A Pump’s Pressure
In the shown figure, a 80% efficient pump uses 3,000 kW of power to push water upwards at 10 cubic meters per second. If the measured outlet pressure is 200 kPa, what is the inlet pump pressure in kPa? Assume 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
How pressure changes with elevation in a fluid can be expressed as:
$$$\Delta P=\gamma \times \Delta h$$$
where
$$\Delta P$$
is the change is pressure,
$$\Delta h$$
is the change in height, and
$$\gamma $$
is the fluid’s specific weight (
$$density \times gravity$$
).
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.
$$$3,000,000W=\frac{(10m^3)(9,800N) h}{(sec)(m^3)(0.8)}$$$
$$$3,000,000\times\frac{N\cdot m}{(sec)}=122,500(h)\times \frac{N}{(sec)}$$$
$$$h=\frac{3,000,000}{122,500}m=24.489\:m$$$
How pressure changes with elevation in a fluid can be expressed as:
$$$\Delta P=\gamma \times \Delta h$$$
where
$$\Delta P$$
is the change is pressure,
$$\Delta h$$
is the change in height, and
$$\gamma $$
is the fluid’s specific weight (
$$density \times gravity$$
).
$$$P_{in}-200kPa=9.8\frac{kN}{m^3}\times 24.489m$$$
$$$P_{in}=[9.8\frac{kN}{m^2}\times 24.489]+200kPa=240kPa+200kPa=440\:kPa$$$
440 kPa
Time Analysis
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