Isentropic nozzle flow describes the movement of a gas or fluid through a narrowing opening without an increase or decrease in entropy.
Whenever a gas is forced through a tube, the gaseous molecules are deflected by the tube's walls. If the speed of the gas is much less than the speed of sound, the gas density will remain constant and the velocity of the flow will increase. However, as the speed of the flow approaches the speed of sound, compressibility effects on the gas are to be considered. The density of the gas becomes position dependent. While considering flow through a tube, if the flow is very gradually compressed (i.e. area decreases) and then gradually expanded (i.e. area increases), the flow conditions are restored (i.e. return to its initial position). So, such a process is a reversible process. According to the Second Law of Thermodynamics, whenever there is a reversible and adiabatic flow, constant value of entropy is maintained. Engineers classify this type of flow as an isentropic flow of fluids. Isentropic is the combination of the Greek word "iso" (which means - same) and entropy. When the change in flow variables is small and gradual, isentropic flows occur. The generation of sound waves is an isentropic process. A supersonic flow that is turned while there is an increase in flow area is also isentropic. Since there is an increase in area, therefore we call this an isentropic expansion. If a supersonic flow is turned abruptly and the flow area decreases, the flow is irreversible due to the generation of shock waves. The isentropic relations are no longer valid and the flow is governed by the oblique or normal shock relations.
Below are nine equations commonly used when evaluating isentropic flow conditions.[1] These assume the gas is calorically perfect; i.e. the ratio of specific heats is a constant across the temperature range. In typical cases the actual variation is only slight.
M=
v | |
cs |
cs=\sqrt
\gammap | |
\rho |
=\sqrt{\gammaRT}
p | |
\rho\gamma |
=
pt | ||||||
|
p | |
pt |
=\left(
\rho | |
\rhot |
\right)\gamma =\left(
T | |
Tt |
| ||||
\right) |
q=
1 | |
2 |
\rhov2=
1 | |
2 |
\gammapM2
p | |
pt |
=\left(1+
\gamma-1 | |
2 |
M2\right)
| ||||
T | |
Tt |
=\left(1+
\gamma-1 | |
2 |
M2\right)-1
\rho | |
\rhot |
=\left(1+
\gamma-1 | |
2 |
M2\right)
| ||||
A | |
A* |
=
1 | \left( | |
M |
\gamma-1 | |
2 |
\right)
| |||||
\left(1+ |
\gamma-1 | |
2 |
M2\right)
| ||||
t
pt
M
v
cs
\gamma
p
\rho
T
A
A*
In fluid dynamics, a stagnation point is a point in a flow field where the local velocity of the fluid is zero. The isentropic stagnation state is the state a flowing fluid would attain if it underwent a reversible adiabatic deceleration to zero velocity. There are both actual and the isentropic stagnation states for a typical gas or vapor. Sometimes it is advantageous to make a distinction between the actual and the isentropic stagnation states. The actual stagnation state is the state achieved after an actual deceleration to zero velocity (as at the nose of a body placed in a fluid stream), and there may be irreversibility associated with the deceleration process. Therefore, the term "stagnation property" is sometimes reserved for the properties associated with the actual state, and the term total property is used for the isentropic stagnation state. The enthalpy is the same for both the actual and isentropic stagnation states (assuming that the actual process is adiabatic). Therefore, for an ideal gas, the actual stagnation temperature is the same as the isentropic stagnation temperature. However, the actual stagnation pressure may be less than the isentropic stagnation pressure. For this reason the term "total pressure" (meaning isentropic stagnation pressure) has particular meaning compared to the actual stagnation pressure.
The isentropic efficiency is
h1-h2a | |
h1-h2 |
There are numerous applications where a steady, uniform, isentropic flow is a good approximation to the flow in conduits. These include the flow through a jet engine, through the nozzle of a rocket, from a broken gas line, and past the blades of a turbine.
