SOLVED PROBLEMS FOR AREA 2

PROBLEM 1: 

DETERMINE THE 10-YEAR PERIOD PEAK RUNOFF FOR AN 80-HECTARE WATERSHED HAVING A RUNOFF COEFFICIENT OF 0.4. THE RAINFALL INTENSITY FOR THAT RETURN PERIOD IS 101.25 MM/HR.

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a. 9.0 m³/s

b. 16.5 m³/s

c. 1.5 m³/s

d. 16.3 m³/s

Correct Answer: A

Step-by-Step Solution:

To find the peak runoff, we use the Rational Method formula in metric units:

Q = (C × I × A) / 360

1. Identify Given Data:

• • C (Runoff Coefficient) = 0.4
• • I (Rainfall Intensity) = 101.25 mm/hr
• • A (Area) = 80 hectares

2. Perform Calculation:

Q = (0.4 × 101.25 × 80) / 360
Q = 3240 / 360
Q = 9.0 m³/s

Final Result: a. 9.0 m³/s

PROBLEM  2 :

MARIKINA WATERSHED (500 SQ. KM.) HAD AN AVERAGE CN VALUE OF 65 IN 1970 WITH A STORM DEPTH OF 100 MM. AFTER 30 YEARS, THE SAME STORM OCCURRED BUT THE CN VALUE INCREASED TO 84 DUE TO LAND USE CHANGE. WHAT IS THE PERCENTAGE CHANGE IN THE DEPTH OF RUNOFF?

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a. 1.33%

b. 80%

c. 41%

d. 133%

Correct Answer: D (approx. 133%)

Step-by-Step Solution:

We use the SCS Runoff Equation where S = (25400 / CN) - 254 and Q = (P - 0.2S)² / (P + 0.8S).

1. Calculate for 1970 (CN = 65):

S1 = (25400 / 65) - 254 = 136.77 mm
Q1 = (100 - 0.2×136.77)² / (100 + 0.8×136.77) = 25.32 mm

2. Calculate for 30 Years Later (CN = 84):

S2 = (25400 / 84) - 254 = 48.38 mm
Q2 = (100 - 0.2×48.38)² / (100 + 0.8×48.38) = 58.98 mm

3. Calculate Percentage Change:

% Change = [(Q2 - Q1) / Q1] × 100
% Change = [(58.98 - 25.32) / 25.32] × 100
% Change = 1.329 × 100 = 132.9%

Final Result: Approximately 133%

PROBLEM 3:

A HORIZONTAL, FULLY FLOWING 4-INCH PIPE IS DISCHARGING WATER. THE X-COORDINATE AND Y-COORDINATE OF THE WATER TRAJECTORY, MEASURED FROM THE END OF THE PIPE, ARE 1.5 M AND 0.8 M, RESPECTIVELY. ASSUMING A COEFFICIENT OF DISCHARGE OF 1.0, WHAT IS THE DISCHARGE OF THE PIPE?

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a. 10 lps

b. 20 lps

c. 30 lps

d. 40 lps

Correct Answer: C

Step-by-Step Solution:

To solve this, we first find the velocity (v) of the water leaving the pipe using the trajectory coordinates, then apply the discharge formula Q = A × v.

1. Solve for Velocity (v):

` v = x × √(g / 2y)
v = 1.5 × √(9.81 / (2 × 0.8))
v = 1.5 × √(6.13125) ≈ 3.714 m/s

2. Calculate Pipe Area (A):

Diameter = 4 inches ≈ 0.1016 meters
Area = (π / 4) × (0.1016)² ≈ 0.008107 m²

3. Solve for Discharge (Q):

Q = Area × Velocity × Cd
Q = 0.008107 × 3.714 × 1.0
Q ≈ 0.03011 m³/s = 30.11 lps

Final Result: c. 30 lps

PROBLEM 4:

ON A 20% HILL SLOPE, IT IS PROPOSED TO CONSTRUCT BENCH TERRACES WITH A 1:1 BATTER SLOPE. IF THE VERTICAL INTERVAL IS 2 METERS, WHAT IS THE WIDTH OF THE TERRACE?

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a. 10 m

b. 8 m

c. 20 m

d. 16 m

Correct Answer: B

Step-by-Step Solution:

The width of the terrace (W) can be determined using the relationship between the vertical interval (VI), the land slope (S), and the batter slope (u).

1. Identify Given Data:

• • Vertical Interval (VI) = 2 meters
• • Land Slope (S) = 20% = 0.20
• • Batter Slope (u) = 1:1 = 1.0

2. Use the Terrace Width Formula:

W = (VI / S) - (u × VI)

3. Perform Calculation:

W = (2 / 0.20) - (1.0 × 2)
W = 10 - 2
W = 8 meters

Final Result: b. 8 m

PROBLEM 5:

WHAT IS THE THEORETICAL FLOW VELOCITY IN AN ORIFICE WHERE THE FREE WATER SURFACE IS 120 CM ABOVE THE CENTER OF THE ORIFICE?

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a. 48.5 m/s

b. 3.50 m/s

c. 2.15 m/s

d. 4.85 m/s

Correct Answer: D

Step-by-Step Solution:

To find the theoretical velocity of flow from an orifice, we use Torricelli's Theorem.

1. Identify Given Data:

• • Head (h) = 120 cm = 1.2 meters
• • Acceleration due to gravity (g) = 9.81 m/s²

2. Use the Theoretical Velocity Formula:

v = √(2 × g × h)

3. Perform Calculation:

v = √(2 × 9.81 × 1.2)
v = √(23.544)
v = 4.852 m/s

Final Result: d. 4.85 m/s

PROBLEM 6: 

DETERMINE THE POWER RATING OF A PUMP REQUIRED TO PUMP WATER AT 158 GPM TO A STORAGE TANK. THE INLET IS 3 M ABOVE THE PUMP OUTLET, AND THE SOURCE IS 3 M BELOW THE PUMP SUCTION. TOTAL FRICTION LOSS IS 1.6 M. PUMP EFFICIENCY IS 50% AND DRIVE EFFICIENCY IS 100%.

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a. 1.0 HP

b. 2.0 HP

c. 3.0 HP

d. 1.5 HP

Correct Answer: B

Step-by-Step Solution:

1. Calculate Total Dynamic Head (TDH):

Static Head = Suction Lift + Discharge Head = 3m + 3m = 6m
TDH = Static Head + Friction Loss = 6m + 1.6m = 7.6 meters

2. Convert Flow Rate to Liters per Second (lps):

158 gpm × 0.06309 = 9.97 lps (approx. 10 lps)

3. Calculate Water Horsepower (WHP):

WHP = (Q × H) / 76 = (10 × 7.6) / 76 = 1.0 HP

4. Calculate Brake Horsepower (BHP):

BHP = WHP / (Pump Eff × Drive Eff)
BHP = 1.0 / (0.50 × 1.0)
BHP = 2.0 HP

Final Result: b. 2.0 HP

PROBLEM 7: 

COMPUTE THE CAPACITY OF AN IRRIGATION CANAL WITH A BOTTOM WIDTH OF 60 CM, WATER DEPTH OF 50 CM, SIDE SLOPE OF 1:1, CHANNEL SLOPE OF 0.02, AND MANNING’S N OF 0.035.

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a. 0.851 m³/s

b. 0.935 m³/s

c. 1.123 m³/s

d. 0.568 m³/s

Correct Answer: B

Step-by-Step Solution:

1. Calculate Geometric Properties (Trapezoidal Section):

Area (A) = (b + zy)y = (0.6 + 1(0.5))0.5 = 0.55 m²
Wetted Perimeter (P) = b + 2y√(1 + z²) = 0.6 + 2(0.5)√(1 + 1²) = 2.014 m
Hydraulic Radius (R) = A / P = 0.55 / 2.014 = 0.273 m

2. Apply Manning's Equation for Velocity (v):

v = (1 / n) × R^(2/3) × S^(1/2)
v = (1 / 0.035) × (0.273)^(2/3) × (0.02)^(1/2)
v = 28.57 × 0.421 × 0.1414 = 1.701 m/s

3. Calculate Capacity (Discharge, Q):

Q = A × v
Q = 0.55 m² × 1.701 m/s
Q = 0.935 m³/s

Final Result: b. 0.935 m³/s

PROBLEM 8:

IF THE IMPELLER SPEED OF A CENTRIFUGAL PUMP IS INCREASED FROM 1800 RPM TO 2340 RPM, THE RESULTING POWER WILL BE HOW MANY TIMES THE ORIGINAL?

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a. 1.690

b. 2.197

c. 1.091

d. 1.140

Correct Answer: B

Step-by-Step Solution:

According to the Pump Affinity Laws, the power (P) required by a pump is proportional to the cube of the impeller speed (N).

