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What power is required to 10liters of water to a height of 6m, in one minute through a pipe of 1.25cm diameter ?

Asked by anirbanbag81 4th April 2018, 5:19 AM
Answered by Expert
Answer:
As per Bernouli's equation, energy of fluid  P+(1/2)ρv2 + ρgh = constant.
 
P - pressure, ρ - density, v - velocity, g - acceleration due to gravity, h is height.
 
velocity at inlet of pipe  = Volumetric flowrate / crosssection Area = begin mathsize 12px style 10 space fraction numerator L i t r e over denominator m i n end fraction fraction numerator 1000 space c c over denominator 1 space L i t r e end fraction fraction numerator 10 to the power of negative 6 end exponent m cubed over denominator c c end fraction fraction numerator 1 space m i n over denominator 60 s end fraction cross times fraction numerator 4 over denominator pi cross times 1.25 cross times 1.25 cross times 10 to the power of negative 4 end exponent end fraction space equals space 1.358 space m divided by s end stylebegin mathsize 12px style 10 space fraction numerator L i t r e over denominator m i n end fraction fraction numerator 1000 space c c over denominator 1 space L i t r e end fraction fraction numerator 10 to the power of negative 6 end exponent m cubed over denominator c c end fraction fraction numerator 1 space m i n over denominator 60 s end fraction cross times fraction numerator 4 over denominator pi cross times 1.25 cross times 1.25 cross times 10 to the power of negative 4 end exponent end fraction space equals space 1.358 space m divided by s end style
 
Let us assume pressure drop from pump outlet to end of 6 metres pipe is negligible. Also velocity of water at outle of 6 m pipe is very less.
 
hence required energy per unit volume E = (1/2)ρv2 + ρgΔh 
power is obtained by multiplying the above equation by volumetric flow rate.
 
hence power P =  begin mathsize 12px style open parentheses 1 half cross times 1000 cross times 1.358 cross times 1.358 space plus 1000 cross times 9.8 cross times 6 close parentheses cross times 10 space fraction numerator L over denominator m i n end fraction fraction numerator 1000 space c c over denominator L end fraction fraction numerator 10 to the power of negative 6 end exponent m cubed over denominator c c end fraction fraction numerator 1 m i n over denominator 60 end fraction end style
we get P = 9.954 Watts

Answered by Expert 4th April 2018, 12:57 PM
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