There are 400kf/min of water being handled by a pump. The lift us firm 20-m deep well and the delivery velocity ks 15m/s. Find (a) the change in potential energy, (b) kinetic energy, (c) the required power of the pumping unit; g= 9.81m/s²
a)
"S=\\frac Qv=0 .44\\cdot 10^{-3}~m^2,"
"V=Sh=8.9\\cdot 10^{-3}~m^3,"
"m=\\rho V=8.9~kg,"
"U=mgh=1780~J,"
b)
"E=\\frac{mv^2}2=1000~J,"
c)
"P=mgv=133~W."
Comments
Leave a comment