Respuesta :
Answer:
Pump power is 23.09 kW
Explanation:
Data
gravitational constant, [tex] g = 9.81 m/s^2 [/tex]
mass flow, [tex] \dot{m} = 60.6 kg/s [/tex]
flow density, [tex] \rho = 1000 kg/m^3 [/tex]
pump efficiency, [tex] \eta = 0.74 [/tex]
output gage pressure, [tex] p_o = 344.75 kPa [/tex]
input gage pressure, [tex] p_i = 68.95 kPa [/tex]
output pipe area, [tex] A_o = 0.069 m^2 [/tex]
input pipe area, [tex] A_i = 0.093 m^2 [/tex]
output height, [tex] z_o = 1.22 m - 0.61 m = 0.61 m [/tex] (considering that pump is at the maximum height, i.e., 1.22 m)
input height, [tex] z_i = 0 m [/tex]
pump hydraulic power,[tex] P = ? kW [/tex]
First of all, volumetric flow (Q) must be computed
[tex] Q = \frac{\dot{m}}{\rho}[/tex]
[tex] Q = \frac{60.6 kg/s}{1000 kg/m^3} [/tex]
[tex] Q = 0.0606 m^3/s[/tex]
Then, velocity (v) must be computed for both input and output
[tex] v_o = \frac{Q}{A_o}[/tex]
[tex] v_o = \frac{0.0606 m^3/s}{0.069 m^2}[/tex]
[tex] v_o = 0.88 m/s [/tex]
[tex] v_i = \frac{Q}{A_i}[/tex]
[tex] v_i = \frac{0.0606 m^3/s}{0.093 m^2}[/tex]
[tex] v_i = 0.65 m/s [/tex]
Now, total head (H) can be calculated
[tex] H = (z_o - z_i) + \frac{v_o^2 - v_i^2}{2 \, g} + \frac{p_o - p_i}{\rho \, g} [/tex]
[tex] 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} [/tex]
[tex] H = 28.74m [/tex]
Finally, pump power is computed as
[tex] P = \frac{Q \, \rho \, g \, H}{\eta}[/tex]
[tex] P = \frac{0.0606 m^3/s \, 1000 kg/m^3 \, 9.81 m/s^2 \, 28.74m}{0.74}[/tex]
[tex] P = 23.09 kW [/tex]
The hydraulic power in kW is mathematically given as
P = 23.09 kW
What is the hydraulic power?
Generally the equation for the volumetric flow (Q) is mathematically given as
Q=m/p
Therefore
Q=60/1000
Q=0.0606
Generally the equation for the velocity (v) is mathematically given as
v=Q/A
Hence for input
[tex]v_i = \frac{Q}{A_i}\\\\v_i = \frac{0.0606 }{0.093 }[/tex]
v_i = 0.65 m/s
For input
[tex]v_o = \frac{Q}{A_o}\\\\v_o = \frac{0.0606 }{0.069}[/tex]
v_o = 0.88 m/s
Therefore, Total head (H)
[tex]H = (z_o - z_i) + \frac{v_o^2 - v_i^2}{2 \, g} + \frac{p_o - p_i}{\mu g}[/tex]
[tex]H = (0.61 0 ) + \frac{{0.88 }^2 - {0.65 }^}{2 *9.81 } + \frac{(344.75 Pa-68.95 Pa)*10^3}{1000* 9.81}[/tex]
H = 28.74m
Generally the equation for the pump power P is mathematically given as
[tex]P = \frac{Q * \rho*g*H}{\eta}[/tex]
[tex]P = \frac{0.0606 * 100*9.81*28.74}{0.74}[/tex]
P = 23.09 kW
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