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kiruha [24]
3 years ago
13

A multipurpose transformer has a secondary coil with several points at which a voltage can be extracted, giving outputs of 5.60,

12.0, and 480 V. (a) The input voltage is 140 V to a primary coil of 220 turns. What are the numbers of turns in the parts of the secondary used to produce the output voltages
Engineering
1 answer:
Rudik [331]3 years ago
4 0

Answer:

for 5.6V 9 turns, for 12.0V 19 turns, for 480V 755 turns

Explanation:

Vp/Vs= Np/Ns

Vp: Primary voltage

Vs: Secondary Voltage

Np: number of turns on primary side

Ns: number of turns on secondary side

for output 5.6V

140/5.6= 220/Ns

Ns= 8.8 or 9 Turns

for output 12.0V

140/12= 220/Ns

Ns= 18.9 or 19 turns

for output 480V

140/480= 220/Ns

Ns= 754.3 or 755 turns

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The wall of drying oven is constructed by sandwiching insulation material of thermal conductivity k = 0.05 W/m°K between thin me
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Answer:

86 mm

Explanation:

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\frac {T_ \infty, i-T_{i}}{ R^{"}_{cv, i}} + \frac {T_{o}-T_{i}}{ R^{"}_{cd}} + q_{rad} = 0 Equation 1

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R^{"}_{cv, o}= \frac {1}{h_{o}}= \frac {1}{10}= 0.100 m^{2}K/w Equation 4

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R^{"}_{cd}= \frac {L}{K}= \frac {L}{0.05} m^{2}K/w Equation 5

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Substituting the values in equations 3,4 and 5 into equations 1 and 2 we obtain

\frac {300-T_{i}}{0.033} +\frac {40-T_{i}}{L/0.05} +100=0  Equation 6

\frac {T_{ i}-40}{L/0.05}+ \frac {25-40}{0.100}=0

T_{i}-40= \frac {L}{0.05}*150

T_{i}-40=3000L

T_{i}=3000L+40 Equation 7

From equation 6 we can substitute wherever there’s T_{i} with 3000L+40 as seen in equation 7 hence we obtain

\frac {300- (3000L+40)}{0.033} + \frac {40- (3000L+40)}{L/0.05}+100=0

The above can be simplified to be

\frac {260-3000L}{0.033}+ \frac {(-3000L)}{L/0.05}+100=0

\frac {260-3000L}{0.033}=50

-3000L=1.665-260

L= \frac {-258.33}{-3000}=0.086*10^{-3}m= 86mm

Therefore, insulation thickness is 86mm

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