This question is so annoying been trying to work it out for ages here it is. A seawater pumping station is to pump 15.6 t/h of seawater (relative density 1.03). If it is desired that the velocity in the intake is not to exceed 3/ms and the velocity in the outlet pipe not to exceed 5m/s determine suitable inlet and outlet pipe diameters.
2007-01-03 03:59:19 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Engineering
Assuming you mean metric tons per hour for the flow rate, we have:
( 15.6 t/h ) * ( 1h/ 3600s ) = .004333 t/s
Now we know the density of the water is 1.03 t/m^3. This means that the volume flow rate is:
( .004333 t/s ) * ( 1 / ( 1.03 t/m^3 ) ) = .004207 m^3/s
Now we know the cross sectional area of the intake pipe is π*(d^2)/4 so we can solve for the diameter by:
( .004207 m^3/s ) * ( 4 / ( π*d^2 m^2)) = 3 m/s
d = sqrt (.00407*4 / ( π*3 )) = .042 m or 4.2 cm
A similar procedure for the outtake side leads to a diameter of:
d = .033 m or 3.3 cm.
2007-01-03 04:49:51 · answer #1 · answered by 1,1,2,3,3,4, 5,5,6,6,6, 8,8,8,10 6 · 0⤊ 0⤋
Q=pi(D1)^2*3*3600
Q=pi(D2)^2*5*3600
Q=15.6 *1/1.03 m*3/hr. Here you have 2 equations in 2 unknowns D1,D2 the two diameters of the inlet and outlet pipes in m.Watch you units and solve good luck.I assumed one cubic meter of fresh water equals one ton.
2007-01-03 05:08:05 · answer #2 · answered by Mesab123 6 · 0⤊ 0⤋
15.6 teaspoons per hour? Heck, the pipes could be capillaries. ;-)
2007-01-03 05:30:18 · answer #3 · answered by gebobs 6 · 0⤊ 0⤋