In a hydroelectric power station, $1000 k g$ of water is allowed to drop a height of $100 m$ in $1 s$ . If the conversion of potential energy to electric energy is $60 %$ , Calculate the power output. [Take $g = \frac{10 m}{s^{2}}$ ]

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Solution

Step 1: Given data,
Efficiency= $60 %$
Height= $100 m$
Mass of water = $1000 k g$
Time= $1 s$
Acceleration due to gravity $g = \frac{10 m}{s^{2}}$
Step 2: To find force and potential energy,
Formula used : $f o r c e = w e i g h t = m g$
$f o r c e = 1000 \times 10 = 10000 N$
The potential energy in terms of mass and height is,
Formula used : $p o t e n t i a l e n e r g y = m g h$
$E_{P} = 10000 \times 100 = 1000000 J$
Step 3: To find time,
Formula used : $p o w e r = \frac{e n r g y}{t i m e}$
$p o w e r = \frac{1000000}{1}$
$= 1000000 w a t t s$
Step 4: To find the power that is efficiently used,
P' = $60 % o f 1000000$
$= \frac{60}{100} \times 1000000 = 600000 w a t t s$
Thus, the Power output is $600000 w a t t s$ $= 6 \times 10^{5} w a t t s .$

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In a hydroelectric power station, 1000kg of water is allowed to drop a height of 100m in 1s. If the conversion of potential energy to electric energy is 60%, Calculate the power output. [Take g=10ms2]

In a hydroelectric power station, $1000 k g$ of water is allowed to drop a height of $100 m$ in $1 s$ . If the conversion of potential energy to electric energy is $60 %$ , Calculate the power output. [Take $g = \frac{10 m}{s^{2}}$ ]