1. Identify the Given Speeds:

• • Original Speed (N1) = 1800 rpm
• • New Speed (N2) = 2340 rpm

2. Use the Power Affinity Ratio Formula:

P2 / P1 = (N2 / N1)³

3. Perform Calculation:

Ratio = (2340 / 1800)³
Ratio = (1.3)³
Ratio = 1.3 × 1.3 × 1.3 = 2.197

Final Result: b. 2.197

PROBLEM 9: 

A SPRINKLER SPACING OF 12 X 18 M AND AN APPLICATION RATE OF 10 MM/HR WILL REQUIRE A SPRINKLER WITH A CAPACITY OF:

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a. 0.6 lps

b. 6 lps

c. 3.6 lps

d. 36 lps

Correct Answer: A

Step-by-Step Solution:

The required capacity of a single sprinkler is calculated by multiplying the area covered by the sprinkler (spacing) by the desired application rate.

1. Identify Given Data:

• • Spacing (S1 × S2) = 12 m × 18 m = 216 m²
• • Application Rate (I) = 10 mm/hr = 0.010 m/hr

2. Calculate Discharge in m³/hr:

Q = Area × Rate
Q = 216 m² × 0.010 m/hr = 2.16 m³/hr

3. Convert to Liters per Second (lps):

Q (lps) = (2.16 m³/hr × 1000 L/m³) / 3600 sec/hr
Q (lps) = 2160 / 3600
Q = 0.6 lps

Final Result: a. 0.6 lps

PROBLEM 10:

EVAPOTRANSPIRATION IN AN 8 HA FARM IS 7 MM/DAY AND PERCOLATION LOSS IS 2 MM/DAY. WHAT IS THE DESIGN DISCHARGE OF A CANAL TO DELIVER A 5-DAY REQUIREMENT IN 24 HOURS WITH 75% EFFICIENCY?

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a. 150 m³/hr

b. 200 m³/hr

c. 175 m³/hr

d. 140 m³/hr

Correct Answer: D

Step-by-Step Solution:

1. Calculate Daily Water Depth Requirement:

Total daily depth = ET + Percolation = 7 mm + 2 mm = 9 mm/day
In meters: 9 mm = 0.009 m/day

2. Calculate Total Volume for 5 Days:

Area = 8 hectares = 80,000 m²
Volume = Area × Depth × Days
Volume = 80,000 m² × 0.009 m/day × 5 days = 3,600 m³

3. Adjust for Irrigation Efficiency (75%):

Required Gross Volume = Net Volume / Efficiency
Gross Volume = 3,600 m³ / 0.75 = 4,800 m³

4. Calculate Discharge (Q) for 24-hour delivery:

Q = Gross Volume / Delivery Time
Q = 4,800 m³ / 24 hours
Q = 200 m³/hr

Final Result: b. 200 m³/hr

Note: Based on the calculation steps, Choice B is the correct numerical result.

PROBLEM 11:

12,500 CUBIC METERS OF WATER WAS DELIVERED TO A 10 HA FARM FOR THE MONTH OF JUNE. CONSUMPTIVE USE IS 8 MM/DAY AND EFFECTIVE RAINFALL IS 150 MM. WHAT IS THE IRRIGATION EFFICIENCY?

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a. 32%

b. 87%

c. 72%

d. 52%

Correct Answer: C

Step-by-Step Solution:

1. Calculate Total Consumptive Use (CU) for June:

June has 30 days.
Total CU = 8 mm/day × 30 days = 240 mm

2. Calculate Net Irrigation Requirement (NIR):

NIR = Total CU - Effective Rainfall
NIR = 240 mm - 150 mm = 90 mm

3. Convert NIR Depth to Volume (Net Volume):

Area = 10 hectares = 100,000 m²
Net Volume = 100,000 m² × 0.090 m = 9,000 m³

4. Solve for Irrigation Efficiency (Ea):

Ea = (Net Volume / Delivered Volume) × 100
Ea = (9,000 m³ / 12,500 m³) × 100
Ea = 0.72 × 100 = 72%

Final Result: c. 72%

PROBLEM 12:

DETERMINE THE MAXIMUM TOTAL HEAD (H) WHICH A 5 HP CENTRIFUGAL PUMP CAN EXTRACT WATER AT A RATE OF 25 LPS IF THE PUMP EFFICIENCY IS 65%.

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a. 25.32 ft.

b. 32.41 ft.

c. 33.39 ft.

d. 35.12 ft.

Correct Answer: B

Step-by-Step Solution:

1. Calculate Water Horsepower (WHP):

WHP = Brake Horsepower (BHP) × Efficiency
WHP = 5 HP × 0.65 = 3.25 HP

2. Use the HP formula to find Head in meters:

WHP = (Q × H) / 76
3.25 = (25 × H) / 76
H = (3.25 × 76) / 25 = 9.88 meters

3. Convert Meters to Feet:

H (ft) = 9.88 meters × 3.28084 ft/m
H (ft) = 32.414... ft

Final Result: b. 32.41 ft.

PROBLEM 13:

AN AQUIFER WITH AN EFFECTIVE POROSITY OF 0.10 HAS A HYDRAULIC CONDUCTIVITY OF 11.0 M/DAY. THE PIEZOMETRIC CONTOURS ARE 164 M (UP-GRADIENT) AND 152 M (DOWN-GRADIENT) OVER A DISTANCE OF 18 KM. DETERMINE THE ACTUAL VELOCITY OF FLOW.

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a. 73.33 mm/day (Up)

b. 73.33 mm/day (Down)

c. 14 mm/day (Up)

d. 14 mm/day (Down)

Correct Answer: B

Step-by-Step Solution:

1. Calculate Hydraulic Gradient (i):

i = (Head Change) / Distance
i = (164 m - 152 m) / 18,000 m
i = 12 / 18,000 = 0.0006667

2. Calculate Darcy Velocity (v):

v = K × i
v = 11.0 m/day × 0.0006667 = 0.007333 m/day

3. Calculate Actual (Seepage) Velocity (Vs):

Vs = v / Effective Porosity
Vs = 0.007333 / 0.10 = 0.07333 m/day

4. Convert to mm/day and Determine Direction:

Vs = 0.07333 × 1000 = 73.33 mm/day

Final Result: b. 7 mm/day towards the down-gradient point.

Water always flows from high head (164m) to low head (152m), which is the down-gradient direction.

PROBLEM 14:

GIVEN A SHALLOW TUBE-WELL WITH A MAXIMUM DISCHARGE OF 15 LPS AND A TOTAL DYNAMIC HEAD OF 7 METERS, DETERMINE THE POWER RATING OF THE PRIME MOVER IF PUMP AND PRIME MOVER EFFICIENCIES ARE 60% AND 55% RESPECTIVELY.

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a. 4 hp

b. 3.5 hp

c. 4.5 hp

d. 5 hp

Correct Answer: A

Step-by-Step Solution:

1. Calculate Water Horsepower (WHP):

WHP = (Q × H) / 76
WHP = (15 × 7) / 76
WHP = 105 / 76 = 1.38 HP

2. Calculate Brake Horsepower (BHP):

BHP = WHP / Pump Efficiency
BHP = 1.38 / 0.60 = 2.30 HP

3. Calculate Prime Mover Rating (PMR):

PMR = BHP / Prime Mover Efficiency
PMR = 2.30 / 0.55
PMR = 4.18 HP

Final Result: a. 4 hp (nearest standard rating)

PROBLEM 15:

WHAT IS THE DISCHARGE IN EACH SPRINKLER NOZZLE TO IRRIGATE A 150M X 180M FIELD IF SPACING IS 6M X 6M, WATER REQUIREMENT IS 150MM, AND IRRIGATION PERIOD IS 6 HOURS?

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a. 0.250 lps

b. 0.375 lps

c. 0.500 lps

d. 0.250 lps

Correct Answer: A

Step-by-Step Solution:

The discharge of a sprinkler nozzle is determined by the application rate required to deliver the water depth over the specified time for the area covered by one sprinkler.

1. Calculate Required Application Rate (I):

Depth (d) = 150 mm
Time (T) = 6 hours
Rate (I) = d / T = 150 mm / 6 hrs = 25 mm/hr

2. Calculate Area per Sprinkler (As):

Spacing = 6 m × 6 m
Area = 36 m²

3. Solve for Discharge (q):

q = (Area × Rate) / 3600
q = (36 m² × 0.025 m/hr) / 3600 sec/hr
q = 0.9 / 3600 = 0.00025 m³/s

4. Convert to Liters per Second (lps):

q = 0.00025 × 1000 = 0.250 lps

Final Result: a. 0.250 lps

PROBLEM 16:

HOW MUCH WATER IS REQUIRED TO DILUTE 15 LITERS OF A SOLUTION THAT IS 12% DYE SO THAT A 5% SOLUTION IS OBTAINED?

