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Fluid Mechanics 07
Hydraulic and Energy Grade Lines
EGL:- Energy Grade line indicate the total head at
any point in the system.
𝐸𝐺𝐿
= π‘‰π‘’π‘™π‘œπ‘π‘–π‘‘π‘¦ β„Žπ‘’π‘Žπ‘‘ + π‘ƒπ‘Ÿπ‘’π‘ π‘ π‘’π‘Ÿπ‘’ β„Žπ‘’π‘Žπ‘‘
+ πΈπ‘™π‘’π‘£π‘Žπ‘‘π‘–π‘œπ‘› π»π‘’π‘Žπ‘‘
𝑣2
𝑝
𝐸𝐺𝐿 =
+ +𝑍
2βˆ—π‘” Ι£
HGL:- Hydraulic Grade Line indicate the
piezometric head at any point in the system
𝐻𝐺𝐿 = π‘ƒπ‘Ÿπ‘’π‘ π‘ π‘’π‘Ÿπ‘’ β„Žπ‘’π‘Žπ‘‘ + πΈπ‘™π‘’π‘£π‘Žπ‘‘π‘–π‘œπ‘› β„Žπ‘’π‘Žπ‘‘
𝑝
𝐻𝐺𝐿 = + 𝑧
Hydraulic and Energy Grade Lines
Pump Add head to the System
Turbine
Nozzle
Nozzle increase the velocity and if discharge to
atmospheric the term of pressure head will be
zero
Change in pipe diameter
Negative Pressure
Example
A pump draws water (50°F) from a reservoir,
where the water-surface elevation is 520 ft, and
forces the water through a pipe 5000 ft long and
1 ft in diameter. This pipe then discharges the
water into a reservoir with water-surface
elevation of 620 ft. The flow rate is 7.85 cfs, and
the head loss in the pipe is given by
Determine the head supplied by the pump, hp,
and the power supplied to the flow, and draw
the HGL and EGL for the system. Assume that
the pipe is horizontal and is 510 ft in elevation.
Solution
𝑝1
Ι£
+ 𝑧1
𝑣12
+
2βˆ—π‘”
+ β„Žπ‘ =
𝑝2
+
Ι£
𝑧2
𝑣22
+
2βˆ—π‘”
+ β„Žπ‘‘ + β„ŽπΏ
Where
P1=P2=Patm=zero
V1=v2=zero
Ht=zero, z1=520 ft, z2=620 ft
𝑄
7.85
𝑣= =
= 10𝑓𝑑/𝑠
2
𝐴 (Ξ /4) βˆ— (1 )
hL= 0.01 βˆ—
=77.6 ft
𝐿
𝐷
βˆ—
𝑣2
( )=0.01
2βˆ—π‘”
βˆ—
5000
1
βˆ—
102
(
)
2βˆ—32.2
β„Žπ‘ = 𝑧2 βˆ’ 𝑧1 + β„ŽπΏ = 620 βˆ’ 510 + 77.6
=178 ft
π‘ƒπ‘œπ‘€π‘’π‘Ÿ = Ι£ βˆ— 𝑄 βˆ— β„Žπ‘ = 62.4 βˆ— 7.85 βˆ— 178
= 159β„Žπ‘