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alukav5142 [94]
3 years ago
11

For an applied transformer with a primary winding of 220, a secondary winding of 113 turns, a core of 45 cm2 cross-section and a

rated supply voltage of 220 V, frequency f = 50 Hz, number system Core style K = 1,2.
one. Calculated from the sensor in the core transformer.
NS. Calculate the rated capacity of the transformer application.
C. Calculating no-load voltage and current levels.
NS. If the power point is reduced by 7% from the voltage level; voltage scortation feature.
e. If use the core machine variable on random form; powerpress same as the host variable on; compare the capacity of the random transformer with the capacity of the transformer application.
Engineering
1 answer:
Damm [24]3 years ago
8 0

Explanation:

https://www.chegg.com/homework-help/questions-and-answers/iron-core-wish-design-transformer-part-dc-power-supply-moderately-high-current-rating-volt-q53186459?trackid=83dbc34cec2e&strackid=3efc9f324415

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Give me some examples of fragile structures.
Anvisha [2.4K]

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i don't know if this help tell me if i am wrong

Explanation:

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3 0
3 years ago
Need answers for these please ​
Step2247 [10]

Answer:

Following are given the answers one-by-one with explanation:

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7 0
3 years ago
A 400 kg machine is placed at the mid-span of a 3.2-m simply supported steel (E = 200 x 10^9 N/m^2) beam. The machine is observe
Alenkinab [10]

Answer:

moment of inertia = 4.662 * 10^6 mm^4

Explanation:

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Mass of machine = 400 kg = 400 * 9.81 = 3924 N

length of span = 3.2 m

E = 200 * 10^9 N/m^2

frequency = 9.3 Hz

Wm ( angular frequency ) = 2 \pi f = 58.434 rad/secs

also Wm = \sqrt{\frac{g}{t} }  ------- EQUATION 1

g = 9.81

deflection of simply supported beam

t = \frac{wl^3}{48EI}

insert the value of t into equation 1

Wm^2 = \frac{g*48*E*I}{WL^3}   make I the subject of the equation

I ( Moment of inertia about the neutral axis ) = \frac{WL^3* Wn^2}{48*g*E}

I = \frac{3924*3.2^3*58.434^2}{48*9.81*200*10^9}  = 4.662 * 10^6 mm^4

6 0
3 years ago
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