A natural-draft cooling tower receives 250,000 ft3/min of air at standard atmospheric pressure, 70oF, and 45 percent relative hu
midity. The air leaves saturated at 100oF. 3500 gallons per minute of water at 104oF is cooled. Calculate
(a) the temperature of the water at the tower exit, and
(b) the necessary height of the tower if the total pressure losses are 2.00 lbf/ft2
(a) the temperature of the water at the tower exit, and
(b) the necessary height of the tower if the total pressure losses are 2.00 lbf/ft2
1 answer:
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Answer:
capacity = 0.555 mAh
capacity = 3600 mAh
Explanation:
given data
battery = 1800 mAh
OCV = 3.9 V
solution
we get here capacity when it is in series
so here Q = 2C
capacity = 2 × ampere × second ...............1
put here value and we get
and 1 Ah = 3600 C
capacity =
capacity = 0.555 mAh
and
when it is in parallel than capacity will be
capacity = Q1 +Q2 ...............2
capacity = 1800 + 1800
capacity = 3600 mAh