A pump draws water from a tank which is 2.5; | vedclass

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(B) The power $P$ required by the pump is given by the rate of change of mechanical energy: $P = \left(gh + \frac{v^{2}}{2}\right) \frac{dm}{dt}$.Firs

Vedclass · Accepted Answer

$185$

Vedclass · Answer

(B) The power $P$ required by the pump is given by the rate of change of mechanical energy: $P = \left(gh + \frac{v^{2}}{2}ight) \frac{dm}{dt}$.First,calculate the mass flow rate $\frac{dm}{dt} = ho A v$.Given $ho = 1000\; kg/m^{3}$,$A = 10\; cm^{2} = 10 	imes 10^{-4}\; m^{2}$,and $v = 5\; m/s$:$\frac{dm}{dt} = 1000 	imes 10 	imes 10^{-4} 	imes 5 = 5\; kg/s$.Now,substitute the values into the power equation using $g = 10\; m/s^{2}$,$h = 2.5\; m$,and $v = 5\; m/s$:$P = \left(10 	imes 2.5 + \frac{5^{2}}{2}ight) 	imes 5$.$P = (25 + 12.5) 	imes 5$.$P = 37.5 	imes 5 = 187.5\; W$.Rounding to the nearest provided option,the correct value is $185\; W$.

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