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The Procedure for Calculating The Maximum Deflection Point

$\;$

The maximum is obtained when . When the right terms defined in (13.20)-(13.21), (13.29), and (13.30) are substituted into this equation and utilizing the trigonometrical $\sin^2\delta + \cos^2\delta = 1$ and other trigonometrical identities results in Maximum Deflection Mach Number's equation in which is

$\displaystyle {M_1}^2 \left( k + 1 \right) ({M_{1n}}^2+1) = 2(k{M_{1n}}^4 +2{M_{1n}}^2 - 1)$

(13.36)

This equation and its twin equation can be obtained by an alternative procedure proposed by someone^13.16who suggested another way to approach this issue. It can be noticed that in equation (13.12), the deflection angle is a function of the Mach angle and the upstream Mach number,

. Thus, one can conclude that the maximum Mach angle is only a function of the upstream Much number,

. This can be shown mathematically by the argument that differentiating equation (13.12) and equating the results to zero creates relationship between the Mach number,

and the maximum Mach angle, $\theta$ . Since in that equation there appears only the heat ratio

, and Mach number,

, $\theta_{max}$ is a function of only these parameters. The differentiation of the equation (13.12) yields

$\displaystyle {d \tan \delta \over d\theta} = {k {M_1}^4 \sin^4\theta + \left(2... ...heta - \left[(k-1)+ {(k+1)^2 {M_1}^2 \over 4 } \right] {M_1}^2\sin^2\theta -1 }$

(13.37)

Because $\tan$ is a monotonous function, the maximum appears when $\theta$ has its maximum. The numerator of equation (13.37) is zero at different values of the denominator. Thus, it is sufficient to equate the numerator to zero to obtain the maximum. The nominator produces a quadratic equation for $\sin^2\theta$ and only the positive value for $\sin^2\theta$ is applied here. Thus, the $\sin^2\theta$ is

$\displaystyle \sin ^2 \theta_{max} = { -1 + { k + 1 \over 4}{M_1}^2+ \sqrt{(k+1... ...ver 2} {M_1}^2 + \left( {k+1 \over 2} {M_1} \right)^4 \right]} \over k {M_1}^2}$

(13.38)

Equation (13.38) should be referred to as the maximum's equation. It should be noted that both the Maximum Mach Deflection equation and the maximum's equation lead to the same conclusion that the maximum $M_{1n}$ is only a function of upstream the Mach number and the heat ratio

. It can be noticed that the Maximum Deflection Mach Number's equation is also a quadratic equation for ${M_{1n}}^2$ . Once $M_{1n}$ is found, then the Mach angle can be easily calculated by equation (13.8). To compare these two equations the simple case of Maximum for an infinite Mach number is examined. It must be pointed out that similar procedures can also be proposed (even though it does not appear in the literature). Instead, taking the derivative with respect to $\theta$ , a derivative can be taken with respect to

. Thus,

$\displaystyle {d \tan \delta \over dM_1} = 0$

(13.39)

and then solving equation (13.39) provides a solution for $M_{max}$ .

A simplified case of the Maximum Deflection Mach Number's equation for large Mach number becomes

$\displaystyle {M_{1n}} = \sqrt{ k+1\over 2k } M_{1} \quad \hbox{for} \quad M_{1} » 1$

(13.40)

Hence, for large Mach numbers, the Mach angle is $\sin\theta = \sqrt{ k+1\over 2k }$ (for k=1.4), which makes $\theta = 1.18$ or $\theta = 67.79^{\circ}$ .

With the value of $\theta$ utilizing equation (13.12), the maximum deflection angle can be computed. Note that this procedure does not require an approximation of $M_{1n}$ to be made. The general solution of equation (13.36) is

$\displaystyle M_{1n} = {{\sqrt{\sqrt{\left(k+1\right)^2 {M_1}^4+8 \left(k^2-1... ...2+16 \left(k+1\right)}+\left(k+1\right) {M_1}^2-4}}\over{2 \sqrt{k}}} %\\$

(13.41)

Note that Maximum Deflection Mach Number's equation can be extended to deal with more complicated equations of state (aside from the perfect gas model).

This typical example is for those who like mathematics.
$\begin{examl} Derive the perturbation of Maximum Deflection Mach Number's equati... ...ue:eq:menikoff} and neglect all the terms that are relatively small. \end{examl}$
Solution

The solution can be done by substituting ( $M_1 = 1+\epsilon$ ) into equation (13.36) and it results in

$\displaystyle M_{1n}= {\sqrt{\sqrt{\epsilon (k) } + {\epsilon^2} +2 \epsilon - 3 + k \epsilon^2+2 k\epsilon+k \over 4k}}$

(13.42)

where the epsilon function is

$\displaystyle \epsilon(k) =$	$\displaystyle (k^2+2k+1 ) \epsilon^4+(4 k^2+8 k+4) \epsilon^3 +$
	$\displaystyle (14 k^2+12 k - 2) \epsilon^2+( 20 k^2+8 k-12) \epsilon + 9 \left(k+1\right)^2$	(13.43)

Now neglecting all the terms with $\epsilon$ results for the epsilon function in

$\displaystyle \epsilon(k) \sim 9 \left(k+1\right)^2$

(13.44)

And the total operation results in

$\displaystyle M_{1n}= \sqrt{3 \left(k+1\right) -3 + k \over 4k} = 1$

(13.45)

Interesting to point out that as a consequence of this assumption the maxinum shock angle, $\theta$ is a normal shock. However, taking the second term results in different value. Taking the second term in the expantion results in

$\displaystyle M_{1n}= \sqrt{\sqrt{9 \left(k+1\right)^2 +( 20 k^2+8 k-12) \epsilon} -3 + k + 2 (1 + k) \epsilon \over 4k}$

(13.46)

Note this equation (13.46) produce un realistic values and additional terms are required to obtained to produced realistic values.

Next: The case of D Up: Upstream Mach number, M1, Previous: The simple procedure Index

Created by:Genick Bar-Meir, Ph.D.
On: 2007-11-21