Conservation of Energy: Control Volume Approach Video Lecture | Fluid Mechanics for Mechanical Engineering

In this lesson, we will derive the conservation
of energy equation for control volumes.
Often we want to know how much energy is required
for a process to occur or how much energy
can be extracted from a device.
For example: How much power is required to
pump water from one storage tank to another?
How much power is required to create a sustained
30-foot high jet of water from a nozzle in
a fountain?
How much power can be extracted by a steam
turbine in a coal-fired power plant?
These questions all involve energy can only
be answered by applying the conservation of
energy equation.
Here we have a mass, shaded in pink, which
we will call our system.
The system has a total energy Esys.
Two types of processes can affect the energy
of the system.
Heat transfer is the exchange of energy between
a system and its surroundings due to differences
in temperature.
There are three modes of heat transfer – conduction,
convection, and radiation.
We will denote heat as capital Q and the rate
heat is transferred to or from the system
as Q-dot.
Work is the exchange of energy between a system
and its surroundings due to other factors,
such as the movement of boundaries, flow of
electricity across boundaries, flow of mass
across boundaries, and other sources.
We will denote work as capital W and the rate
work is transferred to or from the system
as W-dot.
The first law of thermodynamics for a system
states, the time rate of change of a system’s
energy, DEsys/Dt, is equal to the total rate
of heat flux into the system minus the total
rate of heat flux out of the system, plus
the total rate work is transferred into the
system minus the total rate work is transferred
out of the system.
The summation symbol is used in front of Q-dot
and W-dot because there are potentially many
sources of heat and work transfer.
Capital D is used in DEsys/Dt to emphasize
that we are using the system point of view.
The right side of the equation can be rewritten
as the net rate of heat transfer into the
system, Q net in dot, plus the net rate of
work transferred into the system, W net in
dot.
Let’s imagine a control surface, indicated
by red dashed lines, surrounds the system
at this moment in time.
The control volume is contained inside of
this control surface.
At this time, the net rate of heat and work
transfer into the system is equal to the net
rate of heat and work transfer into the control
volume because they occupy the same region
of space.
We now apply the Reynold’s transport theorem
with the extensive property B being the total
energy E, and the corresponding intensive
quantity b being specific energy e.
The specific energy is the total energy per
mass.
Plug in the total energy for B, and specific
energy for b.
We previously stated that the left side of
the equation, DEsys/Dt, is equal to the net
rate of heat and work transfer into the system,
which is also equal to the net rate of heat
and work transfer into the control volume.
Putting everything together, we have the conservation
of energy equation for a control volume in
its most general form.
The time rate of change of energy in the control
volume plus the net flux of energy through
the control surface due to mass flow, is equal
to the net rate of heat and work transfer
to the control volume.
We can simplify this equation by recognizing
that most devices studied in fluid mechanics
operate under steady flow conditions, which
means the amount of energy in the device is
constant in time.
The first term in the conservation of energy
equation, which contains a time derivative,
becomes zero.
Additionally, most devices do not deform significantly,
which removes the time-dependence on the limits
of integration.
We will now examine the remaining terms.
First, we need an expression for the specific
energy e.
The total energy E is equal to the internal
energy plus kinetic energy plus gravitational
potential energy plus all other forms of energy.
These other forms of energy, which could be
due to electromagnetic effects, surface tension
effects, or some other effects, are usually
negligible and will be ignored in this analysis.
The internal energy will be represented by
U.
The kinetic energy is equal to one-half mass
times velocity squared.
The gravitational potential energy is equal
to mass, times the gravitational acceleration
g, times the elevation z.
The specific energy is the total energy divided
by mass, and is equal to the specific internal
energy u plus V squared over 2 plus g z.
Now we plug in the specific energy into the
conservation of energy equation.
From this point forward, we will drop the
subscript from the Q dot and W dot terms in
the conservation of energy equation to avoid
making the equation too messy.
Now let’s examine the rate of work transferred
to the control volume, W net in dot.
