header ads

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.

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?

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?

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 = (Ï€ / 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?

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?

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%.

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.

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?

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:

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?

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?

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%.

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.

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.

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?

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?

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?

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?

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.

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.

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

Post a Comment

0 Comments