Bernoulli's Equation and Mechanical Energy Balance in Fluid Flow, Apuntes de Ingeniería Industrial

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TOPIC 3.

BASIC EQUATIONS OF

FLUID FLOW

3.1. MACROSCOPIC MASS BALANCE

3.2. MACROSCOPIC ENERGY BALANCE

3.3. MECHANICAL ENERGY BALANCE: BERNOUILLI’S EQUATION

3.4. CONSERVATION OF MOMENTUM

TOPIC 3.

BASIC EQUATIONS OF FLUID FLOW

3.1. MACROSCOPIC MASS BALANCE

o As we mentioned in the former topic, the velocity of real fluid varies with the radius.

o As we mentioned in the former topic, the velocity of real fluid varies with the radius.

o Then, in order to carry out the mass balance, we will use the average velocity of the fluid through the section under study, which is calculated as:

o In the case in which the density may be considered as constant, we will have:

o which establishes the equality between the volumetric flow rates and it is known as the equation of continuity.

v= (^) S^1 ∫vdS

_

v 1 .S 1 =v 2 .S 2 ; Qv1 =Qv2 (m 3 /s)

3.1. MACROSCOPIC MASS BALANCE

o If it were a gas in compressible flow conditions, (ρ≠const., v≠const.), then it is obvious that the volumetric flow-rate is not constant.

o In these last cases and if the transversal sections are equal (cylindrical pipe, etc.) the following applies: kg/m^2 s

o The product v·ρ is known as the mass flow-rate (G) and the mass balance is defined in these cases referred to it.

v (^) 1. 1 2. 2 G 1 G 2 2

2 1

(^1) = ∴ ρ = v ρ ∴ = S

m S

m

o Thus, the macroscopic equation of total energy conservation is obtained:

o φ, e, and k represent the average specific potential energy, the average especific internal energy, and the average specific kinetic energy (J/kg), respectively in any tranversal section of the pipe. o Φ, E, and K are total energies of the fluid in the system considered. o Q: Heat transferred form the environment to the system through the surface S. o Qv: volumetric flow-rate; ρ and P the density and pressure of the fluid. o W: The work interchanged between the fluid and a machine (positive if it is a pump or compressor, negative if it is a turbine).

E K [ e k Q e k Q ] Q [ PQ PQ W ]

dt

d (Φ T (^) + T + T )= (φ 1 + 1 + 1 )ρ 1 v 1 −(φ 2 + 2 + 2 ) ρ 2 v 2 + + 1 v 1 − 2 v 2 +

3.2. MACROSCOPIC ENERGY BALANCE

o All the terms are power terms, watts (SI), J/s.

o It might be simplified using the definition of the mass flow- rate, m=ρ·Qv. And considering for a steady state case:

o Taking into account the definitions of potential and kinetic energy, φ=mgz and k=1/2v (^) e^2 (J/s) and dividing all the terms of the equation by m , the mass flow-rate. The following is obtained:

o Which is in accordance with the first principle of thermodynamics, and where all the terms, including Q and W, are expressed as (J/kg).

[ ( ) ( )] 0

2

2 1

1 1 1 2 2 2 1 = 

  

 (^) + 

  m + e + k − + e + k + Q + m ^ P − P W φ φ ρ ρ

g(z z ) (e e ) 21 (v v ) Q P P W 0 2

2 1 1 2 1 2 e 12 e (^221) + = 

 

 − + − + − + + ρ −ρ

3.2. MACROSCOPIC ENERGY BALANCE

o Using the parameter α that relates the two:

α=v 2 /ve^2

o The expression of conservation of the energy becomes:

o This equation shows the balance of the total energy in the whole system, both the terms of the mechanical energy and those of the heat (H,Q).

g (z z ) (H H ) 2 V 2 V Q W 0 1

22 2

12 1 − 2 + 1 − 2 + α − α + + =

3.2. MACROSCOPIC ENERGY BALANCE

3.3. MECHANICAL ENERGY BALANCE: BERNOUILLI’S EQUATION

o An open system in steady state conditions that only interchanges mechanical energy, with no heat transfer Q=0, is a common case.

o The balance takes into account the theorem of “ life forces ”. According to this theorem, in order to go from state 1 to state 2, the increase in kinetic energy (Figure 3.1) is the work carried out by external forces in the system minus the work carried out by internal forces.

