Compressible Flow Examples - Michigan State University Examples.pdf · Fall 2006 1 ME 440 Aerospace...
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Transcript of Compressible Flow Examples - Michigan State University Examples.pdf · Fall 2006 1 ME 440 Aerospace...
Fall 2006
1
ME 440 Aerospace Engineering Fundamentals
Compressible Flow Examples
Example: Compressible Flow Nozzle Analysis Consider our simple turbojet example of our propulsion unit. We found that entering the nozzle we had T5 = 1492 K P5 = 674 kPa For an ambient pressure of 69.2 kPa and an aircraft velocity of 300 m/s, determine the specific thrust of the engine accounting for compressible flow. Solution: We begin by calculating our critical pressure for this converging nozzle. We have P* = 0.528P5 = (0.528)(674) = 356 kPa Since this is above the ambient pressure we will not have complete expansion in our nozzle. Then P6 = 356 kPa The temperature at the exhaust is the critical temperature
K 124314.1
21492)(
12
T TT 5*
6 =+
=+γ
==
The velocity at the exhaust is the sonic velocity s/m 707)1243)(1005)(14.1( Tc)1( V 6P6 =−=−γ= The specific thrust is then ft = V6-Vac = 707-300 = 407 Nt/(kg/s) When we ignored the affect of compressible flow we found a specific thrust of 897 Nt/(kg/s), which means that by not having complete expansion we loose over half of our thrust. Example: Required Area Ratio for Complete Expansion Determine the area ratio required in a converging-diverging nozzle to have complete expansion in the previous example. Solution: In this case we will follow our 4 step approach.
1. Assume full expansion, Pout = 69.2 kPa
ME 440 Aerospace Engineering Fundamentals Fall 2006
2
2. Calculate Tout from the isentropic relationship
K 779 673.969.2
(1492) P
PT T
4.1/)14.1(/)1(
in
outinout =
=
=−γ−γ
3. Calculate Vout from the 1st law
( ) m/s 1197 779)-922(1005)(14 = T-T2c = V outinPout = 4. Calculate area ratio by applying continuity
in
P
outout
outin1/2-1)-/(
t
out
T)1(c
VP
TP
12
A
A −γ
+γ=
γγ
1.90 )1429(
)14.1)(1005()1197)(2.69()779)(9.673(
14.12
A
A 1/2-1)-4.1.4/(1
t
out =−
+=
Example: Normal Shock Wave Air at 270 K, 50 kPa, and a Mach number of 2.4 undergoes a normal shock. Determine the velocity, temperature and pressure downstream of the shock wave. Solution: We begin by calculating M2, T2, and P2.
0.274 1)4.2()}14.1/()4.1(2{
)14.1/(2)4.2(
1M)}1/(2{
)1/(2M M 2
2
21
212
2 =−−
−+=
−−γγ−γ+
=
M2 = (0.274)(1/2) = 0.523
3.13 1)4.2(1-.41
1.4)(2)4.2(
21-.41
1)4.2(1).41(
1)-.412(
1M1-
2M
21-
1M1)(
1)-2(
TT
2222
21
212
12
1
2
=
−
++
=
−γ
γ
γ
++γ
γ=
T2 = (3.13)(270) = 846 K
55.614.11-.41
- 2.4)(14.1
1.4)(2
11-
- M1
2
P
P 221
1
2 =++
=+γ
γ+γγ
=
P2 = (6.55)(50) = 328 kPa Then s/m305)846)(287)(4.1()523.0(RTMV 222 ==γ=
ME 440 Aerospace Engineering Fundamentals Fall 2006
3
Example: Oblique Shock Wave Consider a supersonic flow with M1 = 1.8 P1 = 90 kPa T1 = 268 K that is turned to produce a shock wave. Determine the wall angle and resulting shock angle for a sonic downstream condition. Also determine the downstream temperature and pressure. Solution: Reading from Figure 1 for M1 = 1.8 and M2 = 1, we find θ = 18º β = 61º Then 1.574 )1(1.8)sin(6 )sin(M M 11N =°=β= 0.682)18-1(1.0)sin(6 )sin(M M 22N =°°=θ−β= Using the normal shock relations we find
1.3694 1)574.1(1-.41
1.4)(2)574.1(
21-.41
1)4.2(1).41(
1)-.412(
1M1-
2M
21-
1M1)(
1)-2(
TT
2222
21
212
12
1
2
=
−
++
=
−γ
γ
γ
++γ
γ=
T2 = (1.3694)(270) = 367 K
722.214.11-.41
- )574.1(14.1
1.4)(2
11-
- M1
2
P
P 221
1
2 =++
=+γ
γ+γγ
=
P2 = (2.722)(90) = 245 kPa