modeling and analysis of dynamic systems - lecture 7

Modeling and Analysis of Dynamic Systems

Lecture 7: Thermodynamics part II - Application to Fluiddynamics

Dr. Guillaume Ducard

Fall 2018

Institute for Dynamic Systems and Control

ETH Zurich, Switzerland

G. Ducard c© 1 / 26

Outline

1 Recall on Thermodynamics PrinciplesThermodynamics Principles 1 & 2Entropy Variation

2 Fluiddynamic SystemsValves (for incompressible and compressible fluids)


Recall on Thermodynamics PrinciplesFluiddynamic Systems

Thermodynamics Principles 1 & 2Entropy Variation

Outline






First principle of thermodynamics

is about the “conservation of energy”. All system may bedescribed by its total energy. Its energy variation is given by:

For closed systems:

The variation of the total energy of a closed system is given by

dU = δW + δQ

dU : internal energy variationδW : mechanical energy: work of pressure forces at the systemsurfaces + work of gravity forces in the volume of the system(usually negligible).δQ: thermal energy exchanged with the surrounding

For a closed system (ex: cylinder): the work of the pressure forces isgiven by: δW = −PextdV




First principle of thermodynamics

In many industrial applications, a continuous flow of fluids goes throughthe (thermodynamic) system. This is no longer a closed system, we talkabout “open systems” (valve, turbine, compressor, etc.)

For open systems

We rather use an other state function: Enthalpy defined as

H = U + PV ,

the term PV takes into account the work (energy) of fluidtransport.

The total energy(Etot) variation of an open system is :

dEtot = dH + dK = δτ + δQ

where dK is the variation of the kinetic energyand δτ is the “useful” work (production of thrust, or torque, etc.)




Second Principle

Second principle: intuition

The second principle of thermodynamics is about how (sense, direction)the energy exchanges (or transformation) take place during a process, theassociated variable is called Entropy.

Example of use: to predict the direction of

heat flow (in between 2 medium at different temperatures),

particles flow (in between systems with different pressures).




Second Principle

Rudolf Clausius (1865)original definition; how to use it in practiceEntropy stays the same for reversible processes but increases for

irreversible processes.




Second Principle of Thermodynamics

Second principle: statement

All system is characterized by a state function S called Entropy.If in an isolated system, irreversible transformations (processes)take place,→ the Entropy of the system increases and is maximal atequilibrium.




Outline






Entropy variation w.r.t. Internal energy variation at constant volume(∂S

∂U

)

V

=1

T

Entropy variation w.r.t. volume variation at constant Internal energy(∂S

∂V

)

U

=P

T

Entropy variation w.r.t. variation in U and V

dS =(∂S∂U

)V

dU +(∂S∂V

)U

dV

=(1T

)dU +

(PT

)dV

= mCvdTT

+mR dVV

(ideal gas)

= mCvdTT

−mR dPP

(ideal gas)




A process is said to be “isentropic” when dS = 0. It is the case for

an adiabatic (no heat transfer)

and reversible process.

Isentropic expansion relations for a perfect gas

PV γ = cstγ =

Cp,m

Cv,m=

Cp

Cv, dimensionless

γ = 5/3 for mono-atomic gas, γ = 7/5 for di-atomic gas

(P, V ): P V γ = cst

(T, V ): T V γ−1 = cst

(P, T ): P 1−γ T γ = cst

Isentropic processes will be considered later in the lecture, whenmodeling valves and gas turbines.



Valves (for incompressible and compressible fluids)

Outline






Valves

Figure: Example of a gate valve (top), butterfly valve (bottom)(http://www.ctgclean.com)




Valves

From a modeling point of view: this system can be viewed as theinterface between 2 reservoirs (with associated level variables):

fluid upstream (upstream pressure pin)

fluid downstream (downsteam pressure pout)

This is due to the pressure difference that the fluid flows through theorifice (valve). Two modeling approaches considered:

1 for incompressible fluids

2 for compressible fluidsG. Ducard c© 14 / 26



Valve and Incompressible Fluids

Assumptions

Flow frictions modeled through a correcting factor: determinedexperimentally cd: discharge coefficient

Fluid assumed to be incompressible (constant density). This is thecase for liquids and fluids at low Mach numbers.

