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11111nata11111 [884]

2 years ago

10

Do the coil resistances have any effect on the plots?

1 answer:

PolarNik [594]2 years ago

8 0

Because of the skin depth effect, the current at high frequency tends to flow at very low depth from radius. Then at high frequency the effective cross section of the wire is narrower than at DC.

Fro example skin depth at 100 kHz is 0.206 mm (0.008”), a wire more thicker than AWG26 could be a waste of copper, better use a bunch of thin wire (Litz wire) to rise the Q factor.

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A slab-milling operation is performed on a 0.7 m long, 30 mm-wide cast-iron block with a feed of 0.25 mm/tooth and depth of cut

denis23 [38]

Answer:

a) $T_m=1.787min$

b) $MRR=35259.7mm^3/min$

Explanation:

From the question we are told that:

Cast-iron block Dimension:

Length $l=0.7m=>700mm$

Width $w=30mm$

Feed $F=0.25mm/tooth$

Depth $dp=3mm$

Diameter $d=75mm$

Number of cutting teeth $n=8$

Rotation speed $N=200rpm$

Generally the equation for Approach is mathematically given by

$x=\sqrt{Dd-d^2}$

$X=\sqrt{75*3-3^2}$

$X=14.69mm$

Therefore

Effective length is given as

$L_e=Approach +object Length$

$L_e=700+14.69$

$L_e=714.69mm$

a)

Generally the equation for Machine Time is mathematically given by

$T_m=\frac{L_e}{F_m}$

Where

$F_m=F*n*N$

$F_m=0.25*8*200$

$F_m=400$

Therefore

$T_m=\frac{714.69}{400}$

$T_m=1.787min$

b)

Generally the equation for Material Removal Rate. is mathematically given by

$MRR=\frac{L*B*d}{t_m}$

$MRR=\frac{700*30*3}{1.787}$

$MRR=35259.7mm^3/min$

3 0

3 years ago

6.1-2. Diffusion of CO, in a Binary Gas Mixture. The gas CO2 is diffusing at stcady state through a tube 0.20 m long having a di

zzz [600]

Answer:

Heat flux of CO₂ in cgs

= 170.86 x 10⁻⁹ mol / cm²s

SI units

170.86 x 10⁻⁸ kmol/m²s

Explanation:

4 0

3 years ago

The ???? − i relationship for an electromagnetic system is given by ???? = 1.2i1/2 g where g is the air-gap length. For current

Artemon [7]

Answer:

a) The mechanical force is -226.2 N

b) Using the coenergy the mechanical force is -226.2 N

Explanation:

a) Energy of the system:

$\lambda =\frac{1.2*i^{1/2} }{g} \\i=(\frac{\lambda g}{1.2} )^{2}$

$\frac{\delta w_{f} }{\delta g} =\frac{g^{2}\lambda ^{3} }{3*1.2^{2} }$

$f_{m}=- \frac{\delta w_{f} }{\delta g} =-\frac{g^{2}\lambda ^{3} }{3*1.2^{2} }$

If i = 2A and g = 10 cm

$\lambda =\frac{1.2*i^{1/2} }{g} =\frac{1.2*2^{1/2} }{10x10^{-2} } =16.97$

$f_{m}=-\frac{g^{2}\lambda ^{3} }{3*1.2^{2} }=-\frac{16.97^{3}*2*0.1 }{3*1.2^{2} } =-226.2N$

b) Using the coenergy of the system:

$f_{m}=- \frac{\delta w_{f} }{\delta g} =-\frac{1.2*2*i^{3/2} }{3*g^{2} }=-\frac{1.2*2*2^{3/2} }{3*0.1^{2} } =-226.2N$

8 0

3 years ago

The question belongs to Electrical Engineering (Linear System).

-Dominant- [34]

I’m crying looking at that.

5 0

3 years ago

Which starting circuit uses fuses, switches, and smaller wires to energize a relay and solenoid?

zysi [14]

Answer:

Option B (Starter Control Circuit) is the right option.

Explanation:

This same switching is normally put upon this isolated side of something like the transmission Arduino microcontroller throughout the configuration that is using the ignition just to command the broadcast.
It uses a secondary relay isolated to regulate electrical current throughout the solenoid starting system.

All other given options are not related to the given instance. So the above option is correct.

4 0

3 years ago