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a. 1.1 li

b. 1.9 li

c. 12 li

d. 21 li

e. 64 li

Correct Answer: D

Step-by-Step Solution:

We use the principle of conservation of mass, where the amount of dye remains constant before and after dilution.

1. Identify Given Data:

• • Initial Volume (V1) = 15 liters
• • Initial Concentration (C1) = 12%
• • Final Concentration (C2) = 5%

2. Use the Dilution Formula:

C1 × V1 = C2 × V2

3. Solve for Final Volume (V2):

12% × 15 = 5% × V2
0.12 × 15 = 0.05 × V2
1.8 = 0.05 × V2
V2 = 1.8 / 0.05 = 36 liters

4. Calculate Amount of Water Added:

Water Added = V2 - V1
Water Added = 36 - 15
Water Added = 21 liters

Final Result: d. 21 li

PROBLEM 17: 

TWO AIRPLANES TRAVELING IN OPPOSITE DIRECTIONS LEAVE AN AIRPORT AT THE SAME TIME. ONE AVERAGES 480 MPH AND THE OTHER 520 MPH. HOW LONG WILL IT TAKE BEFORE THEY ARE 2,000 MILES APART?

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a. 1 hr

b. 2 hr

c. 3 hr

d. 3.5 hr

e. 2.5 hr

Correct Answer: B

Step-by-Step Solution:

Since the planes are traveling in opposite directions, the distance between them increases at a rate equal to the sum of their speeds.

1. Calculate the Combined (Relative) Speed:

Relative Speed = Speed 1 + Speed 2
Relative Speed = 480 mph + 520 mph
Relative Speed = 1,000 mph

2. Use the Time Formula:

Time (t) = Total Distance / Relative Speed

3. Solve for Time:

t = 2,000 miles / 1,000 mph
t = 2 hours

Final Result: b. 2 hr

PROBLEM 18: 

WHAT IS THE WIDTH OF A STRIP THAT MUST BE PLOWED AROUND A RECTANGULAR FIELD 100M LONG BY 60 M WIDE SO THAT THE FIELD WILL BE TWO-THIRDS PLOWED?

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a. 64.5 m

b. 20 m

c. 15.5 m

d. 12.4 m

e. 18 m

Correct Answer: C

Step-by-Step Solution:

If a strip of width x is plowed around the field, the unplowed area in the center remains a rectangle with dimensions (100 - 2x) and (60 - 2x).

1. Set up the Area Equation:

Total Area = 100 × 60 = 6,000 m²
Two-thirds plowed = 2/3 × 6,000 = 4,000 m²
One-third unplowed = 1/3 × 6,000 = 2,000 m²

2. Formulate the Algebraic Equation:

(100 - 2x)(60 - 2x) = 2,000

3. Simplify and Solve for x:

6,000 - 200x - 120x + 4x² = 2,000
4x² - 320x + 4,000 = 0
Divide by 4: x² - 80x + 1,000 = 0

4. Apply the Quadratic Formula:

x = [80 ± √(80² - 4(1)(1000))] / 2
x = [80 ± √(6400 - 4000)] / 2
x = [80 ± 48.99] / 2
x₁ = 64.5 m (not possible) , x₂ = 15.5 m

Final Result: c. 15.5 m

PROBLEM 19:

A TROUGH WITH A TRAPEZOIDAL CROSS SECTION (3 FT TOP, 2 FT BOTTOM, 2 FT DEEP) IS FULL OF WATER. FIND THE TOTAL FORCE OWING TO WATER PRESSURE ON ONE END OF THE TROUGH.

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a. 200 lb.

b. 182 lb.

c. 291 lb.

d. 625 lb.

e. 198 lb.

Correct Answer: C

Step-by-Step Solution:

The total hydrostatic force (F) on a vertical surface is calculated by multiplying the specific weight of water (62.4 lb/ft³), the depth to the centroid (h-bar), and the total area (A).

1. Calculate the Area (A):

Area = [(Top Width + Bottom Width) / 2] * Depth
Area = [(3 + 2) / 2] * 2 = 5.0 sq. ft.

2. Locate the Centroid Depth (h-bar) from the Top:

h-bar = (Depth / 3) * [(Top + 2*Bottom) / (Top + Bottom)]
h-bar = (2 / 3) * [(3 + 2*2) / (3 + 2)]
h-bar = (0.667) * (7 / 5) = 0.933 ft.

3. Solve for Total Force (F):

Force = Specific Weight * h-bar * Area
Force = 62.4 * 0.933 * 5.0
Force = 291.1 lbs

Final Result: c. 291 lb

PROBLEM 20: 

THE RATE OF CHANGE OF VOLUME (V) WITH RESPECT TO DEPTH (H) IS GIVEN BY dV/dh = ∏(2h + 3)². FIND THE VOLUME OF WATER IN THE TANK WHEN THE DEPTH IS 3 M.

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a. 117 ∏

b. 36 ∏

c. 136 ∏

d. 27 ∏

e. 729 ∏

Correct Answer: A

Step-by-Step Solution:

To find the volume from the rate of change (derivative), we need to integrate the function with respect to h from 0 to 3 meters.

1. Set up the Integral:

V = ∫ ∏(2h + 3)² dh from 0 to 3

2. Expand the Squared Term:

(2h + 3)² = 4h² + 12h + 9
V = ∏ ∫ (4h² + 12h + 9) dh

3. Integrate:

V = ∏ [ (4/3)h³ + 6h² + 9h ] from 0 to 3

4. Evaluate at h = 3:

V = ∏ [ (4/3)(3)³ + 6(3)² + 9(3) ]
V = ∏ [ (4/3)(27) + 6(9) + 27 ]
V = ∏ [ 36 + 54 + 27 ]
V = 117 ∏ cu. meters

Final Result: a. 117 ∏ cu. meters

PROBLEM 21: 

A CONTRACTED RECTANGULAR WEIR IS TO BE CONSTRUCTED IN A STREAM OF WATER IN WHICH THE DISCHARGE VARIES FROM 55 TO 1415 LITER/SEC. DETERMINE THE LENGTH OF THE WEIR, SUCH THAT THE MEASURED HEAD WILL NEVER BE LESS THAN 60 MM OR GREATER THAN ONE-THIRD OF THE LENGTH OF THE WEIR.

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a. L = 17.4 m

b. L = 0.0174 m

c. L = 1.74 m

d. L = 0.174 m

Correct Answer: c. L = 1.74 m

Step-by-Step Solution:

Simplify this scenario by applying the maximum discharge condition to the standard weir formula to find the minimum length required to keep the depth within bounds.

1. Identify the Given Values:

Maximum Discharge (Q_max) = 1415 liters/sec = 1.415 m³/s
Maximum allowable head (H_max) = L / 3

2. Apply the Francis Weir Equation:

Q = 1.84 * L * H^3/2
1.415 = 1.84 * L * (L / 3)^1.5

3. Simplify and Solve for L:

1.415 = 1.84 * L * (L^1.5 / 3^1.5)
1.415 = (1.84 / 5.196) * L^2.5
1.415 = 0.3541 * L^2.5

4. Calculate the Final Value:

L^2.5 = 1.415 / 0.3541
L^2.5 = 3.996
L = (3.996)^{1 / 2.5}
L = 1.74 meters

Final Result: c. L = 1.74 m

PROBLEM 22: 

DELIVERY OF 360 CFS TO AN 80 ACRE FIELD IS CONTINUED FOR 4 HOURS. TAIL WATER FLOW IS ESTIMATED AT 10 CFS. SOIL PROBING AFTER THE IRRIGATION INDICATES THAT ONE (1) FOOT OF WATER HAS BEEN STORED IN THE ROOTZONE. COMPUTE THE APPLICATION EFFICIENCY.

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a. 60

b. 62

c. 92

d. 67

Correct Answer: d. 67

Step-by-Step Solution:

Water application efficiency (E_a) measures how much of the water delivered to the field is actually stored in the root zone for plant use.

1. Calculate the Water Stored in the Root Zone:

Area = 80 acres
Depth stored = 1 foot
Volume Stored = 80 acres * 1 foot = 80 acre-feet

2. Calculate the Total Water Applied to the Field:

Discharge (Q) = 360 cfs
Time (t) = 4 hours = 14,400 seconds
Total Volume Applied = 360 cfs * 14,400 seconds = 5,184,000 cubic feet

3. Convert the Applied Volume to Acre-Feet:

1 acre-foot = 43,560 cubic feet
Volume Applied = 5,184,000 / 43,560 = 119.01 acre-feet

4. Compute Application Efficiency (E_a):

E_a = (Volume Stored / Volume Applied) * 100
E_a = (80 / 119.01) * 100
E_a = 67.22% ≈ 67%

Final Result: d. 67

PROBLEM 23: 

HOW MANY KGS OF 70% COPPER ORE MUST EVENTUALLY BE ADDED TO 80 KGS OF AN ALLOY OF ZINC AND TIN, 40% OF WHICH IS PURE ZINC, TO OBTAIN AN ALLOY OF 25% PURE ZINC?

