written 6.0 years ago by
teamques10
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modified 6.0 years ago
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Data:-
D=0.6m,H=150m
L=5000m
Rate of water drawing=0.1m3s per 300 m
q∗=0.1300=3.33×10−4m3/sm
f=0.04,L0=1200m,vel. at dead end=0
To find:-
hf in last 1200 m of pipe=?
Solution:

Total distance Q0=1200300×0.1=0.4m3/sec
-As the water is drawn at uniform rate, therefore deriving expression for head loss for pipe having uniform water drawing rate.
-Consider a section at distance 'x' from the start of uniform withdrawal at q* per meter length
∴ Discharge from x length
Qx=Q0−(q∗.x)
in a small distance 'dx; head loss is dhf
∴dhf=f.dx.v2x2gD
but vx=QxA
∴vx=ΔQxπD2
∴v2x=[4(Q0−q∗.x)πD2]2
∴dhf=f2gD×16(Q0−q∗x)2π2D2.dx
dhf=8fπ2gD5×(Q0−q∗x)2dx
Integrating
hf=∫L008fπ2gD5(Q0−q∗x)2dx
=−8fπ2.g.D5×13q∗[(Q0−q∗L0)3−Q30]
∴hf=8f3π2.g.D5×1q∗[(Q30−(Q0−q∗L0)3].........(1)
Equation (1) is expression for pipe in which water is drawn at uniform rate
Now, for last 1200 m, D=0.6m,q∗=3.33×10−4m3/sm,Q0=0.4m3/s,L0=1200m & f=0.04
∴ equation (1) becomes
hf=8×0.043π2×9.81×0.65×13.33×10−4[0.43−(3.33×10−4×1200)3]=2.7m