Filetype PDF | Posted on 20 Jan 2023 | 2 years ago
Authentication
digital resources
145x Filetype PDF File size 0.08 MB Source: www.me.iitb.ac.in
1/14 2/14
ME 209 Compressible Fluid Flow-I
Basic Thermodynamics The compressibility refers to the change of density
Introduction to Compressible Flow-I of the fluid
Density can change due to a change in pressure or
temperature of a fluid
In a Liquid, the density is a very weak function of
pressure and but it can change perceptibly with
temperature.
In gas, the density is a strong function of
temperature and pressure
3/14 4/14
Compressible Fluid Flow-II Compressible Fluid Flow-III
Applications of Compressible Flow
– Gas It is a vast and complex subject
Gas and Steam Turbines Under some cases, we can treat the subject purely based
Rocket Nozzles, on thermodynamic laws of mass and energy
I.C. Engine ports, However, some concepts of momentum conservation is
Combustion chambers needed at places, which we shall consciously minimise.
Re-entry vehicles
– Liquids
Hydraulic Penstocks
High pressure hydraulic circuits
In liquids normally, it is only the transient that calls for
compressible flow analysis
In gases both steady and transient flow may call for
compressible analysis
1
5/14 6/14
Conservation of Momentum-I Conservation of Momentum-II
Consider an arbitrary control volume as shown Let us consider the
δmV
through which mass crosses (flowing from ducts) same mass of fluid as i i
P (t)
shown in yellow CV
Fluid At time, t, the fluid fills
out δm V
the control volume and e e
MCV a portion of inlet duct
The same fluid at t+t
Fluid in P (t+∆t)
fills the control volume CV
The aim shall be to Convert Newton’s Second Law and a portion of exit duct.
for a control mass to a flow system
P (t+∆t)=P (t+∆t)+δm V
CM CV e e
We shall now look at two snapshots one at t and
other at t+t P (t)=P (t)+δmV
CM CV i i
7/14 8/14
Conservation of Momentum-III Conservation of Momentum-IV
Subtracting the above two equations, we get
dP
CV
⇒ +mV −mV =F +F
P (t+∆t)−P (t)= dt e e i i S B
CM CM
dP
P (t+∆t)−P (t)+δm V −δmV CV
CV CV e e i i Or =mV−mV +F +F
dt i i e e S B
Dividing both sides by t and then shrinking t to 0, At steady state
we get
0=mV−mV +F +F
i i e e S B
dP dP
CM CV
= +m V −mV
dt dt e e i i If we put the above equation in words, we can
write
Newton’s Ssecond law implies Rate of
Rate of Sum of all
dP momentum - momentum + forces = 0
CM =F=F +F
dt S B entering CV exiting CV
2
9/14 10/14
Conservation of Momentum-V Pressure Pulse Propagation-I
The equation derived above can be extended to a Pressure pulses propagate in a compressible fluid
steadily moving control volume as follows with a characteristic speed.
This is what we commonly call as speed of sound
0=m V −m V +F +F
i−Rel i−Rel e−Rel e−Rel S B This speed is a property of the medium
In the above equation all quantities refer to Consider a cylinder piston filled with a
quantities with respect to relative frame of compressible fluid
reference. Let the piston be moved instantly
Its application will make it clear in the following This will set a pressure wave moving at a speed c
derivation
V=∆V,p+∆p c
T+∆T,ρ+∆ρ V=0, p,T,ρ
Undisturbed fluid
11/14 12/14
Pressure Pulse Propagation-II Pressure Pulse Propagation-III
To derive a relation between the speed of Momentum balance
propagation and system properties, let an observer mc
m(c−∆V) Positive direction
ride on the wave. In this moving coordinate the p+∆p p
fluid will be in steady state
For the moving coordinate the properties are as
Momentum balance⇒mc−m(c−∆V)+(−∆pA)=0 No friction
shown in out force
V=c−∆V V=c, ∆pA ∆pA ∆p
⇒m∆V−∆pA=0 ⇒∆V= = =
ρ+∆ρ ρ m ρAc ρc
⇒∆V=∆p
Mass balance ⇒(ρ+∆ρ)A(c−∆V)=ρAc ρC 2
⇒ρc+∆ρc−ρ∆V−∆ρ∆V)=ρc Eqs. (1) and (2) c∆ρ = ∆p c2 = ∆p = dp
c∆ρ ρ ρc ∆ρ dρ
Second order ∴∆V= 1
ρ
3
13/14 14/14
Pressure Pulse Propagation-IV Pressure Pulse Propagation-V
For ideal gas dp = ∂p dρ+ ∂p ds Assuming the process
∂ρ ∂s to be adiabatic →ds = 0 For Solids and Liquids
s ρ
∴dp = ∂p =c2 Newton had assumed the
dρ ∂ρ Bulk Modulus E = dp ⇒dp=EV =c2 ⇒c= EV 4
s process to be Isothermal v dρ/ρ dρ ρ ρ
s = constan t ⇒ p = constan t ⇒ln(p)−γln(ρ)=constant
ργ
o 9 2 3
For Water 20 C, E = 2.24 x 10 N/m , ρ = 998 kg/m
dp dρ dp p v
⇒ −γ =0 ⇒ =γ ∴c2=γp=γRT ⇒c= γRT 3
p ρ dρ ρ ρ 2.2x109
∴c= ≈1500 m/s
998
At 300 K c = 1.4X287X300 =347 m/s 9 2 3
o
For Steel 20 C, E = 200 x 10 N/m , ρ = 7830 kg/m
v
Note that c is independent of p and depends only on T 200x109
∴c= ≈5050 m/s
7830
4
no reviews yet
Please Login to review.
