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[7] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [7] The Energy Equation1
The Energy Equation[7]
Dr. Mohammad N. Almasrihttp://sites.google.com/site/mohammadnablus/Home
Fluid Mechanics
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[7] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [7] The Energy Equation
Introduction
There are various types of devices and components that are utilized in flow systems
They occur in most fluid flow systems and they either:
add energy to the fluid (in the case of pumps)
remove energy from the fluid (in the case of turbines), or
cause undesirable losses of energy from the fluid (in the case of friction)
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[7] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [7] The Energy Equation
Pumps
A pump is a common example of a mechanical device that adds energy to a fluid
An electric motor or some other prime power device drives a rotating shaft in the pump. The pump then takes this kinetic energy and delivers it to the fluid, resulting in fluid flow and increased fluid pressure
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[7] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [7] The Energy Equation
Pumps
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[7] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [7] The Energy Equation
Turbines
Turbines are examples of devices that take energy from a fluid and deliver it in the form of work, causing the rotation of a shaft
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[7] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [7] The Energy Equation
Turbines
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[7] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [7] The Energy Equation
Turbines
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[7] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [7] The Energy Equation
Turbines
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[7] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [7] The Energy Equation
Fluid Friction
A fluid in motion suffers losses in energy into thermal energy (heat) due to friction
Heat is dissipated through the walls of the pipe in which the fluid is flowing
The magnitude of the energy loss (called headloss) is dependent on
the properties of the fluid
the flow velocity
the pipe size
the smoothness of the pipe wall
the length of the pipe9
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[7] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [7] The Energy Equation
Headloss in pipes due to Friciton
Generally, headloss in the pipes due to friction is given in the following general form (later we will see different forms):
where hL is the headloss due to friction, L is the length, D is the diameter, v is the velocity, and the constant depends on the status of the internal surface of the pipe, its type, viscosity of the fluid,
Apparently, the hL increases linearly with pipe length
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[7] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [7] The Energy Equation
Valves and Fittings
Elements that control the direction or flow rate of a fluid in a system typically set up local turbulence in the fluid, causing energy to be dissipated as heat
Whenever there is a restriction, a change in flow velocity, or a change in the direction of flow, these energy losses occur
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[7] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [7] The Energy Equation
Headloss due to Abrupt Expansion
The headloss caused by s sudden expansion is given by the following relationship:
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[7] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [7] The Energy Equation
Headloss due to Abrupt Contraction
The headloss caused by a sudden contraction is given by the following relationship:
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[7] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [7] The Energy Equation
The Energy Equation
Now what we need to do is to update Bernoulli equation to account for the addition,
removal, and losses in energy
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[7] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [7] The Energy Equation
The Energy Equation
Take a control volume and do an energy balance
Energy can enter the control volume in two ways:
Energy transported by the flowing fluid
By a pump
Energy can leave the control volume in three ways:
Energy transported by the flowing fluid
By a turbine
By headloss (to heat)
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[7] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [7] The Energy Equation
The Energy Equation
However, we will deal with head rather than energy or work
Head is the energy of the fluid per weight of fluid
This implies that:
hp: head added by the pump (pump head)
ht: head extracted by the turbine (turbine head)
hL: head loss due to friction
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[7] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [7] The Energy Equation
The Energy Equation
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Head carried by flow into the CV
Head added
by pumps
=
+
Head carried by flow out of the CV
Head extracted
by turbines
+ Head loss+
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[7] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [7] The Energy Equation
The Energy Equation
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[7] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [7] The Energy Equation
The Energy Equation
It is essential that the general energy equation be written in the direction of flow
After the fluid leaves point 1 it enters the pump, where energy is added (hp). A motor drives the pump, and the impeller of the pump transfers the energy to the fluid (hp)
Then the fluid flows through a piping system composed of a valve, elbows, and the lengths of pipe, in which energy is dissipated from the fluid and is lost (hp)
Before reaching point 2, the fluid flows through a fluid motor, which removes some of the energy to drive an external device (ht)
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[7] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [7] The Energy Equation
The Energy Equation
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where α is the kinetic energy correction factor. This factor accounts for the cases when the velocity profile is not uniformly distributed. However, in all the cases that we will come across, we will use α = 1
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[7] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [7] The Energy Equation
Example
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[7] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [7] The Energy Equation
Example
A horizontal pipe carries cooling water at 10°C from a reservoir. The head loss in the pipe is:
where L is the length of the pipe from the reservoir to the point in question, V is the mean velocity in the pipe, and D is the diameter of the pipe
If the pipe diameter is 20 cm and the rate of flow is 0.06 m3/s, what is the pressure in the pipe at L = 2000 m. Assume α2 = 1
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[7] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [7] The Energy Equation
Exam
ple
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[7] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [7] The Energy Equation24
Exam
ple
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[7] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [7] The Energy Equation
Power Required by Pumps
Power is defined as the rate of doing work
The unit of power is watt (W)
This is equivalent to N.m/s or J/s
P = γ Q hp
where P is the power added by the pump to the fluid, γ is the specific weight of the fluid, and Q is the volume flow rate
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[7] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [7] The Energy Equation
Power Required by Pumps
Both pumps and turbines lose energy due to mechanical factors
To account for these losses, we use the term pump efficiency η or e
The efficiency is the ratio of power output from the pump (Pout) to power input to the pump (Pin)
η = Pout/Pin
e = Pout/Pin
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[7] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [7] The Energy Equation
Power Required by Pumps
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γ: Specific weight (N/m3)
Q: Flow rate (m3/s)
H: Head added by pump (m)
e: [-]
Added
Required
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[7] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [7] The Energy Equation
Power Required by Pumps
Each one horsepower (HP) is equivalent to 746 Watts
where Q in ft3/sec, hp in ft, γ in lb/ft3, and the power is in HP
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[7] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [7] The Energy Equation
Example
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[7] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [7] The Energy Equation
Example
A pipe 50 cm in diameter carries water (10°C) at a rate of 0.5 m3/s. A pump in the pipe is used to move the water from an elevation of 30 m to 40 m
The pressure at section 1 is 70 kPa gage and the pressure at section 2 is 350 kPa gage
What power in kilowatts and in horsepower must be supplied to the flow by the pump?
