Chute Spillway Discharge

HSWPE 12

Determine the discharge over a Chute spillway with ‘Ogee Crest’ using the following data: The length of the spillway is 250 m, Height of the spillway crest in front of at upstream approach channels is 10 m. The width of the approach channel is 250 m and the depth of water over the spillway crest is 5 m.

Answer :

Given
Length of spillway L = 250 m
Width of channel =250 m 
Cd = 2.2 for Ogee spillway design
h= 10m 
He =5m
h/He =10/5 =2

 from the graph 

C/Cd =1  

C/2.2 =1

C = 2.2

capacity of the Hydropower plant

HSWPE 10

Three generators each of capacity 9000 KW have been installed at a power station. During a certain period of load, the load on the plant varies from 14000 KW to 24000 KW. Determine: (6 mark)          Dec 2018

1. Total installed capacity

2. Load factor

3. Plant factor

4. Utilization factor

Answer :

HSWPE 11

A runoff river plant is installed on a river having a minimum flow of 15 cumecs. Head available at the plant is 16 m and the plant efficiency may be assumed as 80%. If the plant is used as a peak load plant operating for 6 hours daily, compute the firm capacity of the plant: (6 mark) May 2018
i.  Without pondage
ii. With pondage but allowing 8% of water to be lost in evaporation and other losses

Answer :

Spillway Discharge

HSWPE 6

Compute the discharge over an ogee weir with a coefficient of discharge equal to 2.4 and a head of 2 m. The length of the spillway is 100 m. The Weir crest is 8 m above the bottom of the approach channel having the same width as that of the spillway. (7 mark) 

Answer :

HSWPE 7

The Siphon spillway of the rectangular cross-section had the following dimensions at its throat. Height of throat = 1.5 m. Width of throat = 4 m. At the design flow, the tailwater elevation is 7 m below the summit of the siphon, and the headwater elevation is 2 m above the summit. Taking the coefficient of discharge as 0.6, determine the discharge capacity of the siphon. Also, determine the head that would be required on an ogee spillway 3.8 m long to discharge this flow if the coefficient of discharge is 2.25. (7 mark)

Answer :

HSWPE 8

A siphon spillway had the following cross-section at its throat. Height of the throat = 1.5 m. Width of throat = 4 m. At the design flow, the tail water elevation is 2 m above the summit.
i.   Taking a coefficient of discharge as 0.6, determine the capacity of the siphon.
ii. Determine the head that would be required on an ogee spillway 3.8 m long to discharge this flow, if coefficient of discharge is 2.25.
iii. What length of the ogee weir would be required to discharge the same flow with a head of 2.2 m on the crest? (7 mark)

Answer :

HSWPE 9

A saddle siphon spillway has the following data. Full reservoir level = 485 m, Level of the center of siphon outlet = 479.6 m, Highest flood level = 485.9 m, Highest flood discharge = 570 cumecs. If the dimensions of the throat of the siphon are: width = 4.2 m and height = 1.9 m, determine the number of siphon units required to pass the flood safely. The siphon is to discharge freely in the air. Assume the coefficient of discharge = 0.65. (6 mark)

Answer :

Slip Circle Methods

HSWPE 5

In order to find the factor of safety of downstream, slope during steady seepage, the section of the dam was drawn to scale 1 cm = 4 m. The following results were obtained on a critical slip circle.