We will consider three forms of work: Work
transferred through a shaft, work due to normal
stresses, and work due to shear stresses.
In fluid mechanics, devices often utilize
a shaft for work transfer.
For example, in a coal-fired power plant,
steam flows through a turbine and interacts
with a rotor, causing the rotor to rotate.
The rotor is connected to a shaft, which is
connected to a generator, which produces electricity.
If we identify the control volume as just
the turbine, notice that the shaft cuts through
the control surface.
The rate that work is transferred through
a shaft is equal to the shaft torque T, times
the angular speed of the shaft, omega.
Turbines transfer energy out of a control
volume through a rotating shaft.
Other devices, such as pumps, compressors,
and fans, require energy input from a motor
through a shaft.
The net shaft power into a control volume,
is the power required to run a pump or similar
device, which will call W shaft pump dot,
minus the power extracted by a turbine, which
we will call W shaft turbine dot.
Fluid at control surfaces are subject to normal
and shear stresses.
Let’s examine a small section of a control
surface with area dA.
Fluid flows through the control surface with
a velocity V and a small force dF acts on
the surface.
This force, which causes normal and shear
stresses, can be broken into three components:
A component in the normal direction, dFn,
and two components in the tangential directions,
dFt1 and dFt2.
The rate of work transfer, which we call power,
due to a force is equal to the dot product
of the force and flow velocity.
The small amount of power due the normal component
of the force acting on the small surface area
is denoted as dW normal dA dot, which is equal
to dot product of the normal component of
the force dFn n-hat and the velocity vector
V.
The normal component of the force is assumed
to be the negative pressure times area dA.
The negative sign appears because the pressure
always points inward on control surfaces.
We can determine the total power due to pressure
forces on the entire control surface by integrating
dW normal dA dot over the control surface.
The total power due to normal stresses becomes
the integral of negative P, times V dot n-hat,
times dA.
Now we want to determine the power due to
shear stresses or viscous effects.
The small amount of power due the tangential
component of the force in direction t1 acting
on area dA, is the dot product of dFt1 t1-hat
and the velocity vector V.
Similarly, the small amount of power due the
tangential component of the force in direction
t2 acting on area dA, is the dot product of
dFt2 t2-hat and the velocity vector V.
For most devices in fluid mechanics, fluid
will cross the control surface though a well-defined
inlet or outlet in which the flow direction
will be approximately parallel to the outward
normal vector n-hat.
This means the components of the velocity
vector in the t1-hat and t2-hat directions
are small, making the dot products small as
well.
Therefore, we will neglect the power due to
shear stresses in this analysis.
Returning to the conservation of energy equation,
we plug in the expression W net in dot, then
move the integral containing pressure to the
left side of the equation, and multiply and
divide pressure by the density rho.
Since both integrals on the left side of the
equation have the same limits of integration
-- that is the entire control surface -- we
can combine them into a single integral.
This integral can be simplified greatly by
examining steady flow through a generic device
that has multiple inlets and outlets, and
does not deform.
The device is surrounded by a control surface,
which is marked by the red dashed lines.
The control volume is inside this control
surface.
The direction of flow is indicated by blue
velocity vectors, and inlets are labeled 1
and 2 while the outlets are labeled 3 and
4.
The direction of the normal vectors are shown
in orange and point outward from the control
surface.
Heat transfer may occur between the device
and surroundings and work may be transferred
between the device and surroundings through
a rotating shaft.
Notice that the integral is over the entire
control surface.
We can break up the control surface into multiple
sections, and integrate over each of these
sections individually.
The first integral will cover the area for
inlet 1, the second integral will cover the
area for inlet 2, the third integral will
cover the area for outlet 3, and the fourth
integral will cover the area for outlet 4.
We also need a fifth integral to account for
the walls of the device.
The dot products can be simplified by noticing
that the angle between the velocity vector
and normal vectors are parallel at the inlets
and outlets.
Recall that the dot product of any two vectors
can be calculated by multiplying the magnitude
of the two vectors by the cosine of the angle
between them.
