1. Solve. The diameter of the pipe at the sections 1 and 2 are 3m and 6m respectively. Find the discharge through the pipe if the velocity of flowing through the pipe at section 1 is 2 m/s. Determine the velocity at section 2.
Answers
Answer:
Diameter of pipe 1 d_1=15 cm=0.15\ md
1
=15cm=0.15 m
C/S Area of pipe 1 A_1 =\dfrac{\pi(d_1)^2}{4}=0.005625\pi\ m^2A
1
=
4
π(d
1
)
2
=0.005625π m
2
Diameter of pipe 2 d_2=20\ cm=0.2\ md
2
=20 cm=0.2 m
C/S Area of pipe 2 A_2 =\dfrac{\pi(d_1)^2}{4}=0.01\pi\ m^2A
2
=
4
π(d
1
)
2
=0.01π m
2
Velocity at pipe 1 v_1=4\ m/sv
1
=4 m/s
Velocity at pipe 2 v_2=?v
2
=?
Discharge through pipe 1 Q_1=A_1v_1=0.005625\pi\ m^2 (4\ m/s)Q
1
=A
1
v
1
=0.005625π m
2
(4 m/s)
=0.0225\pi\ m^3/s=0.0707\ m^3/s=0.0225π m
3
/s=0.0707 m
3
/s
Q_1=0.0707\ m^3/sQ
1
=0.0707 m
3
/s
From continuity Q_1=Q_2+Q_3Q
1
=Q
2
+Q
3
Discharge through pipe 2 Q_2=A_2v_2=0.01\pi\ m^2 (v_2)Q
2
=A
2
v
2
=0.01π m
2
(v
2
)
=Q_1=0.0225\pi\ m^3/s=Q
1
=0.0225π m
3
/s
Velocity at pipe 2 v_2=\dfrac{Q_2}{A_2}=\dfrac{0.0225\pi\ m^3/s}{0.01\pi\ m^2 }=2.25\ m/sv
2
=
A
2
Q
2
=
0.01π m
2
0.0225π m
3
/s
=2.25 m/s
v_2=2.25\ m/sv
2
=2.25 m/s
HERE IS YOUR ANSWER...
HOPE IT HELPS UHH...
THANKS....