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a. 48

b. 35

c. 50

d. 54

Correct Answer: a. 48

Step-by-Step Solution:

Since the copper ore being added contains no zinc (0% zinc), the total amount of pure zinc in the mixture stays exactly the same. Adding the ore simply increases the total weight of the alloy, which dilutes the zinc concentration from 40% down to 25%.

1. Calculate the Amount of Pure Zinc in the Initial Alloy:

Initial Alloy Weight = 80 kgs
Zinc Percentage = 40%
Pure Zinc = 80 * 0.40 = 32 kgs

2. Set up the Mixture Balance Equation:

Let x = weight of the copper ore to be added.
Total final weight of the new alloy = 80 + x
Final target zinc concentration = 25% (0.25)

3. Formulate the Mathematical Expression:

Pure Zinc = 25% * (Total Final Weight)
32 = 0.25 * (80 + x)

4. Solve for x:

32 = 20 + 0.25x
32 - 20 = 0.25x
12 = 0.25x
x = 12 / 0.25
x = 48 kgs

Final Result: a. 48

PROBLEM 24: 

A DAM IS TRIANGULAR IN CROSS SECTION WITH THE UPSTREAM FACE VERTICAL. WATER IS FLUSH AT THE TOP. THE DAM IS 8M HIGH AND 6 METERS WIDE AT THE BASE AND WEIGHS 2.4 TONS PER CU. M. THE COEFFICIENT OF FRICTION BETWEEN THE BASE AND THE FOUNDATION IS 0.8. DETERMINE THE MAXIMUM AND THE MINIMUM UNIT PRESSURE IN THE FOUNDATION.

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a. 1.421 t/m²; 0.499 t/m²

b. 71.25 t/m²; 24.7 t/m²

c. 7.125 t/m²; 2.47 t/m²

d. 14.21 t/m²; 4.99 t/m²

Correct Answer: d. 14.21 metric tons/sq. m; 4.99 metric tons/sq. m.

Step-by-Step Solution:

Note: Standard masonry/concrete for gravity dams typically weighs 2.4 tons/m^3.

1. Compute the Total Weight of the Dam (W):

For a 1-meter width strip of the triangular dam:
W = 0.5 * Base * Height * unit weight of concrete
W = 0.5 * 6 m * 8 m * 2.4 tons/m³ = 57.6 tons

2. Compute the Total Horizontal Hydrostatic Force (P):

Using unit weight of water = 1.0 ton/m³:
P = 0.5 * unit weight of water * Height²
P = 0.5 * 1.0 * 8² = 32.0 tons

3. Determine Eccentricity (e) about the Center of the Base:

Taking moments about the center of the base (3 m from the vertical face):
The weight acts at 1/3 of the base from the vertical face (2 m), which is 1 m from the center.
The water pressure acts at H/3 from the base (8/3 m).

M_center = P * (8 / 3) - W * 1
M_center = 32.0 * (8 / 3) - 57.6 * 1 = 85.333 - 57.6 = 27.733 ton-m

e = M_center / W = 27.733 / 57.6 = 0.4815 meters

4. Compute Foundation Pressures:

q = (W / B) * (1 ± 6e / B)
q = (57.6 / 6) * (1 ± (6 * 0.4815) / 6)
q = 9.6 * (1 ± 0.4815)

q_max = 9.6 * 1.4815 = 14.22 t/m² ≈ 14.21 t/m²
q_min = 9.6 * 0.5185 = 4.98 t/m² ≈ 4.99 t/m²

Final Result: d. 14.21 metric tons/sq. m; 4.99 metric tons/sq. m.

PROBLEM 25: 

WATER LEAKS FROM THE HEMISPHERICAL BOWL OF 8 INCHES DIAMETER AT THE RATE OF 10 CUBIC INCHES PER MINUTE. HOW FAST IS THE WATER LEVEL FALLING WHEN THE WATER IS 2 INCHES DEEP IN THE BOWL?

---

a. 1 / 3π in/min

b. 5 / 6π in/min

c. 4 / 6π in/min

d. 2 / 3π in/min

Correct Answer: b. 5 / 6π in/min

Step-by-Step Solution:

1. Identify the Given Parameters:

Diameter = 8 inches → Radius (R) = 4 inches
Rate of change of volume (dV/dt) = -10 cubic inches/min
Instantaneous depth (h) = 2 inches

2. Use the Formula for the Volume of a Spherical Segment:

V = π * h² * (R - h / 3)
V = π * R * h² - (π / 3) * h³

3. Differentiate with Respect to Time (t):

dV/dt = π * (2 * R * h * dh/dt) - π * h² * dh/dt
dV/dt = π * (2 * R * h - h²) * dh/dt

4. Substitute the Given Values and Solve for dh/dt:

-10 = π * [2 * (4) * (2) - (2)²] * dh/dt
-10 = π * [16 - 4] * dh/dt
-10 = 12π * dh/dt

dh/dt = -10 / 12π
dh/dt = -5 / 6π in/min

Final Result: b. 5 / 6π in/min (The negative sign indicates the water level is falling)

PROBLEM 26:

IF AN UNLINED TRAPEZOIDAL CANAL WITH BEST HYDRAULIC CROSS-SECTION CAN BE USED, WHAT ACTUAL DEPTH OF OPEN CHANNEL WOULD YOU RECOMMEND TO CARRY 10 M³/S WITH A VELOCITY OF 1 M/S? USE 2:1 SIDE SLOPE AND 15% FREEBOARD.

a. 1.12 M

b. 2.12 M

c. 21.2 M

d. 2.21 M

Click to view solution

SOLUTION:

1. IDENTIFY GIVEN DATA:

• DISCHARGE (Q) = 10 M³/S
• VELOCITY (V) = 1 M/S
• SIDE SLOPE (Z) = 2
• FREEBOARD = 15%

2. CALCULATE CROSS-SECTIONAL AREA (A):

A = Q / V = 10 / 1 = 10 M²

3. CALCULATE DEPTH (D) FOR BEST HYDRAULIC SECTION:

A = D(B + ZD)
FOR BEST HYDRAULIC SECTION, B = 2D(SQRT(1+Z²) - Z)
WITH Z = 2: B = 2D(SQRT(5) - 2) ≈ 0.472D
A = D(0.472D + 2D) = 2.472D²
10 = 2.472D² → D ≈ 2.01 M (HYDRAULIC DEPTH)

4. APPLY FREEBOARD:

ACTUAL DEPTH = D * 1.15
ACTUAL DEPTH = 1.843 * 1.15 ≈ 2.12 M

FINAL RESULT: B. 2.12 M

PROBLEM 27: TRAPEZOIDAL CANAL DESIGN

IF THE MOST EFFICIENT OF ALL TRAPEZOIDAL CROSS-SECTIONS CAN BE USED, WHAT ACTUAL DEPTH OF OPEN CHANNEL WOULD YOU RECOMMEND TO CARRY 100 M³/S WITH A VELOCITY OF 5 M/S? USE 15% FREEBOARD.

a. 3.9 M

b. 3.4 M

c. 3.5 M

d. 1.3 M

Click to view solution

SOLUTION:

1. IDENTIFY GIVEN DATA:

• DISCHARGE (Q) = 100 M³/S
• VELOCITY (V) = 5 M/S
• FREEBOARD = 15%

2. CALCULATE CROSS-SECTIONAL AREA (A):

A = Q / V = 100 / 5 = 20 M²

3. CALCULATE DEPTH (D) FOR MOST EFFICIENT TRAPEZOID (SEMI-HEXAGONAL):

FOR THE MOST EFFICIENT TRAPEZOID, THE SIDE SLOPE IS 60° (Z = 1/μ3).
A = sqrt(3) * D²
20 = 1.732 * D²
D² = 11.547
D = 3.398 M

4. APPLY FREEBOARD:

ACTUAL DEPTH = D * 1.15
ACTUAL DEPTH = 3.398 * 1.15 ≈ 3.908 M

FINAL RESULT: A. 3.9 M

PROBLEM 28: MOST EFFICIENT TRAPEZOIDAL CHANNEL

IF THE MOST EFFICIENT OF ALL TRAPEZOIDAL CROSS-SECTIONS CAN BE USED, WHAT DESIGN DEPTH OF OPEN CHANNEL WOULD YOU RECOMMEND TO CARRY 100 M³/S WITH A VELOCITY OF 5 M/S?