For the inlets, the angle between the velocity
vector and normal vector is always 180 degrees,
which means the dot products in integrals
1 and 2 become negative V.
For the outlets, the angle between the velocity
vector and normal vector is 0 degrees, which
means the dot products for integrals 3 and
4 become positive V.
The no-slip boundary condition means that
the velocity at every wall is zero.
We now have four integrals with integrands
consisting of the quantity, u plus V squared
over 2 plus g z plus P over rho, times rho
V dA.
We need to know how specific internal energy,
velocity, pressure, and density vary over
each inlet and outlet in order to solve these
integrals.
On the right we have a generic inlet or outlet.
There are a few possible velocity profiles
that could occur.
If the flow speed is very slow or the inlet
diameter is very small or the fluid viscosity
is very high, the velocity profile will be
laminar and parabolic in shape.
The velocity will be zero at the walls and
a maximum value at the center.
More often, the velocity profile will be turbulent
and flatter.
Near the walls, the velocity drops rapidly
to zero.
Under certain conditions, the velocity profile
will be appear almost completely uniform and
flat.
In order to simplify the integrals in the
conservation of energy equation, we make a
couple assumptions.
First, we will replace the actual velocity
profile with an average velocity profile that
is uniform and produces the same mass flow
rate as the actual velocity profile.
Next, we assume that the density and specific
internal energy are uniform over each individual
cross section.
We also assume that the sum of g z plus P
over rho is constant over each cross section.
As we descend through an inlet or outlet,
pressure increases and quantity P over rho
increases, but the quantity g z decreases
by the same amount.
This result comes from analyzing the the hydrostatic
pressure distribution for incompressible fluids.
These assumptions allow us to remove all terms
from the integral except dA, and the integral
of dA is simply the area A.
rho times Vavg times A is the mass flux, m-dot.
So the assumptions have allowed us to convert
the integral into a simple algebraic expression:
The quantity u plus Vavg squared over 2 plus
g z plus P over rho, times m-dot, which has
units of energy per time.
If we had evaluated the integral using velocity
profiles for real flows, we would find that
the portion of the integral containing V squared
over 2, times rho V dA is not exactly equal
to Vavg squared over 2 times m-dot.
We would need to introduce a kinetic energy
correction factor, alpha.
Alpha is a measure of how much the actual
flow deviates from the energy flux calculated
using the uniform flow approximation.
For the parabolic, laminar flow on the right,
the value of alpha is 2.
For the flatter, turbulent flow in the middle,
the value of alpha is typically between 1.05
and 1.10.
For the uniform flow on the left, alpha is
exactly 1.
α is often not included in the conservation
of energy equation because the omission of
alpha usually produces insignificant error
compared to typical uncertainties in a problem.
For the rest of this derivation, alpha will
be omitted.
The integral over the control surface has
been replaced by four algebraic expressions.
Notice that a negative sign appears in front
of the inlet terms, while positive signs appear
in front of the outlet terms.
We can express this equation more compactly
by writing the sum of m-dot times the quantity
(u plus Vavg squared over 2 plus g z plus
P over rho) for all outlets, minus the sum
of m-dot times the quantity (u plus Vavg squared
over 2 plus g z plus P over rho) for all inlets.
To avoid making the equations too messy, we
will drop the j subscript from the summations.
Let’s return to the conservation of energy
equation, and plug in the expression for the
surface integral.
At this point, we can define a new quantity
called the specific enthalpy, which is equal
to the specific internal energy plus pressure
divided by density.
We will denote specific enthalpy by lowercase
h.
Specific enthalpy, like specific internal
energy, has units of energy per mass and is
a property of the fluid.
The conservation of energy equation then becomes
the sum of m-dot times the quantity (h plus
Vavg squared over 2 plus g z) for all outlets,
minus the same quantity for all inlets, equals
the rate of heat transfer into the control
volume plus rate of work transfer into the
control volume due to rotating shafts.
