Question

The rate of flow of water in a pump installation is 60.6 kg/s. The intake static gage is 1.22 m below the pump centreline and reads 68.95 kPa gage;the discharge static gage is 0.61 m below the pump centre line and reads 344.75 kPagage. The gages are located close to the pump as much as possible. The area of the intake and discharge pipes are; 0.093 m2 and 0.069 m2 respectively. The pump efficiency is 74%. Take density of water equals 1000 kg/m3. What is the hydraulic power in kW

Answer 1

Answer:

Pump power is 23.09 kW

Explanation:

Data

gravitational constant, $g = 9.81 m/s^2$

mass flow, $\dot{m} = 60.6 kg/s$

flow density, $\rho = 1000 kg/m^3$

pump efficiency, $\eta = 0.74$

output gage pressure, $p_o = 344.75 kPa$

input gage pressure, $p_i = 68.95 kPa$

output pipe area, $A_o = 0.069 m^2$

input pipe area, $A_i = 0.093 m^2$

output height, $z_o = 1.22 m - 0.61 m = 0.61 m$ (considering that pump is at the maximum height, i.e., 1.22 m)

input height, $z_i = 0 m$

pump hydraulic power,$P = ? kW$

First of all, volumetric flow (Q) must be computed

$Q = \frac{\dot{m}}{\rho}$

$Q = \frac{60.6 kg/s}{1000 kg/m^3}$

$Q = 0.0606 m^3/s$

Then, velocity (v) must be computed for both input and output

$v_o = \frac{Q}{A_o}$

$v_o = \frac{0.0606 m^3/s}{0.069 m^2}$

$v_o = 0.88 m/s$

$v_i = \frac{Q}{A_i}$

$v_i = \frac{0.0606 m^3/s}{0.093 m^2}$

$v_i = 0.65 m/s$

Now, total head (H) can be calculated

$H = (z_o - z_i) + \frac{v_o^2 - v_i^2}{2 \, g} + \frac{p_o - p_i}{\rho \, g}$

$H = (0.61 m - 0 m) + \frac{{0.88 m/s}^2 - {0.65 m/s}^2}{2 \, 9.81 m/s^2} + \frac{(344.75 Pa-68.95 Pa)\times 10^3}{1000 kg/m^3 \, 9.81 m/s^2}$

$H = 28.74m$

Finally, pump power is computed as

$P = \frac{Q \, \rho \, g \, H}{\eta}$

$P = \frac{0.0606 m^3/s \, 1000 kg/m^3 \, 9.81 m/s^2 \, 28.74m}{0.74}$

$P = 23.09 kW$