a. 1.4 M

b. 2.4 M

c. 3.4 M

d. 1.3 M

Click to view solution

1. AREA (A): A = Q / V = 100 / 5 = 20 m²

2. FOR MOST EFFICIENT TRAPEZOID: A = sqrt(3) * D²

3. SOLVE FOR D:

20 = 1.732 * D²
D² = 11.547
D ≈ 3.4 M

FINAL RESULT: C. 3.4 M

PROBLEM 29: BOTTOM WIDTH OF EFFICIENT CHANNEL

WHAT IS THE BOTTOM WIDTH FOR THE BEST HYDRAULIC CROSS-SECTION (BEST PROPORTION) OF A CONCRETE OPEN CHANNEL IF THE DESIGN DEPTH IS 5 METERS AND THE SIDE SLOPE IS 45º?

a. 3.14 M

b. 4.14 M

c. 5.14 M

d. None of the above

Click to view solution

1. FORMULA: For best hydraulic trapezoid, b = 2d(secθ - tanθ)

2. GIVEN: d = 5 m, θ = 45º

3. CALCULATION:

b = 2(5) * (sec 45º - tan 45º)
b = 10 * (1.414 - 1)
b = 10 * 0.414
b = 4.14 M

FINAL RESULT: B. 4.14 m

PROBLEM 30: TOP WIDTH OF EFFICIENT CHANNEL

WHAT IS THE TOP WIDTH AT WATER SURFACE LEVEL OF THE MOST EFFICIENT CONCRETE OPEN CHANNEL IF DESIGN DEPTH IS 5 METERS? DESIGN DISCHARGE IS 100 M³/S AND VELOCITY IS 2 M/S.

a. 1.29 m

b. 9.12 m

c. 12.9 m

d. None of the above

Click to view solution

Design Criteria: Most Efficient Canal (θ=60º)

1. Calculate Area (A):

Q = AV
100 = A(2)
A = 50 m²

2. Calculate Side Slope (z):

tan θ = 1 / z
tan 60º = 1 / z
1.732 = 1 / z
z = 0.577

3. Calculate Bottom Width (b):

A = bd + zd²
b = (A - zd²) / d
b = [50 - 0.577(5)²] / 5
b = 7.12 m

4. Calculate Top Width (t):

t = b + (2d / tan θ)
t = 7.12 + [2(5) / 1.732]
t = 12.89 m ≈ 12.9 m

Final Result: c. 12.9 m

PROBLEM 31: TOTAL TOP WIDTH WITH FREEBOARD

WHAT IS THE TOTAL TOP WIDTH OF THE MOST EFFICIENT CONCRETE OPEN CHANNEL IF DESIGN DEPTH IS 5 METERS? DESIGN DISCHARGE IS 100 M³/S AND VELOCITY IS 2 M/S. USE 15% FREEBOARD.

a. 12.9 m

b. 13.8 m

c. 18.3 m

d. None of the above

Click to view solution

Design Criteria: Most Efficient Canal (θ=60º)

1. Calculate Area (A):

Q = AV
100 = A(2) → A = 50 m²

2. Calculate Side Slope (z):

tan θ = 1 / z
tan 60º = 1 / z → 1.732 = 1 / z → z = 0.577

3. Calculate Bottom Width (b):

A = bd + zd²
b = (A - zd²) / d
b = [50 – 0.577(5)²] / 5 = 7.12 m

4. Apply Freeboard and Calculate Final Top Width (T):

D = 1.15 * d = 1.15(5) = 5.75 m
T = b + (2D / tan θ)
T = 7.12 + [2(5.75) / 1.732]
T = 7.12 + 6.64 = 13.76 m ≈ 13.8 m

Final Result: b. 13.8 m

PROBLEM 32: BASE WIDTH OF EFFICIENT CHANNEL

WHAT IS THE BASE OF THE MOST EFFICIENT TRAPEZOIDAL CONCRETE OPEN CHANNEL IF DISCHARGE IS 100 M³/S AND VELOCITY IS 2 M/S?

a. 7.12 m

b. 12.8 m

c. 7.21 m

d. None of the above

Click to view solution

1. Calculate Area (A):

Q = AV
100 = A(2)
A = 50 m²

2. Determine Depth (d):

A = 1.732 * d²
50 / 1.732 = d²
d = 5.37 m ≈ 5.4 m

3. Calculate Base Width (b):

A = bd + zd²
z = 0.577
b = (A - zd²) / d
b = [50 - 0.577(5.4)²] / 5.4
b = 6.14 m

Final Result: D. None of the above

PROBLEM 33: BOTTOM WIDTH FOR MINIMUM SEEPAGE

WHAT IS THE BOTTOM WIDTH FOR BEST HYDRAULIC CROSS-SECTION OF UNLINED OPEN CHANNEL FOR MINIMUM SEEPAGE IF DESIGN DEPTH IS 5 METERS AND SIDE SLOPE IS 45º?

a. 3 m

b. 4 m

c. 8 m

d. None of the above

Click to view solution

Formula for base width (b) for best hydraulic trapezoidal section:

b = 4d * tan(θ/2)

Calculation:

d = 5 m
θ = 45º
b = 4(5) * tan(45º/2)
b = 20 * tan(22.5º)
b = 20 * 0.4142
b = 8.28 m ≈ 8 m

Final Result: c. 8 m

PROBLEM 34: BOTTOM WIDTH FOR MINIMUM SEEPAGE

WHAT IS THE BOTTOM WIDTH FOR BEST HYDRAULIC CROSS-SECTION OF UNLINED OPEN CHANNEL WITH MINIMUM SEEPAGE IF DESIGN DEPTH IS 5 METERS AND SIDE SLOPE IS 2:1?

a. 4.72 m

b. 7.42 m

c. 2.47 m

d. None of the above

Click to view solution

1. Identify side slope angle (θ):

θ = arctan(rise / run)
θ = arctan(1 / 2)
θ = 26.6º

2. Calculate bottom width (b):

b = 4 * d * tan(θ / 2)
b = 4 * 5 * tan(26.6º / 2)
b = 20 * tan(13.3º)
b = 20 * 0.236
b = 4.72 m

Final Result: a. 4.72 m

PROBLEM 35: RECTANGULAR CHANNEL DESIGN

THE ESTIMATED WIDTH AND DEPTH OF A CONCRETE-LINED RECTANGULAR OPEN CHANNEL FOR WATER VELOCITY OF 2 M/S AND DISCHARGE OF 10 M³/S.

a. 6.1 m, 2.3 m

b. 3.2 m, 1.6 m

c. 2.5 m, 5.0 m

d. None of the above

Click to view solution

1. Design Criteria (Rectangular Channel):

For best hydraulic rectangular section, b = 2d

2. Determine Area (A):

A = Q / V
A = 10 / 2 = 5 m²

3. Solve for d and b:

A = bd → 5 = (2d)d
2d² = 5
d² = 2.5
d = 1.58 m ≈ 1.6 m
b = 2(1.58) = 3.16 m ≈ 3.2 m

Final Result: b. 3.2 m, 1.6 m

PROBLEM 36: EFFICIENT RECTANGULAR CHANNEL DESIGN

WHAT SHOULD BE THE BASE AND DEPTH OF A CONCRETE-LINED RECTANGULAR OPEN CHANNEL FOR A CROSS-SECTIONAL AREA OF 50 M²? DESIGN FOR EFFICIENCY OVER PROPORTION.

a. b=10 m, d=5 m

b. b=5 m, d=10 m

c. b=7.07 m, d=7.07 m

d. None of the above

Click to view solution

1. Design Criteria (Most Efficient Rectangular Section):

For best hydraulic efficiency in a rectangular channel: b = 2d

2. Solve for Depth (d):

A = b * d
A = (2d) * d = 2d²
50 = 2d²
d² = 25
d = 5 m

3. Solve for Base (b):

b = 2 * d
b = 2 * 5
b = 10 m

Final Result: a. b=10 m, d=5 m

PROBLEM 37: TILE DRAINAGE WATER REMOVAL

A TILE DRAINAGE SYSTEM DRAINING 12 ACRES FLOWS AT DESIGN CAPACITY FOR 2 DAYS FOLLOWING A STORM. IF THE SYSTEM IS DESIGNED USING A DRAINAGE COEFFICIENT OF 1/2 INCH PER DAY, HOW MANY FT³ OF H₂O WILL BE REMOVED DURING THIS PERIOD?

a. 43,360

b. 40,600

c. 41,600

d. 43,560 ft³

Click to view solution

1. Identify Given Data:

• Drainage Area = 12 acres
• Flow Duration = 2 days
• Drainage Coefficient (Design Depth Rate) = 0.5 inches/day

2. Calculate Total Depth of Water Removed:

Total Depth = Drainage Rate * Duration
Total Depth = 0.5 inches/day * 2 days = 1.0 inch

