## Pump Work

What is the external work done by a pump that delivers 3,000 cubic meters of water per hour against a total head of 10 m? Ignore all losses and note 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**

Convert the volumetric flow rate into
$$m^3/s$$
.

First, letâ€™s convert the volumetric flow rate into
$$m^3/s$$
.

$$$3,000\frac{m^3}{hr}\cdot \frac{1hr}{60min}\cdot \frac{1min}{60sec}=\frac{3,000m^3}{3600sec}=0.83\:m^3/s$$$

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. Since we are ignoring all losses, we can assume efficiency is 100%. Thus,

$$$\dot{W}=\frac{(0.83m^3)(9,800N)(10m)}{(sec)(m^3)1}=81,340\frac{N\cdot m}{s}=81.3\:kW$$$

81.3 kW