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Final Exam.- 1 -
El Minia University Advanced Water Structures II
Faculty of Engineering Date: June 4, 2016
PostgraduateExam. Time: 3 hours
 Answer all the following questions,
 Assume any missing data reasonably,
 Net sketches and clear dimensions are required.
‫ورقتني‬ ‫يف‬ ‫ئةل‬‫س‬‫أ‬ ‫مخسة‬ ‫من‬ ‫مكون‬ ‫الامتحان‬
‫وهجان‬ ‫حدههام‬‫إ‬ ،
&
" ‫إلعظمي‬ ‫إدلرجة‬
0
7
“
Question No 1 (15 Marks) “Net sketches are required in your answer”
A.
 What are the factors affecting selection of type of dams?
 What are the typical forces, which are considered for the analysis of an
elementary profiles of a gravity dam under filled upstream reservoir and
downstream water?
 Why are locks and dams necessary for navigation?
 How is the river diverted during construction?
B. Neglecting ice, and earthquake forces, analyze the stability of the given
gravity dam (see the attached Figure) for the following conditions:
 Friction coefficient between concrete-foundation is 0.70, respectively.
 Allowable shear stress at the foundation level is 2200 kN/m2
,
Take specific weights of concrete and water as 24 kN/m3
, and 9.810 kN/m3
,
respectively.
Question No 2 (12 Marks)
a) What is the purpose of diffusers for wastewater?
b) A round jet of wastewater is discharging upward from near the bottom of a bay
that is 120 m deep. The jet is 0.3 m in diameter, and the discharge rate is
2 m3
/s.
 What are the velocities along the axis of the jet at distances of 30 m and
60 m from the origin of the jet (Z = 30 m and Z = 60 m?),
Final Exam.- 2 -
 What are the velocities at these same Z-direction but at a radial distance
of 6 m from the axis of the jet?
(The density of the wastewater is 1000 kg/ m3
and the density of the water
in the bay is 1026 kg/ m3
).
Question No 3 (20 Marks)
a) How is the following defined (beside net sketches):
Porosity & Permeability,
Drawdown & Specific yield,
Confined & Unconfined aquifer,
Static water table &Hot springs.
b) How can we protect groundwater recharge areas and groundwater
reservoir from contamination and depletion (‫?)نضوب‬
c) What we mean by well rehabilitation?
d) Steady-state to an abstraction well in an unconfined aquifer is given by the
following formula
𝑄
2 𝜋 𝑘
ln (
𝑟2
𝑟1
) =
ℎ2
2
− ℎ1
2
2
Define each term in this equation and show how this formula can be re-
arranged to be expressed in terms of corrected drawdown.
e) A well in an unconfined aquifer is pumped over a long period at a rate of
0.05 m3
/s until approximate steady-state conditions are achieved. Two
observation wells at distances of 25 m and 50 m from the abstraction well
give observed drawdowns below the initial water table level of 15.3 m and
9.5 m respectively. If the saturated thickness of the aquifer prior to pumping
is 80 m, determine the hydraulic conductivity of the aquifer material.
Question No 4 (15 Marks)
 A rectangular swimming pool 1.0 m deep at one end and increases uniformly
in depth to 2.6 m at the other end. The pool is 8 m wide and 32 m long and is
emptied through an orifice of area 0.225 m2
, at the lowest point in the side of
the deep end (see the shown sketch). Taking Cd for the orifice as 0.60, find
from the first principles:
Final Exam.- 3 -
 The time for the depth to fall by 1.0 m,
 The time to empty the pool completely.
Question No 5 (15 Marks)
A pump on test at its designed speed of 1450 rpm gave the following results:
H (ft) 27 25 21.5 17.5 13.5 8 3
Q (gal/min) 0 250 500 750 1000 1250 1375
 (%) 0 20 42 60 69 66 61
The pump is to run continuously at its design speed and is to deliver water through
40 ft of pipe against a static lift of 5 ft. The only pipes available are 6in, 8in and
10in diameter all with f =0.028.
i. Which will be the most suitable pipe for this duty?
ii. Estimate the horse power required to drive the pump and calculate the
specific speed of the pump at maximum efficiency?
iii. Plot on the square paperan estimated head-quantity (Q) curve for the pump
running at1000 rpm.
With Best
Wishes
and Good Luck

Final Exam.- 4 -
 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)
F
w, h
F
Uplif
t
Final Exam.- 5 -
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)
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
Final Exam.- 6 -
∑ 𝑀 = 257040 + 315840 + 1134000 + 101495 = 1808375
𝑜
𝑘𝑁. 𝑚
∑ 𝐹𝑉 = 5040 + 6720 + 37800 + 1915− 17535 = 33940 𝑘𝑁
∑ 𝐹ℎ = 20725 𝑘𝑁
Final Exam.- 7 -
Final Exam.- 8 -
Final Exam.- 9 -
Final Exam.- 10 -
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.
Final Exam.- 11 -
Final Exam.- 12 -
Final Exam.- 13 -
Final Exam.- 14 -
‫الجوفية‬ ‫المياه‬ ‫منسوب‬
‫تطفو‬ ‫التي‬
Final Exam.- 15 -
The vadose zone

