Pump Power
Calculate the power required for a pump to lift water against a head of 25 meters flowing at 5 cubic meters per second. Neglect all friction losses. Assume a pump efficiency of .77, and the specific weight of water is 9.8 kN/m^3.
Expand Hint
The pump power equation:
$$$\dot{W}=\frac{Q\gamma h}{\eta_t}$$$
where
$$Q$$
is the volumetric flow,
$$h$$
is the fluid head needed to be lifted,
$$\eta_t$$
is the total efficiency (
$$\eta_{pump} \times \eta_{motor}$$
),
$$\dot W$$
is the power, and
$$\gamma$$
is the specific weight of the fluid.
Hint 2
Solve for
$$\dot{W}$$
.
The pump power equation:
$$$\dot{W}=\frac{Q\gamma h}{\eta_t}$$$
where
$$Q$$
is the volumetric flow,
$$h$$
is the fluid head needed to be lifted,
$$\eta_t$$
is the total efficiency (
$$\eta_{pump} \times \eta_{motor}$$
),
$$\dot W$$
is the power, and
$$\gamma$$
is the specific weight of the fluid.
Solving for power:
$$$\dot{W}=\frac{(5m^3)(9.8kN)(25m)}{sec(m^3)(.77)}$$$
$$$\dot{W}=1591\:\frac{kN\cdot m}{sec}=1591\:kW$$$
1591 kW
Time Analysis
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