Mach number M

M =u

c[−]

where

u is the local flow velocity [m/s],

c is speed of sound in the medium [m/s].Primary use: to determine whether a fluid flow can be treated as incompressible. If M < 0.2 (up to 0.3) and theflow is quasi-steady and isothermal, → incompressible flow can be assumed.





Modeling objective

Express the fluid mass flow∗m through the valve, with the following

terms:

cd: discharge coefficient (takes into account flow restrictions,flow friction and other losses);

A: open area of the valve;

ρ: density of the fluid (constant: incompressibility assumption);

pin: pressure upstream of the valve;

pout: pressure downstream of the valve.





If the fluid is incompressible, then the valve can be modeled usingBernoulli’s law (conservation of energy), leading to

∗m (t) = cd A

√2ρ

√pin(t)− pout(t)

Proof shown in class.




Valve and Compressible Fluids

If the fluid is compressible, the valve can be modeled using theconcept of “Isenthalpic throttle”.

Why do we call it “Isenthalpic throttle”?

Isenthalpic process:

A fluid circulates in a tube with

1 no moving wall (no work from pressure forces),

2 no heat exchange

→ dH = 0, no enthalpy variation in the system.




Valve and Compressible Fluids

The behavior of the flow may be separated into 2 distinct parts :1) Valve upstream and 2) valve downstream.

pin(t)pout(t)

∗min (t), ϑin(t), pin(t)

∗mout (t), ϑout(t), pout(t)

ϑin(t) ϑout(t)




replacements

pin(t)pout(t)



ϑin(t) ϑout(t)

Valve upstream: up to the narrowest point

Fluid accelerating, laminar flow,

Pressure decrease, temperature decrease,

potential energy stored in the flow (level variable = pressure)is converted isentropically (without losses) into kinetic energy.




replacements

pin(t)pout(t)



ϑin(t) ϑout(t)

Valve downstream: from narrowest point onward

Fluid decelerating,

Turbulent flow

Kinetic energy is dissipated in thermal energy (no pressurerecuperation takes place).




replacements

pin(t)pout(t)



ϑin(t) ϑout(t)

Consequences of assumptions made:

1 The pressure in the narrowest part of the valve is almost equalto downstream pressure.

2 The temperature of the flow before and after the valve isapproximately the same (almost isothermal process).




If we deal with gases, in conditions of perfect gases:

∗m (t) = cdA(t)

pin(t)√Rϑin(t)

Ψ(pin(t), pout(t))

where Ψ(.) is

Ψ(pin(t), pout(t)) =

√κ(

2

κ+1

) κ+1

κ−1

for pout < pcr

(pout

pin

)1/κ

√2κκ−1

[1−

(pout

pin

)κ−1

κ

]for pout ≥ pcr

and where

pcr =

[2

κ+ 1

] κκ−1

pin

Derivations shown in class.




Approximation (air and many other gases OK):

Ψ(pin(t), pout(t)) =

1√2

for pout < 0.5pin√

2poutpin

[1− pout

pin

]for pout ≥ 0.5pin




Approximation (unrealistically large threshold: πtr =poutpin

= 0.9)

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

0.8

0.7

0.6

0.5

0.4

0.3

0.2

0.1

0

Ψ exact (solid) and approximated (dashed)(κ = 1.4), laminar part (dash-dot)Ψ(Π

)(-)

Π (-)




A laminar flow condition can be assumed for very small pressureratios

Πtr :=poutpin

< 1

If larger pressure ratios occur, then use a smooth approximation

Ψ̃(Π) = a · (Π− 1)3 + b · (Π− 1)

with

a =Ψ′

tr · (Πtr − 1)−Ψtr

2 · (Πtr − 1)3, b = Ψ′

tr − 3 · a · (Πtr − 1)2

is used (Ψtr is the value of Ψ and Ψ′tr the value of the gradient of

Ψ at the threshold Πtr).


modeling and analysis of dynamic systems - lecture 7

Documents