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Post graduate exam in hydraulics

  • 1. Final Exam.- 1 - El Minia University Advanced Water Structures II Faculty of Engineering Date: June 4, 2016 PostgraduateExam. Time: 3 hours  Answer all the following questions,  Assume any missing data reasonably,  Net sketches and clear dimensions are required. ‫ورقتني‬ ‫يف‬ ‫ئةل‬‫س‬‫أ‬ ‫مخسة‬ ‫من‬ ‫مكون‬ ‫الامتحان‬ ‫وهجان‬ ‫حدههام‬‫إ‬ ، & " ‫إلعظمي‬ ‫إدلرجة‬ 0 7 “ Question No 1 (15 Marks) “Net sketches are required in your answer” A.  What are the factors affecting selection of type of dams?  What are the typical forces, which are considered for the analysis of an elementary profiles of a gravity dam under filled upstream reservoir and downstream water?  Why are locks and dams necessary for navigation?  How is the river diverted during construction? B. Neglecting ice, and earthquake forces, analyze the stability of the given gravity dam (see the attached Figure) for the following conditions:  Friction coefficient between concrete-foundation is 0.70, respectively.  Allowable shear stress at the foundation level is 2200 kN/m2 , Take specific weights of concrete and water as 24 kN/m3 , and 9.810 kN/m3 , respectively. Question No 2 (12 Marks) a) What is the purpose of diffusers for wastewater? b) A round jet of wastewater is discharging upward from near the bottom of a bay that is 120 m deep. The jet is 0.3 m in diameter, and the discharge rate is 2 m3 /s.  What are the velocities along the axis of the jet at distances of 30 m and 60 m from the origin of the jet (Z = 30 m and Z = 60 m?),
  • 2. Final Exam.- 2 -  What are the velocities at these same Z-direction but at a radial distance of 6 m from the axis of the jet? (The density of the wastewater is 1000 kg/ m3 and the density of the water in the bay is 1026 kg/ m3 ). Question No 3 (20 Marks) a) How is the following defined (beside net sketches): Porosity & Permeability, Drawdown & Specific yield, Confined & Unconfined aquifer, Static water table &Hot springs. b) How can we protect groundwater recharge areas and groundwater reservoir from contamination and depletion (‫?)نضوب‬ c) What we mean by well rehabilitation? d) Steady-state to an abstraction well in an unconfined aquifer is given by the following formula 𝑄 2 𝜋 𝑘 ln ( 𝑟2 𝑟1 ) = ℎ2 2 − ℎ1 2 2 Define each term in this equation and show how this formula can be re- arranged to be expressed in terms of corrected drawdown. e) A well in an unconfined aquifer is pumped over a long period at a rate of 0.05 m3 /s until approximate steady-state conditions are achieved. Two observation wells at distances of 25 m and 50 m from the abstraction well give observed drawdowns below the initial water table level of 15.3 m and 9.5 m respectively. If the saturated thickness of the aquifer prior to pumping is 80 m, determine the hydraulic conductivity of the aquifer material. Question No 4 (15 Marks)  A rectangular swimming pool 1.0 m deep at one end and increases uniformly in depth to 2.6 m at the other end. The pool is 8 m wide and 32 m long and is emptied through an orifice of area 0.225 m2 , at the lowest point in the side of the deep end (see the shown sketch). Taking Cd for the orifice as 0.60, find from the first principles:
  • 3. Final Exam.- 3 -  The time for the depth to fall by 1.0 m,  The time to empty the pool completely. Question No 5 (15 Marks) A pump on test at its designed speed of 1450 rpm gave the following results: H (ft) 27 25 21.5 17.5 13.5 8 3 Q (gal/min) 0 250 500 750 1000 1250 1375  (%) 0 20 42 60 69 66 61 The pump is to run continuously at its design speed and is to deliver water through 40 ft of pipe against a static lift of 5 ft. The only pipes available are 6in, 8in and 10in diameter all with f =0.028. i. Which will be the most suitable pipe for this duty? ii. Estimate the horse power required to drive the pump and calculate the specific speed of the pump at maximum efficiency? iii. Plot on the square paperan estimated head-quantity (Q) curve for the pump running at1000 rpm. With Best Wishes and Good Luck 
  • 4. Final Exam.- 4 -  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) F w, h F Uplif t
  • 5. Final Exam.- 5 - 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) 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
  • 6. Final Exam.- 6 - ∑ 𝑀 = 257040 + 315840 + 1134000 + 101495 = 1808375 𝑜 𝑘𝑁. 𝑚 ∑ 𝐹𝑉 = 5040 + 6720 + 37800 + 1915− 17535 = 33940 𝑘𝑁 ∑ 𝐹ℎ = 20725 𝑘𝑁
  • 10. Final Exam.- 10 - 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.
  • 14. Final Exam.- 14 - ‫الجوفية‬ ‫المياه‬ ‫منسوب‬ ‫تطفو‬ ‫التي‬
  • 15. Final Exam.- 15 - The vadose zone