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Question by R.: A small isolated farm uses a windmill and a gas generator?
A small isolated farm uses a windmill and a gas generator tor power. During a 10 day period, they produced 3010kW/h of power, with the windmill operating at 45% of capacity and the generator at capacity. During the following 10 day period, they produced 2900kW/h with the windmill at 72% of capacity and the generator down 60 hours for repair (at capacity otherwise). what is the capacity (in kW) of each?
yes, I meant 3010 kW-h. Thanks.
Give your answer to this question below!
Thee is no such thing as “kW/h”. Do you mean 3010 kW-h, which is a unit of energy, not power?
60 hours / 10 days = 60 hours/240 hours = 1/4 = 0.25
3010 kW-h x 1 day/24 hr / 10d = 12.54 kW over 10 days
2900 kW-h x 1 day/24 hr / 10d = 12.08 kW over 10 days
0.45W + G = 12.54 kW
0.72W + 0.25G = 12.08 kW
45W + 100G = 1254
72W + 25G = 1208
–288W – 100G = –4832
243W = 3578
W = 14.72 kW
72(14.72) + 25G = 1208
25G = 147
G = 5.91 kW
but chck the arith.
.
billrussell42
October 23, 2011 at 5:34 am