2.4 Steady One-Dimensional Compressible Flow
2.4.4 Diffusers and Nozzles
Diffusers and nozzles are commonly utilized in jet engines, rockets, and spacecrafts.
A diffuser is a device that increases the pressure of a fluid by slowing it down, while a nozzle is a device that increases the velocity of a fluid at the expense of pressure. That is, diffusers and nozzles perform opposite tasks. Diffusers and nozzles involve no work (W( "0) and negligible changes in potential energy (ΔPE"0). Moreover, the rate of heat transfer between the fluid flowing through a diffuser or a nozzle and the surroundings is usually very small (Q( "0). This is due to the very short time air (or gas) spends in either duct (few or fraction of milliseconds) which is insufficient for a significant heat transfer to take place. However, fluid passing through diffusers and nozzles experiences large changes in velocity. Therefore, the kinetic energy changes must be accounted for (ΔKE6¼0). The shape of both diffuser and nozzle may be convergent or divergent depending on the velocity of flowing fluid. Rockets and military high supersonic aircrafts normally haveconvergent–divergent or CD nozzles. In a CD rocket nozzle, the hot exhaust leaves the combustion chamber and converges down to the minimum area, orthroat,of the nozzle. The throat size is chosen tochokethe flow andset the mass flow ratethrough the system. The flow in the throat is sonic which means the Mach number is equal to one in the throat.
Downstream of the throat, the geometry diverges, and the flow is isentropically expanded to a supersonic Mach number that depends on thearea ratioof the exit to the throat. The expansion of a supersonic flow causes the static pressure and temper- ature to decrease from the throat to the exit, so the amount of the expansion also determines the exit pressure and temperature. The exit temperature determines the exitspeed of sound, which determines the exit velocity. The exit velocity, pressure, and mass flow through the nozzledeterminethe amount of thrust produced by the nozzle.
2.4.4.1 Variation of Fluid Velocity with Flow Area
We begin with theconservation of mass equation:
_
m ¼ρVA¼constant
wherem_:is the mass flow rate,ρis the gasdensity,Vis the gas velocity, andAis the cross-sectional flow area. If we differentiate this equation, we obtain
VAdρþρAdVþρVdA¼0 Divide by (ρVA) to get
dρ ρ þdV
V þdA A ¼0 Now we use theconservation of momentum equation:
ρVdV¼ %dP and anisentropic flow relation:Tds¼dh%vdP
dP P ¼γdρ
ρ
whereγis theratio of specific heats. Rewrite the above equation to obtain dP¼γP
ρdρ and use theequation of state(Pρ¼RT) to get
dP¼γRTdρ
Since (γRT) is the square of thespeed of sound(a), then dP¼a2dρ
Combining this equation for the change in pressure with the momentum equation, we obtain
ρVdV¼ % a2dρ V
a2dV¼ %dρ ρ
using the definition of theMach numberM¼V=a, then
%M2dV V ¼dρ
ρ ð2:53Þ
Now we substitute this value of (dρ/ρ) into the mass flow equation to get
%M2dV V þdV
V þdA A ¼0 1%M2
$ %dV
V ¼ %dA
A ð2:54Þ
Equation (2.59) tells us how the velocity (V) changes when the area (A) changes and the results depend on the Mach number (M) of the flow.
If the flow issubsonicthen (M<1:0)—the term multiplying the velocity change is positive [ 1$ %M2%
>0 ]—then an increase in the area (dA>0 ) produces a decrease in the velocity (dV<0), which is the case of a diffuser. On the contrary a decrease in the area produces an increase in velocity, which is the case of a nozzle.
For a supersonic flow (M>1:0 ), the term multiplying velocity change is negative [ 1$ %M2%
<0 ]. Then an increase in the area (dA>0 ) produces an increase in the velocity (dV>0) or a nozzle. The decrease in the area leads to a decrease in velocity or a diffuser.
Table2.5summarizes this behavior.
Figure 2.12 illustrates the geometry of diffusers and nozzles in subsonic and supersonic speeds.
