2.A swimming pool with plan and section shown in Fig. G is filled with water. It has two short tubes both with diameter 20cm and C= 0.80 located at the lowest point. Find the time needed to empty the pool through these tubes. Given: d₁= 20cm or 0.20m Find: d₂= 20cm or 0.20m C = 0.80 T-?

Structural Analysis
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Chapter2: Loads On Structures
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Need asap. Thanks. Why the h is missing on the next highligted solution? And why on the 30/1.8 has multiplied with h? Why it become like that?
12.A swimming pool with plan and section shown in Fig. G is filled with
water. It has two short tubes both with diameter 20cm and C= 0.80
located at the lowest point. Find the time needed to empty the pool
through these tubes.
Given: d₁ =
20cm or 0.20m
d₂= 20cm or 0.20m
C = 0.80
Find:
T =?
15m
ㅗ
1.20 m
Plan
30 m
Section
3m
2 tubes
Transcribed Image Text:12.A swimming pool with plan and section shown in Fig. G is filled with water. It has two short tubes both with diameter 20cm and C= 0.80 located at the lowest point. Find the time needed to empty the pool through these tubes. Given: d₁ = 20cm or 0.20m d₂= 20cm or 0.20m C = 0.80 Find: T =? 15m ㅗ 1.20 m Plan 30 m Section 3m 2 tubes
Solution: for rectangular part
ch1 Adh
t₁ = Sh2 2CA√2gh
=
t₁ = 2021.125 ₁h
for triangular part
x
30
30
→ X=
-h
h
1.8
1.8
30
-1.8 15 )(h)dh
t₂ = So 2(0.8) (π)(0.1)^2√19.62h
= 1122.847 h ¹1/2dh
t₂ = 1807.749 s
T = t₁ + t₂
= 1578.135 s + 1807.749 s
= 3385.88 s
450dh
1₁.8
1.8 2(0.8) (π) (0.1)^2√19.62h
-1/2dh = 1578.135 s
Transcribed Image Text:Solution: for rectangular part ch1 Adh t₁ = Sh2 2CA√2gh = t₁ = 2021.125 ₁h for triangular part x 30 30 → X= -h h 1.8 1.8 30 -1.8 15 )(h)dh t₂ = So 2(0.8) (π)(0.1)^2√19.62h = 1122.847 h ¹1/2dh t₂ = 1807.749 s T = t₁ + t₂ = 1578.135 s + 1807.749 s = 3385.88 s 450dh 1₁.8 1.8 2(0.8) (π) (0.1)^2√19.62h -1/2dh = 1578.135 s
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