A 180m long pipeline runs (through three points ABC) at an upward slope of 1 in
ID: 1844101 • Letter: A
Question
A 180m long pipeline runs (through three points ABC) at an upward slope of 1 in 60. The length of the portion AB is 90m and its diameter is 0.15m. At B, the pipe section suddenly enlarges to 0.3m diameter and remains so for the remainder of its length BC, which is 90m. A flow of 50 L/sec is pumped into the pipe at its lower end A and is discharged at the upper end C into a closed tank. The pressure at the supply end A is 137.34 kN/m^2. Sketch the hydraulic grade line and energy grade line and also find the pressure at the discharge end C. Take f=0.02.
Explanation / Answer
Given discharge Q=50lit/sec
=0.05m3/sec
D1=0.15m D2=0.30m (replace 3.14 with pie for clear understanding)
V1 =Q/(3.14/4)*0.152
=2.83m/sec
V2 =Q/(3.14/4)*0.32
=0.71m/sec
Loss of head in section AB
hf1 = f *L1*v2/2gD1
=0.02*90*2.832/(2*9.81*0.15)
=4.9m
Loss of head in section BC
hf2 = f *L2*v2/2gD2
=0.02*90*0.712/(2*9.81*0.3)
=0.153m
Loss of head at B due to sudden enlargement =(v1-v2)2/2*g
=(2.83-0.71)2/(2*9.81)
=0.229m
Section AB:
Assuming datum to be pass through A,different heads at A,
ZA=0,
?=103*9.81
PA/? =137.34*103/(9.81*103) =14m
VA2/2g=2.832/2*9.81 =0.408m
Different
ZB=90/60
=1.5m
VB2/2g =(2.83)2/2*9.81 =0.408m
By bernoullis equation between A and B
PA/?+ VA2/2g+ ZA = PB/?+ VB2/2g+ ZB+hf1
14+0.408+0= PB/?+0.408+1.5+4.9
PB/?=7.6m
The hydraulic gradient of a line(H.G.L)= P/?
H.G.L= PA/?=14m and PB/?=7.6m
Total energy line(T.E.L)=(P/?)+ V2/2g
T.E.Lat point A=14+0.408=14.408m
T.E.Lat point B=7.6+0.408=8.008m
Both lines(H.G.L and T.E.L)will run parallel to each other
Section BC:
?=103*9.81
Different heads at B:
By bernoullis equation between A and B
PB
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