In the figure shown, solve for the theoretical velocity of the jet, the actual velocity and the discharge through the orifice considering the following: a) p₁ =P₂ =0 h = 5 m of oil Do = 5 cm b) p₁ = h = 5 m of water P₂ = 0 Do = 5 cm c) p₁ = 70 kPa P2=0 C=1.00 (rounded) C₁ = 0.97 C=0.62 (sharp-edged) C₁ = 0.98 C=C 0.95 (rounded) Do D₁ = 7.5 cm h=1.60 m (liquid with sp. gr. = 3) h L.S. P2 2 L.S.
In the figure shown, solve for the theoretical velocity of the jet, the actual velocity and the discharge through the orifice considering the following: a) p₁ =P₂ =0 h = 5 m of oil Do = 5 cm b) p₁ = h = 5 m of water P₂ = 0 Do = 5 cm c) p₁ = 70 kPa P2=0 C=1.00 (rounded) C₁ = 0.97 C=0.62 (sharp-edged) C₁ = 0.98 C=C 0.95 (rounded) Do D₁ = 7.5 cm h=1.60 m (liquid with sp. gr. = 3) h L.S. P2 2 L.S.
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
Related questions
Question
In the figure shown, solve for the theoretical velocity of the jet, the
actual velocity and the discharge through the orifice considering the
following:
a) p1 = p2 = 0 Cc = 1.00 (rounded)
h = 5 m of oil Cv = 0.97
Do = 5 cm
b) p1 = p2 = 0 Cc = 0.62 (sharp-edged)
h = 5 m of water Cv = 0.98
Do = 5 cm
c) p1 = 70 kPa C = Cv = 0.95 (rounded)
p2 = 0 Do = 7.5 cm
h = 1.60 m (liquid with sp. gr. = 3)
![In the figure shown, solve for the theoretical velocity of the jet, the
actual velocity and the discharge through the orifice considering the
following:
a) p₁ =P₂ =0
h = 5 m of oil
Do = 5 cm
b) p₁ =
h = 5 m of water
P₂ = 0
Do = 5 cm
c) p₁ = 70 kPa
P2=0
C=1.00 (rounded)
C₁ = 0.97
C=0.62 (sharp-edged)
C₁ = 0.98
C=C
0.95 (rounded)
Do
D₁ = 7.5 cm
h=1.60 m (liquid with sp. gr. = 3)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5586eeb1-0fc7-46bf-a995-14da4c6931ee%2Fff68376a-4bbb-4747-8aec-33bf4d5b16cb%2F5ql9ohqe_processed.png&w=3840&q=75)
Transcribed Image Text:In the figure shown, solve for the theoretical velocity of the jet, the
actual velocity and the discharge through the orifice considering the
following:
a) p₁ =P₂ =0
h = 5 m of oil
Do = 5 cm
b) p₁ =
h = 5 m of water
P₂ = 0
Do = 5 cm
c) p₁ = 70 kPa
P2=0
C=1.00 (rounded)
C₁ = 0.97
C=0.62 (sharp-edged)
C₁ = 0.98
C=C
0.95 (rounded)
Do
D₁ = 7.5 cm
h=1.60 m (liquid with sp. gr. = 3)
![h
L.S.
P2
2
L.S.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5586eeb1-0fc7-46bf-a995-14da4c6931ee%2Fff68376a-4bbb-4747-8aec-33bf4d5b16cb%2F7xy8yew_processed.png&w=3840&q=75)
Transcribed Image Text:h
L.S.
P2
2
L.S.
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