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Diffuser Efficiency

**Figure:** `Description to clarify the definition of diffuser efficiency`
$\begin{figure}\centerline{\includegraphics {cont/shockVariable/diffuserDiagram}} \end{figure}$

The efficiency of the diffuser is defined as the ratio of the enthalpy change that occurred between the entrance to exit stagnation pressure to the kinetic energy.

$\displaystyle \eta = { 2(h_3 - h_1) \over {U_1}^2 } = { h_3 - h_1 \over {h_{0}}_1 - h_1 }$

(6.6)

For perfect gas equation (6.6) can be converted to

$\displaystyle \eta = { 2C_p(T_3 - T_1) \over {U_1}^2 }$

(6.7)

And further expanding equation (6.7) results in

$\displaystyle \eta = {2 {k R \over k -1} T_1 \left( {T_3 \over T_1} - 1\right) ... ...ver {M_1}^2 (k -1)} \left( \left(T_3 \over T_1\right)^{k -1 \over k} - 1\right)$

(6.8)

$\begin{examl} \index{supersonic tunnel} A wind tunnel combined from a nozzle and... ....4$ can be assumed. Assume that a shock occurs in the test section. \end{examl}$

**Figure:** `Schematic of a supersonic tunnel in a continuous region (and also for example (6.3)`
$\begin{figure}\centerline{\includegraphics {cont/shockVariable/superSonicTunnel}} \end{figure}$

Solution

The condition at

is summarized in following table

$\rule[-0.1in]{0.pt}{0.3 in}\mathbf{M}$	$\mathbf{T \over T_0}$	$\mathbf{\rho \over \rho_0}$	$\mathbf{A \over A^{\star}}$	$\mathbf{P \over P_0}$	$\mathbf{A\times P \over A^{*} \times P_0}$	$\mathbf{F \over F^{*}}$
3.0000	0.35714	0.07623	4.2346	0.02722	0.11528	0.65326

The nozzle area can be calculated by

$\displaystyle {A^{*}}_n = {A^{\star} \over A} A = 0.02/4.2346 = 0.0047[m^2]$

In this case, $P_0 A^{*}$ is constant (constant mass flow). First the stagnation behind the shock will be

$\rule[-0.1in]{0.pt}{0.3 in}\mathbf{M_x}$	$\mathbf{M_y}$	$\mathbf{T_y \over T_x}$	$\mathbf{\rho_y \over \rho_x}$	$\mathbf{P_y \over P_x}$	$\mathbf{{P_0}_y \over {P_0}_x }$
3.0000	0.47519	2.6790	3.8571	10.3333	0.32834

$\displaystyle {A^{*}}_{d} = {{P_0}_n \over {P_0}_d} {A^{*}}_{n} \sim {1 \over 0.32834} 0.0047 \sim 0.0143[m^3]$

$\begin{examl} A shock is moving at 200 [m/sec] in pipe with gas with $k=1.3$, p... ...nd temperature of $350K$. Calculate the conditions after the shock. \end{examl}$
Solution

This is a case of completely and suddenly open valve with the shock velocity, temperature and pressure ``upstream'' known. In this case Potto-GDC provides the following table

$\rule[-0.1in]{0.pt}{0.3 in}\mathbf{M_x}$	$\mathbf{M_y}$	$\mathbf{{M_x}^{'}}$	$\mathbf{{M_y}^{'}}$	$\mathbf{{T_y} \over {T_x} }$	$\mathbf{{P_y} \over {P_x}}$	$\mathbf{{P_0}_y \over {P_0}_x }$
5.5346	0.37554	0.0	1.989	5.479	34.50	0.021717

The calculations were carried as following: First calculate the as

$\displaystyle Mx = U_s / \sqrt{k * 287.* T_x}$

Then calculate the

by using Potto-GDC or utilize the Tables. For example Potto-GDC (this code was produce by the program)

