## 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.

##
__
__**Hint**

**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**

**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