Mid year Exam post graduate 2018

1. Mid-Year Exam.- 1 -
El Minia University Advanced Water Structures II
Faculty of Engineering Date: May 21, 2018
PostgraduateExam. Time: one and half hours
 Answer all the following questions,
 Assume any missing data reasonably,
 Net sketches and clear dimensions are required.
‫من‬ ‫مكون‬ ‫الامتحان‬
‫ثالثة‬
‫ورقتني‬ ‫يف‬ ‫ئةل‬‫س‬‫أ‬
،
ً
‫لك‬
‫مهنام‬
‫وهج‬
‫ني‬
&
" ‫العظمي‬ ‫ادلرجة‬
25
“
Question No 1 (5 Marks)
“Net sketches are required in your answer”
 What is the tractive force? State the forces acting on a particles to check the
bank Stability in a Trapezoidal Channel?
 Why are locks and dams necessary for navigation?
 Functions of Surge Tanks and surge tank types,
 Braiding Channel and when it occurs,

2. Mid-Year Exam.- 2 -
Question No 2 (10 Marks)
a) What is the purpose of diffusers for wastewater?
b) A round jet of water is discharging upward from near the bottom of a bay 100
ft deep. The jet is 1 ft diameter, and the discharge rate is 30 cfs.
 What is the velocity at the points (shown by the Xs) in the diagram shown
below?
“Assume the density of the jet is the same as the density of the water in the
bay”

3. Mid-Year Exam.- 3 -
Question No 3 (10 Marks)
a) How is the following defined (beside net sketches):
Suppose the pump with
characteristic curve shown in the
figure, is situated between two
large water tanks with their free
surfaces open to the atmosphere.
i. Explain qualitatively what would happen to the pump performance curve if the
free surface of the outlet tank were raised in elevation, all else being equal.
ii. “Repeat for the system curve”. What would happen to the operating point-
would the volume flow rate at the operating point decrease, increase, or
remain the same? Indicate the change on a qualitative plot of H versus Q,
and discuss.

4. Mid-Year Exam.- 4 -
iii. Explain qualitatively what would happen to the pump performance curve if a
valve in the piping system were changed from 100 percent open to 50 percent
open, all else being equal. Repeat for the system curve.
iv. What would happen to the operating point-would the volume flow rate at the
operating point decrease, increase, or remain the same? Indicate the change
on a qualitative plot of H versus Q, and discuss.
b) A very long pipe connects two reservoirs whose water levels are the same.
Identical pumps “A” and “B” are connected to the pipe in parallel. The pumps
are operated at 1450 rpm. The operating characteristics of each of these
pumps are as follows:
Q (gpm) 200 400 600 800 1000
H (ft) 156 147 130 104 62
 48 72 83 83 68
When only one pump is operating, the flow rate was found to be 910 gpm.
i. How much flow will there be when both pumps operate?
(Assume friction factor “f” remains constant)
ii. Find also the rates of delivery if the pump speed is reduced to 1200
rpm.

5. Mid-Year Exam.- 5 -
With Best Wishes and
Good Luck

6. Mid-Year Exam.- 6 -
Forces and loads acting the dam:
Fi : Ice Load (for cold climates and
Fi1, and Fi2 for reservoir and tail water in the downstream, respectively)
Fw: Water force produced by earthquake (Fw1, and Fw2 for reservoir and tail
water in the downstream, respectively)
Fwh: Hydrostatic force produced by water in the reservoir and tail water in the
downstream (Fwh1, and Fwh2)
Fwv: Water load produced by water weight (Fwv1, and Fwv2 for reservoir and tail
water in the downstream, respectively)
Fu : Uplift force produced by groundwater (since the tail water in the downstream,
the diagram of uplift force will be in trapezoidal shape)
W : The weight of the dam (W1, W2, W3…Wn)
Fd : Earthquake forces (Fdh1 and Fdv1: horizontally and vertically)
F
w, h
F
Uplif
t