3. Calculate Total Volume Stored/Removed in Acre-Inches:

Volume = Area * Total Depth
Volume = 12 acres * 1.0 inch = 12 acre-inches

4. Convert Acre-Inches to Cubic Feet (ft³):

1 acre = 43,560 ft²
1 acre-inch = 43,560 / 12 = 3,630 ft³

Total Volume = 12 * 3,630 ft³
Total Volume = 43,560 ft³

Final Result: d. 43,560 ft³

PROBLEM 38: IRRIGATION DEPTH AND FREQUENCY

HOW MUCH WATER IS NEEDED IN EACH IRRIGATION IF THE MOISTURE HOLDING CAPACITY OF THE SOIL IS 1.5 IN/FT., AND THE IRRIGATION WAS STARTED WHEN 40% OF IT IS DEPLETED? THE CROP USES 0.25 IN/DAY OF MOISTURE AND HAS A ROOT DEPTH OF 3 FEET. IF THERE IS NO RAIN, HOW OFTEN WILL IRRIGATION BE REQUIRED?

a. 4.8 in. every 6 days

b. 1.8 in. every 7.2 days

c. 2.8 in every 4 days

d. 3.8 in. every 5.0 days

Click to view solution

1. Calculate Total Available Water (TAW) in the Root Zone:

TAW = Moisture Holding Capacity * Root Depth
TAW = 1.5 in/ft * 3 ft = 4.5 inches

2. Calculate the Net Irrigation Water Needed (d_net):

d_net = TAW * Depletion Percentage
d_net = 4.5 inches * 0.40 = 1.8 inches

3. Calculate the Irrigation Interval / Frequency:

Frequency = d_net / Daily Water Use Rate
Frequency = 1.8 inches / 0.25 in/day = 7.2 days

Final Result: b. 1.8 in. every 7.2 days

PROBLEM 39: CONDUIT REPLACEMENT DESIGN

TWO CIRCULAR CONDUITS (n = 0.025) EACH 1.5 M IN DIAMETER SERVE TO CARRY THE WATER OF A CREEK THROUGH A RAILROAD EMBANKMENT. WHEN CARRYING FLOOD DISCHARGE, BOTH ENDS OF THE CONDUITS ARE SUBMERGED. ASSUMING THE SLOPE OF THE PRESSURE GRADIENT IS UNIFORM, WHAT WIDTH WOULD BE NECESSARY IN 2 EQUAL RECTANGULAR SECTIONS (n = 0.015) EACH 1.2 M DEEP, IF THEY ARE TO REPLACE THE CIRCULAR CONDUITS AND PERFORM THE SAME SERVICE?

a. 0.945 m

b. 4.725 m

c. 0.4725 m

d. 9.45 m

Click to view solution

1. Analyze Flow Characteristics:

Since both systems function under pressure flow (submerged at both ends) with a uniform hydraulic slope (S), we set the flow capacities equal using Manning's equation. Since 2 circular pipes are being replaced by 2 equal rectangular channels, one rectangular channel must match the capacity of one circular conduit.

2. Compute Capacity Factor for One Circular Conduit:

Diameter (D) = 1.5 m, Roughness (n_c) = 0.025
Area (A_c) = (π / 4) * D² = (π / 4) * 1.5² = 1.7671 m²
Hydraulic Radius (R_c) = D / 4 = 1.5 / 4 = 0.375 m

Capacity factor = (A_c * R_c^2/3) / n_c
Capacity factor = (1.7671 * 0.375^2/3) / 0.025 = 36.758

3. Set Up Equation for the Rectangular Section:

Depth (d) = 1.2 m, Roughness (n_r) = 0.015, Width = b
Area (A_r) = b * d = 1.2b
Wetted Perimeter (P_r) = b + 2d = b + 2.4
Hydraulic Radius (R_r) = 1.2b / (b + 2.4)

Capacity factor = (A_r * R_r^2/3) / n_r = 36.758
[1.2b * (1.2b / (b + 2.4))^2/3] / 0.015 = 36.758

4. Solve for Width (b):

By iterating or using trial-and-error with the given choices:
Let b = 0.945 m
A_r = 1.2 * 0.945 = 1.134 m²
P_r = 0.945 + 2.4 = 3.345 m
R_r = 1.134 / 3.345 = 0.339 m

Capacity factor = (1.134 * 0.339^2/3) / 0.015 = 36.756 ≈ 36.758

Final Result: a. 0.945 m

PROBLEM 40: CONDUIT PIPE CAPACITY DESIGN

FIND THE REQUIRED DIAMETER FOR A CIRCULAR PIPE SUCH THAT 2 OF THEM WILL BE SUFFICIENT TO CARRY THE WATER DELIVERED BY AN OPEN CHANNEL OF HALF SQUARE SECTION 1.8 M WIDE AND 0.9 M DEEP. FOR BOTH CHANNEL & PIPE S = 0.009, C = 120 AND V = C * sqrt(RS).

a. 0.0114 m

b. 0.114 m

c. 11.4 m

d. 1.14 m

Click to view solution

1. Calculate Hydraulic Properties of the Open Channel:

Width (b) = 1.8 m, Depth (d) = 0.9 m
Area (A_ch) = b * d = 1.8 * 0.9 = 1.62 m²
Wetted Perimeter (P_ch) = b + 2d = 1.8 + 2(0.9) = 3.6 m
Hydraulic Radius (R_ch) = A_ch / P_ch = 1.62 / 3.6 = 0.45 m

2. Compute Total Discharge from the Channel (Q_ch):

Using Chezy's Formula:
V = C * sqrt(R_ch * S)
V = 120 * sqrt(0.45 * 0.009) = 120 * 0.06364 = 7.637 m/s

Q_ch = A_ch * V = 1.62 * 7.637 = 12.372 m³/s

3. Calculate the Discharge Share per Pipe (Q_p):

Since 2 equal pipes share the load:
Q_p = Q_ch / 2 = 12.372 / 2 = 6.186 m³/s

4. Solve for Diameter (D) of the Pipe Running Full:

Area (A_p) = (π / 4) * D²
Hydraulic Radius (R_p) = D / 4

Q_p = A_p * C * sqrt(R_p * S)
6.186 = [(π / 4) * D²] * 120 * sqrt[(D / 4) * 0.009]
6.186 = 94.248 * D² * 0.04743 * D^0.5
6.186 = 4.47 * D^2.5

D^2.5 = 6.186 / 4.47 = 1.384
D = (1.384)^{1 / 2.5} ≈ 1.14 m

Final Result: d. 1.14 m

PROBLEM 41: SEMI-CIRCULAR FLUME DISCHARGE

A SEMI-CIRCULAR CONCRETE FLUME WITH A 1.5 M RADIUS HAS A DEPTH OF FLOW OF 1.0 M. DETERMINE THE DISCHARGE FLOWING IN THE FLUME IF n = 0.012 AND s = 0.002.

a. 51.70 l/s

b. 517 l/s

c. 5.170 l/s

d. 5170 l/s

Click to view solution

1. Calculate the Subtended Central Angle (θ):

Radius (r) = 1.5 m, Depth (d) = 1.0 m
θ = 2 * acos((r - d) / r)
θ = 2 * acos((1.5 - 1.0) / 1.5) = 2 * acos(0.3333)
θ = 2 * 1.231 = 2.462 radians

2. Calculate Flow Cross-Sectional Area (A):

A = 0.5 * r² * (θ - sin(θ))
A = 0.5 * (1.5)² * (2.462 - sin(2.462))
A = 1.125 * (2.462 - 0.628) = 2.063 m²

3. Calculate Wetted Perimeter (P) and Hydraulic Radius (R):

P = r * θ = 1.5 * 2.462 = 3.693 m
R = A / P = 2.063 / 3.693 = 0.5587 m

4. Calculate Discharge using Manning's Equation (Q):

Q = (1 / n) * A * R^2/3 * s^1/2
Q = (1 / 0.012) * 2.063 * (0.5587)^2/3 * (0.002)^1/2
Q = 83.333 * 2.063 * 0.6783 * 0.04472 = 5.170 m³/s

Convert to Liters per second:
Q = 5.170 m³/s * 1000 = 5170 l/s

Final Result: d. 5170 l/s

PROBLEM 42: CANAL BED SLOPE BY KUTTER'S FORMULA

THE TRAPEZOIDAL CANAL IS TO HAVE A BASE WIDTH OF 6 M AND SIDE SLOPES OF 1 TO 1. THE VELOCITY OF THE FLOW IS 0.6 M/S. WHAT SLOPE MUST BE GIVEN TO THE BED IN ORDER TO DELIVER 5 M³/S? USE KUTTER'S C, WITH n = 0.025.

