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-?
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-?
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
Related questions
Question
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?

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

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
Expert Solution
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