Area of N-rectangle = 14.4 sqcm

Area of T-rectangle = 6.4 sqcm

Area of U-rectangle = 4.9 sqcm

Length of arc = 12.6 cm 

Laboratory tests have furnished values 260 for effective angle of friction and 19.5 kN/m2 for cohesion, unit weight of soil = 19 kN/m3. Determine the factor of safety of the slope.            (7 mark) 

Answer :

∴𝐹𝑎𝑐𝑡𝑜𝑟 𝑜𝑓 𝑠𝑎𝑓𝑒𝑡𝑦 𝑜𝑓 𝑠𝑙𝑜𝑝𝑒 𝑖𝑠 1.41

Seepage Analysis

HSWPE 4

A flow net was constructed for a homogeneous earth dam 52 m high and 2 m freeboard, and the following results were obtained. The number of potential drops = 25, Number of flow channels = 4. The dam has a horizontal filter of 40 m in length at its downstream end. Calculate the seepage discharge per meter length of the dam. Assume the dam material's permeability coefficient as 3x10^-3 cm/sec. (7 mark)

Answer: 

Given

K = 3x10^-3 cm/sec = 3x10^-5 m/sec

H = 52-2 = 50 m

Nf = 4

Nd = 25

There seepage discharge per meter length of the dam is 2.4×10^(−4) 𝑐𝑢𝑚𝑒𝑐𝑠/𝑚

HSWPE 3

 The figure shows the section of a gravity dam (non-overflow portion) built of concrete. Calculate (neglecting earthquake effects). (13 mark)       Dec 2016

i. The maximum vertical stresses at the toe and heel of the dam

ii. The major principal stresses at the toe of the dam

iii. Intensity of shear stress on a horizontal plane near the toe

Assume the unit weight of concrete = 23.5 kN/m2.

Allowable compressive stress in concrete = 2500 kN/m².

Allowable shear stress in concrete = 420 kN/m².

Assume that reservoir is full of water up to M.W.L.

ANSWER

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HSWPE 2

HSWPE 2 

As impounding reservoir had an original storage capacity of 738 ha-m. The drainage area of the reservoir is 80 sq. km. from which, annual sediment discharges into the reservoir at the rate of 0.1153 ha-m per sq. km. of the drainage area. Assuming the trap efficiency as 80 percent, find the annual capacity loss of the reservoir in percent per year.

ANS 

Given

Reservoir capacity = 738 ha-m

Drainage area = 80 sq.km

Annual sediment discharge = 0.1153 ha-m per sq.km

Trap efficiency = 80% = 0.80

We know

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HSWPE 1

HSWPE 1

The amounts of water flowing from a certain catchment area at the proposed dam site are tabulated as follows. Determine: 

• The minimum capacity of the reservoir if the water is to be used to feed the turbines of the hydropower plant at a uniform rate and no water is to be spilled over.

• The initial storage required to maintain the uniform demand as above.

Month	Jan	Feb	Mar	Apr	May	Jun	Jul	Aug	Sep	Oct	Nov	Dec
Inflow x105m3	2.83	4.25	5.66	18.4	22.64	22.64	19.81	8.49	7.10	7.10	5.65	5.66

ANS:

The computation is done in a tabular form below

This value has been entered in column (3) of the below table.

Month	Inflow (x105m3)	Av. Demand (x105m3)	Deficit (x105m3)	Surplus (x105m3)
1	2	3	4 = 3 - 2	5 = 2 - 3
January	2.83	10.8533	8.0233
February	4.25	10.8533	6.6033
March	5.66	10.8533	5.1933
April	18.40	10.8533		7.5467
May	22.64	10.8533		11.7867
June	22.64	10.8533		11.7867
July	19.81	10.8533		8.9567
August	8.49	10.8533	2.3633
September	7.10	10.8533	3.7533
October	7.10	10.8533	3.7533
November	5.66	10.8533	5.1933
December	5.66	10.8533	5.1933
Sum	130.24x105	130.24x105		40.0766x105

Since no water is to be spilled, the minimum capacity will be equal to the sum of the surplus water.

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EM 3:

Engineering Mechanics Problem

Find the resultant of the force system shown in and locate it from a point along the x-axis

ANS :

Fx = 1000 cos (30) - 800 = 66.02 N

Fy = 600 + 1000 sin (30) = 1100 N

R= (Fx^2 +Fy^2)^0.5  

R =1101.98 N

ፀx = 86.85 Degree

X distance from Origin = 91.18 m

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