This test should be performed on all cord sets, receptacles that aren't part of a building or structure's permanent wiring, and
1 answer:
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Answer:
The total system active power P = P_1 + P_2 + P_3 = 34.91 KW
Explanation:
Load 1: Active power P_1 = 20 HP = 14.91kW;
Reactive power
Load 2: Active power P_2 = 20 kW;
Reactive power Q2 = 0 since the load is purely resistive.
Load 3: Active power = 0 due to purely capacitiveload
Reactive power = -20 Var
a) since all three loads are connected in parallel therefore
The total system active power P = P_1 + P_2 + P_3 = 34.91 KW
Total system reactive power Q = Q_1 + Q_2 + Q_3 = 11.18 + 0 -20 = -8.82 kVar
Since Q = 0, the power factor is unity.
Supply current per phase is given by
Answer:
See the attached file for the answer.
Explanation:
See the attached file for the explanation
Answer:
The electrical power is 96.5 W/m^2
Explanation:
The energy balance is:
Ein-Eout=0
if:
Gsky=oTsky^4
Eb=oTs^4
qc=h(Ts-Tα)
if Gl≈El(l,5800)
lt= 2*5800=11600 um-K, at this value, F=0.941
The hemispherical emissivity is equal to:
lt=2*333=666 K, at this value, F=0
The hemispherical absorptivity is equal to: