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2 years ago

Applications of fleming hand rule

Engineering

1 answer:

Eva8 [605]2 years ago

4 0

Answer:

Fleming hand rule represents the direction of current in a generator's windings and induced current as a conductor is attached to a circuit such that it moves in a magnetic field.

Explanation:

Fleming hand rule represents the direction of current in a generator's windings and induced current as a conductor is attached to a circuit such that it moves in a magnetic field.

Fleming hand rule is used in the case of electric motors and electric generators.

Fleming hand rule is used to determine the following:

1. Direction of torque

2. Angular velocity

3. Angular acceleration

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2 years ago

A heat engine is coupled with a dynamometer. The length of the load arm is 900 mm. The spring balance reading is 16. Applied wei

miss Akunina [59]

Answer:

P = 80.922 KW

Explanation:

Given data;

Length of load arm is 900 mm = 0.9 m

Spring balanced read 16 N

Applied weight is 500 N

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we know that power is given as

$P = T\times \omega$

T Torque = (w -s) L = (500 - 16)0.9 = 435.6 Nm

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3 years ago

An induced-draft cooling tower cools 90,000 gallons per minute of water from 84 to 68oF. Air at 14.61 psia, 70oF dry bulb and 60

belka [17]

Answer:

a. V = 109.64 × 10⁵ ft/min

b. Mw = 654519.54 kg/hr

Explanation:

Given Parameters

mass flow rate of water, Mw = 90000g/min = 6607.33 kg/s

inlet temperature of water, T1 = 84 F = 28.89 C

outlet temperature of water, T2 = 68 F = 20 C

specific heat capacity of water, c = 4.18kJ/kgK

rate of heat remover from water, Qw is given by

Qw = 6607.33[28.89 - 20] * 4.18

Qw = 245529.545kw

For air, inlet condition

DBT = 70 F hi = 43.43 kJ/kg

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oulet condition,

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check the attached file for complete solution

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2 years ago

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Ostrovityanka [42]

Answer:

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2 years ago

A 1-kW electric resistance heater submerged in 10-kg water is turned on and kept on for 15 min. During the process, 400 kJ of he

hichkok12 [17]

Answer:

ΔT= 11.94 °C

Explanation:

Given that

mass of water = 10 kh

Time t= 15 min

Heat lot from water = 400 KJ

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Heat input the water= 1 x 15 x 60

=900 KJ

By heat balancing

Heat supply - heat rejected = Heat gain by water

As we know that heat capacity of water

$C_p=4.187 \frac{KJ}{kg-K}$

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Now by putting the values

900 - 400 = 10 x 4.187 x ΔT

So rise in temperature of water ΔT= 11.94 °C

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2 years ago