Pressure Control and Speed Regulation
Chap.
10
h'~2
=
1.453
X
106 lb-ft2
ANSWER
(b) It is necessary to characterize the turbine and find specific speed
11,.
Using Eq. (4.17) gives us
To keep the relative speed
change
Anln
less than 0.10,
Eq.
(1 0.63) will apply.
n
d-
Tm
From Table 10.2 for
n,
=
150.9, Kz0.66. For Tg=30,
TWITg
=
7.46130
=
0.25,
and using Fig. 10.20 for
TWITg
=
0.25,
h
=
0.27.-Then
=
135 sec
ANSWER
Using Eq.
(1 0.62) again, we obtain
1.6
X
IO~P,T,,
-
(1.6
X
10~)(2680)(135)
WR~
=
(25712 (257)'
=
8.76
X
lo6
lb-ff2
ANSWER
REFERENCES
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L.,
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E.
Halmes, trans.,
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Society of
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Engineers,
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M.,
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T.
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Jordan, V., "Reverse Water Hammer in Turbine Draft Tubes,"
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Joukovsky,
N.,
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0.
Simin, trans.,
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S.
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and
E.
B.
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:hap.
10
Problems 199
:ovalev,
N.
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1961,
M.
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Jerusalem: Israel Program for Scientific Translation Ltd., 1965.
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J.
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11. Rouse,
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3,
1975.
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M.,
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Prague: Artia/
London: Constable
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J.,
Waterhammer Analysis.
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R.
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'ROBLEMS
10.1.
A
turbine installation has a normal design head of 50 ft, a velocity in the pen-
stock of 6
ftlsec, and a length of penstock from closure valvc to an open
water surface of 500 ft. Assume that the pressure wave velocity,
a.
is 3000
ftl
sec. Determine the relative preskre rise if closure is less than 0.1 scc. What is
the critical time of pressure wave travel? If the time of closure
1s
3
sec. what
will be the expected pressure rise and pressure drop?
10.2.
Visit a hydropower plant and determine the method of controlling water
hammer. Report on closure time of closing vaives or gates.
0.3.
A
simplified problem for a hydro plant involves a penstock of length 2500 ft,
the velocity of the penstock is 6
ftlsec, the penstock area is 100 ft2, the surge
tank area is proposed to be 2000 ft2, and the friction hcad loss in the pen-
stock
is
3
ft. Determine the approximate maximum pressure surgc
in
the
surge tank and time from beginning of flow until
maxiniurn surge if the gate
valve is suddenly closed.