Question

Water in pipe AB (Figure 2), diameter (DAB) 1.2 m, is flowing with the velocity (VAB) of 3 m/second. Then it continues to flow through pipe BC, with a diameter (DBC) of 1.5 m and finally flows through branch pipe of CD, diameter (DCD) of 0.8 m and pipe of CE (DCE), with a velocity (VCE) of 2.1 m/second. Calculate:
1. Discharge of pipe AB (QAB)
2. Flow velocity of pipe BC (VBC)
3. Flow velocity of pipe BC (VBC)
4. Diameter of pipe CE (VCE)

Answer 1

Based on the continuity equation, total discharge is constant.

discharge, Q=AV

A=area

V=velocity

A1= Area of pipe AB=((DAB)2)/4=((1.2)2)/4=1.13m2/s

V1= Velocity of pipe AB=VAB=3m/s

A2= Area of pipe BC=((DBC)2)/4=((1.5)2)/4=1.767m2/s

V2= Velocity of pipe BC=VBC

A3= Area of pipe CD=((DCD)2)/4=((0.8)2)/4=0.502m2/s

V3= Velocity of pipe CD=VCD

A4= Area of pipe CE=((DCE)2)/4

V4= Velocity of pipe CE=VCE=2.1m/s

Q= A1V1  = A2V2 =( A3V3+A4V4)

discharge of pipe AB, Q= A1V1  = 1.13*3=3.39m3/s

A1V1  = A2V2  =1.767m2/s*V2

3.39=1.767m2/s*V2

V2= Velocity of pipe BC=VBC=1.92m/s

3.39m3/s=( A3V3+A4V4)=0.502*V3+A4*2.1

FROM THIS EQUATION SUBTITUTING VALUES OF V3 WE GET DIAMETER DCE OR VICE VERSA.

V3= Velocity of pipe CD(m/s)	A4= Area of pipe CE	The diameter of pipe CE=DCE(m)
1.5	1.258	1.27
2	1.39	1.20
2.5	1.02	1.14
3	0.9	1.07
3.5	0.782	1
4	0.663	0.92