Fundamentals of Water Engineering

This document presents solutions to a series of problems related to fundamentals of water engineering. The problems cover various aspects of fluid mechanics, including friction in hydraulic hoists, buoyancy forces on submerged objects (specifically a cone-shaped plug), the principle of continuity, flow analysis in open channels, and the determination of normal and critical depths. Solutions involve applying relevant formulas and principles, such as Manning's equation and the equation of continuity, along with calculations involving pressure, force, velocity, and volume. The solutions demonstrate the application of fluid mechanics principles to practical engineering problems, including analyzing forces on weirs and understanding the relationship between normal and critical flow depths in open channels. References to relevant textbooks and publications are included.

RUNNING HEAD: FUNDAMENTALS OF WATER ENGINEERING
NAME
UNIT
ASSIGNMENT
ASSESSMENT NUMBER

RUNNING HEAD: FUNDAMENTALS OF WATER ENGINEERING
Qn. 1:
a) Friction on hydraulic hoist:
Ram Diameter = 32cm,
Outer Diameter = 32.02cm
Viscosity = 0.14Pa.s
Density = 820Kg/m3.
Speed = 0.20m/s
Length = 4m
Ram x-sec = 0.162 * 3.142 = 0.08m2.
Volume = 0.08m2 * 4 = 0.322m3.
Mass = 0.322 * 820 = 263.79Kg.
Friction force = 263.73 * 9.81 * 0.14
= 326.29N
Qn. 3
a) Buoyancy force arises when an object is submerged into a fluid with a heavier density. The fluid
with a heavier density will push up the object due to action of gravity which has a stronger effect
on objects or fluids with relatively more weight.
b) Resultant force on plug only due to water:
Radius of plug = 0.28m
Depth at the top = 0.35m
Depth at bottom = 0.50m
Density = 1000kg/m3
Pressure at the top of cone: ρgh1A1 = 1000 * 9.81 * 3.142 * 0.282 * 0.35
= 895.67N
Pressure at the bottom of cone: ρgh2A2 = 1000 * 9.81 * (total surface area of plug-surface
area of plugged part) * sin θ
= 246.43N
Total force exerted by the water = 895.67 – 246.43 = 649.2N
c) Depth of water that will result in no resultant force:
In this case, pressure at top = pressure at bottom.
Pressure at bottom = 246.43
246.43N = ρghA1 = 1000 * 9.81 * h
h = 246.43 / (1000 * 9.81 * 0.246) = 10.21cm
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RUNNING HEAD: FUNDAMENTALS OF WATER ENGINEERING
d) Force per meter around the rim:
Volume of plug = 2.67 E-2 m3.
Density of Plug = 2800kg/m3.
Wt of Plug = 2800 * 0.0267 * 9.81 = 733.40N
Wt of water – wt of plug = 733.40 – 649.2 = 81.2
Circumference = 3.142 * 32 = 100.53m
Force per meter = 81.2/100.53 = 0.84N/m
Qn. 5:
a) Equation of continuity:
Basic principle = Q = A1V1 = A2V2.
Integral calculus = Q = AV =∫
a
b
udA
= AV =∫
a
b
udA where A = w.h
= w.V (h) = ∫
a
b
w u dy
= V (h) =∫
a
b
udy
Qn. 6:
a) The sections represent regions of change of flow which in turn affects the velocity and hence the
total head of water. If the flow in the sections is straight and parallel to the bed, relatively little
head losses occur leading energy conservation. The head losses would occur where a change in
direction of flow is observed as this as energy will be lost in the formation of Eddies currents
leading to lower head pressure.
b) N/A
c) Flow at section 3
Critical flow formula = Q = CLH3/2 where L is 3m and C = 1.6
Q = h * w * vel. = 1.5 * 3 * 0.3 = 1.35m3/s
Q = 1.6 * 3 * h = 1.35
h = 1.35 / (1.6 * 3) = 0.28m
d) Determine magnitude and direction of total force on weir:
Volume striking the wall per second = 1.5 * 3 * 0.3 = 1.35m3/s
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RUNNING HEAD: FUNDAMENTALS OF WATER ENGINEERING
Mass striking wall per second = 1.35 * 1000 = 1350Kg/s
Force = Mass * acceleration = Mass/time * speed = 1350 * 0.3 = 405N
Velocity of water over weir = Q / A = 1.35/ (0.28 * 3) = 1.61m/s
Volume of water per second = 1.61 * 1.5 * 3 = 7.25m3/s
Mass per second = 7.25 * 1000 = 7250Kg/s
Force = Mass * acceleration = Mass/time * speed = 7250 * 1.61 = 11672N
Magnitude of force = (116722 + 4052) ^ 0.5 = 11679.5N = 11.679kN
Direction of force = Tan-1 (11672/405) = 88°
Qn. 8:
a) Normal Depth:
Q = AV where Manning’s equation states V = (R2/3 * S1/2)/ n
Where:
n = Manning coefficient = 0.024
R = A/P where A = y * 2: P = 2y + 2.
A/P = 2y/ (2y + 2)
6.5m3/s = ((((2y) 2)/ (2y + 2)) * 0.5^0.5)/ 0.024
= y = 0.267m
b) Critical depth
Critical depth = Yc = (Q/ ((b(g)1/2))2/3
= Yc = (6.5/ (2 * (9.81)1/2)2/3
= Yc = 1.02m
c) Normal depth occurs when the flow is uniform, parallel and steady, while critical depth occurs
when energy of the discharge is either at a minimum or maximum.
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RUNNING HEAD: FUNDAMENTALS OF WATER ENGINEERING
REFERENCES:
J.M. McDonough, 2004. Lectures in Elementary Fluid Dynamics: Physics, Mathematics and
Applications, University of Kentucky, Lexington.
Stockstill, R. L., C. E. Kees, and R. C. Berger. 2006. Modeling free-surface flow over a weir. Coastal
and Hydraulics Engineering Technical Note ERDC/CHL CHETN-XIII-1, Vicksburg, MS: U.S. Army
Engineer Research and Development Center.
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