NICOLE IRENE DELA PEÑA
BS ECE 5A
PROBLEMS
1. What is the watts output of a 25-hp motor at full load?
Solution:
2. A 150-kw generator has an efficiency of 91 per cent at full load. Calculate the input in
kilowatts and power loss.
Given:
𝑃𝑙 = 150 watts
EFF= 91% at full load
Solution:
Load power is the power output
Since:
EFF=
𝑃𝑖𝑛 =
𝑃𝑜𝑢𝑡
𝑃𝑖𝑛
X 100%
150 𝐾𝑤
91%
X 100%
𝑃𝑖𝑛 = 164.835 Kw
Power losses= 𝑃𝑖𝑛 - 𝑃𝑜𝑢𝑡
= (164.835 – 150) Kw
=14.835 Kw
3. A 15-hp motor operates at an efficiency of 87.5 percent at full load. If the stray power
loss is approximately one-fourth of the total loss, calculate the copper loss.
Given:
𝑝𝑙 = 15 hp motor
EFF = 87.5%, at full load
Stray power loss = 1⁄4 total loss
Solution:
Let:
Pw = the copper loss
Spl = stray power loss
Pt = total power loss
𝑃𝑖𝑛 =
𝑃𝑖𝑛 =
𝑊
15 ℎ𝑝 𝑋 746
ℎ𝑝
0.875
11190 𝑤𝑎𝑡𝑡𝑠
0.875
𝑃𝑖𝑛 = 12789 watts
𝑃𝑡 = 𝑃𝑖𝑛 - 𝑃𝑜𝑢𝑡
= 12789 – 11190
𝑃𝑡 = 1599 watts
Since:
S.P.L. = 1⁄4 Pt
= 1⁄4 (1599)
S.P.L. = 399.75 watts
Thus:
Pcu = Pt – S.P.L.
= 1599 – 399.75
Pcu = 1199.25 ≈ 1200 watts
4. The rotational loss in a generator was found to be 780 watts when the generated emf
was 132 volts. Determine the rotational loss for the voltages of 138 and 126 volts.
Given:
S.P.L. or rotational loss = 780 watts
Eg = 132 volts
Solution:
S.P.L. = Eg X Ia
Ia =
780 𝑤𝑎𝑡𝑡𝑠
132 𝑣
Ia = 5.9 amp
a) At Eg = 138 volts
S.P.L. = (138)(5.9)
= 815.5 watts
b) At Eg = 126 volts
S.P.L. = (126)(5.9)
= 743.4 watts