Energy and Head of Flow

Energy is defined as ability to do work. Both energy and work are measured in Newton-meter (or pounds-foot in English). Kinetic energy and potential energy are the two commonly recognized forms of energy. In a flowing fluid, potential energy may in turn be subdivided into energy due to position or elevation above a given datum, and energy due to pressure in the fluid. Head is the amount of energy per Newton (or per pound) of fluid.
 

Kinetic Energy and Velocity Head
Kinetic energy is the ability of a mass to do work by virtue of its velocity. The kinetic energy of a mass M having a velocity v is ½Mv2. Since M = W/g,

$K.E. = W \dfrac{v^2}{2g}$

$\text{Velocity head} = \dfrac{K.E.}{W} = \dfrac{v^2}{2g}$

 

Elevation Energy and Elevation Head
In connection to the action of gravity, elevation energy is manifested in a fluid by virtue of its position or elevation with respect to a horizontal datum plane.

$\text{Elevation energy} = Wz$

$\text{Elevation head} = \dfrac{\text{Elevation energy}}{W} = z$

 

behavior-of-egl-and-hgl.gif

 

Pressure Energy and Pressure Head
A mass of fluid acquires pressure energy when it is in contact with other masses having some form of energy. Pressure energy therefore is an energy transmitted to the fluid by another mass that possesses some energy.

$\text{Pressure energy} = W \dfrac{p}{\gamma}$

$\text{Pressure head} = \dfrac{\text{Pressure energy}}{W} = \dfrac{p}{W}$

 

01 How to calculate the discharge and the velocity of flow

Problem 1
Compute the discharge of water through 75 mm pipe if the mean velocity is 2.5 m/sec.
 

Problem 2
The discharge of air through a 600-mm pipe is 4 m3/sec. Compute the mean velocity in m/sec.
 

Problem 3
A pipe line consists of successive lengths of 380-mm, 300-mm, and 250-mm pipe. With a continuous flow through the line of 250 Lit/sec of water, compute the mean velocity in each size of pipe.
 

02 Graph of flow velocity in tapering pipe

Problem
The diameter of a 6-m length pipe decreases uniformly from 450 mm to 150 mm. With a flow of 0.15 m3/sec of oil, compute the mean velocity at cross section 1 m apart. Plot the velocity as ordinate against length as abscissa.
 

Graph of velocity of flow versus length of pipe (plotted in MS Excel)

02-graph-of-velocity.gif

 

Discharge or Flow Rate

Discharge (also called flow rate)
The amount of fluid passing a section of a stream in unit time is called the discharge. If v is the mean velocity and A is the cross sectional area, the discharge Q is defined by Q = Av which is known as volume flow rate. Discharge is also expressed as mass flow rate and weight flow rate.

Volume flow rate, $Q = Av$

Mass flow rate, $M = \rho Q$

Weight flow rate, $W = \gamma Q$

 

fundamentals-fluid-flow.gif

 

Consistency of Soil (Atterberg Limits)

Consistency is the term used to describe the ability of the soil to resist rupture and deformation. It is commonly describe as soft, stiff or firm, and hard.
 

Water content greatly affects the engineering behavior of fine-grained soils. In the order of increasing moisture content (see Figure 2 below), a dry soil will exist into four distinct states: from solid state, to semisolid state, to plastic state, and to liquid state. The water contents at the boundary of these states are known as Atterberg limits. Between the solid and semisolid states is shrinkage limit, between semisolid and plastic states is plastic limit, and between plastic and liquid states is liquid limit.
 

atterberg-limits.gif

 

Atterberg limits, then, are water contents at critical stages of soil behavior. They, together with natural water content, are essential descriptions of fine-grained soils.
 

Unit Weights and Densities of Soil

Symbols and Notations
γ, γm = Unit weight, bulk unit weight, moist unit weight
γd = Dry unit weight
γsat = Saturated unit weight
γb, γ' = Buoyant unit weight or effective unit weight
γs = Unit weight of solids
γw = Unit weight of water (equal to 9810 N/m3)
W = Total weight of soil
Ws = Weight of solid particles
Ww = Weight of water
V = Volume of soil
Vs = Volume of solid particles
Vv = Volume of voids
Vw = Volume of water
S = Degree of saturation
w = Water content or moisture content
G = Specific gravity of solid particles
 

Physical Properties of Soil

Phase Diagram of Soil
Soil is composed of solids, liquids, and gases. Liquids and gases are mostly water and air, respectively. These two (water and air) are called voids which occupy between soil particles. The figure shown below is an idealized soil drawn into phases of solids, water, and air.
 

soil-properties.gif

 

Weight-Volume Relationship from the Phase Diagram of Soil
total volume = volume of soilds + volume of voids
$V = V_s + V_v$

volume of voids = volume of water + volume of air
$V_v = V_w + V_a$

total weight = weight of solids + weight of water
$W = W_s + W_w$
 

Example 02: Finding the Number of Steel Bars of Doubly-reinforced Concrete Beam

Problem
A propped beam 8 m long is to support a total load of 28.8 kN/m. It is desired to find the steel reinforcements at the most critical section in bending. The cross section of the concrete beam is 400 mm by 600 mm with an effective cover of 60 mm for the reinforcements. f’c = 21 MPa, fs = 140 MPa, n = 9. Determine the required number of 32 mm ø tension bars and the required number of 32 mm ø compression bars.
 

wsd-example-02-propped-beam.gif

 

Example 01: Finding the Number of Steel Bars of Singly-reinforced Concrete Beam

Problem
A reinforced concrete cantilever beam 4 m long has a cross-sectional dimensions of 400 mm by 750 mm. The steel reinforcement has an effective depth of 685 mm. The beam is to support a superimposed load of 29.05 kN/m including its own weight. Use f’c = 21 MPa, fs = 165 MPa, and n = 9. Determine the required number of 28 mm ø reinforcing bars using Working Stress Design method.
 

wsd-example-01-cantilever-beam.gif

 

Example 05: Stresses of Steel and Concrete in Doubly Reinforced Beam

Problem
A 300 mm × 600 mm reinforced concrete beam section is reinforced with 4 - 28-mm-diameter tension steel at d = 536 mm and 2 - 28-mm-diameter compression steel at d' = 64 mm. The section is subjected to a bending moment of 150 kN·m. Use n = 9.

1. Find the maximum stress in concrete.
2. Determine the stress in the compression steel.
3. Calculate the stress in the tension steel.
 

wsd-example-05-doubly-reinforced-beam.gif

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