Answer:

Explanation:
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In this case, we compute the heat output from coal, given its heating value and the mass flow:

Next, since the work done by the power plant is 230 MW, we compute the efficiency as shown below:

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Answer:
q=39.15 W/m²
Explanation:
We know that
Thermal resistance due to conductivity given as
R=L/KA
Thermal resistance due to heat transfer coefficient given as
R=1/hA
Total thermal resistance

Now by putting the values


We know that
Q=ΔT/R


So heat transfer per unit volume is 39.15 W/m²
q=39.15 W/m²
Answer:
Explanation:
Inductance = 250 mH = 250 / 1000 = 0.25 H
capacitance = 4.40 µF = 4.4 × 10⁻⁶ F ( µ = 10⁻⁶)
ΔVmax = 240, f frequency = 50Hz and I max = 110 mA = 110 /1000 = 0.11A
a) inductive reactance = 2πfl = 2 × 3.142 × 50 × 0.25 H =78.55 ohms
b) capacitive reactance =
= 1 / ( 2 × 3.142× 50 × 4.4 × 10⁻⁶ ) = 723.34 ohms
c) impedance =
= 240 / 0.11 = 2181.82 ohms
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