#### Transcript Document

```Fluid Mechanics 07
any point in the system.
𝐸𝐺𝐿
= 𝑉𝑒𝑙𝑜𝑐𝑖𝑡𝑦 ℎ𝑒𝑎𝑑 + 𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒 ℎ𝑒𝑎𝑑
+ 𝐸𝑙𝑒𝑣𝑎𝑡𝑖𝑜𝑛 𝐻𝑒𝑎𝑑
𝑣2
𝑝
𝐸𝐺𝐿 =
+ +𝑍
2∗𝑔 ɣ
HGL:- Hydraulic Grade Line indicate the
piezometric head at any point in the system
𝐻𝐺𝐿 = 𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒 ℎ𝑒𝑎𝑑 + 𝐸𝑙𝑒𝑣𝑎𝑡𝑖𝑜𝑛 ℎ𝑒𝑎𝑑
𝑝
𝐻𝐺𝐿 = + 𝑧
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
ɣ
+ 𝑧1
𝑣12
+
2∗𝑔
+ ℎ𝑝 =
𝑝2
+
ɣ
𝑧2
𝑣22
+
2∗𝑔
+ ℎ𝑡 + ℎ𝐿
Where
P1=P2=Patm=zero
V1=v2=zero
Ht=zero, z1=520 ft, z2=620 ft
𝑄
7.85
𝑣= =
= 10𝑓𝑡/𝑠
2
𝐴 (Π/4) ∗ (1 )
hL= 0.01 ∗
=77.6 ft
𝐿
𝐷
∗
𝑣2
( )=0.01
2∗𝑔
∗
5000
1
∗
102
(
)
2∗32.2
ℎ𝑝 = 𝑧2 − 𝑧1 + ℎ𝐿 = 620 − 510 + 77.6
=178 ft
𝑃𝑜𝑤𝑒𝑟 = ɣ ∗ 𝑄 ∗ ℎ𝑝 = 62.4 ∗ 7.85 ∗ 178
= 159ℎ𝑝
```