84Kj

b. 2808.99 W/m^2

c. 11.75%

Explanation:

Amount of heat this resistor dissipates during a 24-hour period

= amount of power dissipated * time

= 0.6 * 24 = 14.4 Watt hour

(Note 3.6Watt hour = 1Kj )

=14.4*3.6 = 51.84Kj

Heat flux = amount of power dissipated/ surface area

surface area = area of the two circular end + area of the curve surface

$=2*\frac{\pi D^{2} }{4} + \pi DL\\=2*\frac{\pi *(\frac{0.4}{100} )^{2} }{4} + \pi *\frac{0.4}{100} *\frac{1.5}{100}$

= 2.136 *10^-4 $m^{2}$

Heat flux = $\frac{0.6}{2.136 * 10^{-4} }$ = 2808.99 $W/m^{2}$

fraction of heat dissipated from the top and bottom surface

$=\frac{\frac{2*\pi D^{2} }{4} }{\frac{2*\pi D^{2}}{4} + \pi DL } \\\\=\\\frac{\frac{2*\pi *(\frac{0.4}{100} )^{2} }{4} }{\frac{2*\pi *(\frac{0.4}{100} )^{2} }{4} +\pi *\frac{0.4}{100} *\frac{1.5}{100} } \\\\=\frac{2.51*10^{-5} }{2.136*10^{-4} } \\\\\= 0.1175$

=11.75%

8 0

3 years ago

Read 2 more answers

If a worker currently makes $425.00 per week, what Is the gross amount they’ll earn if they work every week of the year

olga nikolaevna [1]

Answer:

we know that

A worker currently makes \$425.00 per week

remember that

1\ year=52\ weeks

by proportion find the amount that the worker will earn in one year

\frac{425}{1} \frac{\$}{week} =\frac{x}{52} \frac{\$}{weeks} \\ \\x=52*425 \\ \\x=\$22,100

therefore

the answer is

\$22,100

Explanation:

4 0

3 years ago

1. Present the pros and cons of an employer participating in the OSHA Voluntary Protection Program (VPP) or the Star Program. Ba

algol13

Answer:

kohjhfg do gm gm gm tm yl tm yk 9k umm yk to gm I'm y dm b do di Cy do do do do di du t6

3 0

2 years ago

Read 2 more answers

Which of the following types of valves have a valve stem that moves up and down in a valve guide? Select one: a. Poppet valves b

Alja [10]

Answer:c

Explanation:

thats wht it is

5 0

2 years ago

Read 2 more answers

An ideal Rankine cycle operates with turbine inlet steam at 90 bar and 500°C, and a condenser at 40 °C. Calculate the efficiency

lilavasa [31]

Answer:

40.8%

Explanation:

A rankine cycle is a generation cycle using water as a working fluid, when heat enters the boiler the water undergoes a series of changes in state and energy until generating power through the turbine.

This cycle is composed of four main components, the boiler, the pump, the turbine and the condenser as shown in the attached image

To solve any problem regarding the rankine cycle, enthalpies in all states must be calculated using the thermodynamic tables and taking into account the following.

• The pressure of state 1 and 4 are equal

• The pressure of state 2 and 3 are equal

• State 1 is superheated steam

• State 2 is in saturation state

• State 3 is saturated liquid at the lowest pressure

• State 4 is equal to state 3 because the work of the pump is negligible.

Once all enthalpies are found, the following equations are used using the first law of thermodynamics

Wout = m (h1-h2)

Qin = m (h1-h4)

Win = m (h4-h3)

Qout = m (h2-h1)

The efficiency is calculated as the power obtained on the heat that enters

Efficiency = Wout / Qin

Efficiency = (h1-h2) / (h1-h4)

first we calculate the enthalpies in all states

h1=3386kJ/Kg

h2=2073kJ/Kg

h2=h3=167.5kJ/Kg

we use the efficiency ecuation

$Efficiency =\frac{(h1-h2) }{(h1-h4)} =\frac{3386-2073}{3386-167.5} =0.408=40.8%$

8 0

2 years ago