- #1
GSXR-750
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Homework Statement
Calculate head required for the pump and then its power requirement assuming 70% efficiency.
The lower storage vessel is vented to atmosphere (assume 1 bar pressure) .
I have the following given information:
Pipe Area = 0.00636m^3.
Flow(Q)= 0.01m^3/s
Average Velocity = 1.57m/s
Density of Fluid = 960kg/m^-3.
Liquid viscosity = 0.081 Pa/s
I have calculated the head losses of the whole system, using firstly the Equivalent Head (Hm) to be 1.03m and Number Velocity head (Hf) to be 0.54m.
Homework Equations
Bernoullis
The Attempt at a Solution
Pump inlet
## \frac {0} { 960 * 9.81 } + \frac {1.57^2} { 2 * 9.81 \ } + 0 = \frac {P2 }{\ 960 * 9.81 \ } + \frac {1.57^2} { \ 2 * 9.81 \ } + 4 + 0.38 ##
P2 = -41249 Pa
= -0.41Bar
Pump Outlet
## \frac {P1} { 960 * 9.81 } + \frac {1.57^2} { 2 * 9.81 \ } = \frac {200000 }{\ 960 * 9.81 \ } + \frac {1.57^2} { \ 2 * 9.81 \ } + 15 + 1.19 ##
P1= 352470 Pa
P1 = 3.5 Bar + 2 Bar(Pressurised Cylinder)Power = qpgh
= 0.01 x 960 x 9.81 x 19.57
= 1.84 kW
Assuming 70% = ## \frac { 1.84 } { \ 0.7 \ } ##
= 2.63 kW
Is my method correct so I can say the pump required is an 2.63 kW to overcome the 5.5 bar.Any help would be great
Thanks
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