$\rule[-0.1in]{0.pt}{0.3 in}\mathbf{M_x}$	$\mathbf{M_y}$	$\mathbf{T_y \over T_x}$	$\mathbf{\rho_y \over \rho_x}$	$\mathbf{P_y \over P_x}$	$\mathbf{{P_0}_y \over {P_0}_x }$
5.5346	0.37554	5.4789	6.2963	34.4968	0.02172

The calculation of the temperature and pressure ratio also can be obtain by the same manner. The ``downstream'' shock number is

$\displaystyle M_{sy} = {U_s \over \sqrt{k * 287.* T_x * \left( T_y \over T_x \right)}} \sim 2.09668$

Finally utilizing the equation to calculate the following

$\displaystyle {M_y}^{'} = M_{sy} - M_y = 2.09668 - 0.41087 \sim 1.989$

$\begin{examl} An inventor interested in a design of tube and piston so that the ... .... If the steady state is achieved, what will be the piston velocity? \end{examl}$
Solution

This is an open valve case in which the pressure ratio is given. For this pressure ratio of

the following table can be obtained or by using Potto-GDC

$\rule[-0.1in]{0.pt}{0.3 in}\mathbf{M_x}$	$\mathbf{M_y}$	$\mathbf{T_y \over T_x}$	$\mathbf{\rho_y \over \rho_x}$	$\mathbf{P_y \over P_x}$	$\mathbf{{P_0}_y \over {P_0}_x }$
1.3628	0.75593	1.2308	1.6250	2.0000	0.96697

The temperature ratio and the Mach numbers for the velocity of the air (and the piston) can be calculated. The temperature at ``downstream'' (close to the piston) is

$\displaystyle T_y = T_x {T_y \over T_x} = 300 \times 1.2308 = 369.24[\celsius]$

The velocity of the piston is then

$\displaystyle U_y = M_y * c_y = 0.75593 * \sqrt{1.4*287* 369.24} \sim 291.16 [m/sec]$

$\begin{examl} A flow of gas is brought into a sudden stop. The mass flow rate of... ...=1.091$ (Butane?). Calculate the conditions behind the shock wave. \end{examl}$
Solution

This is the case of a close valve in which mass flow rate with the area given. Thus, the ``upstream'' Mach is given.

$\displaystyle {U_x}^{'} = {\dot{m} \over \rho A} = {\dot{m} R T \over P A} = { 2 \times 287 \times 350 \over 200000 \times 0.002} \sim 502.25 [m/sec]$

Thus the static Mach number, ${M_x}^{'}$ is

$\displaystyle {M_x}^{'} = { {U_x}^{'} \over c_x} = { 502.25 \over \sqrt{ 1.091 \times 143 \times 350} } \sim 2.15$

With this value for the Mach number Potto-GDC provides

$\rule[-0.1in]{0.pt}{0.3 in}\mathbf{M_x}$	$\mathbf{M_y}$	$\mathbf{{M_x}^{'}}$	$\mathbf{{M_y}^{'}}$	$\mathbf{{T_y} \over {T_x} }$	$\mathbf{{P_y} \over {P_x}}$	$\mathbf{{P_0}_y \over {P_0}_x }$
2.9222	0.47996	2.1500	0.0	2.589	9.796	0.35101

This table was obtained by using the procedure described in this book. The iteration of the procedure are

$\rule[-0.1in]{0.pt}{0.3 in}\mathbf{i}$	$\mathbf{M_x}$	$\mathbf{M_y}$	$\mathbf{{T_y} \over {T_x} }$	$\mathbf{{P_y} \over {P_x}}$	$\mathbf{{M_y}^{'}}$
0	3.1500	0.46689	2.8598	11.4096	0.0
1	2.940	0.47886	2.609	9.914	0.0
2	2.923	0.47988	2.590	9.804	0.0
3	2.922	0.47995	2.589	9.796	0.0
4	2.922	0.47996	2.589	9.796	0.0
5	2.922	0.47996	2.589	9.796	0.0

Next: Nozzle Flow With External Up: Normal Shock in Variable Previous: Nozzle efficiency Index

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

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