A 2800 m long pipeline conveys water at a veloc- ity of 2.0 m/s. If a valve at the downstream end is closed suddenly, calculate the maximum pressure rise. Sketch the variation of the water hammer pressure with time at (i) the valve end (ii) mid- point B of the pipe and (iii) at point C 2100 m upstream of the valve. The pipe starts from a lake at the upstream end. Assume the pipe to be rigid. For water, take K = 1.956 × 10° Pa.
A 2800 m long pipeline conveys water at a veloc- ity of 2.0 m/s. If a valve at the downstream end is closed suddenly, calculate the maximum pressure rise. Sketch the variation of the water hammer pressure with time at (i) the valve end (ii) mid- point B of the pipe and (iii) at point C 2100 m upstream of the valve. The pipe starts from a lake at the upstream end. Assume the pipe to be rigid. For water, take K = 1.956 × 10° Pa.
Chapter2: Loads On Structures
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![14.21 A 2800 m long pipeline conveys water at a veloc-
ity of 2.0 m/s. If a valve at the downstream end is
closed suddenly, calculate the maximum pressure
rise. Sketch the variation of the water hammer
pressure with time at (i) the valve end (ii) mid-
point B of the pipe and (iii) at point C 2100 m
upstream of the valve. The pipe starts from a lake
at the upstream end. Assume the pipe to be rigid.
For water, take K = 1.956 × 10³ Pa.
Solution: C = √K/P =
Critical time T₁==
Period of pressure fluctuation:
Water hammer pressure
Ph = PCV₁ = 998 x 1400 × 2.0 Pa
(i) Figure 14.16 (a) shows the variation of the water
hammer pressure Ph at the valve V with the static
pressure at V as datum.
Lake
LO
Ph
0
-Ph
700 m 700 m
C
-3
B
1400 m-
V
Valve
1 2 3 4 5 6 7 8 9 10 11
12 13
Time in seconds
(a)
Fig. 14.16(a) Variation of Ph at Valve V
Time taken for P to reach the lake = 2800/1400
= 2s
Time taken for the reflected wave to reach
V=2s
Hence from 0 to 4 s, the pressure at V will be + Ph
and from 4 to 8 s, it will be - Ph
and from 8 to 12 s, it will be + Ph
and so on.
Note the period of the wave is 8 s.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe8ba24fc-cb9e-4dbc-8d18-9934f9dedc0f%2Fb2f306ea-7aed-40c2-a54d-c10380c2b448%2Fc4jj9u_processed.png&w=3840&q=75)
Transcribed Image Text:14.21 A 2800 m long pipeline conveys water at a veloc-
ity of 2.0 m/s. If a valve at the downstream end is
closed suddenly, calculate the maximum pressure
rise. Sketch the variation of the water hammer
pressure with time at (i) the valve end (ii) mid-
point B of the pipe and (iii) at point C 2100 m
upstream of the valve. The pipe starts from a lake
at the upstream end. Assume the pipe to be rigid.
For water, take K = 1.956 × 10³ Pa.
Solution: C = √K/P =
Critical time T₁==
Period of pressure fluctuation:
Water hammer pressure
Ph = PCV₁ = 998 x 1400 × 2.0 Pa
(i) Figure 14.16 (a) shows the variation of the water
hammer pressure Ph at the valve V with the static
pressure at V as datum.
Lake
LO
Ph
0
-Ph
700 m 700 m
C
-3
B
1400 m-
V
Valve
1 2 3 4 5 6 7 8 9 10 11
12 13
Time in seconds
(a)
Fig. 14.16(a) Variation of Ph at Valve V
Time taken for P to reach the lake = 2800/1400
= 2s
Time taken for the reflected wave to reach
V=2s
Hence from 0 to 4 s, the pressure at V will be + Ph
and from 4 to 8 s, it will be - Ph
and from 8 to 12 s, it will be + Ph
and so on.
Note the period of the wave is 8 s.
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