5.8 Prandtl-Meyer Expansion WavesNiklas Andersson
Chalmers University of Technology
Department of Mechanical Engineering
Division of Fluid Dynamics
Figure 5.22: Mach wave flow turning
A single Mach wave has a insignificant effect on the flow passing it but an expansion region constitutes an infinite number of Mach waves and the integrated effect is significant. The net turning of the flow by a single Mach wave is depicted schematically in Fig. Figure 5.22
d θ = M 2 − 1 d V V d\theta=\sqrt{M^2-1}\frac{dV}{V} d θ = M 2 − 1 V d V Since Eqn. (5.1)
To get the integrated effect of all Mach waves in the expansion region, we integrate Eqn. (5.1)
∫ θ 1 θ 2 d θ = ∫ M 1 M 2 M 2 − 1 d V V \int_{\theta_1}^{\theta_2}d\theta=\int_{M_1}^{M_2}\sqrt{M^2-1}\frac{dV}{V} ∫ θ 1 θ 2 d θ = ∫ M 1 M 2 M 2 − 1 V d V To be able to do the integration, we need to rewrite it
V = M a ⇒ ln V = ln M + ln a V=Ma \Rightarrow \ln V=\ln M + \ln a V = M a ⇒ ln V = ln M + ln a Differentiate to get
d V V = d M M + d a a \frac{dV}{V}=\frac{dM}{M}+\frac{da}{a} V d V = M d M + a d a Each Mach wave is isentropic and thus the expansion is an isentropic process, which means that we can use the adiabatic energy equation
T o T = 1 + γ − 1 2 M 2 \frac{T_o}{T}=1+\frac{\gamma-1}{2}M^2 T T o = 1 + 2 γ − 1 M 2 For a calorically perfect gas a = γ R T a=\sqrt{\gamma RT} a = γ RT a o = γ R T o a_o=\sqrt{\gamma RT_o} a o = γ R T o
T o T = ( a o a ) 2 ⇒ \frac{T_o}{T}=\left(\frac{a_o}{a}\right)^2\Rightarrow T T o = ( a a o ) 2 ⇒ ( a o a ) 2 = 1 + γ − 1 2 M 2 \left(\frac{a_o}{a}\right)^2=1+\frac{\gamma-1}{2}M^2 ( a a o ) 2 = 1 + 2 γ − 1 M 2 Solve for a a a
a = a o ( 1 + γ − 1 2 M 2 ) − 1 / 2 a=a_o \left(1+\frac{\gamma-1}{2}M^2\right)^{-1/2} a = a o ( 1 + 2 γ − 1 M 2 ) − 1/2 Differentiate Eqn (5.8)
d a = a o ( − 1 2 ) ( γ − 1 2 ) 2 M ( 1 + γ − 1 2 M 2 ) − 3 / 2 d M da=a_o\left(-\frac{1}{2}\right)\left(\frac{\gamma-1}{2}\right)2M\left(1+\frac{\gamma-1}{2}M^2\right)^{-3/2}dM d a = a o ( − 2 1 ) ( 2 γ − 1 ) 2 M ( 1 + 2 γ − 1 M 2 ) − 3/2 d M Eqn. (5.8) (5.9)
d a a = − ( γ − 1 2 ) ( 1 + γ − 1 2 M 2 ) − 1 M d M \frac{da}{a}=-\left(\frac{\gamma-1}{2}\right)\left(1+\frac{\gamma-1}{2}M^2\right)^{-1}MdM a d a = − ( 2 γ − 1 ) ( 1 + 2 γ − 1 M 2 ) − 1 M d M From Eqn. (5.4)
d V V = d M M + d a a \frac{dV}{V}=\frac{dM}{M}+\frac{da}{a} V d V = M d M + a d a With d a / a da/a d a / a (5.10)
d V V = d M M − ( γ − 1 2 ) ( 1 + γ − 1 2 M 2 ) − 1 M d M \frac{dV}{V}=\frac{dM}{M}-\left(\frac{\gamma-1}{2}\right)\left(1+\frac{\gamma-1}{2}M^2\right)^{-1}MdM V d V = M d M − ( 2 γ − 1 ) ( 1 + 2 γ − 1 M 2 ) − 1 M d M d V V = d M [ ( 1 + γ − 1 2 M 2 ) − ( γ − 1 2 ) M 2 ( 1 + γ − 1 2 M 2 ) M ] = ( 1 + γ − 1 2 M 2 ) − 1 d M M \frac{dV}{V}=dM\left[\frac{\left(1+\dfrac{\gamma-1}{2}M^2\right)-\left(\dfrac{\gamma-1}{2}\right)M^2}{\left(1+\dfrac{\gamma-1}{2}M^2\right)M}\right]=\left(1+\dfrac{\gamma-1}{2}M^2\right)^{-1}\frac{dM}{M} V d V = d M ⎣ ⎡ ( 1 + 2 γ − 1 M 2 ) M ( 1 + 2 γ − 1 M 2 ) − ( 2 γ − 1 ) M 2 ⎦ ⎤ = ( 1 + 2 γ − 1 M 2 ) − 1 M d M Now, insert d V / V dV/V d V / V (5.2)
