Diffuser Efficiency

Figure: Description to clarify the definition of diffuser efficiency

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.

$$\eta = { 2(h_3 - h_1) \over {U_1}^2 } = { h_3 - h_1 \over {h_{0}}_1 - h_1 }$$ (6.6)

For perfect gas equation (6.6) can be converted to

$$\eta = { 2C_p(T_3 - T_1) \over {U_1}^2 }$$ (6.7)

And further expanding equation (6.7) results in

$$\eta = {2 {k R \over k -1} T_1 \left( {T_3 \over T_1} - 1\right) ... ...ver {M_1}^2 (k -1)} \left( \left(T_3 \over T_1\right)^{k -1 \over k} - 1\right)$$ (6.8)

Figure: Schematic of a supersonic tunnel in a continuous region (and also for example (6.3)

Solution

The condition at $M=3$ 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

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The nozzle area can be calculated by

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

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

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

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

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The calculations were carried as following: First calculate the $M_x$ as

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

Then calculate the $M_y$ 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

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

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

Finally utilizing the equation to calculate the following

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

Solution

This is an open valve case in which the pressure ratio is given. For this pressure ratio of $P_y/P_x= 2$ 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

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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

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

The velocity of the piston is then

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

Solution

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

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

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

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

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

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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

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