Given :
Discharge rate Q= 0.650 L/s
Output Pressure, P1 = 2.50 ✕ 10^5 N/m2
Inlet Hose diameter, D1 = 3.00 cm
Fluid height after pump, H = 3.30 m
Loss of fluid height, L = 2.00 m
Exit pipe diameter, D2 = 4.00 cm
As we know that Discharge is equal to Area times fluid velocity.
So let first specify the rate at which fluid being drained = Q L/s (m^3/s)
So the discharge,
where, A1 = Cross section of pipe at inlet, V1 = Speed of flow at inlet
Now looking at the Area
Therefore, V1 = Q/A1 = 0.650/0.706 = 0.92 m/s
Case - A) The water enters a hose with a 3.00 cm inside diameter and rises 3.30 m above the pump. What is its pressure at this point?
Output Pressure, P1 = 2.50 ✕ 10^5 N/m2
Now as there is no change in the cross sectional area of the pipe, the velocity of the flow remains constant due to equation of continuity.
So, the pressure after the flow has has attained a height of 3.30 m after pump is:
P2 + 9.8 * 3.30 = 2.50 ✕ 10^5
Case - B) The hose then loses 2.00 m in height from this point as it goes over the foundation wall, and widens to 4.00 cm diameter. What is the pressure now?
Increase in pipe diameter to = 4 cm
Loss of height = 2 m
As Discharge will be same
Looking at the area
Therefore, V3 = Q/A3 = 0.650/1.256 = 0.517 m/s
Now pressure at this point is:
please only answer of you know it for sure. A sump pump is draining a flooded...
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