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4 years ago

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250 hp, 230 V, and speed of 435 r/min DC shunt motor has a nominal full load current 862 A. a. Calculate the total losses and ef

ficiency at full load. b. Calculate the shunt field exciting current if the shunt field causes 20 percent of the total losses. c. Calculate the value of the armature resistance as well as the counter-emf (Ea), knowing that 50 percent of the total losses at full load are due to armature resistance.

1 answer:

Lostsunrise [7]4 years ago

3 0

Answer:

(a) Losses 11760 watt, Efficiency = 94.06 %

(b) Shunt current 10.21 A

(c) Armature resistance = 0.0084 ohm

Armature voltage = 222.97 volt

Explanation:

Output of the dc motor 2500 hp

As 1 hp = 746 watt

So output power

$P=250\times 746=186500W$

Full load current = 862 A

(A) Input of the motor is equal to

$=VI=230\times 862=198260W$

Therefore losses = input - output

= 192860-186500 = 11760 watt

Efficiency = $\eta =\frac{186500}{192860}=0.9406$ = 94.06 %

(b) Power of shunt field is equal to

$P_{shunt}=11760\times \frac{20}{100}=2352watt$

So shunt current $I_{sh}=\frac{2352}{230}=10.21A$

(C) It is given that 50% of the loss is due to armature current it means 50% loss will be due to shunt current

So loss due to shunt current $=11760\times 0.5=5880W$

Shunt current is equal to $I_{sh}=\frac{5880}{230}=25.56A$

Armature current $I_a=862-25.56=836.43A$

Power loss due to armature = 11760-5880 = 5880 W

Therefore $I_a^2R_a=5880$

$836.43^2\times R_a=5880$

$R_a=0.0084 ohm$

Emf is equal to $E=V-I_aR_a$

$E=230-836.43\times 0.0084=222.97V$

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Answer:

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critical pressure of water Pcr = 22.06 Mpa

and critical temperature of water Tcr = 647.1 K

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reduced pressure will be = $\frac{P}{Pcr}$

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Answer:

The solution code is written in Java.

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