## A Pump’s Efficiency

If a pump requires 2,000 kW of power to lift against a total water head of 30 m at 5 cubic meters per second, what is the pump’s efficiency? Assume no friction losses and the specific weight of water is 9.8 kN/m^3.

##
__
__**Expand Hint**

**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**

**Hint 2**

Solve for
$$\eta_t$$
.

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 efficiency:

$$$2,000kW=\frac{(5m^3)(9.8kN)(30m)}{sec(m^3)\eta_t}$$$

$$$\eta_t=\frac{(5)(9.8kN)(30m)}{sec(2,000\frac{kN\cdot m}{sec})}=\frac{1,470}{2,000}=0.74$$$

0.74

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