7. Mid-Year Exam.- 7 -
The values of the forces (total vertical and total horizontal, and moments):
Forces Moment Arm about “o” Moment “kN.m ”
𝑾, 𝑭 (2) 𝑋𝒘 (3) 𝑊 × 𝑋𝒘
𝑾𝟏 = 𝟎. 𝟓 × 𝟔 × 𝟕𝟎 × 𝟐𝟒
= 𝟓𝟎𝟒𝟎 𝒌𝑵
𝑋𝒘𝟏 =
𝟏
𝟑
× 𝟔 + 𝟒 + 𝟒𝟓
= 𝟓𝟏𝒎
5040 × 51 = 257040 𝑘𝑁
𝑾𝟐 = 𝟒 × 𝟕𝟎 × 𝟐𝟒 = 𝟔𝟕𝟐𝟎 𝒌𝑵
𝑋𝒘𝟐 =
𝟏
𝟐
× 𝟒 + 𝟒𝟓 = 𝟒𝟕𝒎 6720 × 47 = 315840 𝑘𝑁
𝑾𝟑 = 𝟎. 𝟓 × 𝟒𝟓 × 𝟕𝟎 × 𝟐𝟒
= 𝟑𝟕𝟖𝟎𝟎 𝒌𝑵
𝑋𝒘𝟑 =
𝟐
𝟑
× 𝟒𝟓 = 𝟑𝟎𝒎 37800 × 30 = 1134000 𝑘𝑁
𝑭𝑾,𝒉 = 𝟎. 𝟓 × 𝟔𝟓 × (𝟏 × 𝟗. 𝟖𝟏)
× 𝟔𝟓 ≅ 𝟐𝟎𝟕𝟐𝟓 𝒌𝑵
𝑋𝒘,𝒉 =
𝟏
𝟑
× 𝟔𝟓 = 𝟐𝟏.𝟔𝟕𝒎
20725 × 21.67
= 449111 𝑘𝑁
𝑭𝑾,𝒗 = 𝟎. 𝟓 × 𝟔 × (𝟏 × 𝟗. 𝟖𝟏)× 𝟔𝟓
≅ 𝟏𝟗𝟏𝟓 𝒌𝑵
𝑋𝒘,𝒗 =
𝟐
𝟑
× 𝟔 + 𝟒 + 𝟒𝟓
= 𝟓𝟑 𝒎
1915 × 53 = 101495 𝑘𝑁
𝑭𝑼 = 𝟎.𝟓 × 𝟓𝟓 × (𝟏 × 𝟗. 𝟖𝟏)× 𝟔𝟓
≅ 𝟏𝟕𝟓𝟑𝟓 𝒌𝑵
𝑋𝒘,𝒗 =
𝟐
𝟑
× 𝟓𝟓
= 𝟑𝟔.𝟔𝟕 𝒎
17535 × 36.67
= 643010 𝑘𝑁
∑ 𝑀 = 449111 + 643010 = 1092121 𝑘𝑁. 𝑚
𝑜
∑ 𝑀 = 257040 + 315840 + 1134000 + 101495 = 1808375
𝑜
𝑘𝑁. 𝑚
∑ 𝐹𝑉 = 5040 + 6720 + 37800 + 1915− 17535 = 33940 𝑘𝑁
∑ 𝐹ℎ = 20725 𝑘𝑁

8. Mid-Year Exam.- 8 -

9. Mid-Year Exam.- 9 -

10. Mid-Year Exam.- 10 -

11. Mid-Year Exam.- 11 -
Note: The pumping well itself can be taken as one of the observation wells, in
which case r1 equals the radius of the pumping well r.

12. Mid-Year Exam.- 12 -

13. Mid-Year Exam.- 13 -

14. Mid-Year Exam.- 14 -

15. Mid-Year Exam.- 15 -
‫الجوفية‬ ‫المياه‬ ‫منسوب‬
‫تطفو‬ ‫التي‬

16. Mid-Year Exam.- 16 -
The vadose zone