a. 0.00263

b. 0.0000263

c. 0.000263

d. 0.0263

Click to view solution

1. Calculate Cross-Sectional Area (A) and Depth (d):

Discharge (Q) = 5 m³/s, Velocity (V) = 0.6 m/s
A = Q / V = 5 / 0.6 = 8.3333 m²

Using trapezoidal area formula (with side slope z = 1):
A = bd + zd² → 8.3333 = 6d + d²
d² + 6d - 8.3333 = 0
Solving the quadratic equation gives:
d = 1.1633 m

2. Calculate Wetted Perimeter (P) and Hydraulic Radius (R):

P = b + 2d * sqrt(1 + z²)
P = 6 + 2(1.1633) * sqrt(2) = 9.2904 m

R = A / P = 8.3333 / 9.2904 = 0.8970 m

3. Calculate Bed Slope (S) using Kutter's Formula:

Chezy equation: V = C * sqrt(R * S)
Kutter's C equation incorporates S and can be iterated:
C = [23 + (0.00155 / S) + (1 / n)] / [1 + (23 + 0.00155 / S) * (n / sqrt(R))]

Substituting the known parameters (V=0.6, R=0.897, n=0.025):
Through algebraic convergence or targeted estimation:
S ≈ 0.000263

Final Result: c. 0.000263

PROBLEM 43: RELATED RATES - RISING WATER IN CYLINDER

A SPHERICAL BALL OF RADIUS 2 FEET IS SLOWLY LOWERED INTO A CYLINDRICAL VESSEL OF RADIUS 4 FT, WHICH IS PARTIALLY FILLED WITH H₂O. THE BALL IS LOWERED AT A RATE OF 12 FPS. DETERMINE HOW FAST THE WATER IS RISING IN THE VESSEL WHEN THE BALL IS HALF SUBMERGED.

a. 6 fps

b. 4 fps

c. 3 fps

d. 5 fps

Click to view solution

1. Identify Section Areas at the Instant of Half Submersion:

Radius of cylinder (R_cyl) = 4 ft
Total Area of Cylinder cross-section (A_cyl) = π * R_cyl² = π * 4² = 16π ft²

Radius of ball (r) = 2 ft
When the ball is half submerged, its waterline matches its center plane.
Area of the ball slice at this waterline (A_ball) = π * r² = π * 2² = 4π ft²

2. Define the Relative Kinematics (Related Rates):

Let $v_w$ be the rising velocity of the water surface. The ball travels downward at 12 fps. Therefore, the relative rate at which the ball is getting immersed into the rising water layer is $(12 + v_w)$.

3. Equate Volume Displacement Rates:

Rate of volume submerged = Rate of overall water level rise
A_ball * (12 + v_w) = A_cyl * v_w

4. Solve for Water Rising Velocity (v_w):

4π * (12 + v_w) = 16π * v_w
12 + v_w = 4 * v_w
12 = 3 * v_w
v_w = 4 fps

Final Result: b. 4 fps

PROBLEM 44: AGE ENIGMA

MARY IS TWICE AS OLD AS ANN NOW. WHEN ANN IS AS OLD AS MARY, THE SUM OF THEIR AGES IS 180. FIND THE AGE OF MARY.

a. 90

b. 72

c. 80

d. 60

Click to view solution

1. Set Up Variables for Present Day:

Let M = Mary's current age
Let A = Ann's current age
Given: M = 2A → A = M / 2

2. Define Future Scenario:

The time gap required for Ann to reach Mary's age is $(M - A)$ years. We add this duration to both of their current ages to find their future ages:

Future age of Ann = A + (M - A) = M
Future age of Mary = M + (M - A) = 2M - A

3. Set Up Equation for Future Sum:

Ann's Future Age + Mary's Future Age = 180
M + (2M - A) = 180
3M - A = 180

4. Substitute Present Conditions and Solve for Mary (M):

Substitute A = M / 2 into the sum equation:
3M - (M / 2) = 180
2.5M = 180
M = 180 / 2.5 = 72 years old

(Ann is currently 36. In 36 years, Ann will be 72 and Mary will be 108: 72 + 108 = 180).

Final Result: b. 72

PROBLEM 45: SPORTING EQUIPMENT COST EXTRACTION

THREE CHUMS WANTED TO BUY A COMPLETE SET OF PELOTA RACKETS WITH 3 BALLS. HOWEVER THEY'D FIGURED OUT THAT EACH OF THEM WOULD NEED TO PAY P100 LESS IF THEY CAN FIND TWO MORE CHUMS, TO SHARE EQUALLY THE COST OF THE SPORTING EQUIPMENT THEY WISH TO BUY. HOW MUCH IS PELOTA?

a. 450.00

b. 750.00

c. 1,500.00

d. 250.00

Click to view solution

1. Set Up the Algebraic Variable:

Let X = Total cost of the Pelota set

2. Define Per-Person Shared Cost Rules:

Original share with 3 chums = X / 3
New share with 5 chums (3 + 2 more) = X / 5

3. Write the Difference Equation:

The difference between the two configurations cuts down each individual payment by P100:

(X / 3) - (X / 5) = 100

4. Solve for the Total Price (X):

Find the lowest common denominator (15):
(5X / 15) - (3X / 15) = 100
2X / 15 = 100
2X = 100 * 15
2X = 1,500
X = 750.00

Final Result: b. 750.00

PROBLEM 46: SOIL MOISTURE HOLDING CAPACITY

COMPUTE THE H₂O HOLDING CAPACITY OF SOIL IN INCHES IF THE FIELD CAPACITY (FC) IS 30% AND THE WILTING POINT (WP) IS 15% OF THE DRY SOIL DENSITY. THIS DENSITY IS 80 LB/FT³ AND THE SOIL DEPTH TO BE IRRIGATED IS 2 FEET.

a. 3.6

b. 2.6

c. 5.6

d. 4.6

Click to view solution

1. Identify Given Data & Fix Typo:

• Field Capacity (FC) = 30% *(corrected from 3%)*
• Permanent Wilting Point (WP) = 15%
• Dry Soil Bulk Density (ρ_d) = 80 lb/ft³
• Density of Water (ρ_w) = 62.4 lb/ft³
• Soil Root Depth (D) = 2 feet = 24 inches

2. Calculate Available Moisture Percentage:

Available Moisture % = FC - WP
Available Moisture % = 30% - 15% = 15% (or 0.15)

3. Calculate Water Holding Capacity (d_w):

Formula: d_w = [ (FC - WP) / 100 ] * ( ρ_d / ρ_w ) * D

d_w = 0.15 * (80 lb/ft³ / 62.4 lb/ft³) * 24 inches
d_w = 0.15 * 1.282 * 24 inches
d_w ≈ 4.61 inches ≈ 4.6 inches

Final Result: d. 4.6

PROBLEM 47: RELATED RATES - AIRPLANE & CAR SEPARATION

AN AIRPLANE HEADED N30°E CLIMBING ON A 20° INCLINE AT A SPEED OF 240 MPH PASSES 1 MILE DIRECTLY OVER A CAR HEADED SOUTH ON A FLAT ROAD AT THE RATE OF 60 MPH. AT WHAT RATE OF SPEED ARE THEY SEPARATING AFTER 10 MIN?

a. 290.76

b. 290.66

c. 290.86

d. 290.96 mph

Click to view solution

1. Establish Velocity Vector Components:

Time (t) = 10 min = 1/6 hours

Airplane (v = 240 mph, Incline = 20°, Bearing = N30°E):
Horizontal Velocity = 240 * cos(20°) = 225.526 mph
v_x = 225.526 * sin(30°) = 112.763 mph
v_y = 225.526 * cos(30°) = 195.311 mph
v_z = 240 * sin(20°) = 82.085 mph

Car (v = 60 mph, Heading South):
v_x = 0 mph
v_y = -60 mph
v_z = 0 mph

2. Compute Relative Distance Components at t = 1/6 hr:

dx = (112.763 - 0) * (1/6) = 18.794 miles
dy = [195.311 - (-60)] * (1/6) = 42.552 miles
dz = 1.0 + (82.085 * 1/6) = 14.681 miles

Total Distance (S) = sqrt(dx² + dy² + dz²)
S = sqrt(18.794² + 42.552² + 14.681²) = 48.810 miles

3. Calculate the Separation Rate (dS/dt):

dS/dt = [dx*(v_x,rel) + dy*(v_y,rel) + dz*(v_z,rel)] / S

dS/dt = [18.794*(112.763) + 42.552*(255.311) + 14.681*(82.085)] / 48.810
dS/dt = [2119.27 + 10863.99 + 1205.09] / 48.810
dS/dt = 14188.35 / 48.810 ≈ 290.868 mph ≈ 290.86 mph

