Answer to Question #194915 in Classical Mechanics for na su kim

Question

At a bottled water plant, water with a density of 1.00 g/ml passes through a pumping station where it is raised vertically by 5.23 m at the rate of 1,880,000 liters per day. The liquid enters and leaves the pumping station at the same speed and through pipes of equal diameter.

(a)Determine the output mechanical power (in W) of the lift station. Ignore any energy loss due to friction.

  W

(b)Assume an electric motor continuously operating with average power 3.85 kW runs the pump. Find its efficiency.

Answer 1

$P = 3,85 kW= 3850000W$

$t = 24h$

$W_p = Pt$

$\text{work performed by the pump per day}$

$W_p = 3850000*24 = 92400000\ J$

$\text{work required to raise water to a height}$

$\text {this is the difference in potential energies}$

$W_w = E_{p1}-E_{p0}$

$E_{p0} =0$

$E_{p1}=mgh$

$h = 5.23\ m$

$g= 9.8\frac{m}{s^2}$

$m = \rho V$

$V = 1880000 l= 1880m^3$

$\rho= 1\frac{g}{ml}=1000\frac{kg}{m^3}$

$m = 1000*1880=1880000\ kg$

$E_{p1}=mgh= 1880000*9.8*5.23=96357520\ J$

$W_w= 96357520\ J$

$\eta=\frac{W_p}{W_W}=\frac{92400000}{96357520}=0.95=95\%$

$\text{Answer :95\%}$