Assume hL = 3 m and α1 = α2=1
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[7] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [7] The Energy Equation
Example
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[7] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [7] The Energy Equation
Example
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[7] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [7] The Energy Equation
Example
A small hydroelectric power plant takes a discharge of 14.1 m3/s through an elevation drop of 61 m
The head loss through the intakes, penstock, and outlet works is 1.5 m. The combined efficiency of the turbine and electrical generator is 87%
What is the rate of power generation?
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[7] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [7] The Energy Equation
Example
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[7] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [7] The Energy Equation
Example
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[7] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [7] The Energy Equation
Example
Water (10°C) is flowing at a rate of 0.35 m3/s, and it is assumed that hL = 2V2/2g from the reservoir to the gage, where V is the velocity in the 30-cm pipe. What power must the pump supply? Assume α = 1.0 at all locations
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[7] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [7] The Energy Equation
Example
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[7] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [7] The Energy Equation
Example
In the pump test shown, the rate of flow is 0.16 m3/s of oil (S = 0.88)
Calculate the horsepower that the pump supplies to the oil if there is a differential reading of 120 cm of mercury in the U-tube manometer. Assume α = 1.0 at all locations
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[7] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [7] The Energy Equation
Example
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[7] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [7] The Energy Equation
Example
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[7] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [7] The Energy Equation
Example
Neglecting head losses, determine what horsepower the pump must deliver to produce the flow as shown. Here the elevations at points A, B, C, and D are 35 m, 60 m, 35 m, and 30 m, respectively. The nozzle area is 90 cm2
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[7] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [7] The Energy Equation
Example
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[7] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [7] The Energy Equation
Hydraulic and Energy Grade Lines
Engineers find it useful to employ the "energy grade line" (EGL) and the "hydraulic grade line" (HGL) in working with the pipe systems
These imaginary lines help the engineers find the trouble spots in the system (usually points of low pressure)
HGL is the line that indicates the piezometric head at each location in the system (p/γ + z)
EGL is the line that indicates the total head at each location (V2/2g + p/γ + z)
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[7] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [7] The Energy Equation
Hydraulic and Energy Grade Lines
As the velocity goes to zero, the HGL and the EGL approach each other
Thus, in a reservoir,they are identical and lie on the surface
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[7] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [7] The Energy Equation
Hydraulic and Energy Grade LinesEnergy Addition by Pumps
A pump causes an abrupt rise in the EGL and HGL by adding energy to the flow
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[7] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [7] The Energy Equation
Hydraulic and Energy Grade LinesEnergy Removal by Turbines
A turbine causes an abrupt drop in the EGL and HGL by removing energy from the flow
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[7] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [7] The Energy Equation
Hydraulic and Energy Grade LinesDischarge to Atmosphere
When a pipe discharges into the atmosphere the HGL is coincident with the system because P/γ = 0 at these points
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[7] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [7] The Energy Equation
Hydraulic and Energy Grade LinesNegative Pressure
If the HGL falls below the pipe, then P/ γ is negative indicating subatmospheric pressure and a potential location of cavitation
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[7] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [7] The Energy Equation
Hydraulic and Energy Grade LinesDiameter Change
For a steady flow in a pipe of constant diameter and wall roughness, the slope of the EGL and HGL will be constant
However, when there is a change in the diameter there will be a change in the EGL and HGL
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[7] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [7] The Energy Equation
Hydraulic and Energy Grade LinesA general Case
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[7] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [7] The Energy Equation
Example
Water flows from the reservoir through a pipe and then discharges from a nozzle. The head loss in the pipe itself is given as hL = 0.025 (L/D) (V2/2g), where L and D are the length and diameter of the pipe and V is the velocity in the pipe
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What is the discharge of water? Draw the HGL and EGL for the system
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[7] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [7] The Energy Equation
Example
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[7] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [7] The Energy Equation
Example
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[7] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [7] The Energy Equation
Example
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[7] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [7] The Energy Equation
Example
What power must be supplied to the water to pump 0.1 m3/s at 20°C from the lower to the upper reservoir? Assume that the head loss in the pipes is given by hL = 0.018 (L/D) (V2/2g), where L and D are the length and diameter of the pipe and V is the velocity in the pipe. Sketch the HGL and the EGL
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[7] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [7] The Energy Equation
Example
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[7] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [7] The Energy Equation
Example
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[7] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [7] The Energy Equation
Example
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[7] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [7] The Energy Equation
Example
Water flows from the reservoir on the left to the reservoir on the right at a rate of 0.45 m3/s. The formula for the head losses in the pipes is hL = 0.018 (L/D) (V2/2g). What elevation in the left reservoir is required to produce this flow? Sketch the HGL and the EGL for the system
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[7] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [7] The Energy Equation
Exmple
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[7] Fall – 2010 – Fluid Mechanics Dr. Mohammad N. Almasri [7] The Energy Equation
Example
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