For the case of CD nozzle, if the flow in the throat is subsonic, the flow downstream of the throat will decelerate and stay subsonic. So if the converging section is too large and does not choke the flow in the throat, the exit velocity is very slow and does not produce much thrust. On the other hand, if the converging section is small enough so that the flow chokes in the throat, then a slight increase in area causes the flow to go supersonic. This is exactly the opposite of what happens subsonically.
Table 2.5 Variation of duct area with inlet Mach number
Accelerated flow (nozzle) Decelerated flow (diffuser)
Constant velocity
dV>0 dV<0
M<1:0 dA<0 dA>0 dA¼0
M>1:0 dA>0 dA<0 dA¼0
Fig. 2.12 Variation of flow properties in subsonic and supersonic nozzles and diffusers
Example 2.7 Air at 5!C and 80 kPa enters the diffuser of a jet engine steadily with a velocity of 200 m/s. The inlet area of the diffuser is 0.4 m2. The air leaves the diffuser with a velocity that is very small compared with the inlet velocity.
Determine (a) the mass flow rate of the air and (b) the temperature of the air leaving the diffuser.
Solution
We take thediffuseras the system (Fig.2.13). This is acontrol volumesince mass crosses the system boundary during the process. We observe that there is only one inlet and one exit and thusm_1¼m_2¼m_:.
(a) To determine the mass flow rate, we need to find the density of the air first.
This is determined from the ideal gas relation at the inlet conditions:
ρ1¼ P1
RT1¼ 80,103
287,ð273þ5Þ¼1:0027 kg=m3 _
m ¼ρ1V1A1¼1:0027,200,0:4¼79:8 kg=s (b) From the energy equation
dQ d t þ V21
2 þgz1þh1
" #
_ m ¼dWs
dt þ V22
2 þgz2þh2
" #
_ m:
With small exit velocity (V2"0), negligible potential energy variation as well as heat and work exchange (z1 "z2, dQ=dt¼dWs=dt¼0), then, energy equation is reduced to
h2¼h1þV21 2 Fig. 2.13 Diffuser and
control volume
T2¼T1þ V21
2Cp¼278þ 2002
2,1005¼297:9 K
Example 2.8 Gas flows through a converging–diverging nozzle. Points G and H lie between the inlet and outlet of the nozzle. At a point “G,” the cross-sectional area is 500 cm2and the Mach number was measured to be 0.4. At point “H” in the nozzle, the cross-sectional area is 400 cm2. Find the Mach number at point H. Assume that the flow is isentropic and the gas-specific heat ratio is 1:3.
Solution
To obtain the Mach number at point G, apply Eq. (2.46) to find the ratio between the area (AG) to the critical one (A*)
AG
A* ¼ 1 MG
2
γþ1 1þγ%1 2 MG2
" #
) *2ðγγ%1þ1Þ
AG
A* ¼ 1 0:4
" # 2 1:3þ1
" #
1þ1:3%1 2 ð0:4Þ2
" #
) *2 0:3ð2:3Þ
¼1:6023 At point H, the area ratio is evaluated from the relation:
AH
A* ¼AH
AG
AG
A* ¼400
500,1:6023¼1:2818 Again from Eq. (2.46)
AH
A* ¼ 1 MH
2
γþ1 1þγ%1 2 MH2
" #
) *2γγþ1
%1 ð Þ
Rearranging to solve for the Mach numberMH,
MH
AH
A*
" #2ðγþγ%11Þ
%γ%1
γþ1MH2¼ 2 γþ1 MH
AH
A*
" #2ðγγþ%11Þ
%γ%1
γþ1MH2¼ 2 γþ1 1:0669MH0:2609%0:1304MH2¼0:8696 Solving the above equation bytrial and error, we get either
MH¼0:5374 or MH¼1:612
Both solutions are possible, the first is still a subsonic Mach number which may be located in the convergent section, while the second one is supersonic which may be located in the divergent section if the speed at throat is sonic:Mthroat¼1:0.