In other words, the rate energy leaves the
control volume through mass flow, minus the
rate energy enters the control volume through
mass flow, equals the net rate of heat and
work transfer into the control volume.
This form of the conservation of energy equation
for control volumes is more useful for thermodynamics
applications.
We will now obtain a more practical equation
for fluid mechanics applicationa.
First, we recognize that most devices in fluid
mechanics have one inlet and one outlet.
This means the rate of mass flux through the
inlet equals the rate of mass flux through
the outlet, and we can write the mass flux
as simply m-dot.
We will use the subscript “o” for outlet
and the subscript “i” for inlet.
Next, we divide the entire equation by m-dot,
and bring the internal energies to the right
side of the equation.
We now have P over rho plus Vavg squared over
2 plus g z at the outlet, equals the same
expression at the inlet, minus the quantity
specific internal energy at the outlet minus
specific internal energy at the inlet, minus
net heat-in per mass, plus the net shaft work-in
per mass.
Now we divide the equation by the gravitational
acceleration g, and rho and g are combined
into gamma.
In this form, all the terms in the conservation
of energy equation have units of length and
are called “heads”.
Heads make it easier to visualize energy by
relating the amount of energy to a length.
The P over gamma term is called the pressure
head, the Vavg squared over 2g term is called
the velocity head, and the z term is called
the elevation head.
These three terms are related to the total
mechanical energy in the flow.
The two remaining terms in the conservation
of energy equation also have special names.
The quantity (uo minus ui minus the net heat
in per mass), divided by g, is called the
head loss and is denoted with the symbol hL.
This term is related to the loss of mechanical
energy between the inlet and outlet due to
viscous effects in pipes and fittings, and
is always positive.
In later lessons, we will discuss empirical
formulas for calculating the head loss.
The specific shaft work divided by g is called
the shaft work head and is denoted with the
symbol hshaft net in.
This term is related to the gain or loss in
mechanical energy due to shaft work.
Let’s take a closer look at the shaft work
head.
The net shaft work in per mass is equal to
the net power in divided by m-dot.
There are two types of devices that use shaft
power to change the mechanical energy of a
fluid.
Pumps add mechanical energy to the fluid and
are driven by motors.
Power is transferred from the motor to the
pump through a rotating shaft, then the pump
delivers power to the fluid through interactions
with the fluid.
A turbine extracts mechanical energy from
a fluid and is used to run an electric generator
or other device.
Power is transferred from the fluid to the
turbine, then the turbine transfers power
to another device through a rotating shaft.
The net power into the fluid stream due to
a rotating shaft can be rewritten as the sum
of the power delivered by a pump to the fluid,
minus the power extracted from the fluid by
a turbine.
The negative sign appears in front of the
turbine term because the fluid loses mechanical
energy through interactions with the turbine.
We also rewrite m-dot as rho Vavg A.
We separate terms, and combine rho and g into
specific weight gamma, as well as combine
Vavg and A into volumetric flowrate Q.
The first term, which is the power the pump
delivers to the fluid divided by gamma Q,
will be denoted as hP, and the second term,
which is the power delivered from the fluid
to the turbine, will be denoted as hT.
hP is called the pump head and is related
to the gain in mechanical energy by the fluid
due to a pump.
hT is called the turbine head and is related
to the loss of mechanical energy by the fluid
due to a turbine.
We now have an equation for conservation of
energy for a control volume that is more useful
for fluid mechanics applications.
The equation is valid for steady flows with
one inlet and one outlet.
This equation states that the total head at
the outlet is equal to the total head at the
inlet, minus the head loss from fluid friction
in pipes and fittings, plus the head added
by pumps, minus the head removed by turbines.
It is important to not confuse the head loss,
pump head, and turbine head with specific
enthalpy, even though they use the same symbol
h.
hL, hP, and hT have units of length, while
specific enthalpy h has units of energy per
mass.
Throughout this derivation we assumed that
the kinetic energy correction coefficient
alpha was 1.
If we did not make this assumption, the coefficient
would appear in front of the velocity head
terms.