Mach number = M Velocity = V Universal gas constant = R Pressure = p Specific heat ratio = k Temperature = T * = Sonic conditions Density =
\rho
Mm
Energy equation for the steady flow:
qnet+h+
\vec{V | |
2}{2} |
=wnet+ho+
\vec{V | |
o |
2}{2}
To model such situations, consider the control volume in the changing area of the conduit of Fig. The continuity equation between two sections an infinitesimal distance dx apart is
\rhoAV=(\rho+d\rho)(A+dA)(V+dV)
dV | |
V |
+
dA | |
A |
+
d\rho | |
\rho |
=0
V2 | |
2 |
+
k | ⋅ | |
k-1 |
p | |
\rho |
=
\left(V+dV\right)2 | |
2 |
+
k | ⋅ | |
k-1 |
p+dp | |
\rho+d\rho |
This simplifies to, neglecting higher-order terms,:
VdV+
k | ⋅ | |
k-1 |
\rhodp-pd\rho | |
\rho2 |
=0
VdV+
k ⋅ p | |
\rho2 |
⋅ d\rho=0
dV | |
V |
⋅ \left(
\rhoV2 | |
kp |
-1\right)=
dA | |
A |
dV | |
V |
⋅ \left(M2-1\right)=
dA | |
A |
This equation applies to a steady, uniform, isentropic flow.There are several observations that can be made from an analysis of Eq. (9.26).They are:
A nozzle for a supersonic flow must increase in area in the flow direction, and a diffuser must decrease in area, opposite to a nozzle and diffuser for a subsonic flow. So, for a supersonic flow to develop from a reservoir where the velocity is zero, the subsonic flow must first accelerate through a converging area to a throat, followed by continued acceleration through an enlarging area.
The nozzles on a rocket designed to place satellites in orbit are constructed using such converging-diverging geometry. The energy and continuity equations can take on particularly helpful forms for the steady, uniform, isentropic flow through the nozzle. Apply the energy equation with Q_ W_S 0 between the reservoir and some location in the nozzle to obtain
cp ⋅ To=
V2 | |
2 |
+cp ⋅ T
Any quantity with a zero subscript refers to a stagnation point where the velocity is zero, such as in the reservoir. Using several thermodynamic relations equations can be put in the forms:
To | |
T |
=1+
k-1 | |
2 |
M2
po | |
p |
=\left(1+
k-1 | |
2 |
M2
| ||||
\right) |
\right)
\rhoo | |
\rho |
=\left(1+
k-1 | |
2 |
M2
| ||||
\right) |
\right)
If the above equations are applied at the throat (the critical area signified by anAsterisk (*) superscript, where M =1), the energy equation takes the forms
T* | |
To |
=
2 | |
k+1 |
p* | |
po |
=\left(
2 | |
k+1 |
| ||||
\right) |
\right)
\rho* | |
\rhoo |
=\left(
2 | |
k+1 |
| ||||
\right) |
\right)
The critical area is often referenced even though a throat does not exist. For air with k =1.4, the equations above provide T* = 0.833333·To p* = 0.528282·po ρ* = 0.633938·ρo
The mass flux through the nozzle is of interest and is given by:
m |
=\rhoAV=
p | |
RT |
⋅ A ⋅ M\sqrt{kRspecificT}=p\sqrt{kMm\overRT}AM
With the use of Eq. (9.28), the mass flux, after applying some algebra, can beExpressed as
m |
=poMA\sqrt{
kMm | |
RTo |
m |
=poA*\sqrt{
kMm | |
RTo |
A | |
A* |
=
1 | \left( | |
M |
k+1 | |
2+(k-1) ⋅ M2 |
| ||||
\right) |
Consider a converging nozzle connecting a reservoir with a receiver. If the reservoir pressure is held constant and the receiver pressure reduced, the Mach number at the exit of the nozzle will increase until Me=1 is reached, indicated by the left curve in figure 2. After Me =1 is reached at the nozzle exit for
pr=0.5283p0
pr
Me=1
p0
It is interesting that the exit pressure
pe
pr
pe
pr
A