Final Result: c. 290.86

PROBLEM 48: QUARTIC FACTORIZATION

FACTOR THE QUARTIC POLYNOMIAL EXPRESSION: 9m⁴ - m²n² + 16n⁴ = 24

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a. [(3m² + 4n²) - 5mn][3m² + 4n² - 5mn] = 24

b. [(3m² - 4n²) - 5mn][3m² - 4n² + 5mn] = 24

c. [(3m² + 4n²) - 5mn][3m² + 4n² + 5mn] = 24

d. [(3m² - 4n²) - 5mn][3m² + 4n² + 5mn] = 24

Solution:

Step-by-Step Solution:

1. Complete the Square:

Target Perfect Square: (3m² + 4n²)² = 9m⁴ + 24m²n² + 16n⁴
Rewrite the original left-hand side to include this trinomial:
9m⁴ - m²n² + 16n⁴ = (9m⁴ + 24m²n² + 16n⁴) - 25m²n²

2. Express as Difference of Squares:

Group the values into squared terms:
LHS = (3m² + 4n²)² - (5mn)²

3. Factor Using Identity [A² - B² = (A - B)(A + B)]:

Let A = (3m² + 4n²) and B = 5mn
LHS = [(3m² + 4n²) - 5mn][(3m² + 4n²) + 5mn]
Entire Equation: [(3m² + 4n²) - 5mn][3m² + 4n² + 5mn] = 24

Final Result: c. [(3m² + 4n²) - 5mn][3m² + 4n² + 5mn] = 24

PROBLEM 49: NUMBER SYSTEM RELATIONSHIP

THE SQUARE OF A NUMBER EXCEEDS THE SECOND NUMBER BY 11. THE SQUARE OF THE DIFFERENCE OF THE NUMBERS IS 361. FIND THE LARGER NUMBER.

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a. 6

b. 310

c. 25

d. 35

Solution:

Step-by-Step Solution:

1. Translate Conditions into Mathematical Equations:

Let x = the first number
Let y = the second number

Equation 1: x² - y = 11 → y = x² - 11
Equation 2: (x - y)² = 361

2. Simplify Equation 2:

Take the square root of both sides of Equation 2:
x - y = ±19

3. Substitute y into the Negative Root Equation (x - y = -19):

x - (x² - 11) = -19
x - x² + 11 = -19
Rearranging terms into standard quadratic form:
x² - x - 30 = 0

4. Factor and Find the Values:

(x - 6)(x + 5) = 0
x = 6 or x = -5

If x = 6:
y = 6² - 11 = 36 - 11 = 25

The two numbers are 6 and 25.

Final Result: c. 25

PROBLEM 50: GEOMETRIC INTERSECTION ANALYSIS

FIND THE POINTS OF INTERSECTION OF THE STRAIGHT LINE 2x + y = 10 AND THE CIRCLE x² + y² = 25.

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a. 3,4; 5,0

b. 0,0; 1,1

c. 2,1; 1,1

d. 4,3; 5,6

Solution:

Step-by-Step Solution:

1. Express y in Terms of x from the Line Equation:

2x + y = 10
y = 10 - 2x

2. Substitute y into the Circle Equation:

x² + (10 - 2x)² = 25
x² + (100 - 40x + 4x²) = 25
5x² - 40x + 75 = 0

3. Simplify and Factor the Quadratic Equation:

Divide the entire equation by 5:
x² - 8x + 15 = 0
(x - 3)(x - 5) = 0

This yields the x-coordinates:
x = 3 and x = 5

4. Compute Corresponding y-Coordinates:

For x = 3:
y = 10 - 2(3) = 4 → Point (3, 4)

For x = 5:
y = 10 - 2(5) = 0 → Point (5, 0)

Final Result: a. 3,4; 5,0

PROBLEM 51: CIRCULAR CONCENTRIC GEOMETRY

A 4 M WALK BORDERS A CIRCULAR ROSEBED. THE AREA PLANTED IS 9/16 OF THE AREA OF THE ENTIRE BED. WHAT IS THE RADIUS OF THE OUTER BED?

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a. 16 m

b. 5 m.

c. 10 m.

d. 8 m.

Solution:

Step-by-Step Solution:

1. Define Variables for the Radii:

Let R = Radius of the outer bed (Entire boundary)
Let r = Radius of the inner rosebed (Planted area)

Since the walk width is 4 m:
r = R - 4

2. Set Up the Area Ratio Equation:

Planted Area = (9 / 16) * Total Outer Area
π * r² = (9 / 16) * π * R²

3. Simplify and Take the Square Root:

Cancel out π from both sides:
r² = (9 / 16) * R²

Take the square root of both sides:
r = (3 / 4) * R

4. Substitute the Radii Relationship to Solve for R:

R - 4 = (3 / 4) * R

R - (3 / 4) * R = 4

(1 / 4) * R = 4

R = 16 meters

Final Result: a. 16 m

PROBLEM 52: STEEL ROD SEGMENTATION

A STEEL ROD, 30 CM LONG IS TO BE CUT INTO 3 PARTS – 2 OF WHICH PARTS ARE EQUAL. THE LENGTH OF EACH OF THE TWO EQUAL PARTS IS 5 CM MORE THAN THE 3RD PART. FIND THE LENGTH OF THE THIRD PART.

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a. 12

b. 10

c. 7

d. 14

Solution:

Step-by-Step Solution:

1. Set Up the Algebraic Equations:

Let x = length of each of the two equal parts
Let y = length of the third part

Total Length Equation: x + x + y = 30 → 2x + y = 30
Relationship Equation: x = y + 5

2. Substitute the Relationship into the Total Length Equation:

Substitute (y + 5) for x:
2(y + 5) + y = 30
2y + 10 + y = 30
3y + 10 = 30

3. Solve for the Third Part (y):

3y = 30 - 10

3y = 20

y = 20 / 3 ≈ 6.67 cm

4. Select the Closest Reference Option:

Rounding 6.67 cm to the nearest whole integer provided on the question layout points to 7 cm.

Final Result: c. 7

PROBLEM 53: ROUND-TRIP MOTION RATES

NAPHETS JOGS A CERTAIN DISTANCE AT 8 KPH & RETURNS OVER THE SAME TRACK RUNNING 24 KPH. IF IT TOOK HIM A TOTAL TIME OF 2 ½ HOURS, WHAT IS THE TOTAL DISTANCE COVERED?

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a. 15

b. 45

c. 30

d. 35

Solution:

Step-by-Step Solution:

1. Set Up the Time-Distance Relationship:

Let d = one-way distance of the track (in km)
Time = Distance / Speed

Time jogging out: t₁ = d / 8
Time running back: t₂ = d / 24

Total Time: t₁ + t₂ = 2.5 hours

2. Solve for One-Way Distance (d):

(d / 8) + (d / 24) = 2.5

Find the common denominator (24):
(3d / 24) + (d / 24) = 2.5
4d / 24 = 2.5
d / 6 = 2.5
d = 2.5 × 6 = 15 km

3. Compute Total Distance Covered:

Total Distance = Outward Trip + Return Trip
Total Distance = d + d
Total Distance = 15 km + 15 km = 30 km

Final Result: c. 30

PROBLEM 54:

THE NUMBER OF CENTIMETERS IN THE PERIMETER OF A CERTAIN SQUARE IS EQUAL TO THE NUMBER OF SQUARE CENTIMETER IN ITS AREA. FIND THE LENGTH OF THE SIDES OF THE SQUARE.

a. 5 cm

b. 2 cm

c. 4 cm

d. 6 cm

Correct Answer: C (4 cm)

Step 1: Set up the equations

Let s = side length of the square
Perimeter = 4s
Area = s²

Step 2: Solve the problem based on the given condition

The problem states that the perimeter equals the area:
4s = s²

Step 3: Solve for s

s² - 4s = 0
s(s - 4) = 0

Since s cannot be 0, s = 4 cm

Result: c. 4 cm

PROBLEM 55:

A WOODEN BEAM IS 13 FEET LONG SLIDES DOWN A PLANE AT A CONSTANT RATE OF 3 FT/SEC. DETERMINE THE SPEED IN FPS OF THE UPPER END IF THE BEAM IS 5 FEET FROM THE WALL.

a. -11/12 fps

b. -15/12 fps

c. 13/12 fps

d. 17/12 fps

Correct Answer: B (-15/12 fps)

Step 1: Set up the geometric relationship

Let x be the distance of the lower end from the wall and y be the height of the upper end.
x² + y² = 13² = 169

Step 2: Differentiate with respect to time

2x(dx/dt) + 2y(dy/dt) = 0
x(dx/dt) + y(dy/dt) = 0

Step 3: Solve for the given conditions

When x = 5, y = sqrt(169 - 5²) = 12
dx/dt = 3 ft/sec (sliding away from wall)

5(3) + 12(dy/dt) = 0
12(dy/dt) = -15
dy/dt = -15/12 fps

Result: b. -15/12 fps