∫ θ 1 θ 2 d θ = ∫ M 1 M 2 M 2 − 1 ( 1 + γ − 1 2 M 2 ) − 1 d M M \int_{\theta_1}^{\theta_2}d\theta=\int_{M_1}^{M_2}\sqrt{M^2-1}\left(1+\dfrac{\gamma-1}{2}M^2\right)^{-1}\frac{dM}{M} ∫ θ 1 θ 2 d θ = ∫ M 1 M 2 M 2 − 1 ( 1 + 2 γ − 1 M 2 ) − 1 M d M The integral on the right hand side of Eqn. (5.14) ν \nu ν M M M
ν ( M ) = γ + 1 γ − 1 tan − 1 γ − 1 γ + 1 ( M 2 − 1 ) − tan − 1 M 2 − 1 \nu(M)=\sqrt{\frac{\gamma+1}{\gamma-1}}\tan^{-1}\sqrt{\frac{\gamma-1}{\gamma+1}(M^2-1)}-\tan^{-1}\sqrt{M^2-1} ν ( M ) = γ − 1 γ + 1 tan − 1 γ + 1 γ − 1 ( M 2 − 1 ) − tan − 1 M 2 − 1 and thus the net turning of the flow can be calculated as
θ 2 − θ 1 = ν ( M 2 ) − ν ( M 1 ) \theta_2-\theta_1=\nu(M_2)-\nu(M_1) θ 2 − θ 1 = ν ( M 2 ) − ν ( M 1 ) Solving Problems using the Prandtl-Meyer Function ¶ A typical problem is one where we know the net flow turning and the upstream flow conditions and want to calculate the flow conditions downstream of the expansion region. An example of such a problem is given in Fig. Figure 5.23
A problem of that type can be solved as follows:
Calculate ν M 1 \nu{M_1} ν M 1 (5.15)
Calculate ν ( M 2 ) \nu(M_2) ν ( M 2 ) ν ( M 2 ) = θ 2 − θ 1 + ν ( M 1 ) \nu(M_2)=\theta_2-\theta_1+\nu(M_1) ν ( M 2 ) = θ 2 − θ 1 + ν ( M 1 )
Calculate M 2 M_2 M 2 ν M 2 \nu{M_2} ν M 2 (5.15)
Figure 5.23: Expansion corner with known net flow turning
The aim is to derive relations of temperature, pressure and density over an expansion wave. Total temperature upstream and downstream of the expansion wave is calculated as
T o 1 T 1 = 1 + γ − 1 2 M 1 2 \frac{T_{o_1}}{T_1}=1+\frac{\gamma-1}{2}M_1^2 T 1 T o 1 = 1 + 2 γ − 1 M 1 2 T o 2 T 2 = 1 + γ − 1 2 M 2 2 \frac{T_{o_2}}{T_2}=1+\frac{\gamma-1}{2}M_2^2 T 2 T o 2 = 1 + 2 γ − 1 M 2 2 The temperature ratio over the expansion wave may now be calculated as
T 2 T 1 = T 2 T o 2 T o 1 T 1 = 1 + γ − 1 2 M 1 2 1 + γ − 1 2 M 2 2 = 2 + ( γ − 1 ) M 1 2 2 + ( γ − 1 ) M 2 2 \frac{T_2}{T_1}=\frac{T_2}{T_{o_2}}\frac{T_{o_1}}{T_1}=\frac{1+\dfrac{\gamma-1}{2}M_1^2}{1+\dfrac{\gamma-1}{2}M_2^2}=\frac{2+(\gamma-1)M_1^2}{2+(\gamma-1)M_2^2} T 1 T 2 = T o 2 T 2 T 1 T o 1 = 1 + 2 γ − 1 M 2 2 1 + 2 γ − 1 M 1 2 = 2 + ( γ − 1 ) M 2 2 2 + ( γ − 1 ) M 1 2 The expansion is isentropic and thus total temperature and both total pressure are unaffected by the expansion. Therefore, T o 1 = T o 2 T_{o_1}=T_{o_2} T o 1 = T o 2
T 2 T 1 = 2 + ( γ − 1 ) M 1 2 2 + ( γ − 1 ) M 2 2 \frac{T_2}{T_1}=\frac{2+(\gamma-1)M_1^2}{2+(\gamma-1)M_2^2} T 1 T 2 = 2 + ( γ − 1 ) M 2 2 2 + ( γ − 1 ) M 1 2 The pressure and density ratios can be obtained from Eqn. (5.20)
p 2 p 1 = [ 2 + ( γ − 1 ) M 1 2 2 + ( γ − 1 ) M 2 2 ] γ / ( γ − 1 ) \frac{p_2}{p_1}=\left[\frac{2+(\gamma-1)M_1^2}{2+(\gamma-1)M_2^2}\right]^{\gamma/(\gamma-1)} p 1 p 2 = [ 2 + ( γ − 1 ) M 2 2 2 + ( γ − 1 ) M 1 2 ] γ / ( γ − 1 ) ρ 2 ρ 1 = [ 2 + ( γ − 1 ) M 1 2 2 + ( γ − 1 ) M 2 2 ] 1 / ( γ − 1 ) \frac{\rho_2}{\rho_1}=\left[\frac{2+(\gamma-1)M_1^2}{2+(\gamma-1)M_2^2}\right]^{1/(\gamma-1)} ρ 1 ρ 2 = [ 2 + ( γ − 1 ) M 2 2 2 + ( γ − 1 ) M 1 2 ] 1/ ( γ − 1 ) Figure 5.24: Asymptotic behavior of the Prandlt-Meyer function