In order to undertake repairs on a cracked dam face the smal

In order to undertake repairs on a cracked dam face, the small reservoir behind the dam has been mostly emptied. The remaining water is only 15 cm deep and must be maintained at this depth or the repair work will be hampered. Water is continuing to seep into the reservoir at a rate of 8 times 10^-3 m^3/s. You have installed a pump to continually remove this water. If the hose at the pump outlet has a diameter of 10 cm, how fast is the water moving at the outlet?

Solution

1) here to undertake thr repair on dam face ,water in the dam has to maitained below 15 cm and

for that total water added to reservoir should be zero and it is achieved where

flow rate incoming=Q1=8*10^-3 m3/s

flow rate outgoing=Q2

flow rate incoming=flow rate outgoing by pump

2) water added in reservoir=Q1-Q2

0=Q1-Q2

Q1=Q2

hence water pumped is Q2=8*10^-3 m3/s

3) area of hose is

A2=(pi*4)*d2=(pi*4)*.1^2=7.8539*10^-3 m2

hence velocity v2=Q2/A2 from continuity equation

v2=(8*10^-3)/(7.8539*10^-3)=1.018 m/s

water moving at outlet=1.018 m/s