o As the fluid is flowing, we will have to calculate the kinetic energy of the fluid that crosses the S section in a dθ differential time:

o Where m stands for the mass flow-rate [MT-1^ ]

int

2 1 E (^) k − Ek = TFext − T F

2 2

1 Ek = md θ vE

3.3. MECHANICAL ENERGY BALANCE: BERNOUILLI’S EQUATION

o Now let’s analyze the other term of the equation: T (^) Fext-T (^) Fint

o T (^) Fext is the work that external forces carry out in the system. In Figure 3.1, many external forces are shown: pressure, gravity, and friction.

o We will consider the entrance pressure (the one that helps the fluid enter into the system) and the exit pressure (the one that prevents the fluid from leaving). Examples of these would be a pumping station, which pushes the water into the system, and a water tap, which needs to overcome the atmospheric pressure in order to leave the pipe.

o Thus, the work of these forces will be:

o In Figure 3.1, the work would be positive in point 1, and negative in point 2.

PQdθ =Pmρd θ

3.3. MECHANICAL ENERGY BALANCE: BERNOUILLI’S EQUATION

o As the fluid is at different heights in the positions of reference, gravity is another variable that has an influence in the system. If z 2 < z 1 then the work of gravity will be positive. The work that gravity carries out is calculated as:

o Another external force that has an influence in the system is the friction caused by the walls of the pipe on the fluid.

o As we saw before, the fluid layer that is touching the surface of the pipe is still, and therefore, this friction does not carry out any work on the fluid. (This term does not take into account the friction among the different layers of the fluid, as they are internal forces).

TFgrav =mdθg(z 1 −z 2 )

3.3. MECHANICAL ENERGY BALANCE: BERNOUILLI’S EQUATION

o Imagine that the fluid is divided in cylindrical shape elements.

o Then, friction forces and the work that accounts for them need to be taken into account in all the surfaces of the elements.

o We will group all the works generated by this friction forces in the term Tμ.

o If we merge now all the terms that we developed, we obtain the most common equation to describe the mechanical energy balance, Bernouilli’s theorem.

o As we will see now, this expression is different whether the fluid is compressible or not.

+ θ − −^ μ

 

 α = θ ρ −ρ mdθ (v −v ) md P P md g(z z ) T 2

1 11 22 1 2

22 12

3.3. MECHANICAL ENERGY BALANCE: BERNOUILLI’S EQUATION

3.3.1. Incompressible fluids ρ 1 =ρ 2 =ρ T (^) Pi =

o If we remember the definition of incompressible fluid seen in Topic 1, we know that an increase of pressure does not entail a variation of volume and density for this type of fluids.

o Because of the latter, we can assume that the density of the fluid remains constant in any point of the pipe, and even if there are internal pressure forces, these do not carry out any work.

o We can rewrite the previous general equation then as:

Pg vg z Pg vg z h f

g z z F g

P P g

v v

md v v md P P md g z z T

• • = + + +

+ − − 

 

− = −

+ − − 

 

− =  (^) −

∑

2 2 22 1 1 12

(^1212) 22 12

(^1212) 22 12

( ) 2

( ) 1 2

1

( ) 2

1

21 ( ) ( )

ρ α ρ α

α ρ

θ α θ ρ ρ θ μ

3.3. MECHANICAL ENERGY BALANCE: BERNOUILLI’S EQUATION

3.3.2. Compressible fluids ρ 1 ≠ρ 2 T (^) Pi≠ 0

o In order to develop the law of mechanical energy conservation for incompressible fluids, the pressure that each differential element produces in the surrounding differential elements has to be taken into account.

o Thus, internal forces produce work in these cases, and we will have to calculate the amount of that work.

o In this case, the “life forces” law will be like this:

o In order to calculate Tp, it is convenient to define a fluid element of dS surface (Figure 3.3)

μ

μ E E T T T

E E T T T T

k k Fgrav P

k k Fgrav Pext P − = + −

2 1

(^21) int

3.3. MECHANICAL ENERGY BALANCE: BERNOUILLI’S EQUATION

3.3.2. Compressible fluids ρ 1 ≠ρ 2 T (^) Pi≠ 0

Figure 3.3. Definition of the volume element of dS surface.

1

1’

2

2’

dS

h (^) h’

u